INTRODUCTORY ERNEST F. HAEUSSLER JR.

MATHEMATICAL RICHARD S. PAUL

RICHARD J. WOOD

ANALYSIS
FOURTEENTH EDITION
FOR BUSINESS, ECONOMICS, AND
THE LIFE AND SOCIAL SCIENCES
This page intentionally left blank
 ALGEBRA
Algebraic Rules for Exponents Radicals
Real numbers
p
a0 D 1 n
a D a1=n p
p
aCbDbCa . a/n D a; n an D a .a > 0/
p
n
1 p m
ab D ba a!n D .a ¤ 0/ p
n
am D . n a/
p D am=n
an p
a C .b C c/ D .a C b/ C c n n
rab D p a b
n

a.bc/ D .ab/c am an D amCn a n

a
a.b C c/ D ab C ac .am /n D amn n
Dp
.ab/n D an bn p b n
bp
a.b ! c/ D ab ! ac m n
p
a D mn a
.a C b/c D ac C bc # a $n an
.a ! b/c D ac ! bc D n
b b
aC0Da
a m Factoring Formulas
a"0D0 D am!n
a"1Da an
a C .!a/ D 0 ab C ac D a.b C c/
!.!a/ D a a2 ! b2 D .a C b/.a ! b/
.!1/a D !a Special Products a2 C 2ab C b2 D .a C b/2
a ! b D a C .!b/ a2 ! 2ab C b2 D .a ! b/2
a! a3 C b3 D .a C b/.a2 ! ab C b2 /
" DaCb
! .!b/ x.y C z/ D xy C xz
1 .x C a/.x C b/ D x2 C .a C b/x C ab a3 ! b3 D .a ! b/.a2 C ab C b2 /
a D1
a .x C a/2 D x2 C 2ax C a2
a 1 .x ! a/2 D x2 ! 2ax C a2
Da" .x C a/.x ! a/ D x2 ! a2 Straight Lines
b b
.x C a/3 D x3 C 3ax2 C 3a2 x C a3 y2 ! y1
.!a/b D !.ab/ D a.!b/ .x ! a/3 D x3 ! 3ax2 C 3a2 x ! a3 mD (slope formula)
x2 ! x1
.!a/.!b/ D ab y ! y1 D m.x ! x1 / (point-slope form)
y D mx C b (slope-intercept form)
!a a Quadratic Formula x D constant (vertical line)
D
!b b y D constant (horizontal line)
!a a a If ax2 C bx C c D 0, where
D! D a ¤ 0, then
b b !b
p Absolute Value
a b aCb !b ˙ b2 ! 4ac
C D xD
c c c 2a
jabj D jaj " jbj
a b a!b ˇaˇ
! D ˇ ˇ jaj
c c c
Inequalities ˇ ˇD
a c ac b jbj
" D ja ! bj D jb ! aj
b d bd
If a < b, then a C c < b C c. !jaj # a # jaj
a=b ad If a < b and c > 0, then ja C bj # jaj C jbj (triangle inequality)
D
c=d bc ac < bc.
a ac If a < b and c > 0, then
D .c ¤ 0/ a.!c/ > b.!c/. Logarithms
b bc

Summation Formulas logb x D y if and only if x D by

Special Sums logb .mn/ D logb m C logb n
m
P
n P
n logb D logb m ! logb n
cai D c ai P
n n
iDm iDm 1Dn logb mr D r logb m
P
n P
n P
n iD1
.ai C bi / D ai C bi Pn
n.nC1/
logb 1 D 0
iDm iDm iDm iD 2 logb b D 1
P
n P
pCn!m iD1
logb br D r
ai D aiCm!p Pn
n.nC1/.2nC1/
iDm iDp
i2 D 6 blogb p D p .p > 0/
iD1
P
p!1 P
n P
n
Pn
n2 .nC1/2 loga m
ai C ai D ai i3 D logb m D
iDm iDp iDm iD1
4 loga b
 FINITE MATHEMATICS
Business Relations Compound Interest Formulas

Interest D (principal)(rate)(time) S D P.1 C r/n

Total cost D variable cost C fixed cost P D S.1 C r/!n
total cost # r $n
Average cost per unit D re D 1 C !1
quantity n
Total revenue D (price per unit)(number of units sold)
Profit D total revenue ! total cost S D Pert
P D Se!rt
re D er ! 1
Ordinary Annuity Formulas
Matrix Multiplication
1 ! .1 C r/!n
ADR D Ran r (present value)
r X
n
.1 C r/n ! 1 .AB/ik D Aij Bjk D Ai1 B1k C Ai2 B2k C " " " C Ain bnk
SDR D Rsn r (future value)
r jD1

.AB/T D BT AT
A!1 A D I D AA!1
Counting
.AB/!1 D B!1 A!1

nŠ
n Pr D Probability
.n ! r/Š
nŠ
n Cr D #.E/
rŠ.n ! r/Š P.E/ D
C
n 0 C n C1 C " " " C n Cn!1 C n Cn D 2
n #.S/
n C0 D 1 D n Cn #.E \ F/
P.EjF/ D
#.F/
nC1 CrC1 D n Cr C n CrC1
P.E [ F/ D P.E/ C P.F/ ! P.E \ F/
P.E0 / D 1 ! P.E/
Properties of Events P.E \ F/ D P.E/P.FjE/ D P.F/P.EjF/

For E and F events for an experiment with sample space S

For X a discrete random variable with distribution f
E[EDE
E\EDE X
.E0 /0 D E f.x/ D 1
E [ E0 D S x X
E \ E0 D ; " D ".X/ D E.X/ D xf.x/
E[SDS X x
E\SDE Var.X/ D E..X ! "/2 / D .x ! "/2 f.x/
E[;DE p x
E\;D; # D #.X/ D Var.X/
E[FDF[E
E\FDF\E
.E [ F/0 D E0 \ F0 Binomial distribution
.E \ F/0 D E0 [ F0
E [ .F [ G/ D .E [ F/ [ G
E \ .F \ G/ D .E \ F/ \ G f.x/ D P.X D x/ D n Cx px qn!x
E \ .F [ G/ D .E \ F/ [ .E \ G/ " D np
p
E [ .F \ G/ D .E [ F/ \ .E [ G/ # D npq
 CALCULUS
Graphs of Elementary Functions

y y y y

2 4 4 4
f(x) = 1 f(x) = x2
f(x) = x3
2 f(x) = x 2 2

x x x x
-2 -1 1 2 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4
-1 -2 -2 -2

-2 -4 -4 -4

y y y y
2
4 f(x) = x 4 f(x) = 3 x 4 f (x) = x3 4

2 2 2 2 f(x) = x

x x x x
-4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4
-2 -2 -2 -2

-4 -4 -4 -4

y y y y

4 f(x) = 12 4 f(x) = ex 4
x
f(x) = 1
x
f(x) = lnx
2 2 2

x x x x
-4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4
-2 -2 -2 -2

-4 -4 -4 -4

Definition of Derivative of f.x/ Elasticity for Demand q D q.p/

p
0 d f.x C h/ ! f.x/ f.z/ ! f.x/
f .x/ D . f.x// D lim D lim p dq q
dx h!0 h z!x z!x $D " D
q dp dp
Differentiation Formulas dq

d d a du Integration Formulas
.c/ D 0 .u / D aua!1
dx dx dx
Z We assume that u isZa differentiable function
Z of x. Z
d a d 1 du
.x / D axa!1 .ln u/ D k dx D kx C C . f.x/ ˙ g.x// dx D f.x/ dx ˙ g.x/ dx
dx dx u dx
Z Z
d d u du xaC1 uaC1
.cf.x// D cf0 .x/ .e / D eu xa dx D C C; a ¤ !1 ua du D C C; a ¤ 1
dx dx dx Z aC1 Z aC1
d d 1 du ex dx D ex C C eu du D eu C C
. f.x/ ˙ g.x// D f0 .x/ ˙ g0 .x/ .logb u/ D "
dx dx .ln b/u dx Z Z Z
1
d d u du kf.x/ dx D k f.x/ dx du D ln juj C C; u ¤ 0
. f.x/g.x// D f.x/g0 .x/ C g.x/ f0 .x/ .b / D bu .ln b/ u
dx dx dx
(product rule) Consumers’ Surplus for Demand p D f.q/
! " Z q0
d f.x/ g.x/ f0 .x/ ! f.x/g0 .x/ d !1 1
D . f .x// D 0 !1
dx g.x/ .g.x//2 dx f . f .x// CS D Πf.q/ ! p0 % dq
0
(quotient rule)
dy dy du dy 1 Producers’ Surplus for Supply p D g.q/
D " (chain rule) D
dx du dx dx dx R q0
dy PS D 0 Πp0 ! g.q/% dq
INTRODUCTORY ERNEST F. HAEUSSLER JR.
The Pennsylvania State University

RICHARD S. PAUL

MATHEMATICAL
The Pennsylvania State University

RICHARD J. WOOD
Dalhousie University

ANALYSIS
FOURTEENTH EDITION
FOR BUSINESS, ECONOMICS, AND
THE LIFE AND SOCIAL SCIENCES
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7 -0-13-414110-7
10 7654321
Library and Archives Canada Cataloguing in Publication
Haeussler, Ernest F., author
Introductory mathematical analysis for business, economics, and the life
and social sciences Ernest F. Haeussler, r. (The Pennsylvania State
University), Richard S. Paul (The Pennsylvania State University), Richard
. Wood ( alhousie University). Fourteenth edition.
Includes bibliographical references and index.
IS N 7 -0-13-414110-7 (hardcover)
1. athematical analysis. 2. Economics, athematical. 3. usiness
mathematics. I. Paul, Richard S., author II. Wood, Richard ames, author
III. Title.
A300.H32 2017 515 C2017- 035 4-3
For Bronwen
This page intentionally left blank
 Contents
Preface ix

PART I COLLEGE ALGEBRA

CHAPTER 0 Review of Algebra 1
0.1 Sets of Real Numbers 2
0.2 Some Properties of Real Numbers 3
0.3 Exponents and Radicals 10
0.4 Operations with Algebraic Expressions 15
0.5 Factoring 20
0.6 Fractions 22
0.7 Equations, in Particular Linear Equations 28
0.8 Quadratic Equations 39
Chapter 0 Review 45

CHAPTER 1 Applications and More Algebra 47

1.1 Applications of Equations 48
1.2 Linear Inequalities 55
1.3 Applications of Inequalities 59
1.4 Absolute Value 62
1.5 Summation Notation 66
1.6 Sequences 70
Chapter 1 Review 80

CHAPTER 2 Functions and Graphs 83

2.1 Functions 84
2.2 Special Functions 91
2.3 Combinations of Functions 96
2.4 Inverse Functions 101
2.5 Graphs in Rectangular Coordinates 104
2.6 Symmetry 113
2.7 Translations and Reflections 118
2.8 Functions of Several Variables 120
Chapter 2 Review 128

CHAPTER 3 Lines, Parabolas, and Systems 131

3.1 Lines 132
3.2 Applications and Linear Functions 139
3.3 Quadratic Functions 145
3.4 Systems of Linear Equations 152
3.5 Nonlinear Systems 162
3.6 Applications of Systems of Equations 164
Chapter 3 Review 172

CHAPTER 4 Exponential and Logarithmic Functions 175

4.1 Exponential Functions 176
4.2 Logarithmic Functions 188
4.3 Properties of Logarithms 194
4.4 Logarithmic and Exponential Equations 200
Chapter 4 Review 204

v
vi Contents

PART II FINITE MATHEMATICS

CHAPTER 5 Mathematics of Finance 208
5.1 Compound Interest 209
5.2 Present Value 214
5.3 Interest Compounded Continuously 218
5.4 Annuities 222
5.5 Amortization of Loans 230
5.6 Perpetuities 234
Chapter 5 Review 237

CHAPTER 6 Matrix Algebra 240

6.1 Matrices 241
6.2 Matrix Addition and Scalar Multiplication 246
6.3 Matrix Multiplication 253
6.4 Solving Systems by Reducing Matrices 264
6.5 Solving Systems by Reducing Matrices (Continued) 274
6.6 Inverses 279
6.7 Leontief’s Input--Output Analysis 286
Chapter 6 Review 292

CHAPTER 7 Linear Programming 294

7.1 Linear Inequalities in Two Variables 295
7.2 Linear Programming 299
7.3 The Simplex Method 306
7.4 Artificial Variables 320
7.5 Minimization 330
7.6 The Dual 335
Chapter 7 Review 344

CHAPTER 8 Introduction to Probability and Statistics 348

8.1 Basic Counting Principle and Permutations 349
8.2 Combinations and Other Counting Principles 355
8.3 Sample Spaces and Events 367
8.4 Probability 374
8.5 Conditional Probability and Stochastic Processes 388
8.6 Independent Events 401
8.7 Bayes’ Formula 411
Chapter 8 Review 419

CHAPTER 9 Additional Topics in Probability 424

9.1 Discrete Random Variables and Expected Value 425
9.2 The Binomial Distribution 432
9.3 Markov Chains 437
Chapter 9 Review 447
 Contents vii

PART III CALCULUS

CHAPTER 10 Limits and Continuity 450
10.1 Limits 451
10.2 Limits (Continued) 461
10.3 Continuity 469
10.4 Continuity Applied to Inequalities 474
Chapter 10 Review 479

CHAPTER 11 Differentiation 482

11.1 The Derivative 483
11.2 Rules for Differentiation 491
11.3 The Derivative as a Rate of Change 499
11.4 The Product Rule and the Quotient Rule 509
11.5 The Chain Rule 519
Chapter 11 Review 527

CHAPTER 12 Additional Differentiation Topics 531

12.1 Derivatives of Logarithmic Functions 532
12.2 Derivatives of Exponential Functions 537
12.3 Elasticity of Demand 543
12.4 Implicit Differentiation 548
12.5 Logarithmic Differentiation 554
12.6 Newton’s Method 558
12.7 Higher-Order Derivatives 562
Chapter 12 Review 566

CHAPTER 13 Curve Sketching 569

13.1 Relative Extrema 570
13.2 Absolute Extrema on a Closed Interval 581
13.3 Concavity 583
13.4 The Second-Derivative Test 591
13.5 Asymptotes 593
13.6 Applied Maxima and Minima 603
Chapter 13 Review 614

CHAPTER 14 Integration 619

14.1 Differentials 620
14.2 The Indefinite Integral 625
14.3 Integration with Initial Conditions 631
14.4 More Integration Formulas 635
14.5 Techniques of Integration 642
14.6 The Definite Integral 647
14.7 The Fundamental Theorem of Calculus 653
Chapter 14 Review 661
viii Contents

CHAPTER 15 Applications of Integration 665

15.1 Integration by Tables 666
15.2 Approximate Integration 672
15.3 Area Between Curves 678
15.4 Consumers’ and Producers’ Surplus 687
15.5 Average Value of a Function 690
15.6 Differential Equations 692
15.7 More Applications of Differential Equations 699
15.8 Improper Integrals 706
Chapter 15 Review 709

CHAPTER 16 Continuous Random Variables 713

16.1 Continuous Random Variables 714
16.2 The Normal Distribution 721
16.3 The Normal Approximation to the Binomial Distribution 726
Chapter 16 Review 730

CHAPTER 17 Multivariable Calculus 732

17.1 Partial Derivatives 733
17.2 Applications of Partial Derivatives 738
17.3 Higher-Order Partial Derivatives 744
17.4 Maxima and Minima for Functions of Two Variables 746
17.5 Lagrange Multipliers 754
17.6 Multiple Integrals 761
Chapter 17 Review 765

APPENDIX A Compound Interest Tables 769

APPENDIX B Table of Selected Integrals 777

APPENDIX C Areas Under the Standard Normal Curve 780

Answers to Odd-Numbered Problems AN-1

Index I-1
Preface
T
he fourteenth edition of Intr duct ry athematica Ana ysis f r Business Ec n
mics and the ife and S cia Sciences I A continues to provide a mathematical
foundation for students in a variety of fields and ma ors, as suggested by the title.
As begun in the thirteenth edition, the boo has three parts: College Algebra, Chapters 0 4
Finite athematics, Chapters 5 and Calculus, Chapters 10 17.
Schools that have two academic terms per year tend to give usiness students a term
devoted to Finite athematics and a term devoted to Calculus. For these schools we rec-
ommend Chapters 0 through for the first course, starting wherever the preparation of the
students allows, and Chapters 10 through 17 for the second, including as much as the stu-
dents bac ground allows and their needs dictate.
For schools with three quarter or three semester courses per year there are a number
of possible uses for this boo . If their program allows three quarters of athematics, well-
prepared usiness students can start a first course on Finite athematics with Chapter 1
and proceed through topics of interest up to and including Chapter . In this scenario, a
second course on ifferential Calculus could start with Chapter 10 on imits and Continu-
ity, followed by the three differentiation chapters , 11 through 13 inclusive. Here, Section
12.6 on Newton s ethod can be omitted without loss of continuity, while some instructors
may prefer to review Chapter 4 on Exponential and ogarithmic Functions prior to study-
ing them as differentiable functions. Finally, a third course could comprise Chapters 14
through 17 on Integral Calculus with an introduction to ultivariable Calculus. Note that
Chapter 16 is certainly not needed for Chapter 17 and Section 15. on Improper Integrals
can be safely omitted if Chapter 16 is not covered.

A
Intr duct ry athematica Ana ysis f r Business Ec n mics and the ife and S cia
Sciences I A ta es a unique approach to problem solving. As has been the case in ear-
lier editions of this boo , we establish an emphasis on algebraic calculations that sets this
text apart from other introductory, applied mathematics boo s. The process of calculating
with variables builds s ill in mathematical modeling and paves the way for students to use
calculus. The reader will not find a definition-theorem-proof treatment, but there is a sus-
tained effort to impart a genuine mathematical treatment of applied problems. In particular,
our guiding philosophy leads us to include informal proofs and general calculations that
shed light on how the corresponding calculations are done in applied problems. Emphasis
on developing algebraic s ills is extended to the exercises, of which many, even those of
the drill type, are given with general rather than numerical coe cients.
We have refined the organization of our boo over many editions to present the content
in very manageable portions for optimal teaching and learning. Inevitably, that process
tends to put weight on a boo , and the present edition ma es a very concerted effort to
pare the boo bac somewhat, both with respect to design features ma ing for a cleaner
approach and content recognizing changing pedagogical needs.

C F E
We continue to ma e the elementary notions in the early chapters pave the way for their
use in more advanced topics. For example, while discussing factoring, a topic many stu-
dents find somewhat arcane, we point out that the principle ab D 0 implies a D 0 or
b D 0 , together with factoring, enables the splitting of some complicated equations into
several simpler equations. We point out that percentages are ust rescaled numbers via the
p
equation p D 100 so that, in calculus, relative rate of change and percentage rate
of change are related by the equation r D r ! 100 . We thin that at this time, when
negative interest rates are often discussed, even if seldom implemented, it is wise to be
absolutely precise about simple notions that are often ta en for granted. In fact, in the

ix
x Preface

Finance, Chapter 5, we explicitly discuss negative interest rates and as , somewhat rhetor-
ically, why ban s do not use continuous compounding (given that for a long time now
continuous compounding has been able to simplify calculations in practice as well as in
theory).
Whenever possible, we have tried to incorporate the extra ideas that were in the Explore
and Extend chapter-closers into the body of the text. For example, the functions tax rate t.i/
and tax paid .i/ of income i, are seen for what they are: everyday examples of case-defined
functions. We thin that in the process of learning about polynomials it is helpful to include
Horner s ethod for their evaluation, since with even a simple calculator at hand this ma es
the calculation much faster. While doing linear programming, it sometimes helps to thin
of lines and planes, etcetera, in terms of intercepts alone, so we include an exercise to show
that if a line has (nonzero) intercepts x0 and y0 then its equation is given by

x y
C D1
x0 y0

and, moreover, (for positive x0 and y0 ) we as for a geometric interpretation of the equivalent
equation y0 x C x0 y D x0 y0 .
ut, turning to our paring of the previous I A, let us begin with inear Program-
ming. This is surely one of the most important topics in the boo for usiness students. We
now feel that, while students should now about the possibility of u tip e ptimum S u
ti ns and egeneracy and nb unded S uti ns, they do not have enough time to devote
an entire, albeit short, section to each of these. The remaining sections of Chapter 7 are
already demanding and we now content ourselves with providing simple alerts to these
possibilities that are easily seen geometrically. (The deleted sections were always tagged
as omittable .)
We thin further that, in Integral Calculus, it is far more important for Applied athe-
matics students to be adept at using tables to evaluate integrals than to now about Integra
ti n by Parts and Partia Fracti ns. In fact, these topics, of endless oy to some as recre-
ational problems, do not seem to fit well into the general scheme of serious problem solving.
It is a fact of life that an elementary function (in the technical sense) can easily fail to have
an elementary antiderivative, and it seems to us that Parts does not go far enough to rescue
this di culty to warrant the considerable time it ta es to master the technique. Since Par
tia Fracti ns ultimately lead to elementary antiderivatives for all rati na functions, they
are part of serious problem solving and a better case can be made for their inclusion in an
applied textboo . However, it is vainglorious to do so without the inverse tangent function
at hand and, by longstanding tacit agreement, applied calculus boo s do not venture into
trigonometry.
After deleting the sections mentioned above, we reorganized the remaining material of
the integration chapters , 14 and 15, to rebalance them. The first concludes with the Funda-
mental Theorem of Calculus while the second is more properly applied . We thin that the
formerly daunting Chapter 17 has benefited from deletion of Imp icit Partia i erentia
ti n, the Chain Ru e for partial differentiation, and ines f Regressi n. Since ultivariable
Calculus is extremely important for Applied athematics, we hope that this more manage-
able chapter will encourage instructors to include it in their syllabi.

E E
ost instructors and students will agree that the ey to an effective textboo is in the
quality and quantity of the examples and exercise sets. To that end, more than 50 exam-
ples are wor ed out in detail. Some of these examples include a strategy box designed
to guide students through the general steps of the solution before the specific solution
is obtained. (See, for example, Section 14.3 Example 4.) In addition, an abundant num-
ber of diagrams (almost 500) and exercises (more than 5000) are included. f the exer-
cises, approximately 20 percent have been either updated or written completely anew. In
each exercise set, grouped problems are usually given in increasing order of di culty.
In most exercise sets the problems progress from the basic mechanical drill-type to more
 Preface xi

interesting thought-provo ing problems. The exercises labeled with a coloured exercise
number correlate to a Now Wor Problem N statement and example in the section.
ased on the feedbac we have received from users of this text, the diversity of the
applications provided in both the exercise sets and examples is truly an asset of this boo .
any real applied problems with accurate data are included. Students do not need to loo
hard to see how the mathematics they are learning is applied to everyday or wor -related
situations. A great deal of effort has been put into producing a proper balance between
drill-type exercises and problems requiring the integration and application of the concepts
learned.

H F
! Applications An abundance and variety of applications for the intended audience appear
throughout the boo so that students see frequently how the mathematics they are learn-
ing can be used. These applications cover such diverse areas as business, economics,
biology, medicine, sociology, psychology, ecology, statistics, earth science, and archae-
ology. any of these applications are drawn from literature and are documented by
references, sometimes from the Web. In some, the bac ground and context are given
in order to stimulate interest. However, the text is self-contained, in the sense that it
assumes no prior exposure to the concepts on which the applications are based. (See, for
example, Chapter 15, Section 7, Example 2.)
! ow or Problem Throughout the text we have retained the popular rk
Pr b em feature. The idea is that after a wor ed example, students are directed to
an end-of-section problem (labeled with a colored exercise number) that reinforces the
ideas of the wor ed example. This gives students an opportunity to practice what they
have ust learned. ecause the ma ority of these eyed exercises are odd-numbered, stu-
dents can immediately chec their answer in the bac of the boo to assess their level of
understanding. The complete solutions to the odd-numbered exercises can be found in
the Student Solutions anual.
! Cautions Cautionary warnings are presented in very much the same way an instructor
would warn students in class of commonly made errors. These appear in the margin,
along with other explanatory notes and emphases.
! Definitions ey concepts and important rules and formulas These are clearly stated
and displayed as a way to ma e the navigation of the boo that much easier for the
student. (See, for example, the efinition of erivative in Section 11.1.)
! Review material Each chapter has a review section that contains a list of important
terms and symbols, a chapter summary, and numerous review problems. In addition,
ey examples are referenced along with each group of important terms and symbols.
! Inequalities and slac variables In Section 1.2, when inequalities are introduced we
point out that a " b is equivalent to there exists a non-negative number, s, such that
a C s D b . The idea is not deep but the pedagogical point is that s ack ariab es, ey
to implementing the simplex algorithm in Chapter 7, should be familiar and not distract
from the rather technical material in linear programming.
! Absolute value It is common to note that ja # bj provides the distance from a to b. In
Example 4e of Section 1.4 we point out that x is less than ! units from " translates as
jx # "j < ! . In Section 1.4 this is but an exercise with the notation, as it should be, but
the point here is that later (in Chapter ) " will be the mean and ! the standard deviation
of a random variable. Again we have separated, in advance, a simple idea from a more
advanced one. f course, Problem 12 of Problems 1.4, which as s the student to set up
jf.x/ # j < #, has a similar agenda to Chapter 10 on limits.
! Early treatment of summation notation This topic is necessary for study of the defi-
nite integral in Chapter 14, but it is usefu long before that. Since it is a notation that is
new to most students at this level, but no more than a notation, we get it out of the way
in Chapter 1. y using it when convenient, bef re c erage f the de nite integra , it is
not a distraction from that challenging concept.
xii Preface

! Section on sequences This section provides several pedagogical advantages.

The very definition is stated in a fashion that paves the way for the more important and
more basic definition of function in Chapter 2. In summing the terms of a sequence we
are able to practice the use of summation notation introduced in the preceding section.
The most obvious benefit though is that sequences allows us a better organization
in the annuities section of Chapter 5. oth the present and the future values of an annu-
ity are obtained by summing (finite) geometric sequences. ater in the text, sequences
arise in the definition of the number e in Chapter 4, in ar ov chains in Chapter , and
in Newton s method in Chapter 12, so that a helpful unifying reference is obtained.
! Sum of an infinite sequence In the course of summing the terms of a finite sequence,
it is natural to raise the possibility of summing the terms of an infinite sequence. This is
a nonthreatening environment in which to provide a first foray into the world of limits.
We simply explain how certain infinite geometric sequences have well-defined sums and
phrase the results in a way that creates a toehold for the introduction of limits in Chapter
10. These particular infinite sums enable us to introduce the idea of a perpetuity, first
informally in the sequence section, and then again in more detail in a separate section in
Chapter 5.
! Section Functions of Several Variables The introduction to functions of several
variables appears in Chapter 2 because it is a topic that should appear long before Cal-
culus. nce we have done some calculus there are particular ways to use calculus in the
study of functions of several variables, but these aspects should not be confused with the
basics that we use throughout the boo . For example, a-sub-n-angle-r and s-sub-n-
angle-r studied in the athematics of Finance, Chapter 5, are perfectly good functions
of two variables, and inear Programming see s to optimize linear functions of several
variables sub ect to linear constraints.
! Leontief’s input output analysis in Section In this section we have separated vari-
ous aspects of the total problem. We begin by describing what we call the eontief matrix
A as an encoding of the input and output relationships between sectors of an economy.
Since this matrix can often be assumed to be constant for a substantial period of time,
we begin by assuming that A is a given. The simpler problem is then to determine the
production, X, which is required to meet an external demand, , for an economy whose
eontief matrix is A. We provide a careful account of this as the solution of .I#A/X D .
Since A can be assumed to be fixed while various demands, , are investigated, there is
s me ustification to compute .I # A/!1 so that we have X D .I # A/!1 . However, use
of a matrix inverse should not be considered an essential part of the solution. Finally, we
explain how the eontief matrix can be found from a table of data that might be available
to a planner.
! Birthday probability in Section This is a treatment of the classic problem of deter-
mining the probability that at least 2 of n people have their birthday on the same day.
While this problem is given as an example in many texts, the recursive formula that we
give for calculating the probability as a function of n is not a common feature. It is reason-
able to include it in this boo because recursively defined sequences appear explicitly in
Section 1.6.
! Mar ov Chains We noticed that considerable simplification of the problem of finding
steady state vectors is obtained by writing state vectors as columns rather than rows.
This does necessitate that a transition matrix D Œtij $ have tij D probability that next
state is i given that current state is j but avoids several artificial transpositions.
! Sign Charts for a function in Chapter The sign charts that we introduced in the
12th edition now ma e their appearance in Chapter 10. ur point is that these charts
can be made for any real-valued function of a real variable and their help in graph-
ing a function begins prior to the introduction of derivatives. f course we continue to
exploit their use in Chapter 13 Curve S etching where, for each function f, we advo-
cate ma ing a sign chart for each of f, f0 , and f00 , interpreted for f itself. When this is
possible, the graph of the function becomes almost self-evident. We freely ac nowledge
that this is a blac board technique used by many instructors, but it appears too rarely in
textboo s.
 Preface xiii

S
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xiv Preface

A
We express our appreciation to the following colleagues who contributed comments and
suggestions that were valuable to us in the evolution of this text. (Professors mar ed with
an asteris reviewed the fourteenth edition.)

E. Adibi, Chapman ni ersity P. Hune e, he hi State ni ersity

R. . Alliston, Pennsy ania State ni ersity C. Hurd, Pennsy ania State ni ersity
R. A. Alo, ni ersity f ust n . A. iminez, Pennsy ania State ni ersity
. T. Andrews, ak and ni ersity T. H. ones, Bish p s ni ersity
. N. de Arce, ni ersity f Puert Ric W. C. ones, estern entucky ni ersity
E. arbut, ni ersity f Idah R. . ing, Gettysburg C ege
. R. ates, estern I in is ni ersity . . ostreva, ni ersity f aine
S. ec , a arr C ege . A. raus, Gann n ni ersity
. E. ennett, urray State ni ersity . ucera, ashingt n State ni ersity
C. ernett, arper C ege . R. atina, Rh de Is and uni r C ege
A. ishop, estern I in is ni ersity . N. aughlin, ni ersity f A aska Fairbanks
P. lau, Sha nee State ni ersity P. oc wood-Coo e, est exas A ni ersity
R. lute, ni ersity f tta a . F. ongman, i an a ni ersity
S. A. oo , Ca if rnia State ni ersity F. acWilliam, A g ma ni ersity
A. rin , St C ud State ni ersity I. arsha , y a ni ersity f Chicag
R. rown, rk ni ersity . ason, E mhurst C ege
R. W. rown, ni ersity f A aska . atheson, ni ersity f ater
S. . ulman-Fleming, i frid aurier ni ersity F. . ayer, t San Ant ni C ege
. Calvetti, ati na C ege P. c ougle, ni ersity f iami
. Cameron, ni ersity f Akr n F. iles, Ca if rnia State ni ersity
. S. Chung, api ani C mmunity C ege E. ohni e, t San Ant ni C ege
. N. Clar , ni ersity f Ge rgia C. on , ni ersity f Richm nd
E. . Cohen, ni ersity f tta a R. A. oreland, exas ech ni ersity
. awson, Pennsy ania State ni ersity . . orris, ni ersity f isc nsin adis n
A. ollins, Pennsy ania State ni ersity . C. oss, Paducah C mmunity C ege
T. . uda, C umbus State C mmunity C ege . ullin, Pennsy ania State ni ersity
. A. Earles, St C ud State ni ersity E. Nelson, Pennsy ania State ni ersity
. H. Edwards, ni ersity f F rida S. A. Nett, estern I in is ni ersity
. R. Elliott, i frid aurier ni ersity R. H. ehm e, ni ersity f I a
. Fitzpatric , ni ersity f exas at E Pas . . h, Pennsy ania State ni ersity
. . Flynn, Rh de Is and uni r C ege . U. verall, ni ersity f a erne
. . Fuentes, ni ersity f aine . Pace, arrant C unty C ege
. erber, St hn s ni ersity A. Panayides, i iam Patters n ni ersity
T. . oedde, he ni ersity f Find ay . Par er, ni ersity f Paci c
S. . oel, a d sta State ni ersity N. . Patterson, Pennsy ania State ni ersity
. off, k ah ma State ni ersity V. Pedwaydon, a rence echnica ni ersity
. oldman, ePau ni ersity E. Pemberton, i frid aurier ni ersity
E. reenwood, arrant C unty C ege rth est . Per el, right State ni ersity
Campus . . Priest, arding C ege
. T. resser, B ing Green State ni ersity . R. Provencio, ni ersity f exas
. riff, Pennsy ania State ni ersity . R. Pulsinelli, estern entucky ni ersity
R. rinnell, ni ersity f r nt at Scarb r ugh . Racine, ni ersity f tta a
F. H. Hall, Pennsy ania State ni ersity . Reed, a arr C ege
V. E. Han s, estern entucky ni ersity N. . Rice, ueen s ni ersity
T. Harriott, unt Saint incent ni ersity A. Santiago, ni ersity f Puert Ric
R. C. Heitmann, he ni ersity f exas at Austin . R. Schaefer, ni ersity f isc nsin i aukee
. N. Henry, Ca if rnia State ni ersity S. Sehgal, he hi State ni ersity
W. U. Hodgson, est Chester State C ege W. H. Seybold, r., est Chester State C ege
. Hooper, Acadia ni ersity . Shibuya, San Francisc State ni ersity
. C. Horne, r., irginia P ytechnic Institute and State . Shilling, he ni ersity f exas at Ar ingt n
ni ersity S. Singh, Pennsy ania State ni ersity
. Hradnans y, Pennsy ania State ni ersity . Small, s Ange es Pierce C ege
 Preface xv

E. Smet, ur n C ege . . Waits, he hi State ni ersity

. Stein, Ca if rnia State ni ersity ng Beach A. Walton, irginia P ytechnic Institute and State
. Stoll, ni ersity f S uth Car ina ni ersity
T. S. Sullivan, S uthern I in is ni ersity Ed ards i e H. Walum, he hi State ni ersity
E. A. Terry, St seph s ni ersity E. T. H. Wang, i frid aurier ni ersity
A. Tierman, Sagina a ey State ni ersity A. . Weidner, Pennsy ania State ni ersity
. Toole, ni ersity f aine . Weiss, Pennsy ania State ni ersity
. W. Toole, ni ersity f aine N. A. Weigmann, Ca if rnia State ni ersity
. Torres, Athabasca ni ersity S. . Wong, hi State ni ersity
. H. Trahan, a a P stgraduate Sch . Woods, he hi State ni ersity
. P. Tull, he hi State ni ersity C. R. . Wright, ni ersity f reg n
. . Vaughan, r., ni ersity f A abama in C. Wu, ni ersity f isc nsin i aukee
Birmingham . F. Wyman, he hi State ni ersity
. A. Vercoe, Pennsy ania State ni ersity . hang, ashingt n State ni ersity
. Vuilleumier, he hi State ni ersity

Some exercises are ta en from problem supplements used by students at Wilfrid aurier
University. We wish to extend special than s to the epartment of athematics of Wilfrid
aurier University for granting Prentice Hall permission to use and publish this material,
and also to Prentice Hall, who in turn allowed us to ma e use of this material.
We again express our sincere gratitude to the faculty and course coordinators of The
hio State University and Columbus State University who too a een interest in this and
other editions, offering a number of invaluable suggestions.
Special than s are due to PS North America, C. for their careful wor on the solu-
tions manuals. Their wor was extraordinarily detailed and helpful to us. We also appreciate
the care that they too in chec ing the text and exercises for accuracy.

Ernest F aeuss er r
Richard S Pau
Richard d
This page intentionally left blank
0.1
0.2
0
ets of ea
o e Pro ert es of ea
ers
e e of e ra

esley ri th wor ed for a yacht supply company in Antibes, France. ften,

she needed to examine receipts in which only the total paid was reported and
then determine the amount of the total which was French value-added tax .
ers It is nown as TVA for Taxe la Value A out . The French TVA rate was
1 .6 (but in anuary of 2014 it increased to 20 ). A lot of esley s business came
0.3 onents an a ca s
from Italian suppliers and purchasers, so she also had to deal with the similar problem
0.4 erat ons t of receipts containing Italian sales tax at 1 (now 22 ).
e ra c ress ons A problem of this ind demands a formula, so that the user can ust plug in a tax
rate li e 1 .6 or 22 to suit a particular place and time, but many people are able
0.5 actor n
to wor through a particular case of the problem, using specified numbers, without
0.6 ract ons nowing the formula. Thus, if esley had a 200-Euro French receipt, she might have
reasoned as follows: If the item cost 100 Euros before tax, then the receipt total would
0.7 at ons n Part c ar be for 11 .6 Euros with tax of 1 .6, so tax in a receipt t ta f is t as is
near at ons
t . Stated mathematically,
0.8 a rat c at ons tax in 200 1 :6
D ! 0:164 D 16:4
C er 0 e e 200 11 :6
If her reasoning is correct then the amount of TVA in a 200-Euro receipt is about 16.4
of 200 Euros, which is 32. Euros. In fact, many people will now guess that
! "
p
tax in R D R
100 C p
gives the tax in a receipt R, when the tax rate is p . Thus, if esley felt confident about
1
her deduction, she could have multiplied her Italian receipts by 11 to determine the tax
they contained.
f course, most people do not remember formulas for very long and are uncom-
fortable basing a monetary calculation on an assumption such as the one we italicized
above. There are lots of relationships that are more complicated than simple proportion-
ality The purpose of this chapter is to review the algebra necessary for you to construct
your own formulas, ith c n dence as needed. In particular, we will derive esley s
formula from principles with which everybody is familiar. This usage of algebra will
appear throughout the boo , in the course of ma ing genera ca cu ati ns ith ariab e
quantities.
In this chapter we will review real numbers and algebraic expressions and the basic
operations on them. The chapter is designed to provide a brief review of some terms and
methods of symbolic calculation. Probably, you have seen most of this material before.
However, because these topics are important in handling the mathematics that comes
later, an immediate second exposure to them may be beneficial. evote whatever time
is necessary to the sections in which you need review.

1
2 C e e of e ra

Objective S R N
o eco e fa ar t sets n A set is a collection of ob ects. For example, we can spea of the set of even numbers
art c ar sets of rea n ers an
t e rea n er ne between 5 and 11, namely, 6, , and 10. An ob ect in a set is called an element of
that set. If this sounds a little circular, don t worry. The words set and e ement are li e
ine and p int in geometry. We cannot define them in more primitive terms. It is only
with practice in using them that we come to understand their meaning. The situation is
also rather li e the way in which a child learns a first language. Without nowing any
words, a child infers the meaning of a few very simple words by watching and listening
to a parent and ultimately uses these very few words to build a wor ing vocabulary.
None of us needs to understand the mechanics of this process in order to learn how to
spea . In the same way, it is possible to learn practical mathematics without becoming
embroiled in the issue of undefined primitive terms.
ne way to specify a set is by listing its elements, in any order, inside braces. For
example, the previous set is f6; ; 10g, which we could denote by a letter such as A,
allowing us to write A D f6; ; 10g. Note that f ; 10; 6g also denotes the same set, as
does f6; ; 10; 10g. A set is determined by its e ements, and neither rearrangements nor
repetitions in a listing affect the set. A set A is said to be a subset of a set B if and
only if every element of A is also an element of B. For example, if A D f6; ; 10g and
B D f6; ; 10; 12g, then A is a subset of B but B is not a subset of A. There is exactly
one set which contains n elements. It is called the empty set and is denoted by ;.
Certain sets of numbers have special names. The numbers 1, 2, 3, and so on form
the set of positive integers:
set f p siti e integers D f1; 2; 3; : : :g
The three dots are an informal way of saying that the listing of elements is unending
and the reader is expected to generate as many elements as needed from the pattern.
The positive integers together with 0 and the negative integers "1; "2; "3; : : : ;
form the set of integers:
set f integers D f: : : ; "3; "2; "1; 0; 1; 2; 3; : : :g
The set of rational numbers consists of numbers, such as 12 and 53 , that can be
written as a quotient of two integers. That is, a rational number is a number that can
The reason for q ¤ 0 is that we cannot be written as pq , where p and q are integers and q ¤ 0. (The symbol ¤ is read is not
divide by zero.
equal to. ) For example, the numbers 120 , !2 7
, and !6
!2
are rational. We remar that 24 , 12 ,
3 !4
, , 0:5, and 50 all represent the same rational number. The integer 2 is rational,
6 !
Every integer is a rational number. since 2 D 21 . In fact, every integer is rational.
All rational numbers can be represented by decimal numbers that terminate such
as 34 D 0:75 and 32 D 1:5, or by n nterminating repeating decima numbers (composed
of a group of digits that repeats without end), such as 23 D 0:666 : : : ; !4 11
D "0:3636 : : : ;
2
Every rational number is a real number. and 15 D 0:1333 : : : : Numbers represented by n nterminating n nrepeating decimals
are called irrational numbers. An irrational number p cannot be written as an integer
The set of real numbers consists of all divided by an integer. The numbers ! (pi) and 2 are examples of irrational numbers.
decimal numbers. Together, the rational numbers and the irrational numbers form the set of real numbers.
Real numbers can be represented by points on a line. First we choose a point on the
line to represent zero. This point is called the rigin. (See Figure 0.1.) Then a standard
measure of distance, called a unit distance is chosen and is successively mar ed off
both to the right and to the left of the origin. With each point on the line we associate a
directed distance, which depends on the position of the point with respect to the origin.

Some Points and Their Coordinates

1
-r -1.5 2 2 r Positive
-3 -2 -1 0 1 2 3 direction
Origin

FIGURE The real-number line.

 Section 0.2 o e Pro ert es of ea ers 3

Positions to the right of the origin are considered positive .C/ and positions to the left
are negative ."/. For example, with the point 12 unit to the right of the origin there
corresponds the number 12 , which is called the coordinate of that point. Similarly, the
coordinate of the point 1.5 units to the left of the origin is "1:5. In Figure 0.1, the
coordinates of some points are mar ed. The arrowhead indicates that the direction to
the right along the line is considered the positive direction.
To each point on the line there corresponds a unique real number, and to each
real number there corresponds a unique point on the line. There is a ne t ne c r
resp ndence between points on the line and real numbers. We call such a line, with
coordinates mar ed, a real number line. We feel free to treat real numbers as points
on a real-number line and vice versa.

E AM LE I R N

Is it true that 0:151515 : : : is an irrational number

S The dots in 0:151515 : : : are understood to convey repetition of the digit
string 15 . Irrational numbers were defined to be real numbers that are represented by a
n nterminating n nrepeating decimal, so 0:151515 : : : is not irrational. It is therefore a
rational number. It is not immediately clear how to represent 0:151515 : : : as a quotient
5
of integers. In Chapter 1 we will learn how to show that 0:151515 : : : D . ou can
33
chec that this is p ausib e by entering 5 # 33 on a calculator, but you should also thin
5
about why the calculator exercise does not pr e that 0:151515 : : : D .
33
Now ork Problem 7 G

R BLEMS
p
In Pr b ems determine the truth f each statement If the 25 is not a positive integer.
statement is fa se gi e a reas n hy that is s p
p 2 is a real number.
"13 is an integer. 0
"2 is rational.
is rational. 0
7 ! is a positive integer.
"3 is a positive integer. p
0 is to the right of " 2 on the real-number line.
0 is not rational.
p Every integer is positive or negative.
3 is rational.
Every terminating decimal number can be regarded as a
"1 repeating decimal number.
is a rational number. p
0
"1 is a real number.

Objective S R N
o na e strate an re ate We now state a few important properties of the real numbers. et a, b, and c be real
ro ert es of t e rea n ers an
t e r o erat ons numbers.

1 T T E
If a D b and b D c; then a D c.

Thus, two numbers that are both equal to a third number are equal to each other.
For example, if x D y and y D 7, then x D 7.
4 C e e of e ra

2 T C A M
For all real numbers a and b, there are unique real numbers a C b and ab.

This means that any two numbers can be added and multiplied, and the result in
each case is a real number.

3 T C A M
aCbDbCa and ab D ba

This means that two numbers can be added or multiplied in any order. For example,
3 C 4 D 4 C 3 and .7/."4/ D ."4/.7/.

4 T A A M
a C .b C c/ D .a C b/ C c and a.bc/ D .ab/c

This means that, for both addition and multiplication, numbers can be grouped in
any order. For example, 2 C .3 C 4/ D .2 C 3/ C 4 in both cases, the sum is . Simi-
larly, 2x C .x C y/ D .2x C x/ C y, and observe that the right side more obviously sim-
plifies to 3x C y than does the left side. Also, .6 $ 13 / $ 5 D 6. 13 $ 5/, and here the left side
obviously reduces to 10, so the right side does too.

5 T I
There are unique real numbers denoted 0 and 1 such that, for each real number a,
0 C a D a and 1a D a
6 T I
For each real number a, there is a unique real number denoted "a such that
a C ."a/ D 0
The number "a is called the negative of a.

For example, since 6 C ."6/ D 0, the negative of 6 is "6. The negative of a num-
ber is not necessarily a negative number. For example, the negative of "6 is 6, since
."6/ C .6/ D 0. That is, the negative of "6 is 6, so we can write "."6/ D 6.

For each real number a, except 0, there is a unique real number denoted a!1 such
that
a $ a!1 D 1
The number a!1 is called the reciprocal of a.

ero does not have a reciprocal because Thus, all numbers except 0 have a reciprocal. Recall that a!1 can be written 1a . For
there is no number that when multiplied example, the reciprocal of 3 is 13 , since 3. 13 / D 1. Hence, 13 is the reciprocal of 3. The
by 0 gives 1. This is a consequence of
0 $ a D 0 in 7. The istributive Properties. reciprocal of 13 is 3, since . 13 /.3/ D 1. he recipr ca f is n t de ned.

7 T

a.b C c/ D ab C ac and .b C c/a D ba C ca

0$aD0Da$0
 Section 0.2 o e Pro ert es of ea ers 5

For example, although 2.3 C 4/ D 2.7/ D 14, we can also write

2.3 C 4/ D 2.3/ C 2.4/ D 6 C D 14
Similarly,
.2 C 3/.4/ D 2.4/ C 3.4/ D C 12 D 20
and
x.z C 4/ D x.z/ C x.4/ D xz C 4x
The distributive property can be extended to the form
a.b C c C d/ D ab C ac C ad
In fact, it can be extended to sums involving any number of terms.
Subtraction is defined in terms of addition:
a"b means a C ."b/
where "b is the negative of b. Thus, 6 " means 6 C ." /.
In a similar way, we define division in terms of multiplication. If b ¤ 0, then
a#b means a.b!1 /
a 1
Usually, we write either or a=b for a # b. Since b!1 D ,
b ! " b
a !1 1
a D a.b / D a
means a times the reciprocal of b. b b
b
3 a
Thus, 5
means 3 times 15 , where 1
5
is the reciprocal of 5. Sometimes we refer to
as
b
the rati of a to b. We remar that since 0 does not have a reciprocal, division by is
not defined.
The following examples show some manipulations involving the preceding
properties.

E AM LE A R N

a x.y " 3z C 2 / D .y " 3z C 2 /x, by the commutative property of multiplication.

b y the associative property of multiplication, 3.4 $ 5/ D .3 $ 4/5. Thus, the result of
multiplying 3 by the product of 4 and 5 is the same as the result of multiplying the
product of 3 and 4 by 5. In either case, the result is 60.
c Show that a.b $ c/ ¤ .ab/ $ .ac/
S To show the negation of a general statement, it su ces to provide a
c unterexamp e. Here, ta ing a D 2 and b D 1 D c, we see that that a.b $ c/ D 2 while
.ab/ $ .ac/ D 4.
Now ork Problem 9 G

E AM LE A R N
p p
a Show that 2" 2 D " 2 C 2.
p p
S y the definition of subtraction,
p2 " 2pD 2 C ." 2/. However, by the
commutative property of paddition,p2 C ." 2/ D " 2 C 2. Hence, by the transitive
property of equality, 2 " 2 D " 2 C 2. Similarly, it is clear that, for any a and b,
we have
a " b D "b C a
b Show that . C x/"y D C .x " y/.
6 C e e of e ra

S eginning with the left side, we have

. C x/ " y D . C x/ C ."y/ definition of subtraction
D C .x C ."y// associative property
D C .x " y/ definition of subtraction
Hence, by the transitive property of equality,
. C x/ " y D C .x " y/
Similarly, for all a, b, and c, we have
.a C b/ " c D a C .b " c/
c Show that 3.4x C 2y C / D 12x C 6y C 24.
S y the distributive property,
3.4x C 2y C / D 3.4x/ C 3.2y/ C 3. /
ut by the associative property of multiplication,
3.4x/ D .3 $ 4/x D 12x and similarly 3.2y/ D 6y
Thus, 3.4x C 2y C / D 12x C 6y C 24
Now ork Problem 25 G
E AM LE A R N
! "
ab b
a Show that Da , for c ¤ 0.
c c
S The restriction is necessary. Neither side of the equation is defined if
c D 0. y the definition of division,
ab 1
D .ab/ $ for c ¤ 0
c c
ut by the associative property, ! "
1 1
.ab/ $ D a b $
c c
1 b
However, by the definition of division, b $ D . Thus,
c! " c
ab b
Da
c c
ab # a $
We can also show that D b.
c c
aCb a b
b Show that D C f r c ¤ 0.
c c c
S (Again the restriction is necessary but we won t always bother to say so.)
y the definition of division and the distributive property,
aCb 1 1 1
D .a C b/ D a $ C b $
c c c c
However,
1 1 a b
a$ Cb$ D C
c c c c
Hence,
aCb a b
D C
c c c

Now ork Problem 27 G

Finding the product of several numbers can be done by considering products of
numbers ta en ust two at a time. For example, to find the product of x, y, and z, we
 Section 0.2 o e Pro ert es of ea ers 7

could first multiply x by y and then multiply that product by z that is, we find (xy)z.
Alternatively, we could multiply x by the product of y and z that is, we find x(yz). The
associative property of multiplication guarantees that both results are identical, regard-
less of how the numbers are grouped. Thus, it is not ambiguous to write xyz. This con-
cept can be extended to more than three numbers and applies equally well to addition.
Not only should you be able to manipulate real numbers, you should also be aware
of, and familiar with, the terminology involved. It will help you read the boo , follow
your lectures, and most importantly allow you to frame your questions when you
have di culties.
The following list states important properties of real numbers that you should study
thoroughly. eing able to manipulate real numbers is essential to your success in math-
ematics. A numerical example follows each property. A den minat rs are assumed t
be di erent fr m zer (but for emphasis we have been explicit about these restrictions).

Property Example(s)
a " b D a C ."b/ 2 " 7 D 2 C ."7/ D "5
a " ."b/ D a C b 2 " ."7/ D 2 C 7 D
"a D ."1/.a/ "7 D ."1/.7/
a.b C c/ D ab C ac 6.7 C 2/ D 6 $ 7 C 6 $ 2 D 54
a.b " c/ D ab " ac 6.7 " 2/ D 6 $ 7 " 6 $ 2 D 30
".a C b/ D "a " b ".7 C 2/ D "7 " 2 D "
".a " b/ D "a C b ".2 " 7/ D "2 C 7 D 5
"."a/ D a "."2/ D 2
a.0/ D 0 2.0/ D 0
."a/.b/ D ".ab/ D a."b/ ."2/.7/ D ".2 $ 7/ D 2."7/ D "14
."a/."b/ D ab ."2/."7/ D 2 $ 7 D 14
a 7 "2
Da D 7; D "2
1 1 1
! " ! "
a 1 2 1
Da for b ¤ 0 D2
b b 7 7
a a "a 2 2 "2
D" D for b ¤ 0 D" D
"b b b "7 7 7
"a a "2 2
D for b ¤ 0 D
"b b "7 7
0 0
D0 for a ¤ 0 D0
a 7
a 2 "5
D 1 for a ¤ 0 D 1; D1
a 2 "5
! " ! "
b 7
a D b for a ¤ 0 2 D7
a 2
1 1
a$ D 1 for a ¤ 0 2$ D1
a 2
a c ac 2 4 2$4
$ D for b; d ¤ 0 $ D D
b d bd 3 5 3$5 15
! " ! "
ab a b 2$7 2 7
D bDa for c ¤ 0 D $7D2$
c c c 3 3 3
8 C e e of e ra

Property Example(s)
a a 1 1 a 2 2 1 1 2
D $ D $ for b; c ¤ 0 D $ D $
bc b c b c 3$7 3 7 3 7
! "! "
a a c ac 2 2 5 2$5
D $ D for b; c ¤ 0 D D
b b c bc 7 7 5 7$5

a a "a 2 2 "2
D D D D D D
b."c/ ."b/c bc 3."5/ ."3/.5/ 3.5/
"a a "2 2 2
D" for b; c ¤ 0 D" D"
."b/."c/ bc ."3/."5/ 3.5/ 15

a."b/ ."a/b ab 2."3/ ."2/.3/ 2.3/

D D D D D D
c c "c 5 5 "5
."a/."b/ ab ."2/."3/ 2.3/ 6
D" for c ¤ 0 D" D"
"c c "5 5 5

a b aCb 2 3 2C3 5
C D for c ¤ 0 C D D
c c c

a b a"b 2 3 2"3 "1

" D for c ¤ 0 " D D
c c c

a c ad C bc 4 2 4$3C5$2 22
C D for b; d ¤ 0 C D D
b d bd 5 3 5$3 15

a c ad " bc 4 2 4$3"5$2 2
" D for b; d ¤ 0 " D D
b d bd 5 3 5$3 15

a 2
b a c a d ad 3 2 7 2 5 2$5 10
c D b # d D b $ c D bc 7
D # D $ D
3 5 3 7 3$7
D
21
d 5
for b; c; d ¤ 0
a b c ac 2 3 5 2$5 10
D a# D a$ D for b; c ¤ 0 D2# D2$ D D
b c b b 3 5 3 3 3
c 5
a 2
b a a 1 a 3 2 2 1 2 2
D # cD $ D for b; c D 0 D #5D $ D D
c b b c bc 5 3 3 5 3$5 15

Property 23 is particularly important and could be called the fundamental

principle of fractions. It states that mu tip ying r di iding b th the numerat r and
den minat r f a fracti n by the same n nzer number resu ts in a fracti n that is
equa t the rigina fracti n. Thus,

7 7$ 56
D D D 56
1 1 1
$
 Section 0.2 o e Pro ert es of ea ers 9

y Properties 2 and 23, we have

2 4 2 $ 15 C 5 $ 4 50 2 $ 25 2
C D D D D
5 15 5 $ 15 75 3 $ 25 3
We can also do this problem by converting 25 and 15
4
into fractions that have the same
denominators and then using Property 26. The fractions 25 and 154
can be written with
a common denominator of 5 $ 15:
2 2 $ 15 4 4$5
D and D
5 5 $ 15 15 15 $ 5
However, 15 is the east such common denominator and is called the east c mm n
den minat r ( C ) of 25 and 15
4
. Thus,
2 4 2$3 4 6 4 6C4 10 2
C D C D C D D D
5 15 5$3 15 15 15 15 15 3
Similarly,
3 5 3$3 5$2
" D " C D 24
12 $ 3 12 $ 2
10 " 10
D " D
24 24 24
1
D"
24

R BLEMS
In Pr b ems determine the truth f each statement In Pr b ems sh that the statements are true by using
Every real number has a reciprocal. pr perties f the rea numbers

The reciprocal of 6:6 is 0:151515 : : :. 2x.y " 7/ D 2xy " 14x

x z
"1 zDx
The negative of 7 is . y y
7
1.x $ y/ D .1 $ x/.1 $ y/ .x C y/.2/ D 2x C 2y

"x C y D "y C x a.b C .c C d// D a..d C b/ C c/

.x C 2/.4/ D 4x C x..2y C 1/ C 3/ D 2xy C 4x

xC3 x #x$ 3x .1 C a/.b C c/ D b C c C ab C ac
D C3 3 D Show that .x " y C z/ D x " y C z .
5 5 4 4
2.x $ y/ D .2x/ $ .2y/ x.4y/ D 4xy int: b C c C d D .b C c/ C d.
Simp ify each f the f ing if p ssib e
In Pr b ems state hich pr perties f the rea numbers are
being used "2 C ."4/ "a C b 6 C ."4/
7"2 3 "5 " ."13/
2.x C y/ D 2x C 2y
2!1
.x C 5:2/ C 0:7y D x C .5:2 C 0:7y/ "."a/ C ."b/ ."2/. / 7." /
2.3y/ D .2 $ 3/y "1
."1:6/."0:5/ 1 ."1/
a 1 "1
D $a
b b a
5.b " a/ D .a " b/."5/ "."6 C x/ "7.x/ "3.a " b/
y C .x C y/ D .y C x/ C y "."6 C ."y// "3 # 3a " # ."27/
5x " y ."a/ # ."b/ 3 C .3!1 / 3."2.3/ C 6.2//
D 1=7.5x " y/
7 ."a/."b/."1/ ."12/."12/ X.1/
5.4 C 7/ D 5.7 C 4/ "71.x " 2/ 4.5 C x/ ".x " y/
! "
.2 C a/b D 2b C ba 1 X
0."x/
."1/."3 C 4/ D ."1/."3/ C ."1/.4/ 11 1
10 C e e of e ra

14x 2x 2 1 X 3 1 1 3 5
$ p "p " C "
21y "2 3 x 5 5 2 4 6 7
a 5a C .7 " 5a/ "aby "x
.3b/
c "ax 6 y2
x m z
a 1 2 5 1 1
$ $ C y xy
b c x y 2 3
x y 3 7 a c 7 0 0
C " C ; for X ¤ 0
3a a 10 15 b b 0 X 0

Objective E R
o re e os t e nte ra e onents The product x $ x $ x of 3 x s is abbreviated x3 . In general, for n a positive integer, xn is
t e ero e onent ne at e nte ra
e onents rat ona e onents the abbreviation for the product of n x s. The letter n in xn is called the exponent, and
r nc a roots ra ca s an t e x is called the base. ore specifically, if n is a positive integer, we have
roce re of rat ona n t e
eno nator
1 1
1 xn D x„ $ x $ ƒ‚
x $ : : : $ …x 2 x!n D D for x ¤ 0
xn x
„ $ x $ x
ƒ‚ x
$ ::: $…
n factors
Some authors say that 00 is not defined. 1 n factors
However, 00 D 1 is a consistent and 3 D xn for x ¤ 0 4 x0 D 1
often useful definition. x!n

E AM LE E

! "4 ! " ! " ! " ! "

1 1 1 1 1 1
a D D
2 2 2 2 2 16
1 1 1
b 3!5 D 5 D D
3 3$3$3$3$3 243
1 5
c D 3 D 243
3!5
d 20 D 1; ! 0 D 1; ."5/0 D 1
e x1 D x

Now ork Problem 5 G

If rn D x, where n is a positive integer, then r is an nth root of x. Second roots,
the case n D 2, are called square roots and third roots, the case n D 3, are called cube
roots. For example, 32 D , so 3 is a square root of . Since ."3/2 D ; "3 is also a
square root of . Similarly, "2 is a cube root of " , since ."2/3 D " , while 5 is a
fourth root of 625 since 54 D 625.
Some numbers do not have an nth root that is a real number. For example, since
the square of any real number is nonnegative: there is no real number that is a square
root of "4.
The principal nth root of x is the nth root of x that is positive if x is positive
p and is
negative if x is negative and n is odd. We denote the principal nth root of x by n x. Thus,
%
p positive if x is positive
n
x is
negative if x is negative and n is odd
r
p p 1 p
For example, 2
D 3; 3 " D "2, and 3
D 13 . We define n 0 D 0.
27
 Section 0.3 onents an a ca s 11
p
Although both 2 and "2 are square roots
p The symbol p a radical. With principal square roots we usually write
x is called
n

of 4, the principal p
p square root of 4 is 2, x instead of x. Thus,
2
D 3.
not "2. Hence, p 4 D 2. For positive x,
we often write ˙ x to indicate both
If x is positive, the expression xp=q , p
where p and q are integers with no common
p
square roots of x, and ˙ 4 D ˙2 factors and q is positive, is defined to be q xp . Hence,
p is a
convenient
p short way of writing 4D2 p4
p
3
p
and " 4 D "2 , but the only value of x3=4 D x3 I 2=3 D 2 D 3 64 D 4
p
4 is 2. r
p2 1 1
!1=2 !1
4 D 4 D D
4 2
Here are the basic laws of exponents and radicals:
Law Example(s)
xm $ xn D xmCn 23 $ 25 D 2 D 256 x2 $ x3 D x5
x0 D 1 20 D 1
1 1 1
x!n D n 2!3 D 3 D
x 2
1 1 1
D xn !3
D 23 D I !5 D x5
x!n 2 x
12
xm 1 2 x 1
D xm!n D n!m D 24 D 16I 12 D 4
xn x 2 x x
x m 24
D1 D1
xm 24
.xm /n D xmn .23 /5 D 215 I .x2 /3 D x6
.xy/n D xn yn .2 $ 4/3 D 23 $ 43 D $ 64 D 512
! "n ! "3
x xn 2 23
D n D 3 D
y y 3 3 27
! "!n ! "n ! "!2 ! "2
x y 3 4 16
D D D
y x 4 3
p 1=5
p5
x1=n D n x 3 D 3
1 1 1 1 1
x!1=n D 1=n D p 4!1=2 D 1=2 D p D
x n
x 4 4 2
p p p p p3
p
n
x n y D n xy
3
2D 31
p r p r
n
x x
3
0 3 0 p
3
p D n p D D
n y y
3
10 10
q p p p
m p p 3 4
n
x D mn x 2 D 12 2
p p p3
p
2=3 2 D . 3 /2 D 22 D 4
xm=n D xm D . n x/m
n D
p p
. m x/m D x . 7/ D 7

E AM LE E R

When computing
p xm=n , it is often easier a y aw 1,
to first find n x and then raise the result
x6 x D x6C D x14
to the mth power. Thus,
p
."27/4=3 D . 3 "27 /4 D ."3/4 D 1. a3 b2 a5 b D a3 a5 b2 b1 D a b3
x11 x!5 D x11!5 D x6
z2=5 z3=5 D z1 D z
xx1=2 D x1 x1=2 D x3=2
12 C e e of e ra

b y aw 16,
! "3=2 r !3 ! "3
1 1 1 1
D D D
4 4 2
! "4=3 r !4 p !4
3
3 " "
c " D D p aws 16 and 14
27 27 3
27
! "4
"2
D
3
."2/4 16
D 4
D aw
3 1
d .64a3 /2=3 D 642=3 .a3 /2=3 aw
p
3
D . 64/2 a2 aws 16 and 7
D .4/2 a2 D 16a2

Now ork Problem 39 G

Rationalizing the numerator is a similar Rati na izing the den minat r of a fraction is a procedure in which a fraction hav-
procedure. ing a radical in its denominator is expressed as an equal fraction without a radical in
its denominator. We use the fundamental principle of fractions, as Example 3 shows.

E AM LE R
p
2 2 2 $ 51=2 2 $ 51=2 2 5
a p D 1=2 D 1=2 1=2 D D
5 5 5 $5 51 5
2 2 2 2 $ 35=6 x1=6
b p Dp p D 1=6 5=6 D 1=6 5=6 5=6 1=6 for x ¤ 0
6
3x5 6 6
3 $ x5 3 x 3 x $3 x
p
6
2.35 x/1=6 2 35 x
D D
3x 3x

Now ork Problem 63 G

The following examples illustrate various applications of the laws of exponents
and radicals. All denominators are understood to be nonzero.

E AM LE E
x!2 y3
a Eliminate negative exponents in for x ¤ 0, z ¤ 0.
z!2
x!2 y3 1 1 y3 z2
S !2
D x!2 $ y3 $ !2 D 2 $ y3 $ z2 D 2
z z x x
y comparing our answer with the original expression, we conclude that we can
bring a factor of the numerator down to the denominator, and vice versa, by chang-
ing the sign of the exponent.
x2 y7
b Simplify 3 5 for x ¤ 0, y ¤ 0.
xy
x2 y7 y7!5 y2
S D D
x3 y5 x3!2 x
c Simplify .x5 y /5 .
S .x5 y /5 D .x5 /5 .y /5 D x25 y40
 Section 0.3 onents an a ca s 13

d Simplify .x5= y4=3 /1 .

S .x5= y4=3 /1 D .x5= /1 .y4=3 /1 D x10 y24
!5
x1=5 y6=5
e Simplify for z ¤ 0.
z2=5
!5
S x1=5 y6=5 .x1=5 y6=5 /5 xy6
D D
z2=5 .z2=5 /5 z2

x3 x6
f Simplify # for x ¤ 0, y ¤ 0.
y2 y5
x3 x6 x3 y5 y3
S # D $ D
y2 y5 y2 x6 x3

Now ork Problem 51 G

E AM LE E

a For x ¤ 0 and y ¤ 0, eliminate negative exponents in x!1 C y!1 and simplify.

1 1 yCx
S x!1 C y!1 D C D
x y xy
b Simplify x3=2 " x1=2 by using the distributive law.
S x3=2 " x1=2 D x1=2 .x " 1/
c For x ¤ 0, eliminate negative exponents in 7x!2 C .7x/!2 .
7 1 7 1 344
S 7x!2 C .7x/!2 D 2
C 2
D 2C 2
D
x .7x/ x 4 x 4 x2
d For x ¤ 0 and y ¤ 0, eliminate negative exponents in .x!1 " y!1 /!2 .
! " ! "
!1 !1 !2 1 1 !2 y " x !2
S .x " y / D " D
x y xy
! "2
xy x2 y2
D D
y"x .y " x/2
e Apply the distributive law to x2=5 .y1=2 C 2x6=5 /.
S x2=5 .y1=2 C 2x6=5 / D x2=5 y1=2 C 2x =5

Now ork Problem 41 G

E AM LE R
p
4
a Simplify 4 .
p
4
p
4
p
4
p
4
p
4
S 4 D 16 $ 3 D 16 3 D 2 3
p
b Rewrite 2 C 5x without using a radical sign.
p
S 2 C 5x D .2 C 5x/1=2
p5
2
c Rationalize the denominator of p 3
and simplify.
6
p p
15
5
2 21=5 $ 62=3 23=15 610=15 .23 610 /1=15 23 610
S p D D D D
3
6 61=3 $ 62=3 6 6 6
14 C e e of e ra
p
20
d Simplify p .
5
p r
20 20 p
S p D D 4D2
5 5
Now ork Problem 71 G
E AM LE R
p
a Simplify 3
x6 y4 .
p
3
p p p p
S x6 y4 D 3
.x2 /3 y3 y D 3
.x2 /3 $ 3 y3 $ 3 y
p
D x2 y 3 y
r
2
b Simplify .
7
r r r p p
2 2$7 14 14 14
S D D D p D
7 7$7 72 72 7
p p p
c Simplify 250 " 50 C 15 2.
p p p p p p
S 250 " 50 C 15 2 D 25 $ 10 " 25 $ 2 C 15 2
p p p
D 5 10 " 5 2 C 15 2
p p
D 5 10 C 10 2
p
d If x is any real number, simplify x2 .
p %
x if x % 0
S 2
x D
"x if x < 0
p p
Thus, 22 D 2 and ."3/2 D "."3/ D 3.
Now ork Problem 75 G

R BLEMS
In Pr b ems simp ify and express a ans ers in terms f ! "4=5 ! "2=5
1 243
p siti e exp nents "
32 1024
.23 /.22 / x6 x 175 $ 172
In Pr b ems simp ify the expressi ns
x3 x5 p p p
z3 zz2 .x12 /4 50 3
54
3
2x3
y y5 r
! 14 "2 p p x
.a3 /7 13 4x 4 u 4
.2x2 y3 /3 16
.b4 /5 13 r
p p p 3
! 2 3 "2 ! "6 2 " 5 27 C 3 12
s x 2a4 13
y2 x5 7b5 . z4 /1=2 .72 x6 /3=2
! 3 "2=3 ! "
.y3 /4 .x2 /3 .x3 /2 27t 256 !3=4
.y2 /3 y2 .x3 /4 x12

In Pr b ems e a uate the expressi ns In Pr b ems rite the expressi ns in terms f p siti e
p p4
p
7 exp nents n y A id a radica s in the na f rm F r examp e
25 1 "12
r r p x1=2
p5 4 1 3 y!1 x D
0:00243 " y
16 27
.4 /1=2 .64/1=3 13=4 a5 b!3 p
!5=2 !2=5 !1=2
5
x2 y3 z!10 3a!1 b!2 c!3
. / .32/ .0:0 / c2
 Section 0.4 erat ons t e ra c ress ons 15

x C y!1 .3t/!2 .3 " z/!4 In Pr b ems simp ify Express a ans ers in terms f
p p p p siti e exp nents Rati na ize the den minat r here necessary
.X5 !7 /!4
5
5x2 x" y t a id fracti na exp nents in the den minat r
u!2 !6 3 p p 3
a!3 b!2 a5 b!4
4
!5
x2 4 xy!2 z3 2x2 y!3 x4
u5=2 1=2
In Pr b ems re rite the exp nentia f rms using radica s p
243 ...3a3 /2 /!5 /!2
.a " b C c/3=5 .ab2 c3 /3=4 p
3
x!4=5 2x1=2 " .2y/1=2 p
30 s5
3 !3=5 " .3 /!3=5 ..y!2 /1=4 /1=5 !4 2=3 !2 3 p
.3 x y / 3
s2
p p p
In Pr b ems rati na ize the den minat rs 3 2
x yz3 3 xy2 4
. 3/
6 3 4 p
p p4
p 2
3 .32/ !2=5 3
. u3 2 /2=3
5 2x
y 1 2 3
p p p .2x!1 y2 /2 p p
3 y 4 x
2y 5
3b 3 3 y2
p p p p p 2 3p 2 p
12 1 5
2 x x y xy 75k4
p p p
3 2 4
a2 b .a3 b!4 c5 /6 p
p
3
7.4 /
3
3 .a!2 c!3 /!4
p ! 3 "2
.x2 /3 x p
2 # ."6/."6/
x4 .x3 /2

Objective A E
oa s tract t an e If numbers, represented by symbols, are combined by any or all of the operations of
a e ra c e ress ons o e ne a
o no a to se s ec a ro cts addition, subtraction, multiplication, division, exponentiation, and extraction of roots,
an to se on s on to e then the resulting expression is called an algebraic expression.
o no a s
E AM LE A E
r
3
3 3x " 5x " 2
a is an algebraic expression in the variable x.
10 " x
p 5
b 10 " 3 y C is an algebraic expression in the variable y.
7 C y2
.x C y/3 " xy
c C 2 is an algebraic expression in the variables x and y.
y
Now ork Problem 1 G
The algebraic expression 5ax3 "2bxC3 consists of three terms:C5ax3 ; "2bx, and
C3. Some of the factors of the first term, 5ax3 , are 5; a; x; x2 ; x3 ; 5ax, and ax2 . Also,
5a is the coe cient of x3 , and 5 is the numerica c e cient of ax3 . If a and b represent
fixed numbers throughout a discussion, then a and b are called constants.
Algebraic expressions with exactly one term are called monomials. Those having
exactly two terms are binomials, and those with exactly three terms are trinomials.
Algebraic expressions with more than one term are called multinomials. Thus, the
p
multinomial 2x " 5 is a binomial the multinomial 3 y C 2y " 4y2 is a trinomial.
The words p yn mia and mu tin mia A polynomial in x is an algebraic expression of the form
should not be used interchangeably. A
polynomial is a special indpof cn xn C cn!1 xn!1 C $ $ $ C c1 x C c0
multinomial. For example, x C 2 is
a multinomial but not a polynomial. n where n is a nonnegative integer and the coe cients c0 ; c1 ; : : : ; cn are constants with
the other hand, x C 2 is a polynomial and cn ¤ 0. Here, the three dots indicate all other terms that are understood to be included
hence a multinomial. in the sum. We call n the degree of the polynomial. So, 4x3 "5x2 Cx"2 is a polynomial
in x of degree 3, and y5 " 2 is a polynomial in y of degree 5. A nonzero constant is a
polynomial of degree zero thus, 5 is a polynomial of degree zero. The constant 0 is
considered to be a polynomial however, no degree is assigned to it.
16 C e e of e ra

In the following examples, we illustrate operations with algebraic expressions.

E AM LE A A E

Simplify .3x2 y " 2x C 1/ C .4x2 y C 6x " 3/.

S We first remove the parentheses. Next, using the commutative property of
addition, we gather all li e terms together. i e terms are terms that differ only by their
numerical coe cients. In this example, 3x2 y and 4x2 y are li e terms, as are the pairs
"2x and 6x, and 1 and "3. Thus,
.3x2 y " 2x C 1/ C .4x2 y C 6x " 3/ D 3x2 y " 2x C 1 C 4x2 y C 6x " 3
D 3x2 y C 4x2 y " 2x C 6x C 1 " 3
y the distributive property,
3x2 y C 4x2 y D .3 C 4/x2 y D 7x2 y
and
"2x C 6x D ."2 C 6/x D 4x
Hence, .3x y " 2x C 1/ C .4x2 y C 6x " 3/ D 7x2 y C 4x " 2
2

Now ork Problem 3 G

E AM LE S A E

Simplify .3x2 y " 2x C 1/ " .4x2 y C 6x " 3/.

S Here we apply the definition of subtraction and the distributive property:
.3x2 y " 2x C 1/ " .4x2 y C 6x " 3/
D .3x2 y " 2x C 1/ C ."1/.4x2 y C 6x " 3/
D .3x2 y " 2x C 1/ C ."4x2 y " 6x C 3/
D 3x2 y " 2x C 1 " 4x2 y " 6x C 3
D 3x2 y " 4x2 y " 2x " 6x C 1 C 3
D .3 " 4/x2 y C ."2 " 6/x C 1 C 3
D "x2 y " x C 4
Now ork Problem 13 G
E AM LE R G S

Simplify 3f2xŒ2x C 3" C 5Œ4x2 " .3 " 4x/"g.

S We first eliminate the innermost grouping symbols (the parentheses). Then
we repeat the process until all grouping symbols are removed combining similar
terms whenever possible. We have
3f2xŒ2x C 3" C 5Œ4x2 " .3 " 4x/"g D 3f2xŒ2x C 3" C 5Œ4x2 " 3 C 4x"g
D 3f4x2 C 6x C 20x2 " 15 C 20xg
D 3f24x2 C 26x " 15g
D 72x2 C 7 x " 45
bserve that properly paired parentheses are the only grouping symbols needed
3f2xŒ2x C 3" C 5Œ4x2 " .3 " 4x/"g D 3.2x.2x C 3/ C 5.4x2 " .3 " 4x///
but the optional use of brac ets and braces sometimes adds clarity.
Now ork Problem 15 G
 Section 0.4 erat ons t e ra c ress ons 17

The distributive property is the ey tool in multiplying expressions. For example,

to multiply ax C c by bx C d we can consider ax C c to be a single number and then
use the distributive property:
.ax C c/.bx C d/ D .ax C c/bx C .ax C c/d
Using the distributive property again, we have
.ax C c/bx C .ax C c/d D abx2 C cbx C adx C cd
D abx2 C .ad C cb/x C cd
Thus, .ax C c/.bx C d/ D abx2 C .ad C cb/x C cd. In particular, if a D 2; b D 1,
c D 3, and d D "2, then
.2x C 3/.x " 2/ D 2.1/x2 C Œ2."2/ C 3.1/"x C 3."2/
D 2x2 " x " 6
We now give a list of special products that can be obtained from the distributive
property and are useful in multiplying algebraic expressions.

S
x.y C z/ D xy C xz distributive property
.x C a/.x C b/ D x2 C .a C b/x C ab
.ax C c/.bx C d/ D abx2 C .ad C cb/x C cd
.x C a/2 D x2 C 2ax C a2 square of a sum
.x " a/2 D x2 " 2ax C a2 square of a difference
2 2
.x C a/.x " a/ D x " a product of sum and difference
3 3 2 2 3
.x C a/ D x C 3ax C 3a x C a cube of a sum
3 3 2 2 3
.x " a/ D x " 3ax C 3a x " a cube of a difference

E AM LE S

a y Rule 2,
.x C 2/.x " 5/ D .x C 2/.x C ."5//
D x2 C .2 " 5/x C 2."5/
D x2 " 3x " 10
b y Rule 3,
.3z C 5/.7z C 4/ D 3 $ 7z2 C .3 $ 4 C 5 $ 7/z C 5 $ 4
D 21z2 C 47z C 20
c y Rule 5,
.x " 4/2 D x2 " 2.4/x C 42
D x2 " x C 16
d y Rule 6,
p p p
. y2 C 1 C 3/. y2 C 1 " 3/ D . y2 C 1/2 " 32
D .y2 C 1/ "
D y2 "
18 C e e of e ra

e y Rule 7,
.3x C 2/3 D .3x/3 C 3.2/.3x/2 C 3.2/2 .3x/ C .2/3
D 27x3 C 54x2 C 36x C

Now ork Problem 19 G

E AM LE M M

Find the product .2t " 3/.5t2 C 3t " 1/.

S We treat 2t " 3 as a single number and apply the distributive property twice:
.2t " 3/.5t2 C 3t " 1/ D .2t " 3/5t2 C .2t " 3/3t " .2t " 3/1
D 10t3 " 15t2 C 6t2 " t " 2t C 3
D 10t3 " t2 " 11t C 3
Now ork Problem 35 G
aCb a b
In Example 3(b) of Section 0.2, we showed that D C . Similarly,
a"b a b c c c
D " . Using these results, we can divide a multinomial by a monomial
c c c
by dividing each term in the multinomial by the monomial.

E AM LE M M

x3 C 3x x3 3x
a D C D x2 C 3
x x x
4z3 " z2 C 3z " 6 4z3 z2 3z 6
b D " C "
2z 2z 2z 2z 2z
3 3
D 2z2 " 4z C "
2 z
Now ork Problem 47 G
L
To divide a polynomial by a polynomial, we use so-called long division when the
degree of the divisor is less than or equal to the degree of the dividend, as the next
example shows.

E AM LE L

ivide 2x3 " 14x " 5 by x " 3.

S Here 2x3 " 14x " 5 is the dividend and x " 3 is the divisor. To avoid errors,
it is best to write the dividend as 2x3 C 0x2 " 14x " 5. Note that the powers of x are in
decreasing order. We have
2
& 2x C 6x C 4 quotient
3 2
divisor ! x " 3 2x C 0x " 14x " 5 dividend
2x3 " 6x2
6x2 " 14x
6x2 " 1 x
4x " 5
4x " 12
7 remainder
 Section 0.4 erat ons t e ra c ress ons 19

Note that we divided x (the first term of the divisor) into 2x3 and got 2x2 . Then we
multiplied 2x2 by x " 3, getting 2x3 " 6x2 . After subtracting 2x3 " 6x2 from 2x3 C 0x2 ,
we obtained 6x2 and then brought down the term "14x. This process is continued
until we arrive at 7, the remainder. We always stop when the remainder is 0 or is a
polynomial whose degree is less than the degree of the divisor. ur answer can be
written as
7
2x2 C 6x C 4 C
x"3
That is, the answer to the question
dividend
D‹
divisor
has the form
remainder
quotient C
divisor
A way of chec ing a division is to verify that

.quotient/.divisor/ C remainder D dividend

y using this equation, you should be able to verify the result of the example.
Now ork Problem 51 G

R BLEMS
Perf rm the indicated perati ns and simp ify
p p p
. x " 4y C 2/ C .3x C 2y " 5/ . 5x " 2/2 . y " 3/. y C 3/
.4a2 " 2ab C 3/ C .5c " 3ab C 7/ .2s " 1/.2s C 1/ .a2 C 2b/.a2 " 2b/
. t2 " 6s2 / C .4s2 " 2t2 C 6/ .x2 " 3/.x C 4/ .u " 1/.u2 C 3u " 2/
p p p p
. x C 2 x/ C .3 x C 4 x/ .x2 " 4/.3x2 C 2x " 1/ .3y " 2/.4y3 C 2y2 " 3y/
p p p p
. a C 2 3b/ " . c " 3 3b/ t.3.t C 2/.t " 4/ C 5.3t.t " 7///
.3a C 7b " / " .5a C b C 21/ ..2z C 1/.2z " 1//.4z2 C 1/
p p
.7x2 C 5xy C 2/ " .2z " 2xy C 2/ .s " t C 4/.3s C 2t " 1/
p p p p .x2 C x C 1/2 .2a C 3/3
. x C 2 x/ " . x C 3 x/
p p p p .2a " 3/3 .2x " 3/3
. 2 2x C 3 3y/ " . 2 2x C 4 4z/
z2 " 1 z
4.2z " / " 3. " 2z/ .3a C b/3
z
3.3x C 3y " 7/ " 3. x " 2y C 2/
2x3 " 7x C 4 6u5 C u3 " 1
.4s " 5t/ C ."2s " 5t/ C .s C / x 3u2
5.x2 " y2 / C x.y " 3x/ " 4y.2x C 7y/ .3y " 4/ " . y C 5/
.7 C 3.x " 3/ " .4 " 5x// 3y
2.3.3.x2 C 2/ " 2.x2 " 5/// 2
.x C 7x " 5/ # .x C 5/
4.3.t C 5/ " t.1 " .t C 1/// .x2 " 5x C 4/ # .x " 4/
2 2
"2.3u .2u C 2/ " 2.u " .5 " 2u/// .3x3 " 2x2 C x " 3/ # .x C 2/
"."3Œ2a C 2b " 2" C 5.2a C 3b/ " a.2.b C 5/// .x4 C 3x2 C 2/ # .x C 1/
.2x C 5/.3x " 2/ .u C 2/.u C 5/ x3 # .x C 2/
. C 2/. " 5/ .x " 4/.x C 7/ . x2 C 6x C 7/ # .2x C 1/
.2x C 3/.5x C 2/ 2 2
.t " 5t/.3t " 7t/ .3x2 " 4x C 3/ # .3x C 2/
.X C 2 /2 .2x " 1/2 .z3 C z2 C z/ # .z2 " z C 1/
p p
.7 " X/2 . x " 1/.2 x C 5/
20 C e e of e ra

Objective F
o state t e as c r es for factor n If two or more expressions are multiplied together, the expressions are called fact rs of
an a t e to factor e ress ons
the product. Thus, if c D ab, then a and b are both factors of the product c. The process
by which an expression is written as a product of its factors is called fact ring.
isted next are rules for factoring expressions, most of which arise from the special
products discussed in Section 0.4. The right side of each identity is the factored form
of the left side.

R F
xy C xz D x.y C z/ common factor
x2 C .a C b/x C ab D .x C a/.x C b/
abx2 C .ad C cb/x C cd D .ax C c/.bx C d/
x2 C 2ax C a2 D .x C a/2 perfect-square trinomial
x2 " 2ax C a2 D .x " a/2 perfect-square trinomial
x2 " a2 D .x C a/.x " a/ difference of two squares
x3 C a3 D .x C a/.x2 " ax C a2 / sum of two cubes
x3 " a3 D .x " a/.x2 C ax C a2 / difference of two cubes

When factoring a polynomial, we usually choose factors that themselves are

2
polynomials.
p p For example, x " 4 D .x C 2/.x " 2/. We will not write x " 4 as
. x C 2/. x " 2/ unless it allows us to simplify other calculations.
Always factor as completely as you can. For example,

2x2 " D 2.x2 " 4/ D 2.x C 2/.x " 2/

E AM LE C F

a Factor 3k2 x2 C k3 x completely.

S Since 3k2 x2 D .3k2 x/.x/ and k3 x D .3k2 x/.3k/, each term of the orig-
inal expression contains the common factor 3k2 x. Thus, by Rule 1,

3k2 x2 C k3 x D 3k2 x.x C 3k/

Note that although 3k2 x2 C k3 x D 3.k2 x2 C3k3 x/, we do not say that the expression
is completely factored, since k2 x2 C 3k3 x can still be factored.
b Factor a5 x2 y3 " 6a2 b3 yz " 2a4 b4 xy2 z2 completely.
S a5 x2 y3 " 6a2 b3 yz " 2a4 b4 xy2 z2 D 2a2 y.4a3 x2 y2 " 3b3 z " a2 b4 xyz2 /

Now ork Problem 5 G

E AM LE F T

a Factor 3x2 C 6x C 3 completely.

S First we remove a common factor. Then we factor the resulting expres-
sion completely. Thus, we have

3x2 C 6x C 3 D 3.x2 C 2x C 1/
D 3.x C 1/2 Rule 4

b Factor x2 " x " 6 completely.

 Section 0.5 actor n 21

S If this trinomial factors into the form .x C a/.x C b/, which is a

product of two binomials, then we must determine the values of a and b. Since
.x C a/.x C b/ D x2 C .a C b/x C ab, it follows that
x2 C ."1/x C ."6/ D x2 C .a C b/x C ab
It is not always possible to factor a y equating corresponding coe cients, we want
trinomial, using real numbers, even if
the trinomial has integer coe cients. a C b D "1 and ab D "6
We will comment further on this point in If a D "3 and b D 2, then both conditions are met and hence
Section 0. .
x2 " x " 6 D .x " 3/.x C 2/
As a chec , it is wise to multiply the right side to see if it agrees with the left side.
c Factor x2 " 7x C 12 completely.
S x2 " 7x C 12 D .x " 3/.x " 4/

Now ork Problem 9 G

E AM LE F

The following is an assortment of expressions that are completely factored. The num-
bers in parentheses refer to the rules used.
a x2 C x C 16 D .x C 4/2 .4/
b x2 C x C 2 D .3x C 1/.3x C 2/ .3/
c 6y3 C 3y2 " 1 y D 3y.2y2 C y " 6/ .1/
D 3y.2y " 3/.y C 2/ .3/
d x " 6x C D .x " 3/2
2
.5/
e z1=4 C z5=4 D z1=4 .1 C z/ .1/
f x4 " 1 D .x2 C 1/.x2 " 1/ .6/
D .x2 C 1/.x C 1/.x " 1/ .6/
g x2=3 " 5x1=3 C 4 D .x1=3 " 1/.x1=3 " 4/ .2/
h ax2 " ay2 C bx2 " by2 D a.x2 " y2 / C b.x2 " y2 / .1/; .1/
D .x2 " y2 /.a C b/ .1/
D .x C y/.x " y/.a C b/ .6/
i " x D .2/ " .x/ D .2 " x/.4 C 2x C x2 /
3 3 3
. /
x6 " y6 D .x3 /2 " .y3 /2 D .x3 C y3 /.x3 " y3 / .6/
D .x C y/.x2 " xy C y2 /.x " y/.x2 C xy C y2 / .7/; . /

Now ork Problem 35 G

Note in Example 3(f) that x2 " 1 is factorable, but x2 C 1 is not. In Example 3(h),
note that the common factor of x2 " y2 was not immediately evident.
Students often wonder why factoring is important. Why does the prof seem to thin
that the right side of x2 " 7x C 12 D .x " 3/.x " 4/ is better than the left side ften,
the reason is that if a pr duct f numbers is 0 then at east ne f the numbers is 0. In
symbols

If ab D 0 then a D 0 or bD0

This is a useful principle for solving equations. For example, nowing x2 " 7x C
12 D .x " 3/.x " 4/ it follows that if x2 " 7x C 12 D 0 then .x " 3/.x " 4/ D 0 and
from the principle above, x " 3 D 0 or x " 4 D 0. Now we see immediately that either
x D 3 or x D 4. We should also remar that in the displayed principle the word or is
If ab D 0, at east one of a and b is 0. use inclusively. In other words, if ab D 0 it may be that both a D 0 and b D 0.
22 C e e of e ra

R BLEMS
Fact r the f ing expressi ns c mp ete y
5bx C 5b 6y2 " 4y 2x3 C 2x2 " 12x x2 y2 " 4xy C 4
.4x C 2/2 x2 .2x2 " 4x3 /2
10xy C 5xz 3x2 y " x3 y3
x3 y2 " 16x2 y C 64x .5x2 C 2x/ C .10x C 4/
3a3 bcd2 " 4ab3 c2 d2 C 2a3 bc4 d3 .x3 " 4x/ C . " 2x2 / .x2 " 1/ C .x2 " x " 2/
5r2 st2 C 10r3 s2 t3 " 15r2 t2 4ax2 " ay2 C 12bx2 " 3by2
z2 " 4 x2 " x " 6 t3 u " 3tu C t2 2
"3 2

2
p C 4p C 3 2
t " t " 12 b3 C 64 x3 " 1
25y2 " 4 x2 C 2x " 24 x6 " 1 64 C 27t3
a2 C 12a C 35 4t2 " s2 .x C 4/3 .x " 2/ C .x C 4/2 .x " 2/2

y2 C y C 15 t2 " 1 t C 72 .a C 5/3 .a C 1/2 C .a C 5/2 .a C 1/3

P.1 C r/ C P.1 C r/r
5x2 C 25x C 30 3t2 C 12t " 15
.X " 3I/.3X C 5I/ " .3X C 5I/.X C 2I/
3x2 " 3 6x2 C 31x C 35
16u2 " 1 2 2 256y4 " z4
5x2 C 16x C 3 4x2 " x " 3
y "1 t4 " 4
12s3 C 10s2 " s z2 C 30z C 25
4 2
X C 4X " 5 4x4 " 20x2 C 25
a11=3 b " 4a2=3 b3 4x6=5 " 1
a4 b " a2 b C 16b 4x3 " 6x2 " 4x

Objective F
os f a s tract t an Students should ta e particular care in studying fracti ns. In everyday life, numerical
e a e ra c fract ons o rat ona e
t e eno nator of a fract on fractions often disappear from view with the help of calculators. However, manipula-
tion of fractions of algebraic expressions is essential in calculus, and here most calcu-
lators are of no help.

S F
y using the fundamental principle of fractions (Section 0.2), we may be able to sim-
plify algebraic expressions that are fractions. That principle allows us to multiply or
divide both the numerator and the denominator of a fraction by the same nonzero
quantity. The resulting fraction will be equal to the original one. The fractions that
we consider are assumed to have nonzero denominators. Thus, all the factors of the
denominators in our examples are assumed to be nonzero. This will often mean that
certain values are excluded for the variables that occur in the denominators.

E AM LE S F

x2 " x " 6
a Simplify .
x2 " 7x C 12
S First, we completely factor both the numerator and the denominator:
x2 " x " 6 .x " 3/.x C 2/
2
D
x " 7x C 12 .x " 3/.x " 4/
ividing both numerator and denominator by the common factor x " 3, we have
.x " 3/.x C 2/ 1.x C 2/ xC2
D D for x ¤ 3
.x " 3/.x " 4/ 1.x " 4/ x"4
Usually, we ust write
x2 " x " 6 .x " 3/.x C 2/ xC2
2
D D for x ¤ 3
x " 7x C 12 .x " 3/.x " 4/ x"4
 Section 0.6 ract ons 23

The process of eliminating the common factor x " 3 is commonly referred to as

cancellation. We issued a blan et statement before this example that all fractions
are assumed to have nonzero denominators and that this requires excluding certain
values for the variables. bserve that, nevertheless, we explicitly wrote for x ¤ 3 .
xC2
This is because the expression to the right of the equal sign, , is de ned f r
x"4
x D 3. Its value is "5 but we want to ma e it clear that the expressi n t the eft f
the equa sign is n t de ned f r x D 3.
2x2 C 6x "
b Simplify .
" 4x " 4x2
S 2x2 C 6x " 2.x2 C 3x " 4/ 2.x " 1/.x C 4/
2
D 2
D
" 4x " 4x 4.2 " x " x / 4.1 " x/.2 C x/
2.x " 1/.x C 4/
D
The simplified expression is defined for 2.2/Œ."1/.x " 1/".2 C x/
x D 1, but since the original expression
is not defined for x D 1, we explicitly
xC4
D for x ¤ 1
exclude this value. "2.2 C x/

Now ork Problem 3 G

M F
a c
The rule for multiplying by is
b d
a c ac
$ D
b d bd

E AM LE M F

x xC3 x.x C 3/
a $ D
xC2 x"5 .x C 2/.x " 5/
Note that we explicitly excluded the x2 " 4x C 4 6x2 " 6 Œ.x " 2/2 "Œ6.x C 1/.x " 1/"
values that ma e the cancelled b $ D
factors 0. While the final expression is
x2 C 2x " 3 x2 C 2x " Œ.x C 3/.x " 1/"Œ.x C 4/.x " 2/"
defined for these values, the original
6.x " 2/.x C 1/
expression is not. D for x ¤ 1; 2
.x C 3/.x C 4/

Now ork Problem 9 G

a c
To divide by , where b ¤ 0, d ¤ 0, and c ¤ 0, we have
b d
a
In short, to divide by a fraction we invert a c b a d
the divisor and multiply. # D c D $
b d b c
d
E AM LE F

x xC3 x x"5 x.x " 5/

a # D $ D
xC2 x"5 xC2 xC3 .x C 2/.x C 3/
x"5 x"5
x"3 x"3 x"5 1 x"5
b D D $ D
2x 2x x " 3 2x 2x.x " 3/
1
24 C e e of e ra

4x
x2 "1 4x x"1 4x.x " 1/
c D 2 $ 2 D
2
2x C x x " 1 2x C x Œ.x C 1/.x " 1/"Œ2x.x C 4/"
x"1
2
Why did we write for x ¤ 0; 1 D for x ¤ 0; 1
.x C 1/.x C 4/

Now ork Problem 11 G

R
Sometimes
p thepdenominator
p of a fraction has two terms and involves square roots, such
as 2 " 3 or 5 C 2. The denominator may then be rationalized by multiplying by
an expression that ma es the denominator a difference of two squares. For example,
p p
4 4 5" 2
p p D p p $p p
5C 2 5C 2 5" 2
p p p p
4. 5 " 2/ 4. 5 " 2/
D p p D
. 5/2 " . 2/2 5"2
p p
4. 5 " 2/
Rationalizing the numerat r is a similar D
procedure. 3

E AM LE R
pp
x x 2C6
x. 2 C 6/
a p Dp $p D p
2"6 2"6 2C6 . 2/2 " 62
p p
x. 2 C 6/ x. 2 C 6/
D D"
2 " 36 34
p p p p p p
5" 2 5" 2 5" 2
b p p Dp p $p p
5C 2 5C 2 5" 2
p p p p p
. 5 " 2/2 5"2 5 2C2 7 " 2 10
D D D
5"2 3 3

Now ork Problem 53 G

A S F
a b aCb
In Example 3(b) of Section 0.2, it was shown that C D . That is, if we add
c c c
two fractions having a common denominator, then the result is a fraction whose denom-
inator is the common denominator. The numerator is the sum of the numerators of the
a b a"b
original fractions. Similarly, " D .
c c c

E AM LE A S F

p2 " 5 3p C 2 .p2 " 5/ C .3p C 2/

a C D
p"2 p"2 p"2
p2 C 3p " 3
D
p"2
 Section 0.6 ract ons 25

x2 " 5x C 4 x2 C 2x .x " 1/.x " 4/ x.x C 2/

b " D "
x2 C 2x " 3 x2 C 5x C 6 .x " 1/.x C 3/ .x C 2/.x C 3/
x"4 x .x " 4/ " x 4
D " D D" for x ¤ "2; 1
xC3 xC3 xC3 xC3
x2 C x " 5 x2 " 2 "4x C x2 C x " 5 x2 " 2 "4
c " C 2 D " C
x"7 x"7 x " x C 14 x"7 x"7 x"7
.x2 C x " 5/ " .x2 " 2/ C ."4/
D
x"7
x"7
Why did we write for x ¤ 2; 7 D D 1 for x ¤ 2; 7
x"7
Now ork Problem 29 G
To add (or subtract) two fractions with di erent denominators, use the fundamental
principle of fractions to rewrite the fractions as fractions that have the same denomi-
nator. Then proceed with the addition (or subtraction) by the method ust described.
For example, to find
2 3
C
x3 .x " 3/ x.x " 3/2
we can convert the first fraction to an equal fraction by multiplying the numerator and
denominator by x " 3:
2.x " 3/
x3 .x " 3/2
and we can convert the second fraction by multiplying the numerator and denominator
by x2 :
3x2
x3 .x " 3/2
These fractions have the same denominator. Hence,
2 3 2.x " 3/ 3x2
C D C
x3 .x " 3/ x.x " 3/2 x3 .x " 3/2 x3 .x " 3/2

3x2 C 2x " 6
D
x3 .x " 3/2
We could have converted the original fractions into equal fractions with any common
denominator. However, we chose to convert them into fractions with the denominator
x3 .x"3/2 . This denominator is the least common denominator LCD of the fractions
2=.x3 .x " 3// and 3=Œx.x " 3/2 ".
In general, to find the C of two or more fractions, first factor each denomina-
tor completely. he C is the pr duct f each f the distinct fact rs appearing in
the den minat rs each raised t the highest p er t hich it ccurs in any sing e
den minat r.

E AM LE A S F
t 4
a Subtract: " .
3t C 2 t " 1
S The C is .3t C 2/.t " 1/. Thus, we have
t 4 t.t " 1/ 4.3t C 2/
" D "
.3t C 2/ t " 1 .3t C 2/.t " 1/ .3t C 2/.t " 1/

t.t " 1/ " 4.3t C 2/

D
.3t C 2/.t " 1/

t2 " t " 12t " t2 " 13t "

D D
.3t C 2/.t " 1/ .3t C 2/.t " 1/
26 C e e of e ra

4
b Add: C 3.
q"1
S The C is q " 1.
4 4 3.q " 1/
C3D C
q"1 q"1 q"1

4 C 3.q " 1/ 3q C 1
D D
q"1 q"1

Now ork Problem 33 G

E AM LE S F

x"2 xC2
"
x2 C 6x C 2.x2 " /

x"2 xC2
D 2
" ΠC D 2.x C 3/2 .x " 3/"
.x C 3/ 2.x C 3/.x " 3/

.x " 2/.2/.x " 3/ .x C 2/.x C 3/

D 2
"
.x C 3/ .2/.x " 3/ 2.x C 3/.x " 3/.x C 3/

.x " 2/.2/.x " 3/ " .x C 2/.x C 3/

D
2.x C 3/2 .x " 3/

2.x2 " 5x C 6/ " .x2 C 5x C 6/

D
2.x C 3/2 .x " 3/

2x2 " 10x C 12 " x2 " 5x " 6

D
2.x C 3/2 .x " 3/

x2 " 15x C 6
D
2.x C 3/2 .x " 3/
Now ork Problem 39 G

Example is important for later wor . E AM LE C F

Note that we explicitly assume h ¤ 0.
1 1
"
xCh x
Simplify , where h ¤ 0.
h
S First we combine the fractions in the numerator and obtain
1 1 x xCh x " .x C h/
" "
xCh x x.x C h/ x.x C h/ x.x C h/
D D
h h h
"h
x.x C h/ "h 1
D D D"
h x.x C h/h x.x C h/
1

Now ork Problem 47 G

 Section 0.6 ract ons 27

In business applications fractions are often expressed as percentages, which are some-
p
times confusing. We recall that p means . Also p f x simply means
p px 100
$x D . Notice that the p in p is not required to be a number between 0 and
100 100
100. In fact, for any real number r we can write r D 100r . Thus, there are 3100
days in anuary. While this might sound absurd, it is correct and reinforces understand-
ing of the de niti n:
p
p D
100
Similarly, the use of of when dealing with percentages really is ust multiplication.
If we say 5 of 7 , it means five sevens which is 35.
200
If a cost has increased by 200 it means that the cost has increased by D 2.
100
Strictly spea ing, this should mean that the increase is 2 times the old cost so that
new cost D old cost C increase D old cost C 2 $ old cost D 3 $ old cost
but people are not always clear when they spea in these terms. If you want to say that
a cost has doubled, you can say that the cost has increased by 100 .

E AM LE

A restaurant bill comes to 73.5 to which is added Harmonized Sales Tax (HST) of
15 . A customer wishes to leave the waiter a tip of 20 , and the restaurant s credit
card machine calculates a 20 tip by calculating 20 of the after tax total. How much
is charged to the customer s credit card
S
charge D .1 C 20 /.after-tax total/
120
D ..1 C 15 /.bill//
100
! "! "
120 115
D .73:5 /
100 100
13 00
D .73:5 /
10000
D 1:3 .73:5 /
D 101:5542
So, the credit card charge is 101.55.
Now ork Problem 59 G

R BLEMS
In Pr b ems simp ify
x3 C 27 x2 " 3x " 10 x2 " x C 20 ax " b c " x a2 " b2 a2 " 2ab C b2
$ $
x2 C 3x x2 " 4 x2 C x " 20 x " c ax C b a"b 2a C 2b
3x2 " 27x C 24 15x2 C x " 2 6x2 " 1 x " 7 3x C 3 x2 " x
#
2x3 " 16x2 C 14x 3x2 C 20x " 7 15x2 C 11x C 2 x2 C 3x C 2 x2 C x " 2

In Pr b ems perf rm the perati ns and simp ify as much as x2 C 2x x2 " x " 6
#
p ssib e 3x2 " 1 x C 24 x2 " 4x C 4
y2 "1 t2 " t2
$ $
y"3 yC2 t2 C 3t t2 " 6t C
28 C e e of e ra

X2 3x2 15u .x!1 " y/!1 .a C b!1 /2

3
7x 2 xCa
X x 3u 5C
x x
4 14 4
3 a2
x"
2x C y 4x x
x 3 4x 1 x"1 1
3" 2
"
2x " y 2x 3 2x x C 5x C 6 x C 2
x x"7
3x 2x xC 3C
xC2 3
21t5
3
" x t2 2x C 1 In Pr b ems and perf rm the indicated perati ns but d
x "7 2x2 " 5x " 3 n t rati na ize the den minat rs
3 p
x"3 3 3 x x 2
x2 C 6x C 10x3 x2 " x " 6 p "p p Cp
3
xCh 3
x 3Cx x
x x2 " 1 x2 "
xC3 5x x2 " 4 In Pr b ems simp ify and express y ur ans er in a f rm
xC1 x2 C 2x " 3 that is free f radica s in the den minat r
x2 C x C 12 .x C 1/2 4x2 " 1 1
p p
x2 C x C 1 2x " 1 x2 C 3x " 4 aC b 1" 2
p
x2 " 3x " 10 4x C 4 2x " 3 2 5
p p p p
x2 " 2x " 15 1 " 4x2 1 " x2 3" 6 6C 7
p p
6x2 y C 7xy " 3y x2 5x C 6 2 3 a
C p p p p
xy " x C 5y " 5 xC3 xC3 3C 5 b" c
x3 y C 4x2 y 3 x"3 4
p p Cp
xy " x C 4y " 4 tC 7 x"1 x"1
"1 x 4 3 1 Pam Alnwic used to live in Roc ingham, NS, where the
C C 2 "
x"1 x"1 x 5x X3 X2 Harmonized Sales Tax, HST, was 15 . She recently moved
to elbourne, F , where sales tax is 6.5 . When shopping,
x3 4 1 x the tas of comparing American prices with Canadian prices
1" Cs C
x3 " 1 sC4 3x " 1 xC1 was further complicated by the fact that, at the time of her
move the Canadian dollar was worth 0.75US . After thin ing
.x C 1/3 " .x " 1/3 1 1 about it, she calculated a number so that a pre-tax shelf
C 2
.x " 1/.x2 C x " 1/ x2 " 2x " 3 x " price of A US could be sensibly compared with a pre-tax
shelf price of C C N , so as to ta e into account the different
4 x sales tax rates. Her had the property that if A D C then the
"
2x2 " 7x " 4 2x2 " x C 4 after-tax costs in Canadian dollars were the same, while if A
is less (greater) than C then, after taxes, the American
4 "3x2
"3C (Canadian) price is cheaper. Find Pam s multiplier .
x"1 5 " 4x " x2
Repeat the calculation assuming a US tax rate of a and a
xC1 x"1 1 Canadian tax rate of c , when 1 C N R US , so that Pam
" 2 C
2
2x C 3x " 2 3x C 5x " 2 3x " 1 can help her stepson Tom, who moved from Calgary A to
Santa arbara CA.
.1 C x!1 /!1 .x!1 C y!1 /2

Objective E L E
o sc ss e a ent e at ons an E
to e e o tec n es for so n near
e at ons nc n tera e at ons An equation is a statement that two expressions are equal. The two expressions that
as e as fract ona an ra ca ma e up an equation are called its sides. They are separated by the equality sign, D.
e at ons t at ea to near e at ons

E AM LE E E
a xC2D3
b x2 C 3x C 2 D 0
 Section 0.7 at ons n Part c ar near at ons 29
y
c D6
y"4
d D7"z
Now ork Problem 1 G
In Example 1, each equation contains at least one variable. A variable is a symbol
that can be replaced by any one of a set of different numbers. The most popular symbols
for variables are letters from the latter part of the alphabet, such as x, y, z, , and
t. Hence, Equations (a) and (c) are said to be in the variables x and y, respectively.
Equation (d) is in the variables and z. In the equation x C 2 D 3, the numbers 2 and
3 are called c nstants. They are fixed numbers.
We ne er allow a variable in an equation to have a value for which any expression
Here we discuss restrictions on variables.
in that equation is undefined. For example, in
y
D6
y"4

y cannot be 4, because this would ma e the denominator zero while in

p
x"3D

we cannot have x"3 negative because we cannot ta e square roots of negative numbers.
We must have x " 3 % 0, which is equivalent to the requirement x % 3. (We will have
more to say about inequalities in Chapter 1.) In some equations, the allowable values of
a variable are restricted for physical reasons. For example, if the variable q represents
quantity sold, negative values of q may not ma e sense.
To solve an equation means to find all values of its variables for which the equation
is true. These values are called solutions of the equation and are said to satisfy the
equation. When only one variable is involved, a solution is also called a root. The set
of all solutions is called the solution set of the equation. Sometimes a letter representing
an un nown quantity in an equation is simply called an unkn n. Example 2 illustrates
these terms.

E AM LE T E

a In the equation x C 2 D 3, the variable x is the un nown. The only value of x that
satisfies the equation is obviously 1. Hence, 1 is a root and the solution set is f1g.
b "2 is a root of x2 C 3x C 2 D 0 because substituting "2 for x ma es the equation
true: ."2/2 C 3."2/ C 2 D 0. Hence "2 is an element of the solution set, but in
this case it is not the only one. There is one more. Can you find it
c D 7 " z is an equation in two un nowns. ne solution is the pair of values D 4
and z D 3. However, there are infinitely many solutions. Can you thin of another

Now ork Problem 3 G

E E
Two equations are said to be equivalent if they have exactly the same solutions, which
means, precisely, that the solution set of one is equal to the solution set of the other.
Solving an equation may involve performing operations on it. We prefer that any such
operation result in an equivalent equation. Here are three operations that guarantee
equivalence:
Adding (subtracting) the same polynomial to (from) both sides of an equation, where
the polynomial is in the same variable as that occurring in the equation.
30 C e e of e ra

For example, if "5x D 5 " 6x, then adding 6x to both sides gives the equivalent
equation "5x C 6x D 5 " 6x C 6x, which in turn is equivalent to x D 5.

Equivalence is not guaranteed if both ultiplying (dividing) both sides of an equation by the same n nzer constant.
sides are multiplied or divided by an
expression involving a variable. For example, if 10x D 5, then dividing both sides by 10 gives the equivalent equa-
10x 5 1
tion D , equivalently, x D .
10 10 2
Replacing either side of an equation by an equal expression.
For example, if the equation is x.x C 2/ D 3, then replacing the left side by the
equal expression x2 C 2x gives the equivalent equation x2 C 2x D 3.
We repeat: Applying perations 1 3 guarantees that the resulting equation is equiv-
alent to the given one. However, sometimes in solving an equation we have to apply
operations other than 1 3. These operations may n t necessarily result in equivalent
equations. They include the following:

T M N E
E
ultiplying both sides of an equation by an expression involving the variable.
ividing both sides of an equation by an expression involving the variable.
peration 6 includes ta ing roots of both Raising both sides of an equation to equal powers.
sides.
et us illustrate the last three operations. For example, by inspection, the only root
of x " 1 D 0 is 1. ultiplying each side by x ( peration 4) gives x2 " x D 0, which is
satisfied if x is 0 or 1. (Chec this by substitution.) ut 0 d es n t satisfy the rigina
equation. Thus, the equations are not equivalent.
Continuing, you can chec that the equation .x " 4/.x " 3/ D 0 is satisfied when
x is 4 or when x is 3. ividing both sides by x " 4 ( peration 5) gives x " 3 D 0,
whose only root is 3. Again, we do not have equivalence, since in this case a root has
been lost. Note that when x is 4, division by x " 4 implies division by 0, an invalid
operation.
Finally, squaring each side of the equation x D 2 ( peration 6) gives x2 D 4,
which is true if x D 2 or if x D "2. ut "2 is not a root of the given equation.
From our discussion, it is clear that when perations 4 6 are performed, we must
be careful about drawing conclusions concerning the roots of a given equation. pera-
tions 4 and 6 can produce an equation with more roots. Thus, you should chec whether
or not each solution obtained by these operations satisfies the rigina equation.
peration 5 can produce an equation with fewer roots. In this case, any lost root
may never be determined. Thus, avoid peration 5 whenever possible.
In summary, an equation can be thought of as a set of restrictions on any variable
in the equation. perations 4 6 may increase or decrease the number of restrictions,
giving solutions different from those of the original equation. However, perations 1 3
never affect the restrictions.

L E
The principles presented so far will now be demonstrated in the solution of a linear
equation.

A linear equation in the variable x is an equation that is equivalent to one that can
be written in the form
ax C b D 0
where a and b are constants and a ¤ 0.
 Section 0.7 at ons n Part c ar near at ons 31

A linear equation is also called a first-degree equation or an equation of degree

one, since the highest power of the variable that occurs in Equation (1) is the first.
To solve a linear equation, we perform operations on it until we have an equivalent
equation whose solutions are obvious. This means an equation in which the variable is
isolated on one side, as the following examples show.

E AM LE S L E

Solve 5x " 6 D 3x.

S We begin by getting the terms involving x on one side and the constant on
the other. Then we solve for x by the appropriate mathematical operation. We have
5x " 6 D 3x
5x " 6 C ."3x/ D 3x C ."3x/ adding "3x to both sides
2x " 6 D 0 simplifying, that is, peration 3
2x " 6 C6 D 0 C 6 adding 6 to both sides
2x D 6 simplifying
2x 6
D dividing both sides by 2
2 2
xD3
Clearly, 3 is the only root of the last equation. Since each equation is equivalent to the
one before it, we conclude that 3 must be the only root of 5x " 6 D 3x. That is, the
solution set is f3g. We can describe the first step in the solution as moving a term from
one side of an equation to the other while changing its sign this is commonly called
transp sing. Note that since the original equation can be put in the form 2xC."6/ D 0,
it is a linear equation.
Now ork Problem 21 G
E AM LE S L E

Solve 2.p C 4/ D 7p C 2.
S First, we remove parentheses. Then we collect li e terms and solve. We have
2.p C 4/ D 7p C 2
2p C D 7p C 2 distributive property
2p D 7p " 6 subtracting from both sides
"5p D "6 subtracting 7p from both sides
"6
pD dividing both sides by "5
"5
6
pD
5
Now ork Problem 25 G
E AM LE S L E
7x C 3 x"
Solve " D 6.
2 4
S We first clear the equation of fractions by multiplying b th sides by the
C , which is 4. Then we use various algebraic operations to obtain a solution. Thus,
! "
7x C 3 x"
4 " D 4.6/
2 4
32 C e e of e ra

The distributive property requires that 7x C 3 x"

b th terms within the parentheses be 4$ "4$ D 24 distributive property
2 4
multiplied by 4.
2.7x C 3/ " . x " / D 24 simplifying
14x C 6 " x C D 24 distributive property
5x C 14 D 24 simplifying
5x D 10 subtracting 14 from both sides
xD2 dividing both sides by 5

Now ork Problem 29 G

Every linear equation has exactly Each equation in Examples 3 5 has one and only one root. This is true of every
one root. The root of ax C b D 0 is linear equation in one variable.
b
xD" .
a
L E
Equations in which some of the constants are not specified, but are represented by
letters, such as a, b, c, or d, are called literal equations, and the letters are called literal
constants. For example, in the literal equation xCa D 4b, we can consider a and b to be
literal constants. Formulas, such as I D Prt, that express a relationship between certain
quantities may be regarded as literal equations. If we want to express a particular letter
in a formula in terms of the others, this letter is considered the un nown.

E AM LE S L E

a The equation I D Prt is the formula for the simple interest I on a principal of P
dollars at the annual interest rate of r for a period of t years. Express r in terms of I,
P, and t.
S Here we consider r to be the un nown. To isolate r, we divide both sides
by Pt. We have
I D Prt
I Prt
D
Pt Pt
I I
D r so r D
Pt Pt

When we divided both sides by Pt, we assumed that Pt ¤ 0, since we cannot divide
by 0. Notice that this assumption is equivalent to requiring b th P ¤ 0 and t ¤ 0.
Similar assumptions will be made when solving other literal equations.
b The equation S D P C Prt is the formula for the value S of an investment of a
principal of P dollars at a simple annual interest rate of r for a period of t years.
Solve for P.
S S D P C Prt
S D P.1 C rt/ factoring
S
DP dividing both sides by 1 C rt
1 C rt

Now ork Problem 79 G

E AM LE S L E

Solve .a C c/x C x2 D .x C a/2 for x.

 Section 0.7 at ons n Part c ar near at ons 33

S We first simplify the equation and then get all terms involving x on one side:
.a C c/x C x2 D .x C a/2
ax C cx C x2 D x2 C 2ax C a2
ax C cx D 2ax C a2
cx " ax D a2
x.c " a/ D a2
a2
xD for c ¤ a
c"a
Now ork Problem 81 G
E AM LE S T R

We recall esley ri th s problem from the opening paragraphs of this chapter. We

now generalize the problem so as to illustrate further the use of literal equations. esley
had a receipt for an amount R. She new that the sales tax rate was p . She wanted to
now the amount that was paid in sales tax. Certainly,
price C tax D receipt
Writing P for the price (which she did not yet now), the tax was .p=100/P so that she
new
p
PC PDR
100
# p $
P 1C DR
100
! "
100 C p
P DR
100
100R
PD
100 C p
It follows that the tax paid was
! " ! "
100R 100 p
R"PDR" DR 1" DR
100 C p 100 C p 100 C p
where you should chec the manipulations with fractions, supplying more details if
necessary. Recall that the French tax rate was 1 .6 and the Italian tax rate was 1 .
1 :6
We conclude that esley had only to multiply a French receipt by 11 :6
! 0:163 to
determine the tax it contained, while for an Italian receipt she should have multiplied
1
the amount by 11 . With the current tax rates (20 and 22 , respectively) her multi-
20 22
pliers would be 120 and 122 , respectively, but she doesn t have to re-solve the problem.
It should be noted that wor ing from the simple conceptual Equation (2) we have been
able to avoid the assumpti n about proportionality that we made at the beginning of
this chapter.
It is also worth noting that while problems of this ind are often given using per-
p
centages, the algebra may be simplified by writing p D 100 as a decimal. Alge-
braically, we ma e the substitution
p
rD
100
so that Equation (2) becomes
P C rP D R
R
It should be clear that P.1 C r/ D R so that P D and the tax in R is ust
1Cr
Rr Rr p
Pr D . The reader should now replace r in with and chec that this
1Cr ! " 1 C r 100
p
simplifies to R .
100 C p
34 C e e of e ra

oreover, examining this problem side by side with Example 6(b) above, we see
that solving the Tax in a Receipt problem is really the same as determining, from
an investment balance R, the amount of interest earned during the most recent interest
period, when the interest rate per interest period is r.
Now ork Problem 99 G
F E
A fractional equation is an equation in which an un nown is in a denominator. We
illustrate that solving such a nonlinear equation may lead to a linear equation.

E AM LE S F E
5 6
Solve D .
x"4 x"3
S

S We first write the equation in a form that is free of fractions. Then we

use standard algebraic techniques to solve the resulting equation.
An alternative solution that avoids
multiplying both sides by the C is as ultiplying both sides by the C , .x " 4/.x " 3/, we have
follows: ! " ! "
5 6
5 6
" D0 .x " 4/.x " 3/ D .x " 4/.x " 3/
x"4 x"3 x"4 x"3
Assuming that x is neither 3 nor 4 and 5.x " 3/ D 6.x " 4/ linear equation
combining fractions gives
5x " 15 D 6x " 24
"x
D0 Dx
.x " 4/.x " 3/
A fraction can be 0 only when its
numerator is 0 and its denominator is not. In the first step, we multiplied each side by an expression involving the ariab e x.
Hence, x D . As we mentioned in this section, this means that we are not guaranteed that the last
equation is equivalent to the rigina equation. Thus, we must chec whether or not
satisfies the rigina equation. Since
5 5 6 6
D D1 and D D1
"4 5 "3 6
we see that indeed satisfies the original equation.
Now ork Problem 47 G
Some equations that are not linear do not have any solutions. In that case, we say
that the solution set is the empty set, which we denote by ;. Example 10 will illustrate.

E AM LE S F E

3x C 4 3x " 5 12
a Solve " D 2 .
xC2 x"4 x " 2x "
S bserving the denominators and noting that
x2 " 2x " D .x C 2/.x " 4/
we conclude that the C is .x C 2/.x " 4/. ultiplying both sides by the C ,
we have
! "
3x C 4 3x " 5 12
.x C 2/.x " 4/ " D .x C 2/.x " 4/ $
xC2 x"4 .x C 2/.x " 4/
.x " 4/.3x C 4/ " .x C 2/.3x " 5/ D 12
 Section 0.7 at ons n Part c ar near at ons 35

3x2 " x " 16 " .3x2 C x " 10/ D 12

3x2 " x " 16 " 3x2 " x C 10 D 12
" x " 6 D 12
" xD1
x D "2
However, the rigina equation is not defined for x D "2 (we cannot divide by
zero), so there are no roots. Thus, the solution set is ;. Although "2 is a solution of
Equation (3), it is not a solution of the rigina equation.
4
b Solve D 0.
x"5
S The only way a fraction can equal zero is for the numerator to equal
zero (and the denominator to not equal zero). Since the numerator, 4, is not 0, the
solution set is ;.
Now ork Problem 43 G
E AM LE L E
u
If s D , express u in terms of the remaining letters that is, solve for u.
au C
S

S Since the un nown, u, occurs in the denominator, we first clear fractions

and then solve for u.

u
sD
au C
s.au C / D u multiplying both sides by au C
sau C s D u
sau " u D "s
u.sa " 1/ D "s
"s s
uD D
sa " 1 1 " sa

Now ork Problem 83 G

R E
A radical equation is one in which an un nown occurs in a radicand. The next two
examples illustrate the techniques employed to solve such equations.

E AM LE S R E
p
Solve x2 C 33 " x D 3.
S To solve this radical equation, we raise both sides to the same power to
eliminate the radical. This operation does n t guarantee equivalence, so we must chec
any resulting solutions. We begin by isolating the radical on one side. Then we square
both sides and solve using standard techniques. Thus,
p
x2 C 33 D x C 3
x2 C 33 D .x C 3/2 squaring both sides
36 C e e of e ra

x2 C 33 D x2 C 6x C
24 D 6x
4Dx
ou should show by substitution that 4 is indeed a root.
Now ork Problem 71 G
With some radical equations, you may have to raise both sides to the same power
more than once, as Example 13 shows.

E AM LE S R E
p p
Solve y " 3 " y D "3.
The reason we want one radical on each S When an equation has two terms involving radicals, first write the equation
side is to avoid squaring a binomial with so that one radical is on each side, if possible. Then square and solve. We have
two different radicals. p p
y"3D y"3
p
y"3Dy"6 yC squaring both sides
p
6 y D 12
p
yD2
yD4 squaring both sides
p p
Substituting 4 into the left side of the rigina equation gives 1 " 4, which is "1.
Since this does not equal the right side, "3, there is no solution. That is, the solution
set is ;.
Now ork Problem 69 G

R BLEMS
In Pr b ems determine by substituti n hich f the gi en .x C 2/.x C 1/ D .x C 3/.x C 1/ x C 2 D x C 3
numbers if any satisfy the gi en equati n 2x.3x C 1/
D 2x.x C 4/I 3x C 1 D .x C 4/.2x " 3/
x " x2 D 0I 1; 0 2x " 3
1
10 " 7x D "x2 I 2; 4 2x2 " D xI x2 " x D
2 2
z C 3.z " 4/ D 5I 17
4
; 4
In Pr b ems s e the equati ns
x2 C x " 6 D 0I 2; 3
!x D 3:14 0:2x D 7
x.6 C x/ " 2.x C 1/ " 5x D 4I "2; 0
" x D 12 " 20 4 " 7x D 3
x.x C 1/2 .x C 2/ D 0I 0; "1; 2 p3
5x " 3 D 2x C 2 D 11
In Pr b ems determine hat perati ns ere app ied t the 7x C 7 D 2.x C 1/ 4s C 3s " 1 D 41
rst equati n t btain the sec nd State hether r n t the
perati ns guarantee that the equati ns are equi a ent n t 5.p " 7/ " 2.3p " 4/ D 3p
s e the equati ns t D 2 " 2.2t " 3.1 " t//
2x " 3 D 4x C 12I 2x D 4x C 15 2x 5y 6
D 4x " 3 " D 2 " 4y
x " 4 D 16I x " 1
D2 5 7 7
2 4x x x x
x D 5I x4 D 625 7C D "4D
5 7 2 3 5
2x2 C 4 D 5x " 7I x2 C 2 D x" x 1 x x x
2 2 3x C " 5 D C 5x xC C D
x2 " 2x D 0I x " 2 D 0 5 5 2 3 4
a 2y " 3 6y C 7 t 5 7
C x D x2 I a C x.x " b/ D x2 .x " b/ D C t D .t " 1/
x"b 4 3 4 3 2
x2 " 1 t t t 7 C 2.x C 1/ 6x
D 3I x2 " 1 D 3.x " 1/ tC " C D 10 D
x"1 3 4 36 3 5
 Section 0.7 at ons n Part c ar near at ons 37

7 5 .x " 7/3=4 D
.2 " x/ D .x " 2/ p
5 7
y2 " D " y
2x " 7 x" 3x " 5 p p
C D yC yC2D3
3 14 21 p p
x" xC1D1
4
.5x " 2/ D 7Œx " .5x " 2/" p
3 a2 C 2a D 2 C a
.2x " 5/2 C .3x " 3/2 D 13x2 " 5x C 7 r r
1 2
5 " D0
D 25 "1 3 "4
x
4 In Pr b ems express the indicated symb in terms f the
D2
x"1 remaining symb s
5
D0 I D PrtI r
xC3 # p $
3x " 5 P 1C " R D 0I P
D0 100
x"3 p D q " 1I q
3 7
D p D 10 " 2qI q
5 " 2x 2
xC3 2 S D P.1 C rt/I t
D 2mI
x 5 rD II
a c B.n C 1/
D for a ¤ 0 and c ¤ 0
x"b x"d R.1 " .1 C i/!n /
AD I R
2x " 3 i
D6 R..1 C i/n " 1/
4x " 5 SD I R
1 1 3 i
C D S D P.1 C r/n I r
x 7 7
2 3 x"a xCb
D D I x
x"1 x"2 xCb x"a
2mI
2t C 1 3t " 1 rD I n
D B.n C 1/
2t C 3 3t C 4
x"1 x"3 1 1 1
D C D I q
xC2 xC4 p q f
y"6 6 yC6 Geometry Use the formula P D 2 C 2 to find the length
" D of a rectangle whose perimeter P is 660 m and whose width is
y y y"6
160 m.
y"2 y"2
D Geometry Use the formula D !r2 h to find the radius r of
yC2 yC3
an energy drin can whose volume is 355 ml and whose height h
"5 7 11
D C is 16 cm.
2x " 3 3 " 2x 3x C 5
1 2 "6 r
C D
xC1 x"3 3 " 2x
1 3
D
x"2 x"4
x x 3x " 4
" D 2 h = 16
xC3 x"3 x "
p
xC5D4
p
z"2D3
p
2x C 3 " 4 D 0
p Sales ax A salesperson needs to calculate the cost of an
3 " 2x C 1 D 0
r item with a sales tax of 6.5 . Write an equation that represents the
x 2 total cost c of an item costing x dollars.
C1D
2 3 Revenue A day care center s total monthly revenue from
.x C 6/1=2 D 7 the care of x toddlers is given by r D 450x, and its total monthly
p p costs are given by c D 3 0x C 3500. How many toddlers need to
4x " 6 D x
p p be enrolled each month to brea even In other words, when will
x C 1 D 2x " 3 revenue equal costs
38 C e e of e ra

Straight Line Depreciation If you purchase an item for Prey Density In a certain area, the number y of moth larvae
business use, in preparing your income tax you may be able to consumed by a single predatory beetle over a given period of time
spread out its expense over the life of the item. This is called is given by
depreciati n. ne method of depreciation is straight ine
depreciati n in which the annual depreciation is computed by 1:4x
yD
dividing the cost of the item, less its estimated salvage value, by its 1 C 0:0 x
useful life. Suppose the cost is C dollars, the useful life is years, where x is the prey density (the number of larvae per
and there is no salvage value. Then it can be shown that the value unit of area). What prey density would allow a beetle to survive
.n/ (in dollars) of the item at the end of n years is given by if it needs to consume 10 larvae over the given period
# n$
.n/ D C 1 " Store ours Suppose the ratio of the number of hours a
store is open to the number of daily customers is constant. When
If new o ce furniture is purchased for 3200, has a useful life of the store is open hours, the number of customers is 2 less than
years, and has no salvage value, after how many years will it the maximum number of customers. When the store is open
have a value of 2000 10 hours, the number of customers is 46 less than the maximum
Radar Beam When radar is used on a highway to number of customers. Write an equation describing this situation,
determine the speed of a car, a radar beam is sent out and re ected and find the maximum number of daily customers.
from the moving car. The difference F (in cycles per second) in ravel ime The time it ta es a boat to travel a given
frequency between the original and re ected beams is given by distance upstream (against the current) can be calculated by
f dividing the distance by the difference of the speed of the boat and
FD the speed of the current. Write an equation that calculates the time
334:
t it ta es a boat moving at a speed r against a current c to travel a
where is the speed of the car in miles per hour and f is the distance d. Solve your equation for c.
frequency of the original beam (in megacycles per second).
Suppose you are driving along a highway with a speed limit of ireless ower A wireless tower is 100 meters tall. An
65 mi h. A police o cer aims a radar beam with a frequency of engineer determines electronically that the distance from the top
2500 megacycles per second at your car, and the o cer observes of the tower to a nearby house is 2 meters greater than the
the difference in frequency to be 4 5 cycles per second. Can the horizontal distance from the base of the tower to the house.
o cer claim that you were speeding etermine the distance from the base of the tower to the house.
Automobile S idding Police have used the formula
p
s D 30fd to estimate the speed s (in miles per hour) of a car if it
s idded d feet when stopping. The literal number f is the
coe cient of friction, determined by the ind of road (such as
concrete, asphalt, gravel, or tar) and whether the road is wet or
dry. Some values of f are given in Table 0.1. At 5 mi h, about
how many feet will a car s id on a wet concrete road ive your
answer to the nearest foot.

Table 0.1
Savings Theresa wants to buy a house, so she has decided
Concrete Tar
to save one quarter of her salary. Theresa earns 47.00 per hour
and receives an extra 2 .00 a wee because she declined Wet 0.4 0.5
company benefits. She wants to save at least 550.00 each wee .
How many hours must she wor each wee to achieve her goal ry 0. 1.0

Predator Prey Relation Predator prey relations from

biology also apply to competition in economics. To study a
predator prey relationship, an experiment1 was conducted in Interest Earned Cassandra discovers that she has 1257 in
which a blindfolded sub ect, the predator, stood in front of a an off-shore account that she has not used for a year. The interest
3-ft-square table on which uniform sandpaper discs, the prey, rate was 7.3 compounded annually. How much interest did she
were placed. For 1 minute the predator searched for the discs by earn from that account over the last year
tapping with a finger. Whenever a disc was found, it was removed ax in a Receipt In 2006, Nova Scotia consumers paid
and searching resumed. The experiment was repeated for various HST, harm nized sa es tax, of 15 . Tom Wood traveled from
disc densities (number of discs per ft2 ). It was estimated that if y Alberta, which has only federal ST, g ds and ser ices tax of
is the number of discs pic ed up in 1 minute when x discs are on 7 to Nova Scotia for a chemistry conference. When he later
the table, then submitted his expense claims in Alberta, the comptroller was
7
y D a.1 " by/x puzzled to find that her usual multiplier of 107 to determine tax in
a receipt did not produce correct results. What percentage of
where a and b are constants. Solve this equation for y. Tom s Nova Scotia receipts were HST

1
C. S. Holling, Some Characteristics of Simple Types of Predation and
Parasitism, he Canadian Ent m gist, CI, no. 7 (1 5 ), 3 5 .
 Section 0.8 a rat c at ons 39

Objective E
o so e a rat c e at ons To learn how to solve certain classical problems, we turn to methods of solving quadratic
factor n or s n t e a rat c
for a equati ns.

A quadratic equation in the variable x is an equation that can be written in the form
ax2 C bx C c D 0
where a, b, and c are constants and a ¤ 0.

A quadratic equation is also called a sec nd degree equati n or an equati n f

degree t since the highest power of the variable that occurs is the second. Whereas
a linear equation has only one root, a quadratic equation may have two different roots.

S F
A useful method of solving quadratic equations is based on factoring, as the following
example shows.

E AM LE S E F

a Solve x2 C x " 12 D 0.
S The left side factors easily:
.x " 3/.x C 4/ D 0
Thin of this as two quantities, x " 3 and x C 4, whose product is zero. hene er
the pr duct f t r m re quantities is zer at east ne f the quantities must be
zer (We emphasized this principle in Section 0.5 Factoring.) Here, it means that
either
x " 3 D 0 or x C 4 D 0
Solving these gives x D 3 and x D "4, respectively. Thus, the roots of the original
equation are 3 and "4, and the solution set is f"4; 3g.
b Solve 6 2 D 5 .
We do not divide both sides by S We write the equation as
(a variable) since equivalence is not
2
guaranteed and we may lose a root. 6 "5 D0
so that one side is 0. Factoring gives
.6 " 5/ D 0
so we have
D0 or 6 "5D0
D0 or 6 D5
Thus, the roots are D 0 and D 56 . Note that if we had divided both sides of
6 2 D 5 by and obtained 6 D 5, our only solution would be D 56 . That is,
we would lose the root D 0. This is in line with our discussion of peration 5 in
Section 0.7 and sheds light on the problem with peration 5. ne way of approach-
ing the possibilities for a variable quantity, , is to observe that either ¤ 0 r
D 0. In the first case we are free to divide by . In this case, the original equation
is equi a ent to 6 D 5, whose only solution is D 56 . Now turning to the ther
case, D 0, we are obliged to examine whether it is also a solution of the original
equation and in this problem it is.

Now ork Problem 3 G

40 C e e of e ra

E AM LE S E F

Solve .3x " 4/.x C 1/ D "2.

Approach a problem li e this with
caution. If the product of two quantities is S We first multiply the factors on the left side:
equal to "2, it is not true that at least one
of the quantities must be "2. Why 3x2 " x " 4 D "2
Rewriting this equation so that 0 appears on one side, we have
3x2 " x " 2 D 0
.3x C 2/.x " 1/ D 0
2
xD" ; 1
3
Now ork Problem 7 G
Some equations that are not quadratic may be solved by factoring, as Example 3
shows.

E AM LE S H E F

a Solve 4x " 4x3 D 0.

S This is called a third degree equati n. We proceed to solve it as follows:
4x " 4x3 D 0
4x.1 " x2 / D 0 factoring
4x.1 " x/.1 C x/ D 0 factoring
o not neglect the fact that the factor x Setting each factor equal to 0 gives 4 D 0 (impossible), x D 0; 1 " x D 0, or
gives rise to a root. 1 C x D 0. Thus,
x D 0 or x D 1 or x D "1
so that the solution set is f"1; 0; 1g.
b Solve x.x C 2/2 .x C 5/ C x.x C 2/3 D 0.
S Factoring x.x C 2/2 from both terms on the left side, we have
x.x C 2/2 Œ.x C 5/ C .x C 2/" D 0
x.x C 2/2 .2x C 7/ D 0
Hence, x D 0, x C 2 D 0, or 2x C 7 D 0, from which it follows that the solution set
is f" 72 ; "2; 0g.

Now ork Problem 23 G

E AM LE AF E L E

Solve
yC1 yC5 7.2y C 1/
C D 2
yC3 y"2 y Cy"6
S ultiplying both sides by the C , .y C 3/.y " 2/, we get
.y " 2/.y C 1/ C .y C 3/.y C 5/ D 7.2y C 1/
Since Equation (2) was multiplied by an expression involving the variable y, remember
(from Section 0.7) that Equation (3) is not necessarily equivalent to Equation (2). After
simplifying Equation (3), we have
2y2 " 7y C 6 D 0 quadratic equation
.2y " 3/.y " 2/ D 0 factoring
 Section 0.8 a rat c at ons 41

We have shown that if y satisfies the original equation then y D 32 or y D 2. Thus,

3
2
and 2 are the only p ssib e roots of the given equation. ut 2 cannot be a root of
Equation (2), since substitution leads to a denominator of 0. However, you should chec
that 32 does indeed satisfy the rigina equation. Hence, its only root is 32 .
Now ork Problem 63 G
E AM LE S F

Solve x2 D 3.
o not hastily conclude that
p the solution S x2 D 3
of x2 D 3 consists of x D 3 only.
x2 " 3 D 0
Factoring, we obtain
p p
3/.x C 3/ D 0
.x "
p p p
Thus x " 3 D 0 or x C 3 D 0, so x D ˙ 3.
Now ork Problem 9 G
A more general form of the equation x2 D 3 is u2 D k, for k % 0. In the same
manner as the preceding, we can show that
p
If u2 D k for k % 0 then u D ˙ k:

F
Solving quadratic equations
p pby factoring can be di cult, as is evident by trying that
method on 0:7x2 " 2x " 5 D 0. However, there is a formula called the quadratic
formula that gives the roots of any quadratic equation.

F
The roots of the quadratic equation ax2 CbxCc D 0, where a, b, and c are constants
and a ¤ 0, are given by
p
"b ˙ b2 " 4ac
xD
2a

Actually, the quadratic formula is not hard to derive if we first write the quadratic
equation in the form
b c
x2 C x C D 0
a a
and then as
! "
b 2
xC " 2D0
2a
for a number , as yet to be determined. This leads to
! "! "
b b
xC " xC C D0
2a 2a
b b
which in turn leads to x D " 2a C or x D " 2a " by the methods already under
' &
b 2
consideration. To see what is, observe that we require x C 2a " 2 D x2 C ba x C ac
(so that the equation we ust solved is the quadratic equation we started with), which
p
b2 !4ac b
leads to D 2a
. Substituting this value of in x D " 2a ˙ gives the uadratic
Formula.
42 C e e of e ra

From the quadratic formula we see that the given quadratic equation has two real
roots if b2 " 4ac > 0, one real root if b2 " 4ac D 0, and no real roots if b2 " 4ac < 0.
We remar ed in Section 0.5 Factoring that it is not always possible to factor
x2 C bx C c as .x " r/.x " s/ for real numbers r and s, even if b and c are integers. This
is because, for any such pair of real numbers, r and s would be roots of the equation
x2 C bx C c D 0. When a D 1 in the quadratic formula, it is easy to see that b2 " 4c
can be negative, so that x2 C bx C c D 0 can have no real roots. At first glance it might
seem that the numbers r and s can be found by simultaneously solving
r C s D "b
rs D c
for r and s, thus giving another way of finding the roots of a general quadratic. However,
rewriting the first equation as s D "b " r and substituting this value in the second
equation, we ust get r2 C br C c D 0, right bac where we started.
Notice too that we can now verify that x2 C 1 cannot be factored. If we try to solve
2
x C 1 D 0 using the quadratic formula with a D 1, b D 0, and c D 1 we get
p p p p p
"0 ˙ 02 " 4 "4 4 "1 2 "1 p
xD D˙ D˙ D˙ D ˙ "1
2 2 2 2
p p
and "1 is not a real number. It is common to write i D "1 and refer to it as the
imaginary unit. The C mp ex umbers are those of the form a C ib, where a and b
are real. The Complex Numbers extend the Real Numbers, but except for Example
below they will ma e no further appearances in this boo .

E AM LE A E T R R

Solve 4x2 " 17x C 15 D 0 by the quadratic formula.

S Here a D 4; b D "17, and c D 15. Thus,
p p
"b ˙ b2 " 4ac "."17/ ˙ ."17/2 " 4.4/.15/
xD D
2a 2.4/
p
17 ˙ 4 17 ˙ 7
D D

17 C 7 24 17 " 7 10 5
The roots are D D 3 and D D :
4
Now ork Problem 31 G
E AM LE A E R R
p
Solve 2 C 6 2y C y2 D 0 by the quadratic formula.
p
S oo at the arrangement of the terms. Here a D ; b D 6 2, and c D 2.
Hence,
p p p
"b ˙ b2 " 4ac "6 2 ˙ 0
yD D
2a 2. /
Thus,
p p p p
"6 2 C 0 2 "6 2 " 0 2
yD D" or y D D"
1 3 1 3
p
2
Therefore, the only root is " .
3
Now ork Problem 33 G
 Section 0.8 a rat c at ons 43

E AM LE A E N R R

Solve z2 C z C 1 D 0 by the quadratic formula.

S Here a D 1; b D 1, and c D 1. The roots are
p p p p p
"b ˙ b2 " 4ac "1 ˙ "3 "1 ˙ "1 3 1 3
zD D D D" ˙i
2a 2 2 2 2
Neither of the roots are real numbers. oth are complex numbers as described brie y
in the paragraph preceding Example 6.
Now ork Problem 37 G
This describes the nature of the roots of a Examples 6 illustrate the three possibilities for the roots of a quadratic equation:
quadratic equation. either two different real roots, exactly one real root, or no real roots. In the last case
there are two different complex roots, where if one of them is a C ib with b ¤ 0 then
the other is a " ib.

F E
Sometimes an equation that is not quadratic can be transformed into a quadratic equa-
tion by an appropriate substitution. In this case, the given equation is said to have
quadratic form. The next example will illustrate.

E AM LE S F E
1
Solve C 3 C D 0.
x6 x
S This equation can be written as
! "2 ! "
1 1
C C D0
x3 x3

so it is quadratic in 1=x3 and hence has quadratic form. Substituting the variable for
1=x3 gives a quadratic equation in the variable , which we can then solve:
2
C C D0
o not assume that " and "1 are . C /. C 1/ D 0
solutions of the rigina equation.
D" or D "1

Returning to the variable x, we have

1 1
D" or D "1
x3 x3
Thus,

1
x3 D " or x3 D "1

from which it follows that

1
xD" or x D "1
2
Chec ing, we find that these values of x satisfy the original equation.
Now ork Problem 49 G
44 C e e of e ra

R BLEMS
In Pr b ems s e by fact ring 3x C 2 2x C 1 6. C 1/
" D1 C D3
2
x " 4x C 4 D 0 2
t C 3t C 2 D 0 xC1 2x 2" "1
t2 C 4t " 21 D 0 x2 C 3x " 10 D 0 2 rC1 2x " 3 2x
" D0 C D1
x2 " 2x " 3 D 0 x2 " 16 D 0 r"2 rC4 2x C 5 3x C 1
t t"1 t"3 2 3 4
u2 " 13u D "36 5z2 C 14z " 3 D 0 C D 2 C D
t"1 t"2 t " 3t C 2 xC1 x xC2
x2 " 4 D 0 3u2 " 6u D 0
2 1 2 3.x C 3/ 1"x
t2 " 5t D 0 x2 C x D "14 " D 2 5" D
x2 " 1 x.x " 1/ x x2 C 3x x
4x2 " 4x D 3 2z2 C z D 5 p p
.3 " 5/ D "2 2x " 3 D x " 3 xC2DxC1
2 C x " 6x2 D 0 p p
2 2 5 q C 2 D 2 4q " 7 x C 4x " 5 D 0
"x2 C 3x C 10 D 0 u D u p p p p
3 7 z C 3 " 3z " 1 D 0 x " 2x " " 2 D 0
2p2 D 3p 2
"r " r C 12 D 0 p p p p
xC1" xD1 y " 2 C 2 D 2y C 3
x.x C 4/.x " 1/ D 0 . " 3/2 . C 1/2 D 0 p p pp p
xC3C1D3 x t C 2 D 3t " 1
.3t4 " 3t2 /.t C 3/ D 0 x3 " 4x2 " 5x D 0
6x3 C 5x2 " 4x D 0 .x C 1/2 " 5x C 1 D 0 In Pr b ems and nd the r ts r unded t t decima
.x " 3/.x " 4/ D 0 2 2
5.z " 3z C 2/.z " 3/ D 0 p aces
p.p " 3/2 " 4.p " 3/3 D 0 x4 " 3x2 C 2 D 0 0:04x2 " 2:7x C :6 D 0 x2 C .0:2/x " 0:3 D 0
In Pr b ems nd a rea r ts by using the quadratic Geometry The area of a rectangular picture with a width
f rmu a 2 inches less than its length is 4 square inches. What are the
dimensions of the picture
x2 C 2x " 24 D 0 x2 " 2x " 15 D 0
2 emperature The temperature has been rising X degrees
x " 42x C 4 D 0 q2 " 5q D 0 per day for X days. X days ago it was 15 degrees. Today it is 51
p2 " 2p " 7 D 0 2 " 2x C x2 D 0 degrees. How much has the temperature been rising each day
4 " 2n C n2 D 0 2u2 C 3u D 7 How many days has it been rising
4x2 C 5x " 2 D 0 2
"2 C1D0 Economics ne root of the economics equation
2 2 . C 10/
0:02 " 0:3 D 20 0:01x C 0:2x " 0:6 D 0 D
44
p
z2 " z C 1 D 0 "2x2 " 6x C 5 D 0 is "5 C 25 C 44 . Verify this by using the quadratic formula to
In Pr b ems s e the gi en quadratic f rm equati n solve for in terms of . Here is real income and is the level
4
x " 5x C 6 D 0 2
X 4 " 3X 2 " 10 D 0 of money supply.

3 7 Diet for Rats A group of biologists studied the nutritional

" C2D0 x!2 C x!1 " 2 D 0 effects on rats that were fed a diet containing 10 protein.2 The
x2 x protein was made up of yeast and corn our. y changing the
1 percentage P (expressed as a decimal) of yeast in the protein mix,
x!4 " x!2 C 20 D 0 " C D0
x4 x2 the group estimated that the average weight gain g (in grams) of a
rat over a period of time was given by
.X " 5/2 C 7.X " 5/ C 10 D 0
g D "200P2 C 200P C 20
.3x C 2/2 " 5.3x C 2/ D 0
What percentage of yeast gave an average weight gain of
1 7
" C 10 D 0 60 grams
.x " 4/2 x"4
Drug Dosage There are several rules for determining doses
2 7
C C3D0 of medicine for children when the adult dose has been specified.
.x C 4/2 xC4 Such rules may be based on weight, height, and so on. If A is the
In Pr b ems s e by any meth d age of the child, d is the adult dose, and c is the child s dose, then
here are two rules:
x C 3 x 7 5 A
x2 D D " oung s rule: cD d
2 2 x 2 A C 12
3 x"3 2 3 AC1
C D2 C D2 Cowling s rule: cD d
x"4 x 2x C 1 xC2 24

2
Adapted from R. ressani, The Use of east in Human Foods, in R. I. ateles
and S. R. Tannenbaum (eds.), Sing e Ce Pr tein (Cambridge, A: IT Press,
1 6 ).
 Chapter 0 e e 45

At what age(s) are the children s doses the same under both rules Delivered Price of a Good In a discussion of the delivered
Round your answer to the nearest year. Presumably, the child has price of a good from a mill to a customer, eCanio3 arrives at and
become an adult when c D d. At what age does the child become solves the two quadratic equations
an adult according to Cowling s rule According to oung s rule 2
A .2n " 1/ " 2n C 1 D 0
If you now how to graph functions, graph both .A/ D
A C 12 and
AC1
and C.A/ D as functions of A, for A % 0, in the same n 2
" .2n C 1/ C 1 D 0
24
plane. Using the graphs, ma e a more informed comparison of
where n % 1.
oung s rule and Cowling s rule than is obtained by merely
finding the age(s) at which they agree. a Solve the first equation for .
b Solve the second equation for if < 1.
Motion Suppose the height h of an ob ect thrown straight
upward from the ground is given by
h D 3 :2t " 4: t2
where h is in meters and t is the elapsed time in seconds.
a After how many seconds does the ob ect stri e the ground
b When is the ob ect at a height of 6 .2 m

Chapter 0 Review
I T S E
S Sets of Real umbers
set integers rational numbers real numbers coordinates
S Some Properties of Real umbers
commutative associative identity inverse reciprocal distributive Ex. 3, p. 6

S Exponents and Radicals

exponent base principal nth root radical Ex. 2, p. 11

S Operations with Algebraic Expressions

algebraic expression term factor polynomial Ex. 6, p. 1
long division Ex. , p. 1

S Factoring
common factoring perfect square difference of squares Ex. 3, p. 21
S Fractions
multiplication and division Ex. 3, p. 23
addition and subtraction Ex. 5, p. 24
rationalizing denominators Ex. 4, p. 24

S Equations in Particular Linear Equations

equivalent equations linear equations Ex. 5, p. 31
fractional equations Ex. , p. 34
radical equations Ex. 3, p. 31

3
S. . eCanio, elivered Pricing and ultiple asing Point Equilibria: A
Reevalution, uarter y urna f Ec n mics, CI , no. 2 (1 4), 32 4 .
46 C e e of e ra

S Quadratic Equations
solved by factoring Ex. 2, p. 40
quadratic formula Ex. , p. 43

S
There are certainly basic f rmu as that have to be remem- in the form ax C b D 0, where a and b are constants and
bered when reviewing algebra. It is often a good exercise, a ¤ 0. e sure that you understand the derivation of its solu-
while attempting this memory wor , to find the formulas that b
are basic f r y u. For example, the list of properties, each fol- tion, which is x D " . Even though the sub ect matter of
a
lowed by an example, near the end of Section 0.2, has many this boo is Applied athematics, particularly as it pertains
redundancies in it, but all those f rmu as need to be part of to usiness and Economics, it is essential to understand how
your own mathematical tool it. However, Property 2, which to solve itera equati ns in which the coe cients, such as a
says a " ."b/ D a C b, will probably ump out at you as and b in ax C b D 0, are not presented as particular numbers.
a " ."b/ D a C ."."b// D a C b. The first equality here is In Section 0.7 there are many equations, many but not all of
ust the definition of subtraction, as provided by Property 1, which reduce to linear equations.
while the second equality is Property applied to b. If this The general quadratic equation in the variable x is
is obvious to you, you can stri e Property 2 from y ur ist ax2 C bx C c D 0, where a, b, and c are constants with a ¤ 0,
of formulas that y u personally need to memorize. Try to do and it forms the sub ect of Section 0. . Its roots (solutions)
your own treatment of Property 5, perhaps using Property 1 are given by
(twice) and Properties 4 and 10. If you succeed, continue
wor ing through the list, stri ing off what you don t need to p
"b ˙ b2 " 4ac
memorize. All of this is to say that you will remember faster xD
what you need to now if you wor to shorten your personal 2a
list of formulas that require memory. In spite of technology, although, if the coe cients a, b, and c are simple integers,
this is a tas best done with pencil and paper. athematics these roots may be more easily found by factoring. We rec-
is not a spectator sport. ommend simply memorizing Equation 5, which is nown as
The same comments apply to the list of formulas in Sec- the uadratic Formula. The radical in the uadratic Formula
tion 0.3 and the specia pr ducts in Section 0.4. ng di isi n tells us, at a glance, that the nature of the roots of a quadratic
of polynomials (Section 0.3) is a s ill that comes with prac- are determined by b2 " 4ac. If b2 " 4ac is positive, the equa-
tice (and only with practice) as does fact ring (Section 0.4). tion has two real roots. If b2 "4ac is zero, it has one real root.
A inear equati n in the variable x is one that can be written If b2 " 4ac is negative, there are no real roots.

R
p5
Rewrite a!5 b!3 c2 b4 c3 without radicals and using only Interest Earned Emily discovers that she has 5253.14 in a
positive exponents. ban account that has been untouched for two years, with
p interest earned at the rate of 3.5 compounded annually. How
5
Rationalize the denominator of p
7
. much of the current amount of 5253.14 was interest earned
13 Hint: If an amount P is invested for 2 years at a rate r (given
p p
xCh" x as a real number), compounded annually, then the value of the
Rationalize the numerat r of .
h investment after 2 years is given by S D P.1 C r/2 .
3 2 . C 10/
Calculate .3x " 4x C 3x C 7/ # .x " 1/.
We loo ed earlier at the economics equation D ,
1 1 44
" 2
.x C h/2 x where is the level of money supply and is real income. We
Simplify .
h verified that one of its roots is given by
p
Solve S D P.1 C r/n for P. D "5 C 25 C 44 . What is the other root and does it
Solve S D P.1 C r/n for r. have any significance
p
Solve x C 2 x " 15 D 0 by treating it as an equation of
quadratic-form.
 cat ons
an ore e ra

I
n this chapter, we will apply equations to various practical situations. We will
1.1 cat ons of at ons
also do the same with inequalities, which are statements that one quantity is less
1.2 near ne a t es than (<), greater than (>), less than or equal to (!), or greater than or equal to
(") some other quantity.
1.3 cat ons of Here is an example of the use of inequalities in the regulation of sporting
ne a t es
equipment. ozens of baseballs are used in a typical ma or league game and it would
1.4 so te a e be unrealistic to expect that every ball weigh exactly 5 1 ounces. ut it is reasonable to
require that each one weigh no less than 5 ounces and no more than 5 14 ounces, which is
1.5 at on otat on how 1.0 of the cial Rules of a or eague aseball reads. (See
1.6 e ences http://mlb.mlb.com/ and loo up o cial rules .) Note that n ess than means
the same thing as greater than r equa t while n m re than means the same thing
C er 1 e e as ess than r equa t . In translating English statements into mathematics, we recom-
mend avoiding the negative wordings as a first step. Using the mathematical symbols
we have
1
ball weight " 5 ounces and ball weight ! 5 ounces
4
which can be combined to give
1
5 ounces ! ball weight ! 5 ounces
4
and nicely displays the ball weight bet een 5 and 5 14 ounces (where bet een here
includes the extreme values).
Another inequality applies to the sailboats used in the America s Cup race. The
America s Cup Class (ACC) for yachts was defined until 30 anuary, 200 , by
p p
C 1:25 S # : 3 SP
! 24:000 m
0:6 6
The ! signifies that the expression on the left must come out as less than or
equal to the 24.000 m on the right. The , S, and SP were themselves specified by
complicated formulas, but roughly, stood for length, S for sail area, and SP for
displacement (the hull volume below the waterline).
The ACC formula gave yacht designers some latitude. Suppose a yacht had
D 20:2 m, S D 2 2 m2 , and SP D 16:4 m3 . Since the formula is an inequality, the
designer could reduce the sail area while leaving the length and displacement unchanged.
Typically, however, values of , S, and SP were used that made the expression on the
left as close to 24.000 m as possible.
In addition to applications of linear equations and inequalities, this chapter will
review the concept of absolute value and introduce sequences and summation notation.

47
48 C cat ons an ore e ra

Objective A E
o o e s t at ons escr e near In most cases, the solution of practical problems requires the translation of stated rela-
or a rat c e at ons
tionships into mathematical symbols. This is called m de ing. The following examples
illustrate basic techniques and concepts.

n Alcohol: 2n
n
E AM LE M
350 ml
n
n
Acid: 3n A chemist needs to prepare 350 ml of a chemical solution made up of two parts alcohol
n and three parts acid. How much of each should be used

FIGURE Chemical solution S et n be the number of milliliters in each part. Figure 1.1 shows the situa-
(Example 1). tion. From the diagram, we have

Note that the solution to an equation is 2n C 3n D 350

not necessarily the solution to the
problem posed. 5n D 350
350
nD D 70
5
ut n D 70 is n t the answer to the original problem. Each part has 70 ml. The
amount of alcohol is 2n D 2.70/ D 140, and the amount of acid is 3n D 3.70/ D 210.
Thus, the chemist should use 140 ml of alcohol and 210 ml of acid. This example shows
how helpful a diagram can be in setting up a word problem.
Now ork Problem 5 G
E AM LE I

A vehicle inspection pit is to be built in a commercial garage. See Figure 1.2(a).

The garage has dimensions 6 m by 12 m. The pit is to have area 40 m2 and to be
centered in the garage so that there is a uniform wal way around the pit. How wide
will this wal way be
S A diagram of the pit is shown in Figure 1.2(b). et be the width (in meters)
of the wal way. Then the pit has dimensions 12 # 2 by 6 # 2 . Since its area must
be 40 m2 , where area D .length/.width/, we have
.12 # 2 /.6 # 2 / D 40
2
72 # 36 C 4 D 40 multiplying
2
4 # 36 C 32 D 0
2
# C D0 dividing both sides by 4
. # /. # 1/ D 0
D ; 1

w
w w
6 – 2w 12 – 2w 6

12

(a) (b)

FIGURE Pit wal way (Example 2).

 Section . cat ons of at ons 49

Although is a solution of the equation, it is n t a solution to our problem, because

one of the dimensions of the garage itself is only 6 m. Thus, the only possible solution
is that the wal way be 1 m wide.
Now ork Problem 7 G
The ey words introduced here are xed In the next example, we refer to some business terms relative to a manufactur-
c st ariab e c st t ta c st t ta ing firm. Fixed cost is the sum of all costs that are independent of the level of pro-
re enue, and pr t. This is the time to
gain familiarity with these terms because duction, such as rent, insurance, and so on. This cost must be paid whether or not
they recur throughout the boo . output is produced. Variable cost is the sum of all costs that are dependent on the
level of output, such as labor and material. otal cost is the sum of variable cost and
fixed cost:

total cost D variable cost C fixed cost

otal revenue is the money that the manufacturer receives for selling the output:

total revenue D (price per unit) (number of units sold)

Profit is total revenue minus total cost:

profit D total revenue # total cost

E AM LE

The Acme Company produces a product for which the variable cost per unit is 6 and
fixed cost is 0,000. Each unit has a selling price of 10. etermine the number of
units that must be sold for the company to earn a profit of 60,000.
S et q be the number of units that must be sold. (In many business problems,
q represents quantity.) Then the variable cost (in dollars) is 6q. The t ta cost for the
business is therefore 6q C 0;000. The total revenue from the sale of q units is 10q.
Since

profit D total revenue # total cost

our model for this problem is

60;000 D 10q # .6q C 0;000/

Solving gives

60;000 D 10q # 6q # 0;000

4q D 140;000
q D 35;000
Thus, 35,000 units must be sold to earn a profit of 60,000.
Now ork Problem 9 G
E AM LE

Sportcraft manufactures denim clothing and is planning to sell its new line of eans to
retail outlets. The cost to the retailer will be 60 per pair of eans. As a convenience
to the retailer, Sportcraft will attach a price tag to each pair. What amount should be
mar ed on the price tag so that the retailer can reduce this price by 20 during a sale
and still ma e a profit of 15 on the cost
S Here we use the fact that
Note that price D cost C profit. selling price D cost per pair C profit per pair
50 C cat ons an ore e ra

et p be the tag price per pair, in dollars. uring the sale, the retailer actually receives
p # 0:2p. This must equal the cost, 60, plus the profit, (0.15)(60). Hence,
selling price D cost C profit
p # 0:2p D 60 C .0:15/.60/
0: p D 6
p D 6:25
Sportcraft should mar the price tag at 6.25.
Now ork Problem 13 G
E AM LE I

A total of 10,000 was invested in two business ventures, A and . At the end of the first
year, A and yielded returns of 6 and 5 34 , respectively, on the original investments.
How was the original amount allocated if the total amount earned was 5 .75
S et x be the amount (in dollars) invested at 6 . Then 10;000 # x
was invested at 5 34 . The interest earned from A was (0.06).x/ and that from B was
.0:0575/.10;000 # x/ with a total of 5 .75. Hence,
.0:06/x C .0:0575/.10;000 # x/ D 5 :75
0:06x C 575 # 0:0575x D 5 :75
0:0025x D 13:75
x D 5500
Thus, 5500 was invested at 6 , and 10; 000 # 5500 D 4500 was invested at 5 34 .
Now ork Problem 11 G
E AM LE B R

The board of directors of aven Corporation agrees to redeem some of its bonds in two
years. At that time, 1,102,500 will be required. Suppose the firm presently sets aside
1,000,000. At what annual rate of interest, compounded annually, will this money
have to be invested in order that its future value be su cient to redeem the bonds
S et r be the required annual rate of interest. At the end of the first year, the
accumulated amount will be 1,000,000 plus the interest, 1,000,000r, for a total of
1;000;000 C 1;000;000r D 1;000;000.1 C r/
Under compound interest, at the end of the second year the accumulated amount will
be 1;000;000.1 C r/ plus the interest on this, which is 1;000;000.1 C r/r. Thus, the
total value at the end of the second year will be
1;0000;000.1 C r/ C 1;000;000.1 C r/r
This must equal 1,102,500:
1;000;000.1 C r/ C 1;000;000.1 C r/r D 1;102;500
Since 1,000,000.1 C r/ is a common factor of both terms on the left side, we have
1;000;000.1 C r/.1 C r/ D 1;102;500
1;000;000.1 C r/2 D 1;102;500
1;102;500 11;025 441
.1 C r/2 D D D
1;000;000 10;000 400
r
441 21
1CrD˙ D˙
400 20
21
r D #1 ˙
20
 Section . cat ons of at ons 51

Thus, r D #1 C .21=20/ D 0:05, or r D #1 # .21=20/ D #2:05. Although 0.05 and

#2:05 are roots of Equation (1), we re ect #2:05 since we require that r be positive.
Hence, r D 0:05 D 5 is the desired rate.
Now ork Problem 15 G
At times there may be more than one way to model a word problem, as Example 7
shows.

E AM LE A R

A real estate firm owns the Par lane arden Apartments, which consist of 6 apart-
ments. At 550 per month, every apartment can be rented. However, for each 25 per
month increase, there will be three vacancies with no possibility of filling them. The
firm wants to receive 54,600 per month from rents. What rent should be charged for
each apartment
S
et o Suppose r is the rent (in dollars) to be charged per apartment. Then the
r # 550
increase over the 550 level is r # 550. Thus, the number of 25 increases is .
25
ecause each
! 25 increase
" results in three vacancies, the total number of vacancies
r # 550
will be 3 . Hence, the total number of apartments rented will be
25
! "
r # 550
6#3 . Since
25
total rent D (rent per apartment)(number of apartments rented)
we have
! "
3.r # 550/
54;600 D r 6 #
25
! "
2400 # 3r C 1650
54;600 D r
25
! "
4050 # 3r
54;600 D r
25
1;365;000 D r.4050 # 3r/
Thus,
3r2 # 4050r C 1;365;000 D 0
y the quadratic formula,
p
4050 ˙.#4050/2 # 4.3/.1;365;000/
rD
2.3/
p
4050 ˙ 22;500 4050 ˙ 150
D D D 675 ˙ 25
6 6
Hence, the rent for each apartment should be either 650 or 700.
et o Suppose n is the number of 25 increases. Then the increase in rent per
apartment will be 25n and there will be 3n vacancies. Since
total rent D (rent per apartment)(number of apartments rented)
we have
54;600 D .550 C 25n/. 6 # 3n/
54;600 D 52; 00 C 750n # 75n2
75n2 # 750n C 1 00 D 0
n2 # 10n C 24 D 0
.n # 6/.n # 4/ D 0
52 C cat ons an ore e ra

Thus, n D 6 or n D 4. The rent charged should be either 550 C 25.6/ D 700 or

550 C 25.4/ D 650. However, it is easy to see that the real estate firm can receive
54,675 per month from rents by charging 675 for each apartment and that 54,675 is
the maximum amount from rents that it can receive, given existing mar et conditions.
In a sense, the firm posed the wrong question. A considerable amount of our wor in
this boo focuses on a better question that the firm might have as ed.
Now ork Problem 29 G
R BLEMS
Fencing A fence is to be placed around a rectangular plot so Profit A corn refining company produces corn gluten
that the enclosed area is 00 ft2 and the length of the plot is twice cattle feed at a variable cost of 2 per ton. If fixed costs are
the width. How many feet of fencing must be used 120,000 per month and the feed sells for 134 per ton, how
many tons must be sold each month for the company to have a
Geometry The perimeter of a rectangle is 300 ft, and the
monthly profit of 560,000
length of the rectangle is 3 ft more than twice the width. Find the
dimensions of the rectangle. Sales The Pear-shaped Corporation would li e to now the
ent Caterpillars ne of the most damaging defoliating total sales units that are required for the company to earn a profit
insects is the tent caterpillar, which feeds on foliage of shade, of 1,500,000. The following data are available: unit selling price
forest, and fruit trees. A homeowner lives in an area in which the of 550, variable cost per unit of 250, total fixed cost of
tent caterpillar has become a problem. She wishes to spray the 5,000,000. From these data, determine the required sales units.
trees on her property before more defoliation occurs. She needs Investment A person wishes to invest 20,000 in two
145 oz of a solution made up of 4 parts of insecticide A and 5 enterprises so that the total income per year will be 1440. ne
parts of insecticide B. The solution is then mixed with water. How enterprise pays 6 annually the other has more ris and pays
many ounces of each insecticide should be used 7 12 annually. How much must be invested in each
Concrete Mix A builder ma es a certain type of concrete by
mixing together 1 part Portland cement (made from lime and
clay), 3 parts sand, and 5 parts crushed stone (by volume). If
765 ft3 of concrete are needed, how many cubic feet of each
ingredient does he need
omemade Ice Cream nline recipes claim that you can
ma e no-churn ice cream using 7 parts of sweetened condensed Investment A person invested 120,000, part at an interest
mil and parts of cold, heavy whipping cream. How many rate of 4 annually and the remainder at 5 annually. The total
millilitres of whipping cream will you need to ma e 3 litres of interest at the end of 1 year was equivalent to an annual 4 12 rate
ice cream on the entire 120,000. How much was invested at each rate
Forest Management A lumber company owns a forest that Pricing The cost of a product to a retailer is 3.40. If the
is of rectangular shape, 1 mi by 2 mi. If the company cuts a retailer wishes to ma e a profit of 20 on the selling price, at
uniform strip of trees along the outer edges of this forest, how what price should the product be sold
wide should the strip be if 34 sq mi of forest is to remain Bond Retirement In three years, a company will require
Garden Pavement A 10-m-square plot is to have a circular 1,125, 00 in order to retire some bonds. If the company now
ower bed of 60 m2 centered in the square. The other part of the invests 1,000,000 for this purpose, what annual rate of interest,
plot is to be paved so that the owners can wal around the ower compounded annually, must it receive on that amount in order to
bed. What is the minimum width of the paved surface In other retire the bonds
words, what is the smallest distance from the ower bed to the Expansion Program The Pear-shaped Corporation has
edge of the plot planned an expansion program in three years. It has decided to
Ventilating Duct The diameter of a circular ventilating duct invest 3,000,000 now so that in three years the total value of the
is 140 mm. This duct is oined to a square duct system as shown investment will exceed 3,750,000, the amount required for the
in Figure 1.3. To ensure smooth air ow, the areas of the circle and expansion. What is the annual rate of interest, compounded
square sections must be equal. To the nearest millimeter, what annually, that Pear-shaped must receive to achieve its purpose
should the length x of a side of the square section be Business A company finds that if it produces and sells q
p
units of a product, its total sales revenue in dollars is 100 q. If
the variable cost per unit is 2 and the fixed cost is 1200, find the
140 mm values of q for which
total sales revenue D variable cost C fixed cost
(That is, profit is zero.)
Overboo ing A commuter airplane has 1 seats. n the
x average, 0 of those who boo for a ight show up for it. How
many seats should the airline boo for a ight if it wants to fill the
FIGURE Ventilating duct (Problem ). plane
 Section . cat ons of at ons 53

Poll A group of people were polled, and 20 , or 700, of Business Alternatives The band ongeese tried to sell
them favored a new product over the best-selling brand. How its song obra lub to a small label, Epsilon Records, for a
many people were polled lump-sum payment of 50,000. After estimating that future sales
Prison Guard Salary It was reported that in a certain possibilities of obra lub beyond one year are nonexistent,
women s ail, female prison guards, called matrons, received 30 Epsilon management is reviewing an alternative proposal to give
(or 200) a month less than their male counterparts, deputy a lump-sum payment of 5000 to ongeese plus a royalty of
sheriffs. Find the yearly salary of a deputy sheriff. ive your 0.50 for each disc sold. How many units must be sold the first
answer to the nearest dollar. year to ma e this alternative as economically attractive to the
band as their original request int etermine when the
Stri ing urses A few years ago, licensed practical nurses incomes under both proposals are the same.
( PNs) were on stri e for 27 days. efore the stri e, these nurses
earned 21.50 per hour and wor ed 260 eight-hour days a year. Par ing Lot A company par ing lot is 120 ft long and
What percentage increase is needed in yearly income to ma e up 0 ft wide. ue to an increase in personnel, it is decided to double
for the lost time within one year the area of the lot by adding strips of equal width to one end and
one side. Find the width of one such strip.
Brea Even A manufacturer of video games sells each
copy for 21. 5. The manufacturing cost of each copy is 14. 2.
onthly fixed costs are 500. uring the first month of sales of
a new game, how many copies must be sold in order for the
manufacturer to brea even (that is, in order that total revenue
equals total cost)
Investment Club An investment club bought a bond
of an oil corporation for 5000. The bond yields 4 per year.
The club now wants to buy shares of stoc in a windmill
supply company. The stoc sells at 20 per share and earns a
dividend of 0.50 per share per year. How many shares should
the club buy so that its total investment in stoc s and bonds Rentals ou are the chief financial advisor to a corporation
yields 3 per year that owns an o ce complex consisting of 50 units. At 400 per
Vision Care As a fringe benefit for its employees, a month, every unit can be rented. However, for each 20 per
company established a vision-care plan. Under this plan, each month increase, there will be two vacancies with no possibility of
year the company will pay the first 35 of an employee s filling them. The corporation wants to receive a total of 20,240
vision-care expenses and 0 of all additional vision-care per month from rents in the complex. ou are as ed to determine
expenses, up to a maximum t ta benefit payment of 100. For the rent that should be charged for each unit. What is your reply
an employee, find the total annual vision-care expenses covered Investment Six months ago, an investment company had a
by this program. ,500,000 portfolio consisting of blue-chip and glamour stoc s.
Since then, the value of the blue-chip investment increased by 1 ,
1
whereas the value of the glamour stoc s decreased by 12 . The
current value of the portfolio is 10,700,000. What is the current
value of the blue-chip investment
Revenue The monthly revenue of a certain company is
given by R D 00p # 7p2 , where p is the price in dollars of the
product the company manufactures. At what price will the
revenue be 10,000 if the price must be greater than 50
Quality Control ver a period of time, the manufacturer Price Earnings Ratio The price earnings rati , P=E, of
of a caramel-center candy bar found that 3.1 of the bars were a company is the ratio of the mar et value of one share of the
re ected for imperfections. company s outstanding common stoc to the earnings per share.
a If c candy bars are made in a year, how many would the If P=E increases by 15 and the earnings per share decrease by
manufacturer expect to be re ected 10 , determine the percentage change in the mar et value per
b This year, annual consumption of the candy is pro ected to be share of the common stoc .
600 million bars. Approximately how many bars will have to be Mar et Equilibrium When the price of a product is p
made if re ections are ta en into consideration dollars each, suppose that a manufacturer will supply 2p # 10
Business Suppose that consumers will purchase q units of a units of the product to the mar et and that consumers will demand
product when the price is .60 # q/=$3 each. How many units to buy 200 # 3p units. At the value of p for which supply equals
must be sold in order that sales revenue be 300 demand, the mar et is said to be in equilibrium. Find this value
of p.
Investment How long would it ta e to triple an investment
at simple interest with a rate of 4.5 per year int: See Mar et Equilibrium Repeat Problem 33 for the following
Example 6(a) of Section 0.7, and express 4.5 as 0.045. conditions: At a price of p dollars each, the supply is 2p2 # 3p and
the demand is 20 # p2 .
54 C cat ons an ore e ra

Security Fence For security reasons, a company will Compensating Balance C mpensating ba ance refers to
enclose a rectangular area of 7762:5 m2 in the rear of its plant. that practice wherein a ban requires a borrower to maintain on
ne side will be bounded by the building and the other three sides deposit a certain portion of a loan during the term of the loan. For
by fencing. If the bounding side of the building is 130 m long and example, if a firm ta es out a 100,000 loan that requires a
250 m of fencing will be used, what will be the dimensions of the compensating balance of 20 , it would have to leave 20,000 on
rectangular area deposit and would have the use of 0,000. To meet the expenses
Pac age Design A company is designing a pac age for its of retooling, the arber ie Company needs 1 5,000. The Third
product. ne part of the pac age is to be an open box made from National an , with whom the firm has had no prior association,
a square piece of aluminum by cutting out a 2-in. square from requires a compensating balance of 16 . To the nearest thousand
each corner and folding up the sides. (See Figure 1.4.) The box is dollars, what amount of loan is required to obtain the needed
to contain 50 in3 . What are the dimensions of the square piece of funds Now solve the general problem of determining the amount
aluminum that must be used of a loan that is needed to handle expenses E if the ban
requires a compensating balance of p .

2 2
2 2

2 Fold 2
2 2

FIGURE ox construction (Problem 36).

Product Design A candy company ma es the popular

Henney s, whose main ingredient is chocolate. The
rectangular-shaped bar is 10 centimeters (cm) long, 5 cm wide, Incentive Plan A machine company has an incentive
and 2 cm thic . The spot price of chocolate has decreased by plan for its salespeople. For each machine that a salesperson sells,
60 , and the company has decided to reward its loyal customers the commission is 50. The commission for e ery machine sold
with a 50 increase in the volume of the bar The thic ness will will increase by 0.05 for each machine sold over 500. For
remain the same, but the length and width will be increased by example, the commission on each of 502 machines sold is 50.10.
equal amounts. What will be the length and width of the new bar How many machines must a salesperson sell in order to earn
33,000
Product Design A candy company ma es a washer-shaped
candy (a candy with a hole in it) see Figure 1.5. ecause of Real Estate A land investment company purchased a
increasing costs, the company will cut the volume of candy in parcel of land for 7200. After having sold all but 20 acres at a
each piece by 22 . To do this, the firm will eep the same profit of 30 per acre over the original cost per acre, the
company regained the entire cost of the parcel. How many acres
were sold

7.1 Margin of Profit The margin f pr t of a company is

the net income divided by the total sales. A company s margin
2 of profit increased by 0.02 from last year. ast year the
company sold its product at 3.00 each and had a net income of
4500. This year it increased the price of its product by 0.50
each, sold 2000 more, and had a net income of 7140. The
company never has had a margin of profit greater than 0.15. How
FIGURE Washer-shaped candy (Problem 3 ). many of its product were sold last year and how many were sold
this year
thic ness and outer radius but will ma e the inner radius
Business A company manufactures products A and B. The
larger. At present the thic ness is 2.1 millimeters (mm), the
cost of producing each unit of A is 2 more than that of B. The
inner radius is 2 mm, and the outer radius is 7.1 mm. Find the
costs of production of A and B are 1500 and 1000, respectively,
inner radius of the new-style candy. int: The volume of a
and 25 more units of A are produced than of B. How many of each
solid disc is !r2 h, where r is the radius and h is the thic ness of
are produced
the disc.
 Section .2 near ne a t es 55

Objective L I
o so e near ne a t es n one Suppose a and b are two points on the real-number line. Then either a and b coincide,
ar a e an to ntro ce nter a
notat on or a lies to the left of b, or a lies to the right of b. (See Figure 1.6.)
If a and b coincide, then a D b. If a lies to the left of b, we say that a is less than
b and write a < b, where the inequa ity symb < is read is less than. n the other
hand, if a lies to the right of b, we say that a is greater than b, written a > b. The
statements a > b and b < a are equivalent. (If you have trouble eeping these symbols
b
straight, it may help to notice that < loo s somewhat li e the letter for eft and that
a we have a < b precisely when a lies to the eft of b.)
a=b Another inequality symbol ! is read is less than or equal to and is defined as
follows: a ! b if and only if a < b or a D b. Similarly, the symbol " is defined as
follows: a " b if and only if a > b or a D b. In this case, we say that a is greater than
a b or equal to b.
a 6 b, a is less than b We often use the words rea numbers and p ints interchangeably, since there is a
b 7 a, b is greater than a one-to-one correspondence between real numbers and points on a line. Thus, we can
spea of the points #5, #2, 0, 7, and and can write 7 < , #2 > #5, 7 ! 7, and
b a
7 " 0. (See Figure 1.7.) Clearly, if a > 0, then a is positive if a < 0, then a is negative.
a 7 b, a is greater than b
b 6 a, b is less than a a6x6b

FIGURE Relative positions of

-5 -2 0 7 9 a x b
two points.
FIGURE Points on a number line. FIGURE a < x and x < b.

Suppose that a < b and x is between a and b. (See Figure 1. .) Then not only is
a < x, but also, x < b. We indicate this by writing a < x < b. For example, 0 < 7 < .
(Refer bac to Figure 1.7.)

An inequality is a statement that one quantity is less than, or greater than, or less
than or equal to, or greater than or equal to, another quantity.

f course, we represent inequalities by means of inequality symbols. If two inequal-

ities have their inequality symbols pointing in the same direction, then the inequalities
are said to have the same sense. If not, they are said to be pp site in sense. Hence,
a < b and c < d have the same sense, but a < b has the opposite sense of c > d.
Solving an inequality, such as 2.x#3/ < 4, means finding all values of the variable
for which the inequality is true. This involves the application of certain rules, which we
now state.

Ru es f r Inequa ities
If the same number is added to or subtracted from both sides of an inequality, then
another inequality results, having the same sense as the original inequality. Symboli-
cally,
eep in mind that the rules also apply to If a < b; then a C c < b C c and a # c < b # c:
!; >, and ".
For example, 7 < 10 so 7 C 3 < 10 C 3.
If both sides of an inequality are multiplied or divided by the same p siti e num-
ber, then another inequality results, having the same sense as the original inequality.
Symbolically,
a b
If a < b and c > 0; then ac < bc and < :
c c
56 C cat ons an ore e ra

For example, 3 < 7 and 2 > 0 so 3.2/ < 7.2/ and 32 < 72 .
If both sides of an inequality are multiplied or divided by the same negati e number,
then another inequality results, having the opposite sense of the original inequality.
Symbolically,
a b
If a < b and c < 0; then ac > bc and > :
c c
ultiplying or dividing an inequality by For example, 4 < 7 and #2 < 0, so 4.#2/ > 7.#2/ and !2 4 7
> !2 .
a negative number gives an inequality of
the opposite sense. Any side of an inequality can be replaced by an expression equal to it. Symbolically,

If a < b and a D c; then c < b:

For example, if x < 2 and x D y C 4, then y C 4 < 2.

If the sides of an inequality are either both positive or both negative and reciprocals
are ta en on both sides, then another inequality results, having the opposite sense of
the original inequality. Symbolically,
1 1
If 0 < a < b or a < b < 0; then > :
a b
1 1 1 1
For example, 2 < 4 so 2
> 4
and #4 < #2 so !4
> !2
.
If both sides of an inequality are positive and each side is raised to the same positive
power, then another inequality results, having the same sense as the original inequality.
Symbolically,

If 0 < a < b and n > 0; then an < bn :

For n a positive integer, this rule further provides

p pn
If 0 < a < b; then n a < b:
p p
For example, 4 < so 42 < 2 and 4 < .
A pair of inequalities will be said to be equivalent inequalities if when either is
true then the other is true. When any of Rules 1 6 are applied to an inequality, it is easy
to show that the result is an equivalent inequality.
Expanding on the terminology in Section 0.1, a number a is p siti e if 0 < a and
negati e if a < 0. It is often useful to say that a is n nnegati e if 0 ! a.
bserve from Rule 1 that a ! b is equivalent to b # a is nonnegative. Another
simple observation is that a ! b is equivalent to there exists a nonnegative number s
such that a C s D b. The s which does the ob is ust b # a but the idea is useful when
one side of a ! b contains an un nown.
This idea allows us to replace an inequality with an equality at the expense of
introducing a variable. In Chapter 7, the powerful simplex method builds on replace-
ment of inequalities a ! b with equations a C s D b, for nonnegative s. In this context,
s is called a s ack ariab e because it ta es up the slac between a and b.
We will now apply Rules 1 4 to a inear inequa ity.

The definition also applies to !, >,

and ". A linear inequality in the variable x is an inequality that can be written in the form
ax C b < 0
where a and b are constants and a ¤ 0.
We should expect that the inequality will be true for some values of x and false
for others. To s e an inequality involving a variable is to find all values of the
variable for which the inequality is true.
 Section .2 near ne a t es 57
A L IT I
E AM LE S L I
A salesman has a monthly income
given by I D 200 C 0: S, where S Solve 2.x # 3/ < 4.
is the number of products sold in a
month. How many products must he sell S
to ma e at least 4500 a month
S We will replace the given inequality by equivalent inequalities until the
solution is evident.

x65
2.x # 3/ < 4
2x # 6 < 4 Rule 4
FIGURE All real numbers less 2x # 6 C 6 < 4 C 6 Rule 1
than 5. 2x < 10 Rule 4
2x 10
< Rule 2
A L IT I 2 2
x<5 Rule 4
A zoo veterinarian can purchase
four different animal foods with vari- All of the foregoing inequalities are equivalent. Thus, the original inequality is
ous nutrient values for the zoo s grazing true for a real numbers x such that x < 5. For example, the inequality is true for
animals. et x1 represent the number of x D #10; #0:1; 0; 12 , and 4. . We can write our solution simply as x < 5 and present it
bags of food 1, x2 represent the number
geometrically by the colored half-line in Figure 1. . The parenthesis indicates that 5 is
of bags of food 2, and so on. The num-
ber of bags of each food needed can be
n t inc uded in the solution.
described by the following equations: Now ork Problem 9 G
x1 D 150 # x4
In Example 1 the solution consisted of a set of numbers, namely, all real numbers
x2 D 3x4 # 210 less than 5. It is common to use the term interval to describe such a set. In the case
x3 D x4 C 60
of Example 1, the set of all x such that x < 5 can be denoted by the inter a n tati n
.#1; 5/. The symbol #1 is not a number, but is merely a convenience for indicating
Assuming that each variable must be that the interval includes all numbers less than 5.
nonnegative, write four inequalities There are other types of intervals. For example, the set of all real numbers x for
involving x4 that follow from these which a ! x ! b is called a closed interval and includes the numbers a and b, which
equations.
are called endpoints of the interval. This interval is denoted by Œa; b" and is shown in
Figure 1.10(a). The square brac ets indicate that a and b are inc uded in the interval.
(a, b] a6x…b n the other hand, the set of all x for which a < x < b is called an open interval and
a b is denoted by .a; b/. The endpoints are n t inc uded in this set. See Figure 1.10(b).
Extending these concepts and notations, we have the intervals shown in Figure 1.11.
[a, b) a…x6b ust as #1 is not a number, so 1 is not a number but .a; 1/ is a convenient notation
a b
for the set of all real numbers x for which a < x. Similarly, Œa; 1/ denotes all real x for
[a, q) xÚa which a ! x. It is a natural extension of this notation to write .#1; 1/ for the set of
a a real numbers and we will do so throughout this boo .

(a, q) x7a
a
a b a b
(- q, a] x…a Closed interval [a, b] Open interval (a, b)
a (a) (b)

(- q, a) x6a FIGURE Closed and open intervals.

a

(- q, q) -q 6 x 6 q
E AM LE S L I
FIGURE Intervals.
Solve 3 # 2x ! 6.
S 3 # 2x ! 6

ividing both sides by #2 re erses the #2x ! 3 Rule 1

sense of the inequality. 3
x"# Rule 3
2
58 C cat ons an ore e ra

xÚ-2
3 The solution is x " # 32 , or, in interval notation, Œ# 32 ; 1/. This is represented
geometrically in Figure 1.12.
-2
3
Now ork Problem 7 G
FIGURE The interval Œ# 32 ; 1/.
E AM LE S L I

Solve 32 .s # 2/ C 1 > #2.s # 4/.

3
S .s # 2/ C 1 > #2.s # 4/
2
s7
20 ! "
7 3
2 .s # 2/ C 1 > 2.#2.s # 4// Rule 2
20 2
7
3.s # 2/ C 2 > #4.s # 4/
FIGURE The interval . 20
7 ; 1/.
3s # 4 > #4s C 16
7s > 20 Rule 1
20
s> Rule 2
7
20
The solution is . 7 ; 1/ see Figure 1.13.
Now ork Problem 19 G
E AM LE S L I

a Solve 2.x # 4/ # 3 > 2x # 1.

S 2.x # 4/ # 3 > 2x # 1
2x # # 3 > 2x # 1
#11 > #1
Since it is never true that #11 > #1, there is no solution, and the solution set is ;
(the set with no elements).
-q 6 x 6 q b Solve 2.x # 4/ # 3 < 2x # 1.

FIGURE The interval S Proceeding as in part (a), we obtain #11 < #1. This is true for all real
.#1; 1/. numbers x, so the solution is .#1; 1/ see Figure 1.14.
Now ork Problem 15 G

R BLEMS
In Pr b ems s e the inequa ities Gi e y ur ans er in 5 2
x < 40 # x>6
inter a n tati n and indicate the ans er ge metrica y n the 6 3
rea number ine 3y C 1 3y # 2 1
! 5y # 1 "
3x > 21 4x < #2 2 3 4
5x # 11 ! 5x ! 0 #3x C 1 ! #3.x # 2/ C 1 0x ! 0
#4x " 2 2z C 5 > 0 1#t 3t # 7 3.2t C 2/ t#3 t
< > C
5 # 7s > 3 4s # 1 < #5 2 3 2 4 3
3 < 2y C 3 4 ! 3 # 2y 1 1 5
2x C 13 " x # 7 3x # ! x
3 3 2
t C 4 ! 3 C 2t #3 " .2 # x/
2 5 7
3.2 # 3x/ > 4.1 # 4x/ .x C 1/ C 1 < 3.2x/ C 1 r< r t># t
3 6 4 3
2.4x # 2/ > 4.2x C 1/ 7 # .x C 3/ ! 3.3 # x/ y y y 2 # 0:01x
p p p 2y C < C # 0:1x !
xC2< 3#x 2.x C 2/ > .3 # x/ 5 2 3 0:2
 Section .3 cat ons of ne a t es 59

0:1.0:03x C 4/ " 0:02x C 0:434 Geometry In a right triangle, one of the acute angles x is
less than 3 times the other acute angle plus 10 degrees. Solve
3y # 1 5.y C 1/
< for x.
#3 #3
Savings Each month last year, rittany saved more than Spending A student has 360 to spend on a stereo system
50, but less than 150. If S represents her total savings for the and some compact discs. If she buys a stereo that costs 21 and
year, describe S by using inequalities. the discs cost 1 . 5 each, find the greatest number of discs she
can buy.
Labor Using inequalities, symbolize the following
statement: The number of labor hours t to produce a product is at
least 5 and at most 6.

Objective A I
o o e rea fe s t at ons n ter s Solving word problems may sometimes involve inequalities, as the following examples
of ne a t es
illustrate.

E AM LE

For a company that manufactures aquarium heaters, the combined cost for labor and
material is 21 per heater. Fixed costs (costs incurred in a given period, regardless of
output) are 70,000. If the selling price of a heater is 35, how many must be sold for
the company to earn a profit
S

S Recall that
profit D total revenue # total cost
We will find total revenue and total cost and then determine when their difference
is positive.

et q be the number of heaters that must be sold. Then their cost is 21q. The total
cost for the company is therefore 21q C 70;000. The total revenue from the sale of
q heaters will be 35q. Now,
profit D total revenue # total cost
and we want profit > 0. Thus,
total revenue # total cost > 0
35q # .21q C 70;000/ > 0
14q > 70;000
q > 5000
Since the number of heaters must be a nonnegative integer, we see that at least 5001
heaters must be sold for the company to earn a profit.
Now ork Problem 1 G
E AM LE R

A builder must decide whether to rent or buy an excavating machine. If he were to rent
the machine, the rental fee would be 3000 per month (on a yearly basis), and the daily
cost (gas, oil, and driver) would be 1 0 for each day the machine is used. If he were
to buy it, his fixed annual cost would be 20,000, and daily operating and maintenance
costs would be 230 for each day the machine is used. What is the least number of days
each year that the builder would have to use the machine to ustify renting it rather than
buying it
60 C cat ons an ore e ra

S We will determine expressions for the annual cost of renting and the
annual cost of purchasing. We then find when the cost of renting is less than that of
purchasing.

et d be the number of days each year that the machine is used. If the machine is
rented, the total yearly cost consists of rental fees, which are (12)(3000), and daily
charges of 1 0d. If the machine is purchased, the cost per year is 20;000 C 230d.
We want
costrent < costpurchase
12.3000/ C 1 0d < 20;000 C 230d
36;000 C 1 0d < 20;000 C 230d
16;000 < 50d
320 < d
Thus, the builder must use the machine at least 321 days to ustify renting it.
Now ork Problem 3 G
E AM LE C R

The current ratio of a business is the ratio of its current assets (such as cash, merchan-
dise inventory, and accounts receivable) to its current liabilities (such as short-term
loans and taxes payable).
After consulting with the comptroller, the president of the Ace Sports Equipment
Company decides to ta e out a short-term loan to build up inventory. The company
has current assets of 350,000 and current liabilities of 0,000. How much can the
company borrow if the current ratio is to be no less than 2.5 (Note: The funds received
are considered as current assets and the loan as a current liability.)
S et x denote the amount the company can borrow. Then current assets will
be 350;000 C x, and current liabilities will be 0;000 C x. Thus,
current assets 350;000 C x
current ratio D D
current liabilities 0;000 C x
We want
350;000 C x
" 2:5
0;000 C x
Although the inequality that must be Since x is positive, so is 0;000 C x. Hence, we can multiply both sides of the inequality
solved is not apparently linear, it is by 0;000 C x and the sense of the inequality will remain the same. We have
equivalent to a linear inequality.
350;000 C x " 2:5. 0;000 C x/
150;000 " 1:5x
100;000 " x
Consequently, the company may borrow as much as 100,000 and still maintain a cur-
rent ratio greater than or equal to 2.5.
Now ork Problem 8 G
E AM LE C S

In the television show he ire, the detectives declared a homicide to be c eared

if the case was solved. If the number of homicides in a month was > 0 and C
were cleared then the c earance rate was defined to be C= . The section boss, Rawls,
 Section .3 cat ons of ne a t es 61

fearing a smaller clearance rate, is angered when cNulty adds 14 new homicides to
the responsibility of his section. However, the 14 cases are related to each other, so
solving one case will solve them all. If all 14 new cases are solved, will the clearance
rate change, and if so will it be for the better or worse
C
S The question amounts to as ing about the relative size of the fractions
C C 14
and . For nonnegative numbers a and c and positive numbers b, and d, we have
C 14
a c
< if and only if ad < bc
b d
We have 0 < and 0 ! C ! . If C D , a perfect clearance rate, then C C 14 D
C 14 and the clearance rate is still perfect when 14 new cases are both added and
solved.
ut if C < then 14C < 14 and C C 14C < C C 14 shows that
C. C 14/ < .C C 14/ and hence
C C C 14
<
C 14
giving a better clearance rate when 14 new cases are both added and solved.
Now ork Problem 13 G
f course there is nothing special about the positive number 14 in the last example.
Try to formulate a general rule that will apply to Example 4.

R BLEMS
Profit The avis Company manufactures a product received for all magazines sold beyond 100,000. Find the
that has a unit selling price of 20 and a unit cost of 15. If fixed least number of magazines that can be published profitably,
costs are 600,000, determine the least number of units that must if 0 of the issues published are sold.
be sold for the company to have a profit. Production Allocation A company produces alarm cloc s.
Profit To produce 1 unit of a new product, a company uring the regular wor wee , the labor cost for producing one
determines that the cost for material is 1.50 and the cost of cloc is 2.00. However, if a cloc is produced on overtime, the
labor is 5. The fixed cost, regardless of sales volume, is 7000. labor cost is 3.00. anagement has decided to spend no more
If the cost to a wholesaler is .20 per unit, determine the least than a total of 25,000 per wee for labor. The company must
number of units that must be sold by the company to realize a produce 11,000 cloc s this wee . What is the minimum number
profit. of cloc s that must be produced during the regular wor wee
Leasing versus Purchasing A businesswoman wants to Investment A company invests a total of 70,000 of surplus
determine the difference between the costs of owning and leasing funds at two annual rates of interest: 5 and 6 14 . The company
an automobile. She can lease a car for 420 per month (on an wants an annual yield of no less than 5 12 . What is the
annual basis). Under this plan, the cost per mile (gas and oil) is least amount of money that the company must invest at the
0.06. If she were to purchase the car, the fixed annual expense 6 14 rate
would be 4700, and other costs would amount to 0.0 per Current Ratio The current ratio of Precision achine
mile. What is the least number of miles she would have to Products is 3. . If the firm s current assets are 570,000, what
drive per year to ma e leasing no more expensive than are its current liabilities To raise additional funds, what is the
purchasing maximum amount the company can borrow on a short-term basis
Shirt Manufacturer A T-shirt manufacturer produces if the current ratio is to be no less than 2.6 (See Example 3 for an
shirts at a total labor cost (in dollars) of 1.3 and a total material explanation of current ratio.)
cost of 0.4 . The fixed cost for the plant is 6500. If each shirt Sales Allocation At present, a manufacturer has 2500 units
sells for 3.50, how many must be sold by the company to realize of product in stoc . The product is now selling at 4 per
a profit unit. Next month the unit price will increase by 0.50. The
manufacturer wants the total revenue received from the sale of the
2500 units to be no less than 10,750. What is the maximum
number of units that can be sold this month
Revenue Suppose consumers will purchase q units of a

Publishing The cost of publication of each copy of a product at a price of 200

q C 3 dollars per unit. What is the
magazine is 1.30. It is sold to dealers for 1.50 per copy. minimum number of units that must be sold in order that sales
The amount received for advertising is 20 of the amount revenue be greater than 000
62 C cat ons an ore e ra

ourly Rate Painters are often paid either by the hour or yearly salary. ne method pays 50,000 plus a bonus of 2 of
on a per- ob basis. The rate they receive can affect their wor ing your yearly sales. The other method pays a straight 4
speed. For example, suppose they can wor either for .00 per commission on your sales. For what yearly sales amount is it
hour or for 320 plus 3 for each hour less than 40 if they better to choose the second method
complete the ob in less than 40 hours. Suppose the ob will ta e Fractions If a, b, and c are positive numbers, investigate the
t hours. If t " 40, clearly the hourly rate is better. If t < 40, for aCc
what values of t is the hourly rate the better pay scale value of when c is ta en to be a ery large number.
bCc
Acid est Ratio The acid test rati (or quick rati )
of a business is the ratio of its liquid assets cash and securities
plus accounts receivable to its current liabilities. The minimum
acid test ratio for a financially healthy company is around 1.0, but
the standard varies somewhat from industry to industry. If a
company has 450,000 in cash and securities and has 3 ,000
in current liabilities, how much does it need to be carrying as
accounts receivable in order to eep its acid test ratio at or
Compensation Suppose a company offers you a sales above 1.3
position with your choice of two methods of determining your

Objective A
o so e e at ons an ne a t es
n o n a so te a es
A E
n the real-number line, the distance of a number x from 0 is called the absolute value
of x and is denoted by jxj. For example, j5j D 5 and j#5j D 5 because both 5 and #5
are 5 units from 0. (See Figure 1.15.) Similarly, j0j D 0. Notice that jxj can never be
negative that is, jxj " 0.
If x is positive or zero, then jxj is simply x itself, so we can omit the vertical bars
5 units 5 units
and write jxj D x. n the other hand, consider the absolute value of a negative number,
li e x D #5.
-5 0 5
5 = -5 = 5
jxj D j#5j D 5 D #.#5/ D #x
FIGURE Absolute value.

Thus, if x is negative, then jxj is the positive number #x. The minus sign indicates
that we have changed the sign of x. The geometric definition of absolute value as a
distance is equivalent to the following:

The absolute value of a real number x, written jxj, is defined by

#
x if x " 0
jxj D
#x if x < 0

p p
x2 is not necessarily
p x but x2 D jxj.
For example, .#2/2 D j#2j D 2, not bserve that j#xj D jxj follows from the definition.
#2. Applying the definition, we have j3j D 3; j# j D #.# / D , and j 12 j D 1
2
.
Also, #j2j D #2 and #j#2j D #2.
Also, j#xj is not necessarily x and, thus, j#x # 1j is not necessarily x C 1.
For example, if we let x D #3, then j#.#3/j ¤ #3, and

j#.#3/ # 1j ¤ #3 C 1
 Section .4 so te a e 63

E AM LE S A E

a Solve jx # 3j D 2.
S This equation states that x # 3 is a number 2 units from 0. Thus, either
x#3D2 or x # 3 D #2
Solving these equations gives x D 5 or x D 1. See Figure 1.16.
b Solve j7 # 3xj D 5.
S The equation is true if 7 # 3x D 5 or if 7 # 3x D #5. Solving these
equations gives x D 23 or x D 4.
c Solve jx # 4j D #3.
S The absolute value of a number is never negative, so the solution set
is ;.
Now ork Problem 19 G
2 units 2 units
We can interpret ja # bj D j#.b # a/j D jb # aj as the distance between a and b.
x x For example, the distance between 5 and can be calculated via
1 3 5 either j # 5j D j4j D 4
FIGURE The solution of or j5 # j D j#4j D 4
jx # 3j D 2 is 1 or 5. Similarly, the equation jx # 3j D 2 states that the distance between x and 3 is
2 units. Thus, x can be 1 or 5, as shown in Example 1(a) and Figure 1.17.

-3 6 x 6 3
A I
et us turn now to inequalities involving absolute values. If jxj < 3, then x is less than
3 units from 0. Hence, x must lie between #3 and 3 that is, on the interval #3 < x < 3.
0 3
See Figure 1.17(a). n the other hand, if jxj > 3, then x must be greater than 3 units
-3
(a) Solution of x 6 3 from 0. Hence, there are two intervals in the solution: Either x < #3 or x > 3. See
Figure 1.17(b). We can extend these ideas as follows: If jxj ! 3, then #3 ! x ! 3
if jxj " 3, then x ! #3 or x " 3. Table 1.1 gives a summary of the solutions to
x 6 -3 x73
absolute-value inequalities.

-3 0 3 Table 1.1
(b) Solution of x 7 3 Inequality (d > 0) Solution

FIGURE Solutions of jxj < 3 jxj < d #d < x < d

and jxj > 3.
jxj ! d #d ! x ! d
jxj > d x < #d or x > d
jxj " d x ! #d or x " d

E AM LE S A I

a Solve jx # 2j < 4.
S The number x # 2 must be less than 4 units from 0. From the preceding
discussion, this means that #4 < x # 2 < 4. We can set up the procedure for
solving this inequality as follows:
#4 < x # 2 < 4
-2 6 x 6 6 #4 C 2 < x < 4 C 2 adding 2 to each member
-2 6 #2 < x < 6
FIGURE The solution of Thus, the solution is the open interval .#2; 6/. This means that all numbers between
jx # 2j < 4 is the interval .#2; 6/. #2 and 6 satisfy the original inequality. (See Figure 1.1 .)
64 C cat ons an ore e ra

b Solve j3 # 2xj ! 5.
S #5 ! 3 # 2x ! 5
#5 # 3 ! #2x ! 5 # 3 subtracting 3 throughout
# ! #2x ! 2
4 " x " #1 dividing throughout by #2
#1 ! x ! 4 rewriting
Note that the sense of the original inequality was re ersed when we divided by a
negative number. The solution is the closed interval Œ#1; 4".
Now ork Problem 29 G
E AM LE S A I

a Solve jx C 5j " 7.
S Here x C 5 must be at east 7 units from 0. Thus, either x C 5 ! #7 r
x C 5 " 7. This means that either x ! #12 r x " 2. Thus, the solution consists of
two intervals: .#1; #12" and Œ2; 1/. We can abbreviate this collection of numbers
x … -12 or xÚ2 by writing

-12 2 .#1; #12" [ Œ2; 1/

FIGURE The union where the connecting symbol [ is called the uni n symbol. (See Figure 1.1 .)
.#1; #12" [ Œ2; 1/. ore formally, the union of sets A and B is the set consisting of all elements that
are in either A or B (or in both A and B).
The inequalities x ! #12 or x " 2 in (a) b Solve j3x # 4j > 1.
and x < 1 r x > 53 in (b) do not give
rise to a single interval as in Examples 2a
S Either 3x # 4 < #1 r 3x # 4 > 1. Thus, either 3x < 3 r 3x > 5.
and 2b. Therefore, x < 1 r x > 53 , so the solution consists of all numbers in the set
.#1; 1/ [ . 53 ; 1/.
Now ork Problem 31 G
A L IT I E AM LE A N
Express the following statement
using absolute-value notation: The Using absolute-value notation, express the following statements:
actual weight of a box of cereal must a x is less than 3 units from 5.
be within 0.3 oz of the weight stated on
the box, which is 22 oz. S jx # 5j < 3

b x differs from 6 by at least 7.

S jx # 6j " 7

c x < 3 and x > #3 simultaneously.

S jxj < 3

d x is more than 1 unit from #2.

S jx # .#2/j > 1
jx C 2j > 1

e x is less than # (a ree letter read sigma ) units from $ (a ree letter read
mu ).
S jx # $j < #

f x is within % (a ree letter read epsilon ) units from a.

S jx # aj < %
Now ork Problem 11 G
 Section .4 so te a e 65

A
Five basic properties of the absolute value are as follows:

jabj D jaj $ jbj

ˇaˇ
ˇ ˇ jaj
ˇ ˇD
b jbj
ja # bj D jb # aj
#jaj ! a ! jaj
ja C bj ! jaj C jbj

For example, Property 1 states that the absolute value of the product of two numbers
is equal to the product of the absolute values of the numbers. Property 5 is nown as
the triang e inequa ity.

E AM LE A

a j.#7/ $ 3j D j#7j $ j3j D 21

b j4 # 2j D j2 # 4j D 2
c j7 # xj D jx # 7j
ˇ ˇ ˇ ˇ
ˇ #7 ˇ j#7j 7 ˇ #7 ˇ j#7j 7
d ˇˇ ˇˇ D D I ˇˇ ˇˇ D D
3 j3j 3 #3 j#3j 3
ˇ ˇ
ˇx # 3ˇ
e ˇˇ ˇ D jx # 3j D jx # 3j
#5 ˇ j#5j 5
f #j2j ! 2 ! j2j
g j.#2/ C 3j D j1j D 1 ! 5 D 2 C 3 D j#2j C j3j
Now ork Problem 5 G

R BLEMS
In Pr b ems e a uate the abs ute a ue expressi n Find all values of x such that jx # $j < 3#.
!1
j#13j j2 j j3 # 5j
In Pr b ems s e the gi en equati n r inequa ity
j.#3 # 5/=2j j2.# 72 /j j3 # 5j # j5 # 3j ˇxˇ
ˇ ˇ
p jxj D 7 j#xj D 2 ˇ ˇD7
jxj < 4 jxj < #1 j3 # 10j ˇ ˇ 5
ˇ3ˇ
p ˇ ˇD7 jx # 5j D j4 C 3xj D 6
j 5 # 2j ˇxˇ
Using the absolute-value symbol, express each fact. j5x # 2j D 0 j7x C 3j D x j3 # 5xj D 2
a x is less than 3 units from 7.
b x differs from 2 by less than 3. j5 # 3xj D 7 jxj < for >0
ˇxˇ ˇxˇ 1
c x is no more than 5 units from 7. ˇ ˇ ˇ ˇ
j#xj < 3 ˇ ˇ>2 ˇ ˇ>
d The distance between 7 and x is 4. 4 2 3
e x C 4 is less than 2 units from 0. ˇ ˇ
ˇ 1 ˇ 1
f x is between #3 and 3, but is not equal to 3 or #3. jx C 7j < 3 j2x # 17j < #4 ˇx # ˇ >
ˇ 2ˇ 2
g x < #6 or x > 6.
h The number x of hours that a machine will operate e ciently j1 # 3xj > 2 j3 # 2xj ! 2 j3x # 2j " 0
differs from 105 by less than 3. ˇ ˇ ˇ ˇ
ˇ 3x # ˇ ˇx # 7ˇ
ˇ ˇ ˇ ˇ
i The average monthly income x (in dollars) of a family differs ˇ 2 ˇ"4 ˇ 3 ˇ!5
from 50 by less than 100.
Use absolute-value notation to indicate that f .x/ and differ In Pr b ems express the statement using abs ute a ue
by less than %. n tati n
Use absolute-value notation to indicate that the prices p1 and In a science experiment, the measurement of a distance d is
p2 of two products differ by at least 5 dollars. 35.2 m, and is accurate to ˙20 cm.
66 C cat ons an ore e ra

The difference in temperature between two chemicals that are Find those values of x such that jx # $j > h#.
to be mixed must be at least 5 degrees and at most 10 degrees. Manufacturing olerance In the manufacture of widgets,
Statistics In statistical analysis, the Chebyshev inequality the average dimension of a part is 0.01 cm. Using the
asserts that if x is a random variable, $ is its mean, and # is its absolute-value symbol, express the fact that an individual
standard deviation, then measurement x of a part does not differ from the average by more
1 than 0.005 cm.
.probability that jx # $j > h#/ " 2
h

Objective S N
o r te s s n s at on notat on There was a time when school teachers made their students add up all the positive
an e a ate s c s s
integers from 1 to 105 (say), perhaps as punishment for unruly behavior while the
teacher was out of the classroom. In other words, the students were to find
1 C 2 C 3 C 4 C 5 C 6 C 7 C $ $ $ C 104 C 105
A related exercise was to find
1 C 4 C C 16 C $ $ $ C 1 C 100 C 121
The three dots notation is supposed to convey the idea of continuing the tas , using
the same pattern, until the last explicitly given terms have been added, too. With this
notation there are no hard and fast rules about how many terms at the beginning and
end are to be given explicitly. The custom is to provide as many as are needed to ensure
that the intended reader will find the expression unambiguous. This is too imprecise
for many mathematical applications.
Suppose that for any positive integer i we define ai D i2 . Then, for example,
a6 D 36 and a D 64. The instruction, Add together the numbers ai , for i ta ing
on the integer values 1 through 11 inclusive is a precise statement of Equation (2).
It would be precise regardless of the formula defining the values ai , and this leads to
the following:

If, for each positive integer i there is given a unique number ai , and m and n are
positive integers with m ! n, then the sum of the numbers ai ; with i successively
ta ing on all the integer values in the interval Œm; n" is denoted
X
n
ai
iDm

Thus,
X
The -notation on the left side of (3) X
n

eliminates the imprecise dots on the right. ai D am C amC1 C amC2 C $ $ $ C an

iDm
X
The is the ree capital letter read sigma , from which we get the letter S. It stands
Xn
for sum and the expression ai , can be read as the the sum of all numbers ai ,
iDm
i ranging from m to n (through positive integers being understood). The description of
ai may be very simple. For example, in Equation (1) we have ai D i and
X
105
i D 1 C 2 C 3 C $ $ $ C 105
iD1
while Equation (2) is
X
11
i2 D 1 C 4 C C $ $ $ C 121
iD1
 Section .5 at on otat on 67

We have merely defined a notation, which is called summation notation. In Equa-

tion (3), i is the index of summation and m and n are called the bounds of summation.
It is important to understand from the outset that the name of the index of summation
can be replaced by any other so that we have
X
n X
n X
n X
n
ai D aj D a˛ D a
iDm jDm ˛Dm Dm

for example. In each case, replacing the index of summation by the positive integers m
through n successively, and adding gives
am C amC1 C amC2 C $ $ $ C an
We now illustrate with some concrete examples.

E AM LE E S

Evaluate the given sums.

X
7
a .5n # 2/
nD3

S
X
7
.5n # 2/ D Œ5.3/ # 2" C Œ5.4/ # 2" C Œ5.5/ # 2" C Œ5.6/ # 2" C Œ5.7/ # 2"
nD3
D 13 C 1 C 23 C 2 C 33
D 115
X
6
b . j2 C 1/
jD1

S
X
6
. j2 C 1/ D .12 C 1/ C .22 C 1/ C .32 C 1/ C .42 C 1/ C .52 C 1/ C .62 C 1/
jD1

D 2 C 5 C 10 C 17 C 26 C 37
D 7
Now ork Problem 5 G
E AM LE S U S N

Write the sum 14 C 16 C 1 C 20 C 22 C $ $ $ C 100 in summation notation.

S There are many ways to express this sum in summation notation. ne method
is to notice that the values being added are 2n, for n D 7 to 50. The sum can thus be
written as
X50
2n
nD7
Another method is to notice that the values being added are 2k C 12, for k D 1 to 44.
The sum can thus also be written as
X44
.2k C 12/
kD1

Now ork Problem 9 G

68 C cat ons an ore e ra

Since summation notation is used to express the addition of terms, we can use
the properties of addition when performing operations on sums written in summation
notation. y applying these properties, we can create a list of properties and formulas
for summation notation.
y the distributive property of addition,
ca1 C ca2 C $ $ $ C can D c.a1 C a2 C $ $ $ C an /
So, in summation notation,
X
n X
n
cai D c ai
iDm iDm

Note that c must be constant with respect to i for Equation (6) to be used.
y the commutative property of addition,

a1 C b1 C a2 C b2 C $ $ $ C an C bn D a1 C a2 C $ $ $ C an C b1 C b2 C $ $ $ C bn

So we have
X
n X
n X
n
.ai C bi / D ai C bi
iDm iDm iDm

Sometimes we want to change the bounds of summation.

X
n X
pCn!m
ai D aiCm!p
iDm iDp

A sum of 37 terms can be regarded as the sum of the first 17 terms plus the sum of the
next 20 terms. The next rule generalizes this observation.

X
p!1
X
n X
n
ai C ai D ai
iDm iDp iDm

In addition to these four basic rules, there are some other rules worth noting.
X
n
1Dn
iD1
P
This is because niD1 1 is a sum of n terms, each of which is 1. The next follows from
combining Equation (6) and Equation (10).
X
n
c D cn
iD1

Similarly, from Equations (6) and (7) we have

X
n X
n X
n
.ai # bi / D ai # bi
iDm iDm iDm

Formulas (14) and (15) are best established by a proof method called mathematical
induction, which we will not demonstrate here.
X
n
n.n C 1/
iD
iD1
2
X
n
n.n C 1/.2n C 1/
i2 D
iD1
6
X
n
n2 .n C 1/2
i3 D
iD1
4
 Section .5 at on otat on 69

However, there is an easy derivation of Formula (13). If we add the following equations,
vertically, term by term,
X
n
i D 1 C 2 C 3 C $$$ C n
iD1
Xn
i D n C .n # 1/ C .n # 2/ C $ $ $ C 1
iD1
we get
X
n
2 i D .n C 1/ C .n C 1/ C .n C 1/ C $ $ $ C .n C 1/
iD1
and since there are n terms on the right, we have
X
n
2 i D n.n C 1/
iD1
and, finally
X
n
n.n C 1/
iD
iD1
2
bserve that if a teacher assigns the tas of finding
1 C 2 C 3 C 4 C 5 C 6 C 7 C $ $ $ C 104 C 105
as a punishment and if he or she nows the formula given by Formula (13), then a
student s wor can be chec ed quic ly by
X
105
105.106/
iD D 105 $ 53 D 5300 C 265 D 5565
iD1
2

E AM LE A S N

Evaluate the given sums.

X
100 X
100 X
200
a 4 b .5k C 3/ c k2
jD30 kD1 kD1

S
a X
100 X
71
4D 4 by Equation ( )
jD30 jD1
D 4 $ 71 by Equation (11)
D2 4

b X
100 X
100 X
100
.5k C 3/ D 5k C 3 by Equation (7)
kD1 kD1 kD1
! !
X
100 X
100
D5 k C3 1 by Equation (6)
kD1 kD1
! "
100 $ 101
D5 C 3.100/ by Equations (13) and (10)
2
D 25;250 C 300
D 25;550
70 C cat ons an ore e ra

c X
200 X
200
k2 D k2 by Equation (6)
kD1 kD1
! "
200 $ 201 $ 401
D by Equation (14)
6
D 24;1 0;300
Now ork Problem 19 G
R BLEMS
n !
X "
In Pr b ems and gi e the b unds f summati n and the index 1 X
200
f summati n f r each expressi n 5$ .k # 100/
kD1
n
X
17 X
450 kD1
2
. t # 5t C 3/ . m # 4/ X
100 Xn
n 3
tD12 mD3 10k k
kD1
n C 1
kD51
In Pr b ems e a uate the gi en sums
X
20 X
100
3k2 # 200k
X
5 X
3
.3i2 C 2i/
3i 7q iD1 kD1
101
iD1 qD0
X
100 X
50
X X
11
k2 .k C 50/2
.10k C 16/ .2n # 3/ kD51 kD1

"2 ! ! "!
kD3 nD7
X !
In Pr b ems express the gi en sums in summati n n tati n k 1
3#
36 C 37 C 3 C 3 C $ $ $ C 60 kD1
10 10
X
100 ! "2 ! ! "!
1 C C 27 C 64 C 125 1 1
3# j
32 C 33 C 34 C 35 C 36 jD1
100 50
11 C 15 C 1 C 23 C $ $ $ C 71 X
n ! "2 ! !
3 3
2 C 4 C C 16 C 32 C 64 C 12 C 256 5# $k
kD1
n n
10 C 100 C 1000 C $ $ $ C 100;000;000 X
n
k2
In Pr b ems e a uate the gi en sums .n C 1/.2n C 1/
kD1
X
75 X
75
10 10
kD1 kD1

Objective S
o ntro ce se ences art c ar
ar t et c an eo etr c se ences I
an t e r s s
Consider the following list of five numbers:
p p p p
2; 2 C 3; 2 C 2 3; 2 C 3 3; 2 C 4 3
If it is understood that the ordering of the numbers is to be ta en into account, then
such a list is called a sequence of length 5 and it is considered to be different from
p p p p
2; 2 C 3 3; 2 C 3; 2 C 4 3; 2 C 2 3
which is also a sequence of length 5. oth of these sequences are different again from
p p p p
2; 2; 2 C 3; 2 C 2 3; 2 C 3 3; 2 C 4 3
which is a sequence of length 6. However, each of the sequences (1), (2), and (3) ta es
on all the numbers in the 5-element set
oth rearrangements and repetitions o p p p p
affect a sequence. f2; 2 C 3; 2 C 2 3; 2 C 3 3; 2 C 4 3g
 Section .6 e ences 71

In Section 0.1 we emphasized that a set is determined by its e ements, and neither
repetitions nor rearrangements in a listing affect the set . Since both repetitions and
rearrangements do affect a sequence, it follows that sequences are not the same as sets.
We will also consider listings such as
p p p p p
2; 2 C 3; 2 C 2 3; 2 C 3 3; 2 C 4 3; $ $ $ ; 2 C k 3; $ $ $
and
1; #1; 1; #1; 1; $ $ $ ; .#1/kC1 ; $ $ $
oth are examples of what is called an infinite sequence. However, note that the
infinite sequence (4) involves the infinitely many different numbers in the set
p
f2 C k 3jk a nonnegative integerg
while the infinite sequence (5) involves only the numbers in the finite set
f#1; 1g
For n a positive integer, ta ing the first n numbers of an infinite sequence results
in a sequence of length n. For example, ta ing the first five numbers of the infinite
sequence (4) gives the sequence (1). The following more formal definitions are helpful
in understanding the somewhat subtle idea of a sequence.

For n a positive integer, a sequence of length n is a rule that assigns to each element
of the set f1; 2; 3; $ $ $ ; ng exactly one real number. The set f1; 2; 3; $ $ $ ; ng is called
the domain of the sequence of length n. A finite sequence is a sequence of length
n for some positive integer n.

An infinite sequence is a rule which assigns to each element of the set of all posi-
tive integers f1; 2; 3; $ $ $g exactly one real number. The set f1; 2; 3; $ $ $g is called the
domain of the infinite sequence.

The word ru e in both definitions may appear vague but the point is that for any
sequence there must be a definite way of specifying exactly one number for each of the
elements in its domain. For a finite sequence the rule can be given by simply listing the
numbers in the sequence. There is no requirement that there be a discernible pattern
(although in practice there often is). For example,
3
; #!; ; 102:7
5
is a perfectly good sequence of length 4. For an infinite sequence there should be some
sort of procedure for generating its numbers, one after the other. However, the proce-
dure may fail to be given by a simple formula. The infinite sequence
2; 3; 5; 7; 11; 13; 17; 1 ; 23; $ $ $
is very important in number theory, but its rule is not given by a mere formula. (What
is apparent y the rule that gives rise to this sequence In that event, what is the next
number in this sequence after those displayed )
We often use letters li e a, b, c, and so on, for the names of sequences. If the
sequence is called a, we write a1 for the number assigned to 1, a2 for the number
assigned to 2, a3 for the number assigned to 3, and so on. In general, for k in the domain
of the sequence, we write ak for the number assigned to k and call it the kth term of the
sequence. (If you have studied Section 1.5 on summation, you will already be familiar
with this notation.) In fact, rather than listing all the numbers of a sequence by
a1 ; a2 ; a3 ; : : : ; an
72 C cat ons an ore e ra

or an indication of all the numbers such as

a1 ; a2 ; a3 ; : : : ; ak ; : : :
a sequence is often denoted by .ak /. Sometimes .ak /nkD1 is used to indicate that the
sequence is finite, of length n, or .ak /1
kD1 is used to emphasize that the sequence is
infinite. The range of a sequence .ak / is the set
fak jk is in the domain of ag
Notice that
f.#1/kC1 jk is a positive integerg D f#1; 1g
so an infinite sequence may have a finite range. If a and b are sequences, then, by
definition, a D b if and only if a and b have the same domain and, for all k in the
common domain, ak D bk .

A L IT I E AM LE L T S
A fast-food chain had 1012 restau- a ist the first four terms of the infinite sequence .ak /1
kD1 whose kth term is given by
rants in 2015. Starting in 2016 it plans to ak D 2k2 C 3k C 1.
expand its number of outlets by 27 each S We have a1 D 2.12 / C 3.1/ C 1 D 6, a2 D 2.22 / C 3.2/ C 1 D 15,
year for six years. Writing rk for the a3 D 2.3 / C 3.3/ C 1 D 2 , and a4 D 2.42 / C 3.4/ C 1 D 45. So the first four
2
number of restaurants in year k, mea-
terms are
sured from 2014, list the terms in the
sequence .rk /7kD1 . 6; 15; 2 ; 45
" !
kC1 k
b ist the first four terms of the infinite sequence .ek /, where ek D .
k
! " ! "1 ! " ! "2
1C1 1 2 2C1 2 3
S We have e1 D D D 2, e2 D D D
! "3 ! "3 1 1
! "4 ! "4 2 2
3C1 4 64 4C1 5 625
, e3 D D D , e4 D D D .
4 3 3 27 4 4 256
! "
3 6
c isplay the sequence .
2k!1 kD1
S Noting that 20 D 1, we have
3 3 3 3 3
3; ; ; ; ;
2 4 16 32
Now ork Problem 3 G
A L IT I E AM LE G F S
A certain inactive ban account that a Write 41; 44; 47; 50; 53 in the form .ak /5kD1 .
bears interest at the rate of 6 com-
pounded yearly shows year-end bal- S Each term of the sequence is obtained by adding three to the previous
ances, for four consecutive years, of term. Since the first term is 41, we can write the sequence as .41 C .k # 1/3/5kD1 .
.57, 10.14, 10.75, 11.40. Write bserve that this formula is not unique. The sequence is also described by
the sequence of amounts in the form .3 C 3k/5kD1 and by .32 C .k C 2/3/5kD1 , to give ust two more possibilities.
.ak /4kD1 . b Write the sequence 1; 4; ; 16; : : : in the form .ak /.
S The sequence is apparent y the sequence of squares of positive integers,
$ %1
so .k2 / or k2 kD1 would be regarded as the correct answer by most people. ut the
sequence described by (k4 # 10k3 C 36k2 # 50k C 24) also has its first four terms
given by 1, 4, , 16, and yet its fifth term is 4 . The sixth and seventh terms are 156
and 40 , respectively. The point we are ma ing is that an infinite sequence cannot
be determined by finitely many values alone.
n the other hand, it is correct to write
1; 4; ; 16; : : : ; k2 ; : : : D .k2 /
 Section .6 e ences 73

because the display on the left side of the equation ma es it clear that the general
term is k2 .
Now ork Problem 9 G
E AM LE E S

Show that the sequences ..i C 3/2 /1 2 1

iD1 and . j C 6j C /jD1 are equal.

S oth ..i C 3/2 /1 2 1

iD1 and . j C 6j C /jD1 are explicitly given to have the same
domain, namely f1; 2; 3; : : :g, the infinite set of all positive integers. The names i and j
being used to name a typical element of the domain are unimportant. The first sequence
is the same as ..k C 3/2 /1 2
kD1 , and the second sequence is the same as .k C 6k C /kD1 .
1
2
The first rule assigns to any positive integer k, the number .k C 3/ , and the second
assigns to any positive integer k, the number k2 C 6k C . However, for all k, .k C 3/2 D
k2 C 6k C , so by the definition of equality of sequences the sequences are equal.
Now ork Problem 13 G
R S
Suppose that a is a sequence with
a1 D 1 and, for each positive integer k; akC1 D .k C 1/ak
Ta ing k D 1, we see that a2 D .2/a1 D .2/1 D 2, while with k D 2 we have
a3 D .3/a2 D .3/2 D 6. A sequence whose rule is defined in terms of itself evaluated
at smaller values, and some explicitly given small values, is said to be recursively
defined. Thus, we can say that there is a sequence a recursively defined by (6) above.
Another famous example of a recursively defined sequence is the Fibonacci
sequence:
F1 D 1 and F2 D 1 and, for each positive integer k; FkC2 D FkC1 C Fk
Ta ing k D 1, we see that F3 D F2 C F1 D 1 C 1 D 2, F4 D F3 C F2 D 2 C 1 D 3,
F5 D F4 C F3 D 3 C 2 D 5. In fact, the first ten terms of .Fk / are
1; 1; 2; 3; 5; ; 13; 21; 34; 55

E AM LE A R

a Use the recursive definition (6) to determine a5 (without referring to the earlier
calculations).
S We have
a5 D .5/a4
D .5/.4/a3
D .5/.4/.3/a2
D .5/.4/.3/.2/a1
D .5/.4/.3/.2/.1/
D 120
The standard notation for ak as defined by (6) is kŠ and it is read k factorial . We
also define 0Š D 1.
b Use the recursive definition (7) to determine F6 .
S F6 D F5 C F4
D .F4 C F3 / C .F3 C F2 /
D F4 C 2F3 C F2
D .F3 C F2 / C 2.F2 C F1 / C F2
74 C cat ons an ore e ra

D F3 C 4F2 C 2F1
D .F2 C F1 / C 4F2 C 2F1
D 5F2 C 3F1
D 5.1/ C 3.1/
D
Now ork Problem 17 G
In Example 4 we deliberately avoided ma ing any numerical evaluations until a
terms had been expressed using only those terms whose values were given explicitly in
the recursive definition. This helps to illustrate the structure of the recursive definition
in each case.
While recursive definitions are very useful in applications, the computations in
Example 4(b) underscore that for large values of k, the computation of the kth term
may be time-consuming. It is desirable to have a simple formula for ak that does not
refer to a , for < k. Sometimes it is possible to find such a c sed formula. In the case
of (6) it is easy to see that ak D k $ .k # 1/ $ .k # 2/ $ : : : $ 3 $ 2 $ 1. n the other hand, in
the case of (7), it is not so easy to derive
p !k p !k
1 1C 5 1 1# 5
Fk D p #p
5 2 5 2

A S G S

An arithmetic sequence is a sequence .bk / defined recursively by

b1 D a and, for each positive integer k; bkC1 D d C bk
for fixed real numbers a and d.

In words, the definition tells us to start the sequence at a and get the next term
by adding d (no matter which term is currently under consideration). The number a
is simply the first term of the arithmetic sequence. Since the recursive definition gives
bkC1 # bk D d, for every positive integer k, we see that the number d is the difference
between any pair of successive terms. It is, accordingly, called the common difference
of the arithmetic sequence. Any pair of real numbers a and d determines an infinite
arithmetic sequence. y restricting to a finite number of terms, we can spea of finite
arithmetic sequences.

A L IT I E AM LE L A S
In 200 the enrollment at Spring-
field High was 1237, and demographic Write explicitly the terms of an arithmetic sequence of length 6 with first term a D 1:5
studies suggest that it will decline by and common difference d D 0:7.
12 students a year for the next seven
S et us write .bk / for the arithmetic sequence. Then
years. ist the pro ected enrollments of
Springfield High. b1 D 1:5
b2 D 0:7 C b1 D 0:7 C 1:5 D 2:2
b3 D 0:7 C b2 D 0:7 C 2:2 D 2:
b4 D 0:7 C b3 D 0:7 C 2: D 3:6
b5 D 0:7 C b4 D 0:7 C 3:6 D 4:3
b6 D 0:7 C b5 D 0:7 C 4:3 D 5:0
Thus the required sequence is
1:5; 2:2; 2: ; 3:6; 4:3; 5:0

Now ork Problem 21 G

 Section .6 e ences 75

A geometric sequence is a sequence .ck / defined recursively by

c1 D a and, for each positive integer k; ckC1 D ck $ r
for fixed real numbers a and r.

In words, the definition tells us to start the sequence at a and get the next term by
multiplying by r (no matter which term is currently under consideration). The number
a is simply the first term of the geometric sequence. Since the recursive definition gives
ckC1 =ck D r for every positive integer k with ck ¤ 0, we see that the number r is the
ratio between any pair of successive terms, with the first of these not 0. It is, accordingly,
called the ommon ratio of the geometric sequence. Any pair of real numbers a and r
determines an infinite geometric sequence. y restricting to a finite number of terms,
we can spea of finite geometric sequences.

A L IT I E AM LE L G S
The population of the rural area sur- p
rounding Springfield is declining as a Write explicitly the terms of a geometric sequence of length 5 with first term a D 2
result of movement to the urban core. In and common ratio r D 1=2.
200 it was 23,500, and each year, for
S et us write .ck / for the geometric
p sequence. Then
the next four years, it is expected to be
only 2 of the previous year s popula- c1 p D p 2
tion. ist the anticipated annual popula- c2 D .c1 / $ 1=2 D p. 2/1=2 D p2=2
tion numbers for the rural area. c3 D .c2 / $ 1=2 D .p2=2/1=2 D p2=4
c4 D .c3 / $ 1=2 D .p2=4/1=2 D p 2=
c5 D .c4 / $ 1=2 D . 2= /1=2 D 2=16
Thus, the required sequence is
p p p p p
2; 2=2; 2=4; 2= ; 2=16

Now ork Problem 25 G

We have remar ed that sometimes it is possible to determine an explicit formula for
the kth term of a recursively defined sequence. This is certainly the case for arithmetic
and geometric sequences.

E AM LE F k A S

Find an explicit formula for the kth term of an arithmetic sequence .bk / with first term
a and common difference d.
S We have
b1 D a D 0d C a
b2 D d C .b1 / D d C .0d C a/ D 1d C a
b3 D d C .b2 / D d C .1d C a/ D 2d C a
b4 D d C .b3 / D d C .2d C a/ D 3d C a
b5 D d C .b4 / D d C .3d C a/ D 4d C a
It appears that, for each positive integer k, the kth term of an arithmetic sequence .bk /
with first term a and common difference d is given by

bk D .k # 1/d C a

This is true and follows easily via the proof method called mathematical induction,
which we will not demonstrate here.
Now ork Problem 29 G
76 C cat ons an ore e ra

E AM LE F k G S

Find an explicit formula for the kth term of a geometric sequence .ck / with first term a
and common ratio r.
S We have
c1 D a D ar0
c2 D .c1 / $ r D ar0 r D ar1
c3 D .c2 / $ r D ar1 r D ar2
c4 D .c3 / $ r D ar2 r D ar3
c5 D .c4 / $ r D ar3 r D ar4
It appears that, for each positive integer k, the kth term of a geometric sequence .ck /
with first term a and common difference r is given by
ck D ark!1
This is true and also follows easily via mathematical induction.
Now ork Problem 31 G
It is clear that any arithmetic sequence has a unique first term a and a unique com-
mon difference d. For a geometric sequence we have to be a little more careful. From
(11) we see that if any term ck is 0, then either a D 0 or r D 0. If a D 0, then every
term in the geometric sequence is 0. In this event, there is not a uniquely determined r
because r $ 0 D 0, for any r. If a ¤ 0 but r D 0, then every term except the first is 0.

S S
For any sequence .ck / we can spea of the sum of the first k terms. et us call this sum
sk . Using the summation notation introduced in Section 1.5, we can write
X
k
sk D ci D c1 C c2 C $ $ $ C ck
iD1

We can regard the sk as terms of a new sequence .sk /, of sums, associated to the original
sequence .sk /. If a sequence .ck / is finite, of length n, then sn can be regarded as t e
sum o t e sequen e.

E AM LE F S A S
A L IT I
If a company has an annual revenue Find a formula for the sum sn of the first n terms of an arithmetic sequence .bk / with
of 27 in 200 and revenue grows by first term a and common difference d.
1.5 each year, find the total revenue
through 200 2015, inclusive. S Since the arithmetic sequence .bk / in question has, by Example 7, bk D
.k # 1/d C a, the required sum is given by
X
n X
n X
n X
n X
n
sn D bk D ..k # 1/d C a/ D .dk # .d # a// D dk # .d # a/
kD1 kD1 kD1 kD1 kD1

X
n X
n
? n.n C 1/ n
Dd k # .d # a/ 1Dd # .d # a/n D ..n # 1/d C 2a/
kD1 kD1
2 2

Notice that the equality labeled ? uses both (13) and (10) of Section 1.5. We remar
that the last term under consideration in the sum is bn D .n # 1/d C a so that in our
formula for sn the factor ..n#1/dC2a/ is the first term a plus the last term .n#1/dCa.
If we write z D .n # 1/d C a for the last term, then we can summarize with
n n
sn D ..n # 1/d C 2a/ D .a C z/
2 2
 Section .6 e ences 77

Note that we could also have found (13) by the same technique used to find (13) of
Section 1.5. We preferred to calculate using summation notation here. Finally, we
should remar that the sum (13) in Section 1.5 is the sum of the first n terms of the
special arithmetic sequence with a D 1 and d D 1.
Now ork Problem 33 G
E AM LE F S G S
A L IT I
rs. Simpson put 1000 in a spe- Find a formula for the sum sn of the first n terms of a geometric sequence .ck / with first
cial account for art on each of his term a and common ratio r.
first 21 birthdays. The account earned
interest at the rate of 7 compounded S Since the geometric sequence .ck / in question has, by Example ,
annually. We will see in Chapter 5 that ck D ark!1 , the required sum is given by
the amount deposited on art s .22 # X
n X
n
k/th birthday is worth 1000.1:07/k!1 sn D ck D ark!1 D a C ar C ar2 C $ $ $ C arn!1
on art s 21st birthday. Find the total kD1 kD1
amount in the special account on art s
It follows that if we multiply (14) by r we have
21st birthday.
Xn X
n X n
rsn D r ck D r ark!1 D ark D ar C ar2 C $ $ $ C arn!1 C arn
kD1 kD1 kD1
If we subtract (15) from (14) we get
sn # rsn D a # arn so that .1 # r/sn D a.1 # rn /
Thus, we have
a.1 # rn /
sn D for r ¤ 1
1#r
(Note that if r D 1, then each term in the sum is a and, since there are n terms, the
answer in this easy case is sn D na:/
Now ork Problem 37 G
!1
X
k
1
For s me infinite sequences .ck /1
kD1 the sequence of sums .sk /kD1 D ci
iD1 kD1
appears to approach a definite number. When this is indeed the case we write the num-
X
1
ber as ci . Here we consider only the case of a geometric sequence. As we see from
iD1
a.1 # rk /
(16), if ck D ark!1 then, for r ¤ 1, sk D . bserve that only the factor 1 # rk
1#r
depends on k. If jrj > 1, then for large values of k, jrk j will become large, as will j1#rk j.
In fact, for jrj > 1 we can ma e the values j1 # rk j as large as we li e by ta ing k to
a.1 # rk /
be su ciently large. It follows that, for jrj > 1, the sums d n t approach a
1#r
definite number. If r D 1, then sk D ka and, again, the sums do not approach a definite
number.
However, for jrj < 1 (that is for #1 < r < 1), we can ma e the values rk as close
to 0 as we li e by ta ing k to be su ciently large. ( e sure to convince yourself that
this is true before reading further because the rest of the argument hinges on this point.)
Thus, for jrj < 1, we can ma e the values 1 # rk as close to 1 as we li e by ta ing k to
a.1 # rk /
be su ciently large. Finally, for jrj < 1, we can ma e the values as close to
a 1#r
as we li e by ta ing k to be su ciently large. In precisely this sense, an infinite
1#r
geometric sequence with jrj < 1 has a sum and we have
X
1
a
for jrj < 1; ari!1 D
iD1
1#r
78 C cat ons an ore e ra

E AM LE R

In Example 1 of Section 0.1 we stated that the repeating decimal 0:151515 : : : represents
5
the rational number . We pointed out that entering 5 % 33 on a calculator strongly
33
suggests the truth of this assertion but were unable at that point to explain how the
5
numbers 5 and 33 were found. Is it true that 0:151515 : : : D
33
S et us write 0:151515 : : : D 0:15 C 0:0015 C 0:000015 C : : : . We can
recognize this infinite sum as the sum of the infinite geometric sequence whose first
term a D 0:15 and whose common ratio r D 0:01. Since jrj D 0:01 < 1 we have
0:15 0:15 15=100 15 5
0:151515 $ $ $ D D D D D
1 # 0:01 0: =100 33

E AM LE F S I G S

A rich woman would li e to leave 100,000 a year, starting now, to be divided equally
among all her direct descendants. She puts no time limit on this bequeathment and
is able to invest for this long-term outlay of funds at 2 compounded annually. How
much must she invest now to meet such a long-term commitment
S et us write R D 100;000, set the cloc to 0 now, and measure time in
years from now. With these conventions we are to account for payments of R at times
0; 1; 2; 3; : : : ; k; : : : by ma ing a single investment now. (Such a sequence of payments
is called a perpetuity.) The payment now simply costs her R. The payment at time 1 has
a present a ue of R.1:02/!1 . The payment at time 2 has a present value of R.1:02/!2 .
The payment at time 3 has a present value of R.1:02/!3 , and, generally, the payment
at time k has a present value of R.1:02/!k . Her investment n must exactly cover the
present value of a these future payments. In other words, the investment must equal
the sum

R C R.1:02/!1 C R.1:02/!2 C R.1:02/!3 C : : : C R.1:02/!k C : : :

We recognize the infinite sum as that of a geometric series, with first term a D R D
100;000 and common ratio r D .1:02/!1 . Since jrj D .1:02/!1 < 1, we can evaluate
the required investment as
a 100;000 100;000 100;000.1:02/
D D D D 5;100;000
1#r 1 0:02 0:02
1#
1:02 1:02
In other words, an investment of a mere 5,100,000 now will allow her to leave 100,000
per year to her descendants f re er
Now ork Problem 57 G
There is less here than meets the eye. Notice that 2 of 5,000,000 is 100,000.
The woman sets aside 5,100,000 at time 0 and simultaneously ma es her first 100,000
payment. uring the first year of investment, the remaining principal of 5,000,000
earns interest of exactly 100,000 in time for the payment of 100,000 at time 1.
Evidently, this process can continue indefinitely. However, there are
!1other infinite
Xk
1
sequences .ck /1
kD1 for which the sequence of sums .sk /kD1 D ci approaches
iD1 kD1
a definite number that cannot be dismissed by the argument of this paragraph.
 Section .6 e ences 79

R BLEMS
In Pr b ems rite the indicated term f the gi en sequence X
10 X
10
p 3 100.1=2/k!1 50.1:07/k!1
a D 2; # ; 2:3; 57 a3 kD1 kD1
7
5
b D 1; 13; #0: ; ; 100; 3 b6 X
10 X
5
2
7 k
.ak /kD1 D .3 / a4 .ck /kD1 D .3k C k/ c4 50.1:07/1!k 3 $ 2k
kD1 kD1
.ck / D .3 C .k # 5/2/ c15 .bk / D .5 $ 2k!1 / b6
.ak / D .k4 # 2k2 C 1/ a2 In Pr b ems nd the in nite sums if p ssib e r state hy
3 2
.ak / D .k C k # 2k C 7/ a3 this cann t be d ne

In Pr b ems nd a genera term .ak / descripti n that ts the X1 ! "k!1 1 ! "i

X
1 1
disp ayed terms f the gi en sequence 3
kD1
2 iD0
3
#1; 2; 5; 7; 4; 1; #2; : : :
2; #4; ; #16 5 5 5 X
1 X
1
5; ; ; ; ::: 1 2
3 27 .17/k!1 .1:5/k!1
2 3
In Pr b ems determine hether the gi en sequences are kD1 kD1
equa t each ther
X
1 X
1
..i C 3/3 / and . j3 # j2 C j # 27/ 20.1:01/!k 75.1:0 /1!j
2
.k # 4/ and ..k C 2/.k # 2// kD1 jD1
! " ! "1
1 1 3
3 k!1 and k Inventory Every 30 days a grocery store stoc s 0 cans of
5 kD1 5 kD1
elephant noodle soup and, rather surprisingly, sells 3 cans each
. j3 # j2 C 27j # 27/1 3 1
jD1 and ..k # 3/ /kD1 day. escribe the inventory levels of elephant noodle soup at the
In Pr b ems determine the indicated term f the gi en end of each day, as a sequence, and determine the inventory level
recursi e y de ned sequence 1 days after restoc ing.
a1 D 1, a2 D 2, akC2 D akC1 $ ak a7 Inventory If a corner store has 5 previously viewed
a1 D 1, akC1 D aak a17 V movies for sale today and manages to sell 6 each day,
bk write the first 7 terms of the store s daily inventory sequence for
b1 D 1, bkC1 D b6 the V s. How many V s will the store have on hand after
k 10 days
c1 D 0, ckC1 D .k C 2/ C ak c
Chec ing Account A chec ing account, which earns no
In Pr b ems rite the rst e terms f the arithmetic interest, contains 125.00 and is forgotten. It is nevertheless
sequence ith the gi en rst term a and c mm n di erence d sub ect to a 5.00 per month service charge. The account is
a D 22:5, d D 0: a D 0, d D 1 remembered after months. How much does it then contain
a D 6, d D #1:5 a D A, d D Savings Account A savings account, which earns interest
at a rate of 5 compounded annually, contains 25.00 and is
In Pr b ems rite the rst e terms f the ge metric forgotten. It is remembered 7 years later. How much does it then
sequence ith the gi en rst term a and c mm n rati r contain
1 1 a D 50, r D .1:06/!1
aD ,rD# Population Change A town with a population of 50,000 in
2 2
200 is growing at the rate of per year. In other words, at the
1
a D 100, r D 1:05 a D 3, r D end of each year the population is 1.0 times the population at the
3 end of the preceding year. escribe the population sequence and
In Pr b ems rite the indicated term f the arithmetic determine what the population will be at the end of 2020, if this
sequence ith gi en parameters a and d r f the ge metric rate of growth is maintained.
sequence ith gi en parameters a and r
Population Change Each year 5 of the inhabitants of a
27th term, a D 3, d D 2
rural area move to the city. If the current population is 24,000, and
th term, a D 2:5, d D #0:5 this rate of decrease continues, give a formula for the population k
11th term, a D 1, r D 2 7th term, a D 2, r D 10 years from now.
Revenue Current daily revenue at a campus burger
In Pr b ems nd the required sums restaurant is 12,000. ver the next 7 days revenue is expected to
X
7 X increase by 1000 each day as students return for the fall
..k # 1/3 C 5/ .k $ 2 C / semester. What is the pro ected total revenue for the days for
kD1 kD1 which we have pro ected data
X
4 X
34
..k # 1/0:2 C 1:2/ ..k # 1/10 C 5/
kD1 kD1
80 C cat ons an ore e ra

Revenue A car dealership s finance department is going to Perpetuity Rewor Problem 57 under the assumption that
receive payments of 300 per month for the next 60 months to pay rad s estate can ma e an investment at 10 compounded
for art s car. The kth such payment has a present value of annually.
300.1:01/!k . The sum of the present values of all 60 payments
must equal the selling price of the car. Write an expression for the The Fibonacci sequence given in (7) is defined recursively
selling price of the car and evaluate it using your calculator. using addition. Is it an arithmetic sequence Explain.

Future Value Five years from now, rittany will need a The sequence with a1 D 1 and akC1 D kak is defined
new truc . Starting next month, she is going to put 100 in the recursively using multiplication. Is it a geometric sequence
ban each month to save for the inevitable purchase. Five years Explain.
from now the kth ban deposit will be worth 100.1:005/60!k
The recursive definition for an arithmetic sequence .bk /
(due to compounded interest). Write a formula for the
called for starting with a number a and adding a fixed number d to
accumulated amount of money from her 60 ban deposits. Use
each term to get the next term. Similarly, the recursive definition
your calculator to determine how much rittany will have
for a geometric sequence .ck / called for starting with a number a
available towards her truc purchase.
and multiplying each term by a fixed number r to get the next term.
Future Value isa has ust turned 7 years old. She would If instead of addition or multiplication we use exp nentiati n, we
li e to save some money each month, starting next month, so that get two other classes of recursively defined sequences:
on her 21st birthday she will have 1000 in her ban account.
arge told her that with current interest rates her kth deposit will d1 D a and, for each positive integer k; dkC1 D .dk /p
be worth, on her 21st birthday, .1:004/16 !k times the deposited
amount. isa wants to deposit the same amount each month. Write for fixed real numbers a and p and
a formula for the amount isa needs to deposit each month to meet
her goal. Use your calculator to evaluate the required amount.
e1 D a and, for each positive integer k; ekC1 D bek
Perpetuity rad s will includes an endowment to
alhousie University that is to provide each year after his death, for fixed real numbers a and b. To get an idea of how sequences
forever, a 500 prize for the top student in the business can grow in size, ta e each of the parameters a, d, r, p, and b that
mathematics class, ATH 1115. rad s estate can ma e an have appeared in these definitions to be the number 2 and write
investment at 5 compounded annually to pay for this the first five terms of each of the arithmetic sequence .bk /, the
endowment. Adapt the solution of Example 11 to determine how geometric sequence .ck /, and the sequences .dk / and .ek / defined
much this endowment will cost rad s estate. above.

Chapter 1 Review
I T S E
S Applications of Equations
fixed cost variable cost total cost total revenue profit Ex. 3, p. 4
S Linear Inequalities
a<b a!b a>b a"b a<x<b Ex. 1, p. 57
inequality sense of an inequality Ex. 2, p. 57
equivalent inequalities linear inequality Ex. 1, p. 57
interval open interval closed interval endpoint
.a; b/ Œa; b! .#1; b/ .#1; b! .a; 1/ Œa; 1/ .#1; 1/ Ex. 3, p. 5
S Applications of Inequalities
renting versus purchasing Ex. 2, p. 5
current assets current liabilities current ratio Ex. 3, 60
S Absolute Value
distance absolute value, jxj union, [ Ex. 3, p. 64
S Summation
P otation
notation index bounds Ex. 1, p. 67
S Sequences
arithmetic sequence Ex. 5, p. 74
geometric sequence Ex. 6, p. 75
sum of an arithmetic sequence Ex. , p. 76
sum of a geometric sequence Ex. 10, p. 77
 Chapter e e 81

S
With word problems, you may not be given any equations. ultiplying (or dividing) both sides by the same positive
ou may have to construct equations and inequalities (often number.
more than one) by translating natural language statements of ultiplying (or dividing) both sides by the same negative
the word problem into mathematical statements. This pro- number and reversing the sense of the inequality.
cess is mathematica m de ing. First, read the problem more
than once so that you understand what facts are given and The algebraic definition of absolute value is
what you are to find. Next, choose variables to represent the jxj D x if x " 0 and jxj D #x if x < 0
un nown quantities you need to find. Translate each relation-
ship or fact given in the problem into equations or inequal- We interpret ja # bj or jb # aj as the distance between
ities involving the variables. Finally, solve the equations a and b. If d > 0, then the solution to the inequality jxj < d
(respecting any inequalities) and chec that your solution is the interval .#d; d/. The solution to jxj > d consists of the
answers what was as ed. Sometimes solutions to the equa union of two intervals and is given by .#1; #d/ [ .d; 1/.
ti ns will not be answers to the pr b em (but they may help Some basic properties of the absolute value are as follows:
in obtaining the final answers). ˇaˇ
ˇ ˇ jaj
Some basic relationships that are used in solving busi- jabj D jaj $ jbj ˇ ˇD
b jbj
ness problems are as follows:
ja # bj D jb # aj #jaj ! a ! jaj
total cost D variable cost C fixed cost
ja C bj ! jaj C jbj
total revenue D .price per unit/.number of units sold/
Summation notation provides a compact and precise way
profit D total revenue # total cost
of writing sums that have many terms. The basic equations
The inequality symbols <, !, >, and " are used to rep- of summation notation are ust restatements of P the proper-
n
resent an inequality, which is a statement that one number is, ties
Pn of addition. Certain particular sums, such as kD1 k and
2
for example, less than another number. Three basic opera- kD1 k , are memorable and useful.
tions that, when applied to an inequality, guarantee an equiv- oth arithmetic sequences and geometric sequences
alent inequality, are as follows: arise in applications, particularly in business applica-
tions. Sums of sequences, particularly those of geometric
Adding (or subtracting) the same number to (or from) sequences, will be important in our study of the mathematics
both sides. of finance in Chapter 5.

R
In Pr b ems s e the equati n r inequa ity X
11 X
11 X
3

2x C 1 " x # 3 2x # .7 C x/ ! x Evaluate i3 by using i3 # i3 . Explain why this

iD4 iD1 iD1
#.5x C 2/ < #.2x C 4/ #2.x C 6/ > x C 4 wor s, quoting any equations from Section 1.5 that are used.
! Explain why the answer is necessarily the same as that in
3 " Problem 16.
3p.1 # p/ > 3.2 C p/ # 3p2 2 5# q <4
2
Profit A profit of 40 on the selling price of a product is
xC5 1 x x x equivalent to what percent profit on the cost
# !2 # >
3 2 3 4 5 Stoc Exchange n a certain day, there were 1132
1 1 1 1 different issues traded on the New or Stoc Exchange. There
s # 3 ! .3 C 2s/ .t C 2/ " t C 4 were 4 more issues showing an increase than showing a decline,
4 3 4
ˇ ˇ and no issues remained the same. How many issues suffered a
ˇ 5x # 6 ˇ
j2 # 3xj D 7 ˇ ˇ decline
ˇ 13 ˇ D 0
Sales ax The sales tax in a certain province is 16.5 . If a
ˇ ˇ
j2z # 3j < 5 ˇ2 ˇ total of 303 .2 in purchases, including tax, is made in the
4 < ˇ x C 5ˇˇ
ˇ
course of a year, how much of it is tax
3
Production Allocation A company will manufacture a
j3 # 2xj " 4 total of 10,000 units of its product at plants A and . Available
X
7
data are as follows:
Evaluate .k C 5/2 by first squaring the binomial and then
kD1 Plant A Plant
using equations from Section 1.5.
Unit cost for labor and material 6 7.50
Fixed cost 25,000 30,000
82 C cat ons an ore e ra

etween the two plants, the company has decided to allot no more to net sales (i.e., gross sales minus returns and allowances). An
than 115,000 for total costs. What is the minimum number of operating ratio less than 100 indicates a profitable operation,
units that must be produced at plant A while an operating ratio in the 0 0 range is extremely good.
Propane an s A company is replacing two propane tan s If a company has net sales of 236,460 in one period, write an
with one new tan . The old tan s are cylindrical, each 25 ft high. inequality describing the operating costs that would eep the
ne has a radius of 10 ft and the other a radius of 20 ft. The new operating ratio below 0 .
tan is essentially spherical, and it will have the same volume Write the first five terms of the arithmetic sequence with first
as the old tan s combined. Find the radius of the new tan . term 32 and common difference 3.
int: The volume of a cylindrical tan is D "r2 h, where Write the first five terms of the geometric sequence with first
r is the radius of the circular base and h is the height of the tan . term 100 and common ratio 1:02.
The volume of a spherical tan is D 43 "R3 , where R is the
radius of the tan . Find the sum of the first five terms of the arithmetic sequence
with first term 12 and common difference 5.
Operating Ratio The perating rati of a retail business is
the ratio, expressed as a percentage, of operating costs Find the sum of the first five terms of the geometric sequence
(everything from advertising expenses to equipment depreciation) with first term 100 and common ratio 1:02.
 nct ons
an ra s

S
uppose a 1 0-pound man drin s four beers in quic succession. We now that
2.1 nct ons
his blood alcohol concentration, or AC, will first rise, then gradually fall
2.2 ec a nct ons bac to zero. ut what is the best way to describe how quic ly the AC rises,
where it pea s, and how fast it falls again
2.3 Co nat ons of If we obtain measured AC values for this particular individual, we can display
nct ons
them in a table, as follows:
2.4 n erse nct ons
2.5 ra s n ectan ar Time (h) 1 2 3 4 5 6
Coor nates
AC( ) 0.0 20 0.066 0.0516 0.0364 0.0212 0.0060
2.6 etr
2.7 rans at ons an
e ect ons
However, a table can show only a limited number of values and so does not really give
2.8 nct ons of e era the overall picture.
ar a es We might instead relate the AC to time t using a combination of linear and
quadratic equations (recall Chapter 0):
C er 2 e e
AC D !0:1025t2 C 0:1 44t if t " 0: 7
AC D !0:0152t C 0:0 72 if t > 0: 7

As with the table, however, it is hard to loo at the equations and understand quic ly
0.10 what is happening with AC over time.
Probably the best description of changes in the AC over time is given by a graph,
0.08
li e the one on the left. Here we see easily what happens. The blood alcohol concentra-
tion climbs rapidly, pea s at 0.0 3 after about an hour, and then tapers off gradually
BAC (%)

0.06
over the next five-and-a-half hours. Note that for about three hours, this male s AC is
0.04
above 0.05 , the point at which one s driving s ills begin to decline. The curve will
0.02 vary from one individual to the next, but women are generally affected more severely
than men, not only because of weight differences but also because of the different water
0 1 2 3 4 5 6 7 8 contents in men s and women s bodies.
Time (hours) The relationship between time and blood alcohol content is an example of a func
ti n. This chapter deals in depth with functions and their graphs.

83
84 C nct ons an ra s

Objective F
o n erstan at a f nct on s an In the 17th century, ottfried Wilhelm eibniz, one of the inventors of calculus, intro-
to eter ne o a ns an f nct on
a es duced the term functi n into the mathematical vocabulary. The concept of a function
is one of the most basic in all of mathematics. In particular, it is essential to the study
of calculus.
In everyday speech we often hear educated people say things li e (Prime) interest
rates are a function of oil prices or Pension income is a function of years wor ed
or lood alcohol concentration after drin ing beer is a function of time. Sometimes
such usage agrees with mathematical usage but not always. We have to be careful
with our meaning of the word functi n in order to ma e it a good mathematical tool.
Nevertheless, everyday examples can help our understanding. We build the definition
in the next three paragraphs.
A ey idea is to realize that a set, as first mentioned in Section 0.1, need not have
numbers as its e ements. We can spea of a set of interest rates, a set of oil prices, a
set of incomes, and so on. If X and are sets, in that generality, and x is an element
of X and y is an an element of , then we write .x; y/ for what we call the ordered
pair consisting of x and y in the order displayed. We accept that the notation for an
ordered pair of real numbers is the same as that for an open interval, but the practice
is strongly entrenched and almost never causes any confusion. Note that .y; x/ is in
general different from .x; y/. In fact, given two ordered pairs .x; y/ and .a; b/, we have
.x; y/ D .a; b/ if and only if both x D a and y D b. We will write X # for the set of all
ordered pairs .x; y/, where x is an element of X and y is an element of . For example,
if X is the set of oil prices and is the set of interest rates, then an element of X # is
a pair .p; r/, where p is an oil price and r is an interest rate.
A relation R from a set X to a set is a subset of X # . We recall from Section 0.1
that this means any element of R is also an element of X # . If it happens that .x; y/ is
f course .!1; 1/ here is an interval , an element of R, then we say that x is R-related to y and write xRy. Each of <, >, ",
as in Chapter 1, Section 2, not an ordered and $ are relations from the set .!1; 1/ of all real numbers to itself. For example,
pair.
we can define < as that subset of .!1; 1/ # .!1; 1/ consisting of all .a; b/ such
that a < b is true. The use of xRy for x is R-related to y is inspired by the notation
for inequalities. To give another example, let P and denote, respectively, the set of all
points and the set of all lines in a given plane. For an ordered pair .p; / in P # , it is
either the case that p is on or p is not on . If we write p ı for p is on , then
ı is a relation from P to in the sense of this paragraph. Returning to prices and rates,
we might say that oil price p is R-related to interest rate r, and write pRr, if there has
been a time at which both the price of oil has been p and the interest rate has been r .
A function f from a set X to a set is a relation from X to with the special property
that if both xfy and xfz are true, then y D z. (In many boo s, it is also required that for
each x in X there exists a y in , such that xfy. We will not impose this further condition.)
The point is that if x is f-related to anything, then that thing is uniquely determined by x.
After all, the definition says that if two things, y and z, are both f-related to x, then they
are in fact the same thing, so that y D z. We write y D f.x/ for the unique y, if there is
one, such that x is f-related to y.
With this definition we see that the notion of function is not symmetric in x
and y. The notation fW X % is often used for f is a function from X to because it
underscores the directedness of the concept.
We now re-examine the examples from everyday speech of the second paragraph
of this section. The relation R defined by pRr if there has been a time at which both
the price of oil has been p and the (prime) interest rate has been r does n t define a
function from oil prices to interest rates. any people will be able to recall a time when
oil was 30 a barrel and the interest rate was and another time when oil was 30 a
barrel and the interest rate was 1 . In other words, both .30; / and .30; 1/ are ordered
pairs belonging to the R relation, and since ¤ 1, R is not a function. est you thin
that we may be trying to do it the wrong way around, let us write Rı for the relation
from the set of interest rates to the set of oil prices given by rRı p if and only if pRr.
 Section 2. nct ons 85

If you can remember a time when the interest rate was 6 with oil at 30 a barrel and
another time when the interest rate was 6 with oil at 70 a barrel, then you will have
both .6; 30/ and .6; 70/ in the relation Rı . The fact that 30 ¤ 70 shows that Rı is also
not a function.
n the other hand, suppose we bring into a testing facility a person who has ust
drun five beers, and test her blood alcohol concentration then and each hour thereafter,
for six hours. For each of the time values f0; 1; 2; 3; 4; 5; 6g, the measurement of
blood alcohol concentration will produce exact y ne a ue. If we write for the set
of all times beginning with that of the first test and B for the set of all blood alcohol
concentration values, then testing the woman in question will determine a function
b W % B, where, for any time t in , b.t/ is the blood alcohol concentration of the
woman at time t.
It is not true that Pension income is a function of years wor ed. If the value of
years wor ed is 25, then the value of pension income is not yet determined. In most
organizations, a CE and a systems manager will retire with different pensions after
25 years of service. However, in this example we might be able to say that, f r each j b
descripti n in a particu ar rganizati n, pension income is a function of years wor ed.
If 100 is invested at, say, 6 simple interest, then the interest earned I is a function
of the length of time t that the money is invested. These quantities are related by
I D 100.0:06/t
Here, for each value of t, there is exactly one value of I given by Equation (1). In a
situation li e this we will often write I.t/ D 100.0:06/t to reinforce the idea that the
I-value is determined by the t-value. Sometimes we write I D I.t/ to ma e the claim
that I is a function of t even if we do not now a formula for it. Formula (1) assigns the
output 3 to the input 12 and the output 12 to the input 2. We can thin of Formula (1) as
defining a ru e: ultiply t by 100(0.06). The rule assigns to each input number t exactly
one output number I, which is often symbolized by the following arrow notation:
t ‘ 100.0:06/t
A formula provides a way of describing a rule to cover, potentially, infinitely many
cases, but if there are only finitely many values of the input variable, as in the chapter-
opening paragraph, then the ru e, as provided by the observations recorded in the table
there, may not be part of any recognizable f rmu a. We use the word ru e rather than
f rmu a below to allow us to capture this useful generality. The following definition is
sometimes easier to eep in mind than our description of a function as a special ind
of relation:

A function f W X % is a rule that assigns to each of certain elements x of X at

most one element of . If an element is assigned to x in X, it is denoted by f.x/. The
subset of X consisting of all the x for which f.x/ is defined is called the domain of
f. The set of all elements in of the form f.x/, for some x in X, is called the range
of f.

For the interest function defined by Equation (1), the input number t cannot be
negative, because negative time ma es no sense in this example. Thus, the domain
consists of all nonnegative numbers that is, all t $ 0, where the variable gives the
time elapsed from when the investment was made.
A variable that ta es on values in the domain of a function f W X % is sometimes
called an input, or an independent variable for f. A variable that ta es on values in the
range of f is sometimes called an utput, or a dependent variable of f. Thus, for the
interest formula I D 100.0:06/t, the independent variable is t, the dependent variable
is I, and I is a function of t.
As another example, the equation
yDxC2
86 C nct ons an ra s

defines y as a function of x. The equation gives the rule, Add 2 to x. This rule assigns
to each input x exactly one output x C 2, which is y. If x D 1, then y D 3 if x D !4,
then y D !2. The independent variable is x, and the dependent variable is y.
In y2 D x, x and y are related, but the Not all equations in x and y define y as a function of x. For example, let y2 D x. If
relationship does not give y as a function x is , then y2 D , so y D ˙3. Hence, to the input , there are assigned not one but
of x. t output numbers: 3 and !3. This violates the definition of a function, so y is n t a
function of x.
n the other hand, some equations in two variables define either variable as a func-
tion of the other variable. For example, if y D 2x, then for each input x, there is exactly
one output, 2x. Thus, y is a function of x. However, solving the equation for x gives
x D y=2. For each input y, there is exactly one output, y=2. Consequently, x is a func-
tion of y.
Usually, the letters f, g, h, F, G, and so on are used to name functions. For example,
Equation (2), y D x C 2, defines y as a function of x, where the rule is Add 2 to the
input. Suppose we let f represent this rule. Then we say that f is the function. To indicate
that f assigns to the input 1 the output 3, we write f.1/ D 3, which is read f of 1 equals
3. Similarly, f.!4/ D !2. ore generally, if x is any input, we have the following
notation:

input
f.x/, which is read f of x, and which means the #
output, in the range of f, that results when the rule f f.x/
„ƒ‚…
is applied to the input x, from the domain of f. "
output

f .x/ does n t mean f times x. f .x/ is the Thus, the output f.x/ is the same as y. ut since y D x C 2, we can also write
output that corresponds to the input x. f.x/ D y D x C 2 or, more simply,
f.x/ D x C 2
For example, to find f.3/, which is the output corresponding to the input 3, we replace
each x in f.x/ D x C 2 by 3:
f.3/ D 3 C 2 D 5
utputs are also called function values.
For another example, the equation g.x/ D x3 Cx2 defines the function g that assigns
the output x3 C x2 to an input x:
gW x ‘ x3 C x2
In other words, g adds the cube of the input to the square of the input. Some function
values are
g.2/ D 23 C 22 D 12
g.!1/ D .!1/3 C .!1/2 D !1 C 1 D 0
g.t/ D t3 C t2
g.x C 1/ D .x C 1/3 C .x C 1/2
The idea of rep acement, also nown as Note that g.x C 1/ was found by replacing each x in x3 C x2 by the input x C 1. When
substituti n, is very important in we refer to the function g defined by g.x/ D x3 C x2 , we are free to say simply the
determining function values. function g.x/ D x3 C x2 and similarly the function y D x C 2.
Unless otherwise stated, the domain of a function f W X % is the set of all x in
X for which f.x/ ma es sense, as an element of . When X and are both .!1; 1/,
this convention often refers to arithmetical restrictions. For example, suppose
1
h.x/ D
x!6
Here any real number can be used for x except 6, because the denominator is 0 when x
is 6. So the domain of h is understood to be all real numbers except 6. A useful notation
 Section 2. nct ons 87

for this set is .!1; 1/ ! f6g. ore generally, if A and B are subsets of a set X, then
we write A ! B for the set of all x in X such that x is in A and x is n t in B. We note too
that the range of h is the set of all real numbers except 0. Each output of h is a fraction,
and the only way that a fraction can be 0 is for its numerator to be 0. While we do have
1 c
D for all c ¤ 0
x!6 c.x ! 6/
by the fundamenta princip e f fracti ns of Section 0.2, we see that 0 is not a function
1
value for h. ut if y is any nonzero real number, we can solve D y for x and
x!6
1
get x D 6 C as the (unique) input for which h.x/ is the given y. Thus, the range is
y
.!1; 1/ ! f0g, the set of all real numbers other than 0.

E F
To say that two functions f; g W X % are equal, denoted f D g, is to say that
The domain of f is equal to the domain of g
For every x in the domain of f and g, f.x/ D g.x/.

Requirement 1 says that an element x is in the domain of f if and only if x is in the

domain of g. Thus, if we have f.x/ D x2 , with no explicit mention of domain, and
g.x/ D x2 for x $ 0, then f ¤ g. For here the domain of f is the whole real line
.!1; 1/ and the domain of g is Œ0; 1/. n the other hand, if we have f.x/ D .x C 1/2
and g.x/ D x2 C 2x C 1, then, for both f and g, the domain is understood to be
.!1; 1/ and the issue for deciding if f D g is whether, for each real number x, we have
.x C 1/2 D x2 C 2x C 1. ut this is true it is a special case of item 4 in the Special Prod-
ucts of Section 0.4. In fact, older textboo s refer to statements li e .xC1/2 D x2 C2xC1
as identities, to indicate that they are true for any admissible value of the variable and
to distinguish them from statements li e .x C 1/2 D 0, which are true for some values
of x.
iven functions f and g, it follows that we have f ¤ g if either the domain of f is
different from the domain of g r there is some x for which f.x/ ¤ g.x/.

E AM LE E F

etermine which of the following functions are equal.

.x C 2/.x ! 1/
a f.x/ D
.x ! 1/
b g.x/ D x C 2
!
x C 2 if x ¤ 1
c h.x/ D
0 if x D 1
!
x C 2 if x ¤ 1
d k.x/ D
3 if x D 1
S The domain of f is the set of all real numbers other than 1, while that of g is
the set of all real numbers. (For these we are following the convention that the domain
is the set of all real numbers for which the rule ma es sense.) We will have more to
say about functions li e h and k that are defined by cases in Example 4 of Section 2.2.
Here we observe that the domain of h and the domain of k are both .!1; 1/, since for
both we have a rule that ma es sense for each real number. The domains of g, h, and
k are equal to each other, but that of f is different. So by Requirement 1 for equality of
functions, f ¤ g, f ¤ h and f ¤ k. y definition, g.x/ D h.x/ D k.x/ for all x ¤ 1,
so the matter of equality of g, h and k depends on their values at 1. Since g.1/ D 3,
h.1/ D 0 and k.1/ D 3, we conclude that g D k and g ¤ h (and h ¤ k). While this
88 C nct ons an ra s

example might appear to be contrived, it is typical of an issue that arises frequently in

calculus.
Now ork Problem 3 G
E AM LE F
A L IT I
The area of a circle depends on the Find the domain of each function.
length of the radius of the circle. x
a f.x/ D 2
a Write a function a.r/ for the area of a x !x!2
circle when the length of the radius is r.
b What is the domain of this function S We cannot divide by zero, so we must find any values of x that ma e the
out of context denominator 0. These cann t be inputs. Thus, we set the denominator equal to 0 and
c What is the domain of this function in solve for x:
the given context
x2 ! x ! 2 D 0 quadratic equation
.x ! 2/.x C 1/ D 0 factoring
x D 2; ! 1

Therefore, the domain of f is all real numbers except 2 and !1.

p
b g.t/ D 2t ! 1 as a function gW .!1; 1/ % .!1; 1/
p
S 2tp! 1 is a real number if 2t ! 1 is greater than or equal to 0. If 2t ! 1 is
negative, then 2t ! 1 is not a real number, so we must assume that

2t ! 1 $ 0
2t $ 1 adding 1 to both sides
1
t$ dividing both sides by 2
2
Thus, the domain is the interval Π12 ; 1/.
Now ork Problem 7 G
E AM LE F F
A L IT I
The time it ta es to go a given dis- et g.x/ D 3x2 ! x C 5. Any real number can be used for x, so the domain of g is all
tance depends on the speed at which one real numbers.
is traveling. a Find g(z).
a Write a function t.r/ for the time it
S Replacing each x in g.x/ D 3x2 ! x C 5 by z gives
ta es if the distance is 300 miles and the
speed is r.
b What is the domain of this function
g.z/ D 3z2 ! z C 5
out of context
c What is the domain of this function in b Find g.r2 /.
the given context
"x# "x# S Replacing each x in g.x/ D 3x2 ! x C 5 by r2 gives
d Find t.x/, t , and t .
2 4
e What happens to the time if the speed g.r2 / D 3.r2 /2 ! r2 C 5 D 3r4 ! r2 C 5
is divided by a constant c escribe this
situation using an equation. c Find g.x C h/.

on t be confused by notation. In S
Example 3(c), we find g.x C h/ by
replacing each x in g.x/ D 3x2 ! x C 5 by g.x C h/ D 3.x C h/2 ! .x C h/ C 5
the input x C h. g.x C h/, g.x/ C h, and
g.x/ C g.h/ are all different quantities. D 3.x2 C 2hx C h2 / ! x ! h C 5
D 3x2 C 6hx C 3h2 ! x ! h C 5
Now ork Problem 31 G
 Section 2. nct ons 89

E AM LE F
f.x C h/ ! f.x/
If f.x/ D x2 , find .
h
f.x C h/ ! f.x/
S The expression is referred to as a difference quotient. Here
h
the numerator is a difference of function values. We have
f.x C h/ ! f.x/ .x C h/2 ! x2
D
The difference quotient of a function is h h
an important mathematical concept.
x2 C 2hx C h2 ! x2 2hx C h2
D D
h h
h.2x C h/
D
h
D 2x C h for h ¤ 0
If we consider the original difference quotient as a function of h, then it is different
from 2x C h because 0 is not in the domain of the original difference quotient but it is
in the default domain of 2x C h. For this reason, we had to restrict the final equality.
Now ork Problem 35 G
In some cases, the domain of a function is restricted for physical or economic
reasons. For example, the previous interest function I D 100.0:06/t has t $ 0 because
t represents time elapsed since the investment was made. Example 5 will give another
illustration.

E AM LE F
A L IT I
Suppose the wee ly demand func- Suppose that the equation p D 100=q describes the relationship between the price per
tion for large pizzas at a local pizza par- unit p of a certain product and the number of units q of the product that consumers will
q buy (that is, demand) per wee at the stated price. This equation is called a demand
lor is p D 26 ! .
40 equati n for the product. If q is an input, then to each value of q there is assigned at
a If the current price is 1 .50 per
pizza, how many pizzas are sold each
most one output p:
wee 100
q ‘ Dp
b If 200 pizzas are sold each wee , q
what is the current price For example,
c If the owner wants to double the num-
ber of large pizzas sold each wee (to
100
20 ‘ D5
400), what should the price be 20
that is, when q is 20, p is 5. Thus, price p is a function of quantity demanded, q. This
function is called a demand function. The independent variable is q, and p is the depen-
dent variable. Since q cannot be 0 (division by 0 is not defined) and cannot be negative
(q represents quantity), the domain is all q > 0.
Now ork Problem 43 G
We have seen that a function is a rule that assigns to each input in the domain
exactly one output in the range. For the rule given by f.x/ D x2 , some sample assign-
ments are shown by the arrows in Figure 2.1. The next example discusses a rule given
by a finite listing rather than an an algebraic formula.

f
1 = f(1)
1
4 = f(2)
2 Range
Domain x x2 = f(x)

FIGURE Some function values for f .x/ D x2 .

90 C nct ons an ra s

A L IT I E AM LE S S
For the supply function given by the
following table, determine the wee ly The table in Apply It 4 is a supp y schedu e. Such a table lists for each of certain prices
revenue function, assuming that all p of a certain product the quantity q that producers will supply per wee at that price.
units supplied are sold. For each price, the table provides exactly one quantity so that it exhibits q as a function
p q of p.
Price per uantity ut also, for each quantity, the table provides exactly one price so that it also
Unit in Supplied exhibits p as a function of q. If we write q D f.p/, then the table provides
ollars per Wee
f.500/ D 11 f.600/ D 14 f.700/ D 17 f. 00/ D 20
500 11
If we write p D g.q/, then the table also provides
600 14
700 17 g.11/ D 500 g.14/ D 600 g.17/ D 700 g.20/ D 00
00 20 bserve that we have g. f.p// D p, for all p, and f.g.q// D q, for all q. We will
have more to say about pairs of functions of this ind in Section 2.4. oth functions
determined by this table are called supply functions.
Now ork Problem 53 G

R BLEMS
In Pr b ems determine hether the gi en functi ns are equa x!5
p k.x/ D k.5/; k.2x/; k.x C h/
2
f .x/ D x g.x/ D x x2 C 1
p p
G.x/ D . x C 3/2 .x/ D x C 3 k.x/ D x ! 3 k.4/, k.3/, k.x C 1/ ! k.x/
! $ %
jxj 1 if x $ 0 !1
h.x/ D k.x/ D f .x/ D x2=5 f .0/, f .243/, f
x !1 if x < 0 32
8
< x2 ! 4x C 3 g.x/ D x2=5 g.32/; g.!64/; g.t10 /
f .x/ D if x ¤ 3
: x!3
2 if x D 3 f .x C h/ ! f .x/
g.x/ D x ! 1 In Pr b ems nd a f .x C h/ and b
h
In Pr b ems gi e the d main f each functi n simp ify y ur ans ers
x x
6 g.x/ D f .x/ D 4x ! 5 f .x/ D
f .x/ D 5 3
x!1
r f .x/ D x2 C 2x f .x/ D 2x2 ! 5x C 3
x!2 1 2
h.x/ D .z/ D p f .x/ D 3 ! 2x C 4x f .x/ D x3
xC1 z!1 1 xC
2 x2 f .x/ D f .x/ D
f .z/ D 3z C 2z ! 4 .x/ D x!1 x
xC3 f .2 C h/ ! f .2/
x! p If f .x/ D 3x C 7, find .
f .x/ D g.x/ D 2 ! 3x h
2x C 7
4 xC5 f .x/ ! f .2/
g.y/ D 2 !.x/ D 2 If f .x/ D 2x2 ! x C 1, find .
y ! 4y C 4 x Cx!6 x!2
3 ! x2 2 In Pr b ems is y a functi n f x Is x a functi n f y
h.s/ D 2 G.r/ D 2
3x ! 5x ! 2 r C1 y ! 3x ! 4 D 0 x4 ! 1 C y D 0
In Pr b ems nd the functi n a ues f r each functi n
y D 7x2 x3 C y2 D 1
f .x/ D 3 ! 5x f .0/, f .2/, f .!2/
$ %
p 2 The formula for the area of a circle of radius r is A D "r2 . Is
.s/ D 5s2 ! 3 .4/; . 2/; the area a function of the radius
3
G.x/ D 2 ! x2 G.! /; G.u/; G.u2 / Suppose f .b/ D a2 b3 C a3 b2 . a Find f .a/. b Find f .ab/.
F.x/ D !7x C 1I F.s/, F.t C 1/; F.x C 3/ Value of Business A business with an original capital of
2
g.u/ D 2u ! u g.!2/, g.2 /, g.x C a/ 50,000 has income and expenses each wee of 7200 and 4 00,
$ % respectively. If all profits are retained in the business, express the
2 1
h. / D p h.36/, h , h.1 ! x/ value of the business at the end of t wee s as a function of t.
4 4
Depreciation If a 30,000 machine depreciates 2 of its
f .x/ D x2 C 2x C 1 f .1/; f .!1/; f .x C h/ original value each year, find a function f that expresses the
.x/ D .x C 4/2 .0/; .2/; .t ! 4/ machine s value after t years have elapsed.
 Section 2.2 ec a nct ons 91

Profit Function If q units of a certain product are sold (q is the intensity I of the shoc (I in microamperes) and was
nonnegative), the profit P is given by the equation P D 2:57q ! 127. estimated by
Is P a function of q What is the dependent variable the
I 4=3
independent variable R D f .I/ D 500 " I " 3500
2500
Demand Function Suppose the yearly demand function for
Evaluate a f(1000) and b f(2000). c Suppose that I0 and 2I0
1;200;000
a particular actor to star in a film is p D , where q is the are in the domain of f. Express f .2I0 / in terms of f .I0 /. What effect
q does the doubling of intensity have on response
number of films he stars in during the year. If the actor currently
charges 600,000 per film, how many films does he star in each Profit For n $ 3 the profit from selling n items is nown to
p
year If he wants to star in four films per year, what should his be P.n/ D n=2 C n ! 3. Find P.2 / and P.52/.
price be Demand Schedule The following table is called a demand
Supply Function Suppose the wee ly supply function for a schedu e. It gives a correspondence between the price p of a
q product and the quantity q that consumers will demand (that is,
pound of house-blend coffee at a local coffee shop is p D ,
4 purchase) at that price. a If p D f .q/, list the numbers in the
where q is the number of pounds of coffee supplied per wee . domain of f. Find f .2 00/ and f .3000/. b If q D g.p/, list the
How many pounds of coffee per wee will be supplied if the price numbers in the domain of g. Find g.10/ and g.17/.
is .3 a pound How many pounds of coffee per wee will be
supplied if the price is 1 .4 a pound How does the amount Price per Unit, p uantity emanded per Wee , q
supplied change as the price increases
10 3000
ospital Discharges An insurance company examined the
12 2 00
records of a group of individuals hospitalized for a particular
illness. It was found that the total proportion discharged at the end 17 2300
of t days of hospitalization is given by 20 2000

$ %3
200 In Pr b ems use y ur ca cu at r t nd the indicated
f .t/ D 1 ! a ues f r the gi en functi n R und ans ers t t decima
200 C t
p aces
fp.x/ D 2:03x3 ! 5:27x2 ! 13:71 a f .1:73/, b f .!5:7 /,
Evaluate a f .0/, b f .100/, and c f . 00/. d At the end of how c f . 2/
many days was half of the group discharged 14:7x2 ! 3: 5x ! 15:76
Psychology A psychophysical experiment was conducted f .x/ D a f(4), b f .!17=4/,
24:3 ! x3
to analyze human response to electrical shoc s.1 The sub ects c f ."/
received a shoc of a certain intensity. They were told to assign a f .x/ D .20:3 ! 3:2x/.2:25x2 ! 7:1x ! 16/4 a f .0:3/,
magnitude of 10 to this particular shoc , called the standard b f .!0:02/, c f .1: /
stimulus. Then other shoc s (stimuli) of various intensities were sp
given. For each one, the response R was to be a number that 5x2 C 3:23.x C 1/
f .x/ D a f .11:7/, b f .!73/,
indicated the perceived magnitude of the shoc relative to that 7:2
of the standard stimulus. It was found that R was a function of c f .0/

Objective S F
o ntro ce constant f nct ons In this section, we loo at functions having special forms and representations. We begin
o no a f nct ons rat ona
f nct ons case e ne f nct ons t e with perhaps the simplest type of function there is: a c nstant functi n.
a so te a e f nct on an factor a
notat on
E AM LE C F

et h W .!1; 1/ % .!1; 1/ be given by h.x/ D 2. The domain of h is .!1; 1/,

the set of all real numbers. All function values are 2. For example,

h.10/ D 2 h.!3 7/ D 2 h.x C 3/ D 2

1 Adapted from H. ab off, agnitude Estimation of Short Electrocutaneous Pulses, Psych gica Research
3 , no. 1 (1 76), 3 4 .
 92 C nct ons an ra s

A L IT I We call h a c nstant functi n because all the function values are the same. ore gen-
Suppose the monthly health insur- erally, a function of the form h.x/ D c, where c is a c nstant, is called a constant
ance premiums for an individual are function.
125.00. Now ork Problem 17 G
a Write the monthly health insurance
premiums as a function of the num- A constant function belongs to a broader class of functions, called p yn mia func
ber of visits the individual ma es to the ti ns. In general, a function of the form
doctor.
b How do the health insurance premi- f.x/ D cn xn C cn!1 xn!1 C & & & C c1 x C c0
ums change as the number of visits to
the doctor increases where n is a nonnegative integer and cn ; cn!1 ; : : : ; c0 are constants with cn ¤ 0, is called
c What ind of function is this a polynomial function (in x). The number n is called the degree of the polynomial, and
cn is the eading c e cient Thus,
Each term in a polynomial function is f.x/ D 3x2 ! x C
either a constant or a constant times a
positive integral power of x. is a polynomial function of degree 2 with leading coe cient 3. i ewise,
g.x/ D 4!2x has degree 1 and leading coe cient !2. Polynomial functions of degree 1
or 2 are called linear or quadratic functions, respectively. For example, g.x/ D 4!2x
is linear and f.x/ D 3x2 ! x C is quadratic. Note that a nonzero constant function,
such as f.x/ D 5 which can be written as f.x/ D 5x0 , is a polynomial function of
degree 0. The constant function f.x/ D 0, also called the zer functi n, is a polynomial
function but, by convention, the zero function has no degree assigned to it. The domain
of any polynomial function is the set of all real numbers.

E AM LE F
A L IT I
The function d.t/ D 3t2 , for t $ 0, a f.x/ D x3 !6x2 C7 is a polynomial (function) of degree 3 with leading coe cient 1.
represents the distance in meters a car 2x 2
b g.x/ D is a linear function with leading coe cient .
will go in t seconds when it has a con- 3 3
stant acceleration of 6 m per second. 2
a What ind of function is this c f.x/ D 3 is n t a polynomial function. ecause f.x/ D 2x!3 and the exponent
x
b What is its degree for x is not a nonnegative integer,
c What is its leading coe cient p this function does not have the proper form for a
polynomial. Similarly, g.x/ D x is not a polynomial, because g.x/ D x1=2 .
Now ork Problem 3 G
A function that is a quotient of polynomial functions is called a rational function.

E AM LE R F

x2 ! 6x
a f.x/ D is a rational function, since the numerator and denominator are each
xC5
polynomials. Note that this rational function is not defined for x D !5.
2x C 3
b g.x/ D 2x C 3 is a rational function, since 2x C 3 D . In fact, every polyno-
Every polynomial function is a rational 1
mial function is also a rational function.
function.
Now ork Problem 5 G
Sometimes more than one expression is needed to define a function, as Example 4
shows.

E AM LE C F

et
8
< 1 if ! 1 " s < 1
F.s/ D 0 if 1 " s " 2
:s ! 3 if 2 < s "
 Section 2.2 ec a nct ons 93

A L IT I This is called a case defined function because the rule for specifying it is given
by rules for each of several dis oint cases. Here s is the independent variable, and the
To reduce inventory, a department
store charges three rates. If you buy 0 5 domain of F is all s such that !1 " s " . The value of s determines which expression
pairs of soc s, the price is 3.50 per to use.
pair. If you buy 6 10 pairs of soc s, Find F.0/: Since !1 " 0 < 1, we have F.0/ D 1
the price is 3.00 per pair. If you buy Find F.2/: Since 1 " 2 " 2, we have F.2/ D 0
more than 10 pairs, the price is 2.75
Find F.7/: Since 2 < 7 " , we substitute 7 for s in s ! 3.
per pair. Write a case-defined function
to represent the cost of buying n pairs of F.7/ D 7 ! 3 D 4
soc s.
Now ork Problem 19 G
E AM LE A F

The absolute-value function is an The function j!j.x/ D jxj is called the abs ute a ue functi n. Recall that the absolute
example of a case-defined function. value of a real number x is denoted jxj and is defined by
!
x if x $ 0
jxj D
!x if x < 0
Thus, the domain of j!j is all real numbers. Some function values are
j16j D 16
4
& '
j! 3
j D ! ! 43 D 4
3
j0j D 0
Now ork Problem 21 G
In our next examples, we ma e use of fact ria n tati n.

The symbol r , with r a positive integer, is read r factorial . It represents the prod-
uct of the first r positive integers:
rŠ D 1 & 2 & 3 & & & r
We also define
0Š D 1

For each nonnegative integer n, .!/Š.n/ D nŠ determines a unique number, so it

follows that .!/Š is a function whose domain is the set of nonnegative integers.

E AM LE F
A L IT I
a 5Š D 1 & 2 & 3 & 4 & 5 D 120
Seven different boo s are to be
placed on a shelf. How many ways can b 3Š.6 ! 5/Š D 3Š & 1Š D .3 & 2 & 1/.1/ D .6/.1/ D 6
they be arranged Represent the ques- 4Š 1&2&3&4 24
c D D D 24
tion as a factorial problem and give the 0Š 1 1
solution.
Now ork Problem 27 G
E AM LE G

Suppose two blac guinea pigs are bred and produce exactly five offspring. Under cer-
tain conditions, it can be shown that the probability P that exactly r of the offspring
Factorials occur frequently in probability will be brown and the others blac is a function of r, P D P.r/, where
theory. & 'r & 3 '5!r
5Š 14 4
P.r/ D r D 0; 1; 2; : : : ; 5
rŠ.5 ! r/Š
94 C nct ons an ra s

The letter P in P D P.r/ is used in two ways. n the right side, P represents the func-
tion rule. n the left side, P represents the dependent variable. The domain of P is all
integers from 0 to 5, inclusive. Find the probability that exactly three guinea pigs will
be brown.
S We want to find P(3). We have
& 1 '3 & 3 '2 & 1 '& '
5Š 4 4
120 64 16 45
P.3/ D D D
3Š2Š 6.2/ 512

Now ork Problem 35 G

E AM LE I T

The Canadian Federal tax rates for 2015 were given by 15 on the first 44,701 22
on income over 44,701 up to ,401 26 on income over ,401 up to 13 ,5 6
and 2 on income over 13 ,5 6. a Express the Canadian Federal tax rate t as a
case-defined function of income i. b Express Canadian Federal income tax paid as
a function of income i. c Express after-Federal-tax income a as a function of income
i and graph it.
S a Translating the given information directly, in the style of Example 4, we
have
8̂
ˆ 0:15 if 0 " i " 44;701
ˆ
< 0:22 if 44;701 < i " ;401
t.i/ D
ˆ
ˆ 0:26 if ;401 < i " 13 ;5 6
:̂
0:2 if 13 ;5 6 < i

b For 0 " i " 44;701, .i/ D 0:15i. Note that .44;701/ D 6705:15.
For 44;701 < i " ;401, it follows that .i/ D 6705:15 C 0:22.i ! 44;701/.
Note that . ;401/ D 6705:15 C 0:22. ;401 ! 44;701/ D 16;53 :15.
For ;401 < i " 13 ;5 6, .i/ D 16;53 :15 C 0:26.i ! ;401/.
Note that .13 ;5 6/ D 16;53 :15 C 0:26.13 ;5 6 ! ;401/ D 2 ;327:25.
Finally, for 13 ;5 6 < i, we have .i/ D 2 ;327:25 C 0:2 .i ! 13 ;5 6/ and
8̂
ˆ 0:15i if 0 " i " 44;701
ˆ
< 6705:15 C 0:22.i ! 44;701/ if 44;701 < i " ;401
.i/ D
ˆ
ˆ 16;53 :15 C 0:26.i ! ;401/ if ;401 < i " 13 ;5 6
:̂
2 ;327:25 C 0:2 .i ! 13 ;5 6/ if 13 ;5 6 < i

c The function a is given by a.i/ D i ! .i/, another case-defined function, with

the same case rules as those for :
8̂
ˆ 0: 5i if 0 " i " 44;701
ˆ
< 312 :07 C 0:7 i if 44;701 < i " ;401
a.i/ D
ˆ
ˆ 6705:11 C 0:74i if ;401 < i " 13 ;5 6
:̂
10; 62:6 C 0:71i if 13 ;5 6 < i

bserve from Figure 2.2 that whenever i < j for incomes i and j, a.i/ < a.j/.
There is a sort of urban myth that one can end up with a reducti n in after-tax income
by getting an increase in income that puts one in a higher tax brac et. The graph in
Figure 2.2 shows that this myth is false.
 Section 2.2 ec a nct ons 95

140,000
120,000
100,000
80,000
60,000
40,000
20,000
0
0 50,000 100,000 150,000 200,000

FIGURE After-Federal-tax income as a function of income.

Now ork Problem 33 G

E AM LE H M

Consider the tas of evaluating the polynomial f.x/ D 2x4 C 5x3 C 7x2 ! 2x C 5
at 6: 51 2, say, on a very unsophisticated hand-held calculator. If we apply the
operations in the order suggested, we find ourselves entering 6: 51 2 ten times and
stri ing the multiplication ey ten times. Horner s ethod begins by rewriting
f.x/ D ...2x C 5/x C 7/x ! 2/x C 5. How many times does this method of evalua-
tion require us to enter 6: 51 2 and how many multiplications are required
S Simply counting occurrences of x we see that only four entries of 6: 51 2
are required. Similarly, we see that this evaluation requires only four multiplications.
(Entry of coe cients and number of additions subtractions are unaffected by Horner s
ethod).
G

R BLEMS
In Pr b ems determine hether the gi en functi n is a 1
f .x/ D ! 3x5 C 2x6 C x7 f .x/ D
p yn mia functi n "
x3 C 7x ! 3
f .x/ D x2 ! x4 C 4 f .x/ D
3 In Pr b ems nd the functi n a ues f r each functi n
p
5 !3 3 f .x/ D f .2/, f .t C /; f .! 17/
g.x/ D g.x/ D 2 x
3x C 1 g.x/ D j2x C 1j g.20/, g.5/, g.!7/
8
In Pr b ems determine hether the gi en functi n is a < 2 if t > 1
rati na functi n F.t/ D 0 if t D 1 I
x2 C x 3 : !1 if t < 1
f .x/ D 3 f .x/ D $ %
x C4 2x C 1 p 1
! F.12/, F.! 3/, F.1/, F
1 if x < 5 5
g.x/ D g.x/ D 2x!5 !
4 if x $ 5 4 if x $ 0
f .x/ D I
3 if x < 0
In Pr b ems nd the d main f each functi n f .3/; f .!4/; f .0/
p !
k.z/ D 26 f .x/ D " x ! 1 if x $ 3
! ! G.x/ D I
3 ! x2 if x < 3
5x if x > 1 4 if x D 3
f .x/ D f .x/ D 2
4 if x " 1 x if 1 " x < 3 G. /, G.3/, G.!1/, G.1/
!
In Pr b ems state a the degree and b the eading 2$ ! 5 if $ < 2
F.$ / D I
c e cient f the gi en p yn mia functi n $ 2 ! 3$ C 1 if $ > 2
F.x/ D 2x3 ! 32x2 C 5x4 g.x/ D x2 C 2x C 1 F.3/, F.!3/, F.2/
96 C nct ons an ra s

In Pr b ems determine the a ue f each expressi n P D P.r/, where

Š .3 ! 3/Š .4 ! 2/Š & 'r& '3!r
3Š 14 34
nŠ Š P.r/ D ; r D 0; 1; 2; 3
6Š & 2Š rŠ.3 ! r/Š
.n ! 1/Š 4Š. ! 4/Š
Find the probability that exactly two of the children will be
Subway Ride A return subway ride tic et within the city blue eyed.
costs 2.50. Write the cost of a return tic et as a function of a Genetics In Example 7, find the probability that all five
passenger s income. What ind of function is this offspring will be brown.
Geometry A rectangular prism has length three more than Bacteria Growth acteria are growing in a culture. The
its width and height one less than twice the width. Write the time t (in hours) for the bacteria to double in number (the
volume of the rectangular prism as a function of the width. What generation time) is a function of the temperature (in ı C) of the
ind of function is this culture. If this function is given by2
Cost Function In manufacturing a component for a 8̂
1 11
machine, the initial cost of a die is 50 and all other additional ˆ
< 24 C 4 if 30 " " 36
costs are 3 per unit produced. a Express the total cost C (in t D f. / D
dollars) as a linear function of the number q of units produced. ˆ4 175
:̂ ! if 36 < " 3
b How many units are produced if the total cost is 1600 3 4
Investment If a principal of P dollars is invested at a a determine the domain of f and b find f .30/, f .36/, and f .3 /.
simple annual interest rate of r for t years, express the total In Pr b ems use a ca cu at r t nd the indicated functi n
accumulated amount of the principal and interest as a function a ues f r the gi en functi n R und ans ers t t decima
of t. Is your result a linear function of t p aces
(
Capital Gains The Canadian tax rates in Example refer 0:11x3 ! 15:31 if x < 2:57
to personal income other than capital gains. Using the same f .x/ D
0:42x4 ! 12:31 if x $ 2:57
cut-off points as in Example , which define what are called tax
brackets in tax terminology, the capital gains rates for each a f .2:14/ b f .3:27/ c f .!4/
brac et are precisely half of the rates given in Example . Write (
a case-defined function that describes capital gains tax rate c as a 2 :5x4 C 30:4 if x < 3
f .x/ D
function of capital gains income j. 7: x3 ! 2:1x if x $ 3
Factorials The business mathematics class has elected a a f .2:5/ b f .!3:6/ c f .3:2/
grievance committee of five to complain to the faculty about the 8̂
introduction of factorial notation into the course. They decide < 4:07x ! 2:3 if x < !
that they will be more effective if they label themselves as f .x/ D 1 :12 if ! " x < !2
members A, , , N, and S, where member A will lobby faculty :̂ 2
with surnames A through F, member will lobby faculty with x ! 4x!2 if x $ !2
surnames through , and so on. In how many ways can the a f .!5: / b f .!14: / c f(7.6)
committee so label its members 8̂
Genetics Under certain conditions, if two brown-eyed < x=.x C 3/ if x < !5
parents have exactly three children, the probability that there f .x/ D x.x ! 4/2 if ! 5 " x < 0
:̂ p
will be exactly r blue-eyed children is given by the function 2:1x C 3 if x $ 0
p
a f .! 30/ b f .46/ c f .!2=3/

Objective C F
o co ne f nct ons eans of There are several ways of combining two functions to create a new function. Suppose
a t on s tract on t cat on
s on t cat on a constant f and g are the functions given by
an co os t on f.x/ D x2 and g.x/ D 3x
Adding f.x/ and g.x/ gives
f.x/ C g.x/ D x2 C 3x
This operation defines a new function called the sum of f and g, denoted f C g. Its
function value at x is f.x/ C g.x/. That is,
. f C g/.x/ D f.x/ C g.x/ D x2 C 3x

2 Adapted from F. . E. Imrie and A. . Vlitos, Production of Fungal Protein from Carob, in Sing e Ce Pr tein
II ed. S. R. Tannenbaum and . I. C. Wang (Cambridge, A: IT Press, 1 75).
 Section 2.3 Co nat ons of nct ons 97

For example,
. f C g/.2/ D 22 C 3.2/ D 10

In general, for any functions f; g W X % .!1; 1/, we define the sum f C g, the
f
di erence f ! g, the pr duct fg, and the qu tient as follows:
g

. f C g/.x/ D f.x/ C g.x/

. f ! g/.x/ D f.x/ ! g.x/
. fg/.x/ D f.x/ & g.x/
f f.x/
.x/ D for g.x/ ¤ 0
g g.x/

For each of the four new functions, the domain is the set of all x that belong to both the
domain of f and the domain of g, with the domain of the quotient further restricted to
exclude any value of x for which g.x/ D 0. In each of the four combinations, we have
a new function from X to .!1; 1/. For example, we have

f C g W X % .!1; 1/

A special case of fg deserves separate mention. For any real number c and any function
f, we define cf by

.cf /.x/ D c & f.x/

This restricted case of product is calledp

the sca ar pr duct
For f.x/ D x2 , g.x/ D 3x, and c D 2 we have

. f C g/.x/ D f.x/ C g.x/ D x2 C 3x

. f ! g/.x/ D f.x/ ! g.x/ D x2 ! 3x
. fg/.x/ D f.x/ & g.x/ D x2 .3x/ D 3x3
f f.x/ x2 x
.x/ D D D for x ¤ 0
g g.x/ 3x 3
p 2
.cf /.x/ D cf.x/ D 2x

E AM LE C F

If f.x/ D 3x ! 1 and g.x/ D x2 C 3x, find

a . f C g/.x/
b . f ! g/.x/
c . fg/.x/
f
d .x/
g
e ..1=2/f /.x/

a . f C g/.x/ D f.x/ C g.x/ D .3x ! 1/ C .x2 C 3x/ D x2 C 6x ! 1

b . f ! g/.x/ D f.x/ ! g.x/ D .3x ! 1/ ! .x2 C 3x/ D !1 ! x2
c . fg/.x/ D f.x/g.x/ D .3x ! 1/.x2 C 3x/ D 3x3 C x2 ! 3x
98 C nct ons an ra s

f f.x/ 3x ! 1
d .x/ D D 2
g g.x/ x C 3x
e ..1=2/f /.x/ D .1=2/. f.x// D .1=2/.3x ! 1/

Now ork Problem 3(a)--(f) G

C
We can also combine two functions by first applying one function to an input and then
applying the other function to the output of the first. For example, suppose g.x/ D 3x,
f.x/ D x2 , and x D 2. Then g.2/ D 3 & 2 D 6. Thus, g sends the input 2 to the output 6:
g
2 ‘ 6
Next, we let the output 6 become the input for f:
f.6/ D 62 D 36
So f sends 6 to 36:
f
6 ‘ 36
y first applying g and then f, we send 2 to 36:
g f
2 ‘ 6 ‘ 36
To be more general, replace the 2 by x, where x is in the domain of g. (See Figure 2.3.)
Applying g to x, we get the number g.x/, which we will assume is in the domain of
f. y applying f to g.x/, we get f.g.x/), read f of g of x, which is in the range of f.
The operation of applying g and then applying f to the result is called c mp siti n, and
the resulting function, denoted f ı g, is called the c mp site of f with g. This function
assigns the output f.g.x/) to the input x. (See the bottom arrow in Figure 2.3.) Thus,
. f ı g/.x/ D f.g.x//.

Range of f
Domain of f f
g f(g(x))
Domain = ( f 5 g)(x)
of g
x g(x)

f5g

FIGURE Composite of f with g.

For functions g W X % and f W % , the composite of f with g is the function

f ı g W X % defined by
. f ı g/.x/ D f.g.x//
where the domain of f ı g is the set of all those x in the domain of g such that g.x/
is in the domain of f.

For f.x/ D x2 and g.x/ D 3x, we can get a simple form for f ı g:
. f ı g/.x/ D f.g.x// D f.3x/ D .3x/2 D x2
For example, . f ı g/.2/ D .2/2 D 36, as we saw before.
 Section 2.3 Co nat ons of nct ons 99

When dealing with real numbers and the operation of addition, 0 is special in that
for any real number a, we have
aC0DaD0Ca
The number 1 has a similar property with respect to multiplication. For any real number
a, we have
a1 D a D 1a
For reference, in Section 2.4 we note that the function I defined by I.x/ D x satisfies,
for any function f,

fıIDfDIıf

where here we mean equality of functions as defined in Section 2.1. Indeed, for any x,

. f ı I/.x/ D f.I.x// D f.x/ D I. f.x// D .I ı f /.x/

The function I is called the identity function.

A L IT I
A C costs x dollars wholesale. The E AM LE C
price the store pays is given by the func-
tion s.x/ D x C 3. The price the cus- p
et f.x/ D x and g.x/ D x C 1. Find
tomer pays is c.x/ D 2x, where x is the a . f ı g/.x/
price the store pays. Write a composite
function to find the customer s price as b .g ı f /.x/
a function of the wholesale price.
S
a . f ı g/.x/ is f.g.x/). Now g adds 1 to x, and f ta es the square root of the result.
Thus,
p
. f ı g/.x/ D f.g.x// D f.x C 1/ D x C 1

enerally, f ı g and g ı f are different. In The domain of g is all real numbers x, and the domain of f is all nonnegative reals.
Example 2, Hence, the domain of the composite is all x for which g.x/ D x C 1 is nonnegative.
p That is, the domain is all x $ !1, which is the interval Œ!1; 1/.
. f ı g/.x/ D x C 1
b .g ı f /.x/ is g( f.x/).
p Now f ta es the square root of x, and g adds 1 to the result.
but we have
p Thus, g adds 1 to x, and we have
.g ı f /.x/ D xC1 p p
p .g ı f /.x/ D g. f.x// D g. x/ D x C 1
bserve that . f ı g/.1/ D 2, while
.g ı f /.1/ D 2. Also, do not confuse
f .g.x/) with . fg/.x/, which is the product The domain of f is all x $ 0, and the domainpof g is all reals. Hence, the domain of
f .x/g.x/. Here the composite is all x $ 0 for which f.x/ D x is real, namely, all x $ 0.
p
f .g.x// D x C 1 Now ork Problem 7 G
but
p Composition is ass ciati e, meaning that for any three functions f, g, and h,
f .x/g.x/ D x.x C 1/
. f ı g/ ı h D f ı .g ı h/

E AM LE C

If F.p/ D p2 C 4p ! 3, G.p/ D 2p C 1, and .p/ D jpj, find

a F.G.p//
b F.G. .p///
c G.F.1//
100 C nct ons an ra s

S
a F.G.p// D F.2p C 1/ D .2p C 1/2 C 4.2p C 1/ ! 3 D 4p2 C 12p C 2 D .F ı G/.p/
b F.G. .p/// D .F ı .G ı //.p/ D ..F ı G/ ı /.p/ D .F ı G/. .p// D
.F ı G/.jpj/ D 4jpj2 C 12jpj C 2 D 4p2 C 12jpj C 2
c G.F.1// D G.12 C 4 & 1 ! 3/ D G.2/ D 2 & 2 C 1 D 5
Now ork Problem 9 G
In calculus, it is sometimes necessary to thin of a particular function as a com-
posite of two simpler functions, as the next example shows.

E AM LE E F C

Express h.x/ D .2x ! 1/3 as a composite.

S
We note that h.x/ is obtained by finding 2x ! 1 and cubing the result. Suppose we let
g.x/ D 2x ! 1 and f.x/ D x3 . Then
h.x/ D .2x ! 1/3 D .g.x//3 D f.g.x// D . f ı g/.x/
which gives h as a composite of two functions.
Now ork Problem 13 G

R BLEMS
If f .x/ D x C 3 and g.x/ D x C 5, find the following. 2
If F.t/ D t2 C 7t C 1 and G.t/ D , find .F ı G/.t/ and
a . f C g/.x/ b . f C g/.0/ c . f ! g/.x/ t!1
.G ı F/.t/.
f p
d ( fg).x/ e . fg/.!2/ f .x/ If F.t/ D t and G.t/ D 2t2 ! 2t C 1, find .F ı G/.t/ and
g
.G ı F/.t/.
g . f ı g/.x/ h . f ı g/.3/ i .g ı f /.x/
2 p
.g ı f /.3/ If f . / D 2 and g. / D 3 C 1, find . f ı g/. / and
!3
.g ı f /. /.
If f .x/ D 2x and g.x/ D 6 C x, find the following.
a . f C g/.x/ b . f ! g/.x/ c . f ! g/.4/ If f .x/ D x2 C 2x ! 1, find . f ı f /.x/.
f f In Pr b ems nd functi ns f and g such that
d . fg/.x/ e .x/ f .2/ h.x/ D f .g.x//
g g
g . f ı g/.x/ h .g ı f /.x/ i .g ı f /.2/ h.x/ D 11x ! 7
p
If f .x/ D x2 ! 1 and g.x/ D x2 C x, find the following. h.x/ D x2 ! 2
& '
a . f C g/.x/ b . f ! g/.x/ c . f ! g/ ! 12 3
$ % h.x/ D
f f 1 x2 CxC1
d ( fg).x/ e .x/ f !
g g 2 h.x/ D 7.4x2 C 7x/2 ! 5.4x2 C 7x/ C 1
g . f ı g/.x/ h .g ı f /.x/ i .g ı f /.!3/ s
2
4 x ! 1
h.x/ D
If f .x/ D 2x2 C 5 and g.x/ D 3, find the following. xC3
& '
a . f C g/.x/ b . f C g/ 12 c . f ! g/.x/ 2 ! .3x ! 5/
f h.x/ D
d . fg/.x/ e . fg/.2/ f .x/ .3x ! 5/2 C 2
g
g . f ı g/.x/ h . f ı g/.100:003/ i .g ı f /.x/ Profit A coffeehouse sells a pound of coffee for .75.
Expenses are 4500 each month, plus 4.25 for each pound of
If f .x/ D 3x2 C 6 and g.x/ D 4 ! 2x, find f(g(2)) and g(f(2)). coffee sold.
4 p!2 a Write a function r.x/ for the total monthly revenue as a
If f .p/ D and g.p/ D , find both . f ı g/.p/ and function of the number of pounds of coffee sold.
p 3
.g ı f /.p/. b Write a function e.x/ for the total monthly expenses as a
function of the number of pounds of coffee sold.
c Write a function .r ! e/.x/ for the total monthly profit as a
function of the number of pounds of coffee sold.
 Section 2.4 n erse nct ons 101

Geometry Suppose the volume of a sphere is Furthermore, suppose a person s income I is a function of the
.x/ D 43 ".3x ! 1/3 . Express as a composite of two functions, number of years of education E, where
and explain what each function represents.
I D g.E/ D 7202 C 0:2 E3:6
Business A manufacturer determines that the total number
Find . f ı g/.E/. What does this function describe
of units of output per day, q, is a function of the number of
employees, m, where In Pr b ems f r the gi en functi ns f and g nd the
indicated functi n a ues R und ans ers t t decima p aces
.20m ! m2 / f .x/ D .4x ! 13/2 , g.x/ D 0:2x2 ! 4x C 3
q D f .m/ D
2 a . f C g/.4:5/, b . f ı g/.!2/
r
The total revenue r that is received for selling q units is given by x!3
f .x/ D , g.x/ D 11:2x C 5:3
the function g, where r D g.q/ D 24q. Find .g ı f /.m/. What does xC1
this composite function describe f
a .!2/, b .g ı f /.!10/
g
Sociology Studies have been conducted concerning the
statistical relations among a person s status, education, and f .x/ D x4=5 , g.x/ D x2 !
income.3 et S denote a numerical value of status based on annual a . fg/.7/, b .g ı f /.3:75/
income I. For a certain population, suppose 2 1
f .x/ D , g.x/ D 3
xC1 x
S D f .I/ D 0:45.I ! 1000/0:53 a . f ı g/.2:17/, b .g ı f /.2:17/

Objective I F
o ntro ce n erse f nct ons t e r ust as !a is the number for which
ro ert es an t e r ses
a C .!a/ D 0 D .!a/ C a

and, for a ¤ 0, a!1 is the number for which

aa!1 D 1 D a!1 a

so, given a function f W X % , we can inquire about the existence of a function g

satisfying

fıgDIDgıf

where I is the identity function, introduced in the subsection titled Composition of

Section 2.3 and given by I.x/ D x. Suppose that we have g as above and a function h
that also satisfies the equations of (1) so that

fıhDIDhıf

Then

h D h ı I D h ı . f ı g/ D .h ı f / ı g D I ı g D g

shows that there is at most one function satisfying the requirements of g in

(1). In mathematical argon, g is uniquely determined by f and is therefore given a
o not confuse f !1 , the inverse of f,
name, g D f !1 , that re ects its dependence on f. The function f !1 is read as f inverse
1
and , the multiplicative reciprocal of f. and called the in erse of f.
f
The additive inverse !a exists for any number a the multiplicative inverse a!1
Unfortunately, the notation for inverse
functions clashes with the numerical use exists precisely if a ¤ 0. The existence of f !1 places a strong requirement on a function
of .!/!1 . Usually, f !1 .x/ is different f. It can be shown that f !1 exists if and only if, for all a and b, whenever f.a/ D f.b/,
1 1 then a D b. It may be helpful to thin that such an f can be cance ed n the eft .
from .x/ D . For example, I!1 D I
f f .x/
(since I ı I D I) so I!1 .x/ D x, but
1 1 1 3
.x/ D D . R. . ei and . F. ee er, athematica S ci gy (Englewood Cliffs, N :
I I.x/ x Prentice Hall, 1 75).
102 C nct ons an ra s

A function f that satisfies

for all a and b; if f.a/ D f.b/ then a D b
is called a one to one function.

Thus, we can say that a function has an inverse precisely if it is one-to-one. An equiv-
alent way to express the one-to-one condition is
for all a and b; if a ¤ b then f.a/ ¤ f.b/
so that distinct inputs give rise to distinct outputs. bserve that this condition is not
met for many simple functions. For example, if
f.x/ D x2 , then f.!1/ D .!1/2 D 1 D .1/2 D f.1/
and !1 ¤ 1 shows that the squaring function is not one-to-one. Similarly, f.x/ D jxj
is not one-to-one.
In general, the domain of f !1 is the range of f and the range of f !1 is the domain of f.
et us note here that the equations of (1) are equivalent to
f !1 . f.x// D x for all x in the domain of f
and
f. f !1 .y// D y for all y in the range of f
In general, the range of f, which is equal to the domain of f !1 , can be different from
the domain of f.

E AM LE I L F

According to Section 2.2, a function of the form f.x/ D ax C b, where a ¤ 0, is a linear

function. Show that a linear function is one-to-one. Find the inverse of f.x/ D ax C b
and show that it is also linear.
S Assume that f.u/ D f. / that is,
au C b D a C b
To show that f is one-to-one, we must show that u D follows from this assumption.
Subtracting b from both sides of (4) gives au D a , from which u D follows by
dividing both sides by a. (We assumed that a ¤ 0.) Since f is given by first multiplying
by a and then adding b, we might expect that the effect of f can be undone by first
x!b
subtracting b and then dividing by a. So consider g.x/ D . We have
a
x!b
. f ı g/.x/ D f.g.x// D a C b D .x ! b/ C b D x
a
and
.ax C b/ ! b ax
.g ı f /.x/ D g. f.x// D D Dx
a a
Since g satisfies the two requirements of (1), it follows that g is the inverse of f. That
x!b 1 !b
is, f !1 .x/ D D xC and the last equality shows that f !1 is also a linear
a a a
function.
Now ork Problem 1 G
E AM LE I I

Show that
a If f and g are one-to-one functions, the composite f ı g is also one-to-one and
. f ı g/!1 D g!1 ı f !1 .
b If f is one-to-one, then . f !1 /!1 D f.
 Section 2.4 n erse nct ons 103

S
a Assume . f ı g/.a/ D . f ı g/.b/ that is, f.g.a// D f.g.b//. Since f is one-to-one,
g.a/ D g.b/. Since g is one-to-one, a D b and this shows that f ı g is one-to-one.
The equations
. f ı g/ ı .g!1 ı f !1 / D f ı .g ı g!1 / ı f !1 D f ı I ı f !1 D f ı f !1 D I
and
.g!1 ı f !1 / ı . f ı g/ D g!1 ı . f !1 ı f / ı g D g!1 ı I ı g D g!1 ı g D I
show that g!1 ı f !1 is the inverse of f ı g, which, in symbols, is the statement
g!1 ı f !1 D . f ı g/!1 .
b In Equations (2) and (3), replace f by f !1 . Ta ing g to be f shows that Equation (1)
is satisfied, and this gives . f !1 /!1 D f.
G
E AM LE I U S E

any equations ta e the form f.x/ D 0, where f is a function. If f is a one-to-one

function, then the equation has x D f !1 .0/ as its unique solution.

S Applying f !1 to both sides of f.x/ D 0 gives f !1 . f.x// D f !1 .0/,

and f . f.x// D x shows that x D f !1 .0/ is the only possible solution. Since
!1

f. f !1 .0// D 0, f !1 .0/ is indeed a solution.

G
E AM LE R F

It may happen that a function f whose domain is the natural one, consisting of all ele-
ments for which the defining rule ma es sense, is not one-to-one, and yet a one-to-one
function g can be obtained by restricting the domain of f.

S For example, we have shown that the function f.x/ D x2 is not one-to-one
g.x/ D x2 ith d main exp icit y gi en as Œ0; 1/ is one-to-one. Since
but the function p
p 2 p
. x/ D x and x2 D x, for x $ 0, it follows that is the inverse of the restricted
squaring function g. Here is a more contrived example. et f.x/ D jxj (with its natural
domain). et g.x/ D jxj ith d main exp icit y gi en as .!1; !1/[Œ0; 1%. The function
g is one-to-one and hence has an inverse.
G
E AM LE F I F

To find the inverse of a one-to-one function f, solve the equation y D f.x/ for x in terms
of y obtaining x D g.y/. Then f !1 .x/ D g.x/. To illustrate, find f !1 .x/ if f.x/ D .x!1/2 ,
for x $ 1.
p p
S ! 1/2 , for x $ 1. Then x ! 1 D
et y D .x p y and hence x D y C 1. It
!1
follows that f .x/ D x C 1.
Now ork Problem 5 G

R BLEMS
In Pr b ems nd the in erse f the gi en functi n In Pr b ems determine hether r n t the functi n is
f .x/ D 3x C 7 g.x/ D 5x ! 3 ne t ne
F.x/ D 12 x ! 7 f .x/ D .4x ! 5/2 , for x $ 5 f .x/ D 5x C 12 g.x/ D .3x C 4/2
4
h.x/ D .5x C 12/2 , for x $ ! 12 F.x/ D jx C 10j
A.r/ D 4"r2 , for r $ 0 .r/ D 43 "r3 5
104 C nct ons an ra s

In Pr b ems and s e each equati n by nding an in erse is a function of p. Show that the resulting function is inverse to the
functi n function giving p in terms of q.
5
.4x ! 5/2 D 23, for x $ 4
Supply Function The wee ly supply function for a pound
3
2x C 1 D 12 of house-blend coffee at a coffee shop is

Demand Function The function q

p D p.q/ D q>0
1,200,000 4
p D p.q/ D q>0
q where q is the number of pounds of coffee supplied per wee and
expresses an actor s charge per film p as a function of the number p is the price per pound. Express q as a function of p and
of films, q, that she stars in. Express the number of films in which demonstrate the relationship between the two functions.
she stars in terms of her charge per film. Show that the expression oes the function f .x/ D 10x have an inverse

Objective G R C
o ra e at ons an f nct ons A rectangular coordinate system allows us to specify and locate points in a plane. It
n rectan ar coor nates to
eter ne nterce ts to a also provides a geometric way to graph equations in two variables, in particular those
t e ert ca ne test an t e arising from functions.
or onta ne test an to In a plane, two real-number lines, called coordinate axes, are constructed per-
eter ne t e o a n an ran e
of a f nct on fro a ra pendicular to each other so that their origins coincide, as in Figure 2.4. Their point of
intersection is called the origin of the coordinate system. We will call the horizontal
y line the x axis and the vertical line the y axis.
The plane on which the coordinate axes are placed is called a rectangu ar c rdi
nate p ane or simply an x y plane. Every point in the x y-plane can be labeled to indicate
3
its position. To label point P in Figure 2.5(a), we draw perpendiculars from P to the
2 x-axis and y-axis. They meet these axes at 4 and 2, respectively. Thus, P determines
1 Origin two numbers, 4 and 2. We say that the rectangu ar c rdinates of P are given by the
x rdered pair .4; 2/. As we remar ed in Section 2.1, the word rdered is important. In
-4 -3 -2 -1 1 2 3 4 the terminology of Section 2.1, we are labeling the points of the plane by the elements
of the set .!1; 1/ # .!1; 1/. In Figure 2.5(b), the point corresponding to .4; 2/ is
-1
-2 not the same as that corresponding to .2; 4/:
-3
.4; 2/ ¤ .2; 4/

FIGURE Coordinate axes. y y

(2, 4) Z (4, 2)
4 (2, 4)

P(4, 2)
2 2 (4, 2)

x x
4 2 4

(a) (b)

y FIGURE Rectangular coordinates.

In general, if P is any point, then its rectangular coordinates will be given by an

P(a, b) ordered pair of the form .a; b/. (See Figure 2.6.) We call a the x c rdinate of P, and b
b the y c rdinate of P. We accept that the notation for an ordered pair of real numbers is
the same as that for an open interval but the practice is strongly entrenched and almost
a
x never causes any confusion.
Accordingly, with each point in a given coordinate plane, we can associate exactly
one ordered pair .a; b/ of real numbers. Also, it should be clear that with each ordered
pair .a; b/ of real numbers, we can associate exactly one point in that plane. Since there
FIGURE Coordinates of P. is a ne t ne c rresp ndence between the points in the plane and all ordered pairs
 Section 2.5 ra s n ectan ar Coor nates 105

y
of real numbers, we refer to a point P with x-coordinate a and y-coordinate b simply
as the point .a; b/, or as P.a; b/. oreover, we use the words p int and rdered pair f
rea numbers interchangeably.
In Figure 2.7, the coordinates of various points are indicated. For example,
5
(- 2 , 3) (0, 3) the point .1; !4/ is located one unit to the right of the y-axis and four units below the
(3, 2) x-axis. The origin is .0; 0/. The x-coordinate of every point on the y-axis is 0, and
(-3, 0) (0, 0) the y-coordinate of every point on the x-axis is 0.
x The coordinate axes divide the plane into four regions called quadrants
(4, 0)
(Figure 2. ). For example, quadrant I consists of all points .x1 ; y1 / with x1 > 0 and
(0, -2) y1 > 0. The points on the axes do not lie in any quadrant.
Using a rectangular coordinate system, we can geometrically represent equations
(-2, -3) (1, -4) in two variables. For example, let us consider

y D x2 C 2x ! 3
FIGURE Coordinates of
points. A solution of this equation is a value of x and a value of y that ma e the equation true.
For example, if x D 1, substituting into Equation (1) gives
y
y D 12 C 2.1/ ! 3 D 0
Quadrant II Quadrant I Thus, x D 1; y D 0 is a solution of Equation (1). Similarly,
(x2, y2) (x1, y1) if x D !2 then y D .!2/2 C 2.!2/ ! 3 D !3
x2 6 0, y2 7 0 x1 7 0, y1 7 0
and so x D !2; y D !3 is also a solution. y choosing other values for x, we can get
x
more solutions. See Figure 2. (a). It should be clear that there are infinitely many
Quadrant III Quadrant IV
solutions of Equation (1).
(x3, y3) (x4, y4) Each solution gives rise to a point .x; y/. For example, to x D 1 and y D 0 corre-
x3 6 0, y3 6 0 x4 7 0, y4 6 0 sponds .1; 0/. The graph of y D x2 C 2x ! 3 is the geometric representation of all its
solutions. In Figure b , we have plotted the points corresponding to the solutions
in the table.
Since the equation has infinitely many solutions, it seems impossible to determine
FIGURE uadrants. its graph precisely. However, we are concerned only with the graph s general shape. For
this reason, we plot enough points so that we can intelligently guess its proper shape.
(The calculus techniques to be studied in Chapter 13 will ma e such guesses much
more intelligent.) Then, we oin these points by a smooth curve wherever conditions
permit. This gives the curve in Figure 2. (c). f course, the more points we plot, the
better our graph is. Here we assume that the graph extends indefinitely upward, as
indicated by arrows.
The point .0; !3/ where the curve intersects the y-axis is called the y-intercept.
ften, we simply say that the y-intercept The points .!3; 0/ and .1; 0/ where the curve intersects the x-axis are called the
is !3 and the x-intercepts are !3 and 1. x-intercepts. In general, we have the following definition.

x y y y

-4 5
5
-3 0

-2 -3
y = x2 + 2x - 3
-1 -4
x x
0 -3 -4 -2 2

1 0 x-intercept x-intercept

2 5 -4 y-intercept

(a) (b) (c)

FIGURE raphing y D x2 C 2x ! 3.
106 C nct ons an ra s

An x intercept of the graph of an equation in x and y is a point where the graph

intersects the x-axis. A y intercept is a point where the graph intersects the y-axis.

To find the x-intercepts of the graph of an equation in x and y, we first set y D 0 and
then solve the resulting equation for x. To find the y-intercepts, we first set x D 0 and
then solve for y. For example, let us find the x-intercepts for the graph of y D x2 C2x!3.
Setting y D 0 and solving for x gives
0 D x2 C 2x ! 3
0 D .x C 3/.x ! 1/
x D !3; 1
Thus, the x-intercepts are .!3; 0/ and .1; 0/, as we saw before. If x D 0, then
y D 02 C 2.0/ ! 3 D !3
So .0; !3/ is the y-intercept. eep in mind that an x-intercept has its y-coordinate 0, and
a y-intercept has its x-coordinate 0. Intercepts are useful because they indicate precisely
where the graph intersects the axes.

E AM LE I G
A L IT I
Rachel has saved 7250 for college Find the x- and y-intercepts of the graph of y D 2x C 3, and s etch the graph.
expenses. She plans to spend 600 a S If y D 0, then
month from this account. Write an equa-
tion to represent the situation, and iden- 3
tify the intercepts.
0 D 2x C 3 so that xD!
2
Thus, the x-intercept is .! 32 ; 0/. If x D 0, then
y D 2.0/ C 3 D 3
So the y-intercept is .0; 3/. Figure 2.10 shows a table of some points on the graph and
a s etch of the graph.
y

y = 2x + 3

y-intercept
x-intercept
1
x
1
3 1 1
x 0 - 2 2
- 2 1 -1 2 -2

y 3 0 4 2 5 1 7 -1

FIGURE raph of y D 2x C 3.

Now ork Problem 9 G

E AM LE I G
100
etermine the intercepts, if any, of the graph of s D
, and s etch the graph.
t
S For the graph, we will label the horizontal axis t and the vertical axis s
(Figure 2.11). ecause t cannot equal 0 (division by 0 is not defined), there is no
 Section 2.5 ra s n ectan ar Coor nates 107

s-intercept. Thus, the graph has no point corresponding to t D 0. oreover, there

is no t-intercept, because if s D 0, then the equation
100
0D
t
has no solution. Remember, the only way that a fraction can be 0 is by having its
A L IT I numerator 0. Figure 2.11 shows the graph. In general, the graph of s D k=t, where k is
The price of admission to an amuse- a nonzero constant, is called a rectangu ar hyperb a.
ment par is 24. 5. This fee allows
s
the customer to ride all the rides at the
par as often as he or she li es. Write
an equation that represents the relation- 20
ship between the number of rides, x, that s = 100
t
a customer ta es and the cost per ride, 10
y, to that customer. escribe the graph t
of this equation, and identify the inter- 20 40
cepts. Assume x > 0. No intercepts
t 5 -5 10 -10 20 -20 25 -25 50 -50

s 20 -20 10 -10 5 -5 4 -4 2 -2

100
FIGURE raph of s D .
t

Now ork Problem 11 G

y E AM LE I G
3 x=3
x-intercept
etermine the intercepts of the graph of x D 3, and s etch the graph.
x S We can thin of x D 3 as an equation in the variables x and y if we write it
3 as x D 3C0y. Here y can be any value, but x must be 3. ecause x D 3 when y D 0, the
-2 x-intercept is .3; 0/. There is no y-intercept, because x cannot be 0. (See Figure 2.12.)
The graph is a vertical line.
Now ork Problem 13 G
x 3 3 3

y 0 3 -2 Each function f gives rise to an equation, namely y D f.x/, which is a special case of
the equations we have been graphing. Its graph consists of all points .x; f.x//, where x
FIGURE raph of x D 3.
is in the domain of f. The vertical axis can be labeled either y or f.x/, where f is the name
of the function, and is referred to as the function value axis. In this b k e a ays
abe the h riz nta axis ith the independent ariab e but n te that ec n mists abe
the ertica axis ith the independent ariab e bserve that in graphing a function the
solutions .x; y/ that ma e the equation y D f.x/ true are handed to us. For each x
in the domain of f, we have exactly one y obtained by evaluating f.x/. The resulting
pair .x; f.x// is a point on the graph, and these are the only points on the graph of the
equation y D f.x/.
The x-intercepts of the graph of a real-valued function f are all those real numbers
x for which f.x/ D 0. As such they are also nown as roots of the equation f.x/ D 0
and still further as eros of the function f.
A useful geometric observation is that the graph of a function has at most one point
of intersection with any vertical line in the plane. Recall that the equation of a vertical
line is necessarily of the form x D a, where a is a constant. If a is not in the domain
of the function f, then x D a will not intersect the graph of y D f.x/. If a is in the
domain of the function f, then x D a will intersect the graph of y D f.x/ at the point
.a; f.a//, and only there. Conversely, if a set of points in the plane has the property that
any vertical line intersects the set at most once, then the set of points is actually the
graph of a function. (The domain of the function is the set of all real numbers a with
the property that the line x D a does intersect the given set of points, and for such an a
the corresponding function value is the y-coordinate of the unique point of intersection
108 C nct ons an ra s

f(x) of the line x D a and the given set of points.) This is the basis of the vertical line test
that we will discuss after Example 7.

3
2 E AM LE G S R F
f(x) = x
1 p
x raph the function fW .!1; 1/ % .!1; 1/ given by f.x/ D x.
1 4 9
S p The graph is shown in Figure 2.13. We label the vertical
p axis as f.x/. Recall
1
that x denotes the principa square root of x. Thus, f. / D D 3, not ˙3. Also,
x 0 1 4 9
4
the domain of f is Œ0; 1/ because its values
p are declared to be real numbers. et us now
f(x) 0 1
2 1 2 3 consider intercepts. If f.x/ D 0, then x D 0, so that x D 0. Also, if x D 0, then
f.x/ D 0. Thus, the x-intercept and the vertical-axis intercept are the same, namely,
FIGURE raph of .0; 0/.
p
f .x/ D x. Now ork Problem 29 G
E AM LE G A F
A L IT I
raph p D G.q/ D jqj.
rett rented a bi e from a rental
shop, rode at a constant rate of 12 mi h S We use the independent variable q to label the horizontal axis. The function-
for 2.5 hours along a bi e path, and then value axis can be labeled either G.q/ or p. (See Figure 2.14.) Notice that the q- and
returned along the same path. raph the p-intercepts are the same point, .0; 0/. p
absolute-value-li e function that repre-
sents rett s distance from the rental
5
shop as a function of time over the
appropriate domain. q 0 1 -1 3 -3 5 -5 3
p= q
p 0 1 1 3 3 5 5
q
-5 -3 -1 1 3 5
Note: Sharp corner
at origin

FIGURE raph of p D jqj.

Now ork Problem 31 G

f (x) (x, f(x))

Range:
all y 9 0
f(4) = 3

x
4 x

Domain: all real numbers

FIGURE omain, range, and function values.

Figure 2.15 shows the graph of a function y D f.x/. The point .x; f.x// tells us that
corresponding to the input number x on the horizontal axis is the output number f.x/
on the vertical axis, as indicated by the arrow. For example, corresponding to the input
4 is the output 3, so f.4/ D 3.
From the shape of the graph, it seems reasonable to assume that, for any value of
x, there is an output number, so the domain of f is all real numbers. Notice that the set
of all y-coordinates of points on the graph is the set of all nonnegative numbers. Thus,
 Section 2.5 ra s n ectan ar Coor nates 109

the range of f is all y $ 0. This shows that we can ma e an educated guess about
the domain and range of a function by loo ing at its graph. In genera the d main
c nsists f a x a ues that are inc uded in the graph and the range is a y a ues
that arep
inc uded. For example, Figure 2.13 tells us that both the domain and range of
f.x/ D x are all nonnegative numbers. From Figure 2.14, it is clear that the domain
s of p D G.q/ D jqj is all real numbers and the range is all p $ 0.
s = F(t )
1
E AM LE R F
Range: t
-1 … s … 1 1 2 3 4 Figure 2.16 shows the graph of a function F. To the right of 4, assume that the graph
-1 repeats itself indefinitely. Then the domain of F is all t $ 0. The range is !1 " s " 1.
Some function values are
Domain: t Ú 0
F.0/ D 0 F.1/ D 1 F.2/ D 0 F.3/ D !1
FIGURE omain, range, and
function values. Now ork Problem 5 G

E AM LE G C F

raph the case-defined function

8
< x if 0 " x < 3
f.x/ D x ! 1 if 3 " x " 5
: 4 if 5 < x " 7

S The domain of f is 0 " x " 7. The graph is given in Figure 2.17, where the
h d t means that the point is n t included in the graph. Notice that the range of f
is all real numbers y such that 0 " y " 4.
A L IT I
To encourage conservation, a gas f(x)
company charges two rates. ou pay x if 0 … x 6 3
0.53 per therm for 0 70 therms and 4 f(x ) = x - 1 if 3 … x … 5
0.74 for each therm over 70. raph the Range: 4 if 5 6 x … 7
0…y…4 2
case-defined function that represents the
monthly cost of t therms of gas. x
3 5 7

Domain: 0 … x … 7

x 0 1 2 3 4 5 6 7

f(x) 0 1 2 2 3 4 4 4

FIGURE raph of a case-defined

function.

Now ork Problem 35 G

There is an easy way to tell whether a curve is the graph of a function. In
Figure 2.1 (a), notice that with the given x there are associated t values of yW y1
and y2 . Thus, the curve is n t the graph of a function of x. oo ing at it another way,
we have the following general rule, called the vertical line test. If a ertica line, , can
be drawn that intersects a curve in at least two points, then the curve is n t the graph
of a function of x. When no such vertical line can be drawn, the curve is the graph of a
function of x. Consequently, the curves in Figure 2.1 do not represent functions of x,
but those in Figure 2.1 do.
110 C nct ons an ra s

y y y
L Two outputs
y1 for one input L

x x x x
y2

(a) (b) (c)

FIGURE y is not a function of x.

y y y

x x x

FIGURE Functions of x.

E AM LE AG T N R F

raph x D 2y2 .
S Here it is easier to choose values of y and then find the corresponding values
of x. Figure 2.20 shows the graph. y the vertical-line test, the equation x D 2y2 does
not define a function of x.

y
x = 2y2
3
x 0 2 2 8 8 18 18 1
x
y 0 1 -1 2 -2 3 -3 10 20

FIGURE raph of x D 2y2 .

Now ork Problem 39 G

After we have determined whether a curve is the graph of a function, perhaps using
the vertical-line test, there is an easy way to tell whether the function in question is one-
to-one. In Figure 2.15 we see that f.4/ D 3 and, apparently, also f.!4/ D 3. Since the
distinct input values !4 and 4 produce the same output, the function is not one-to-one.
oo ing at it another way, we have the following general rule, called the hori ontal
line test. If a h riz nta line, , can be drawn that intersects the graph of a function in
at least two points, then the function is n t one-to-one. When no such horizontal line
can be drawn, the function is one-to-one.

R BLEMS
In Pr b ems and cate and abe each f the p ints and gi e Figure 2.21(a) shows the graph of y D f .x/.
the quadrant if p ssib e in hich each p int ies a Estimate f .0/, f(2), f(4), and f .!2/.
$ %
2 b What is the domain of f
.!1; !3/, .4; !2/, ! ; 4 , .6; 0/
5 c What is the range of f
.!4; 5/; .3; 0/; .1; 1/; .0; !6/ d What is an x-intercept of f
 Section 2.5 ra s n ectan ar Coor nates 111

Figure 2.21(b) shows the graph of y D f .x/. p 1

s D f .t/ D t2 ! F.r/ D !
a Estimate f .0/ and f(2). r
b What is the domain of f f .x/ D j7x ! 2j D .u/ D ju ! 3j
c What is the range of f 16 2
F.t/ D y D f .x/ D
d What is an x-intercept of f t2 x ! 4
In Pr b ems graph each case de ned functi n and gi e the
y y d main and range
!
p C 1 if 0 " p < 7
c D g.p/ D
3 5 if p $ 7
2 y = f(x)
2 y = f (x) !
1 x if 0 " x < 1
g.x/ D 2
x x x ! 2x C 2 if x $ 1
2 4 2 !
-2
x C 6 if x $ 3
g.x/ D
x2 if x < 3
8
< x C 1 if 0 < x " 3
(a) (b)
f .x/ D 4 if 3 < x " 5
: x ! 1 if x > 5
FIGURE iagram for Problems 3 and 4.
Which of the graphs in Figure 2.23 represent functions of x
Figure 2.22(a) shows the graph of y D f .x/.
a Estimate f .0/, f .1/, and f .!1/.
b What is the domain of f y y
c What is the range of f
d What is an x-intercept of f
Figure 2.22(b) shows the graph of y D f .x/.
x x
a Estimate f .0/, f .2/, f .3/, and f .4/.
b What is the domain of f (a) (b)
c What is the range of f
d What is an x-intercept f
y y
y y
y = f(x)
3
y = f(x) x x
2
-1 1
x 1
(c) (d)
x
1 2 3 4
FIGURE iagram for Problem 3 .

(a) (b) Which of the graphs in Figure 2.24 represent one-to-one

functions of x
FIGURE iagram for Problems 5 and 6.
In Pr b ems determine the intercepts f the graph f each
equati n and sketch the graph Based n y ur graph is y a
y y
functi n f x and if s is it ne t ne and hat are the d main
and range
y D 2x yDxC1
y D 3x ! 5 y D 3 ! 2x x x
2
y D x5 C x yD 2
x (a) (b)
xD0 y D 4x2 ! 16
y D x3 x D 17
x D !jyj x2 D y2 y y
2x C y ! 2 D 0 xCyD1
In Pr b ems graph each functi n and gi e the d main and
range A s determine the intercepts x x
u D f. / D 3 ! 1 f .x/ D 5 ! 2x2
y D h.x/ D 3 g.s/ D !17 (c) (d)
y D h.x/ D x2 ! 4x C 1 y D f .x/ D !2x2 ! 5x C 12
f .t/ D !t3 p D h.q/ D 1 C 2q C q2 FIGURE iagram for Problem 40.
112 C nct ons an ra s

Debt Payments eatrix has charged 700 on her credit Distance Running A alhousie University student training
cards. She plans to pay them off at the rate of 300, plus all for distance running finds that, after running for x hours, her
interest charges, per month. Write an equation to represent the distance traveled, in ilometers, is given by
amount she owes, after she has made n payments, and identify the 8
intercepts, explaining their financial significance. < 10x if 0 " x " 3
Pricing To encourage an even ow of customers, a y D f .x/ D 5x C 15 if 3 < x " 4
: 35 if 4 < x " 5
restaurant varies the price of an item throughout the day. From
6:00 . . to :00 . ., customers pay full price. At lunch,
from 10:30 . . until 2:30 . ., customers pay half price. From Plot this function and determine if she is ready to attempt the
2:30 . . until 4:30 . ., customers get a dollar off the lunch price. luenose arathon. A marathon is 42.2 ilometers.
From 4:30 . . until 6:00 . ., customers get 5.00 off the dinner
price. From :00 . . until closing time at 10:00 . ., customers In Pr b ems use a graphing ca cu at r t nd a rea
get 5.00 off the dinner price. raph the case-defined function that r ts if any f the gi en equati n R und ans ers t t decima
represents the cost of an item throughout the day for a dinner price p aces
of 1 .
5x3 C 7x D 3
Supply Schedule iven the following supply schedule x2 .x ! 3/ D 2x4 ! 1
(see Example 6 of Section 2.1), plot each quantity price pair by
choosing the horizontal axis for the possible quantities. . x C 3:1/2 D 7:4 ! 4x2
Approximate the points in between the data by connecting the data .x ! 2/3 D x2 ! 3
points with a smooth curve. The result is a supp y cur e. From the
graph, determine the relationship between price and supply. (That
In Pr b ems use a graphing ca cu at r t nd a
is, as price increases, what happens to the quantity supplied )
x intercepts f the graph f the gi en functi n R und ans ers
Is price per unit a function of quantity supplied
t t decima p aces
f .x/ D x3 C 3x C 57
uantity Supplied per Wee , q Price per Unit, p
f .x/ D 2x4 ! 1:5x3 C 2
30 10 g.x/ D x4 ! 1:7x2 C 2x
100 20
p
g.x/ D 3x5 ! 4x2 C 1
150 30
1 0 40 In Pr b ems use a graphing ca cu at r t nd (a) the
210 50 maximum a ue f f .x/ and ( ) the minimum a ue f f .x/ f r
the indicated a ues f x R und ans ers t t decima p aces
f .x/ D x4 ! 4:1x3 C x2 C 10 1"x"4
Demand Schedule The following table is called a demand 2 2 3
schedu e. It indicates the quantities of brand that consumers will f .x/ D x.2x C 2/ ! x C 1 !1"x"1
demand (that is, purchase) each wee at certain prices per unit x2 ! 4
(in dollars). Plot each quantity price pair by choosing the vertical f .x/ D 3"x"5
2x ! 5
axis for the possible prices. Connect the points with a smooth p
curve. In this way, we approximate points in between the given From the graph of f .x/ D 2x3 C 1:1x2 C 4, find a the
data. The result is called a demand cur e. From the graph, range and b the intercepts. Round values to two decimal places.
determine the relationship between the price of brand and the
From the graph of f .x/ D 1 ! 4x3 ! x4 , find a the maximum
amount that will be demanded. (That is, as price decreases, what
value of f .x/, b the range of f, and c the (real) zeros of f. Round
happens to the quantity demanded ) Is price per unit a function of
values to two decimal places.
quantity demanded
x3 C 1:1
From the graph of f .x/ D , find a the range of f
uantity emanded, q Price per Unit, p 3: C x2=3
and b the intercepts. c oes f have any real zeros Round
5 20 values to two decimal places.
10 10 x3 C 64
raph f .x/ D 2 for 3 " x " 5. etermine a the
20 5 x !5
25 4 maximum value of f .x/, b the minimum value of f .x/, c the
range of f, and d all intercepts. Round values to two decimal
places.
Inventory S etch the graph of
8
< !100x C 1000 if 0 " x < 7
y D f .x/ D !100x C 1700 if 7 " x < 14
: !100x C 2400 if 14 " x < 21
A function such as this might describe the inventory y of a
company at time x.
 Section 2.6 etr 113

Objective S
o st s etr a o t t e x a s Examining the graphical behavior of equations is a basic part of mathematics. In this
t e y a s an t e or n an to a
s etr to c r e s etc n section, we examine equations to determine whether their graphs have symmetry.
Consider the graph of y D x2 in Figure 2.25. The portion to the left of the y-axis
is the re ection (or mirror image) through the y-axis of that portion to the right of the
y y-axis, and vice versa. ore precisely, if .a; b/ is any point on this graph, then the point
.!a; b/ must also lie on the graph. We say that this graph is symmetric ab ut the y axis.

y = x2

y0 A graph is symmetric about the y axis if and only if .!a; b/ lies on the graph when
(-x0, y0) (x0, y0) .a; b/ does.
x
-x0 x0

FIGURE Symmetry
about the y-axis.
E AM LE y A S

Use the preceding definition to show that the graph of y D x2 is symmetric about the
y-axis.
S Suppose .a; b/ is any point on the graph of y D x2 . Then

b D a2

We must show that the coordinates of .!a; b/ satisfy y D x2 . ut

y
.!a/2 D a2 D b

x = y2 shows this to be true. Thus, we have pr ed with simple algebra what the picture of the
(x, y) graph led us to believe: The graph of y D x2 is symmetric about the y-axis.

x
Now ork Problem 7 G
When one is testing for symmetry in Example 1, .a; b/ can be any point on the
(x, -y)
graph. In the future, for convenience, we write .x; y/ for a typical point on the graph.
This means that a graph is symmetric about the y-axis if replacing x by !x in its equation
results in an equivalent equation.
FIGURE Symmetry Another type of symmetry is shown by the graph of x D y2 in Figure 2.26. Here
about the x-axis. the portion below the x-axis is the re ection through the x-axis of that portion above
the x-axis, and vice versa. If the point .x; y/ lies on the graph, then .x; !y/ also lies on
it. This graph is said to be symmetric ab ut the x axis.

A graph is symmetric about the x axis if and only if .x; !y/ lies on the graph when
(x, y) .x; y/ does.

x
Thus, the graph of an equation in x and y has x-axis symmetry if replacing y by !y
y = x3 results in an equivalent equation. For example, applying this test to the graph of x D y2 ,
(-x, -y) we see that .!y/2 D x if and only if y2 D x, simply because .!y/2 D y2 . Hence, the
graph of x D y2 is symmetric about the x-axis.
A third type of symmetry, symmetry about the origin, is illustrated by the graph
FIGURE Symmetry of y D x3 (Figure 2.27). Whenever the point .x; y/ lies on the graph, .!x; !y/ also lies
about the origin. on it.
114 C nct ons an ra s

A graph is symmetri a out t e ori in if and only if .!x; !y/ lies on the graph when
.x; y/ does.

Thus, the graph of an equation in x and y has symmetry about the origin if simulta-
neously replacing x by !x and y by !y results in an equivalent equation. For example,
applying this test to the graph of y D x3 shown in Figure 2.27 gives
!y D .!x/3
!y D !x3
y D x3
where all three equations are equivalent, in particular the first and last. Accordingly,
the graph is symmetric about the origin.
Table 2.1 summarizes the tests for symmetry. When we now that a graph has
symmetry, we can s etch it by plotting fewer points than would otherwise be needed.

Table .1 T S
Symmetry about x-axis Replace y by !y in given equation. Symmetric if equivalent
equation is obtained.
Symmetry about y-axis Replace x by !x in given equation. Symmetric if equivalent
equation is obtained.
Symmetry about origin Replace x by !x and y by !y in given equation. Symmetric if
equivalent equation is obtained.

E AM LE G I S
1
Test y D for symmetry about the x-axis, the y-axis, and the origin. Then find the
x
intercepts and s etch the graph.
S
ymmetry x axis Replacing y by !y in y D 1=x gives
1 1
!y D equivalently y D !
x x
y
which is not equivalent to the given equation. Thus, the graph is n t symmetric about
the x-axis.
y axis Replacing x by !x in y D 1=x gives
1 1
Symmetric y= 1
x yD equivalently y D !
about origin !x x
1 which is not equivalent to the given equation. Hence, the graph is n t symmetric about
x the y-axis.
1
rigin Replacing x by !x and y by !y in y D 1=x gives
No intercepts 1 1
!y D equivalently y D
!x x
which is the given equation. Consequently, the graph is symmetric about the origin.
nter epts Since x cannot be 0, the graph has no y-intercept. If y is 0, then 0 D 1=x,
and this equation has no solution. Thus, no x-intercept exists.
1 1
x 1 2 4
4 2
is ussion ecause no intercepts exist, the graph cannot intersect either axis. If
y 4 2 1
1
2
1
4
x > 0, we obtain points only in quadrant I. Figure 2.2 shows the portion of the graph
in quadrant I. y symmetry, we re ect that portion through the origin to obtain the
1 entire graph.
FIGURE raph of y D .
x Now ork Problem 9 G
 Section 2.6 etr 115

E AM LE G I S

Test y D f.x/ D 1 ! x4 for symmetry about the x-axis, the y-axis, and the origin. Then
find the intercepts and s etch the graph.
S
ymmetry x axis Replacing y by !y in y D 1 ! x4 gives
!y D 1 ! x4 equivalently y D !1 C x4
which is not equivalent to the given equation. Thus, the graph is n t symmetric about
the x-axis.
y axis Replacing x by !x in y D 1 ! x4 gives
y D 1 ! .!x/4 equivalently y D 1 ! x4
which is the given equation. Hence, the graph is symmetric about the y-axis.
rigin Replacing x by !x and y by !y in y D 1 ! x4 gives
!y D 1 ! .!x/4 equivalently ! y D 1 ! x4 equivalently y D !1 C x4
which is not equivalent to the given equation. Thus, the graph is n t symmetric about
the origin.
nter epts Testing for x-intercepts, we set y D 0 in y D 1 ! x4 . Then,
1 ! x4 D 0
.1 ! x2 /.1 C x2 / D 0
.1 ! x/.1 C x/.1 C x2 / D 0
xD1 or x D !1

The x-intercepts are therefore .1; 0/ and .!1; 0/. Testing for y-intercepts, we set x D 0.
Then y D 1, so .0; 1/ is the only y-intercept.
is ussion If the intercepts and some points .x; y/ to the right of the y-axis are plot-
ted, we can s etch the entire graph by using symmetry about the y-axis (Figure 2.2 ).

y
x y
y-intercept
0 1 y = f(x) = 1 - x 4

1 15
1
2 16

3 175 x-intercept
4 256
x
-1 1
1 0 x-intercept

3
2 - 65
16
y-axis symmetry

FIGURE raph of y D 1 ! x4 .

Now ork Problem 19 G

The constant function f.x/ D 0, for all x, is easily seen to be symmetric about
the x-axis. In Example 3, we showed that the graph of y D f.x/ D 1 ! x4 does not
have x-axis symmetry. For any functi n f, suppose that the graph of y D f.x/ has x-axis
The only functi n whose graph is symmetry. According to the definition, this means that we also have !y D f.x/. This
symmetric about the x-axis is the tells us that for an arbitrary x in the domain of f we have f.x/ D y and f.x/ D !y. Since
function constantly 0. for a function each x-value determines a unique y-value, we must have y D !y, and
116 C nct ons an ra s

this implies y D 0. Since x was arbitrary, it follows that if the graph of a functi n is
symmetric about the x-axis, then the function must be the constant 0.

E AM LE G I S

Test the graph of 4x2 C y2 D 36 for intercepts and symmetry. S etch the graph.
S
nter epts If y D 0, then 4x2 D 36, so x D ˙3. Thus, the x-intercepts are .3; 0/ and
.!3; 0/. If x D 0, then y2 D 36, so y D ˙2. Hence, the y-intercepts are (0, 2) and
.0; !2/.
ymmetry x axis Replacing y by !y in 4x2 C y2 D 36 gives
4x2 C .!y/2 D 36 equivalently 4x2 C y2 D 36
which is the original equation, so there is symmetry about the x-axis.
y axis Replacing x by !x in 4x2 C y2 D 36 gives
4.!x/2 C y2 D 36 equivalently 4x2 C y2 D 36
which is the original equation, so there is also symmetry about the y-axis.
rigin Replacing x by !x and y by !y in 4x2 C y2 D 36 gives
4.!x/2 C .!y/2 D 36 equivalently 4x2 C y2 D 36
which is the original equation, so the graph is also symmetric about the origin.
is ussion In Figure 2.30, the intercepts and some points in the first quadrant are
plotted. The points in that quadrant are then connected by a smooth curve. y symmetry
about the x-axis, the points in the fourth quadrant are obtained. Then, by symmetry
about the y-axis, the complete graph is found. There are other ways of graphing the
equation by using symmetry. For example, after plotting the intercepts and some points
in the first quadrant, we can obtain the points in the third quadrant by symmetry about
the origin. y symmetry about the x-axis (or y-axis), we can then obtain the entire
graph.
y
x y

;3 0 2 4x2 + 9y2 = 36

0 ;2

1 4 2 x
3 -3 3
2 5
2 3 Symmetry about x-axis,
-2 y-axis, and origin.
5 11
2 3

FIGURE raph of 4x2 C y2 D 36.

Now ork Problem 23 G

In Example 4, the graph is symmetric about the x-axis, the y-axis, and the origin. It
This fact can be a time-saving device in
chec ing for symmetry. can be shown that for any graph if any two of the three types of symmetry discussed
so far exist then the remaining type must also exist

E AM LE S L yDx

A graph is symmetric about the line y D x if and only if .b; a/ lies on the graph
when .a; b/ does.
 Section 2.6 etr 117

Another way of stating the definition is to say that interchanging the roles of x and
y in the given equation results in an equivalent equation.
Use the preceding definition to show that x2 C y2 D 1 is symmetric about the line
y D x.
S Interchanging the roles of x and y produces y2 Cx2 D 1, which is equivalent
to x C y D 1. Thus, x2 C y2 D 1 is symmetric about y D x.
2 2

G
The point with coordinates .b; a/ is the mirror image in the line y D x of the point
.a; b/. If f is a one-to-one function, b D f.a/ if and only if a D f !1 .b/. Thus, the graph
of f !1 is the mirror image in the line y D x of the graph of f. It is interesting to note
that for any function f we can form the mirror image of the graph of f. However, the
resulting graph need not be the graph of a function. For this mirror image to be itself
the graph of a function, it must pass the vertical-line test. However, vertical lines and
horizontal lines are mirror images in the line y D x, and we see that for the mirror
image of the graph of f to pass the vertical-line test is for the graph of f to pass the
horizontal-line test. This last happens precisely if f is one-to-one, which is the case if
and only if f has an inverse.

E AM LE S I F

S etch the graph of g.x/ D 2x C 1 and its inverse in the same plane.
S As we shall study in greater detail in Chapter 3, the graph of g is the straight
line with slope 2 and y-intercept 1. This line, the line y D x, and the re ection of
y D 2x C 1 in y D x are shown in Figure 2.31.

y = g(x) = 2x + 1
5 y=x

4
3
2 y = g-1(x ) = 12 x - 12

1
x
-5 -4 -3 -2 -1 1 2 3 4 5
-1
-2
-3
-4
-5

FIGURE raph of y D g.x/ and y D g!1 .x/.

Now ork Problem 27 G

R BLEMS
In Pr b ems nd the x and y intercepts f the graph f the
equati n A s test f r symmetry ab ut the x axis the y axis the 25x2 C 144y2 D 16 y D 57
rigin and the ine y D x n t sketch the graph
y D 5x x D !7 y D j2xj ! 2
y D f .x/ D x2 ! p
!4
x D !y y D x2 ! 36
2x2 C y2 x4 D !y x D y3
118 C nct ons an ra s

x ! 4y ! y2 C 21 D 0 x3 C xy C y3 D 0 jxj ! jyj D 0 x2 C y2 D 25
2 2
x3 ! 2x2 C x x C 4y D 25 x2 ! y2 D 4
y D f .x/ D x2 C xy C y2 D 0
x2 C 1 Prove that the graph of y D f .x/ D 5 ! 1: 6x2 ! "x4 is
2 4
x symmetric about the y-axis, and then graph the function. (a) a e
yD 3 yD
x C 27 xCy use of symmetry, where possible, to find all intercepts. etermine
(b) the maximum value of f .x/ and (c) the range of f. Round all
In Pr b ems nd the x and y intercepts f the graph f the values to two decimal places.
equati n A s test f r symmetry ab ut the x axis the y axis the Prove that the graph of y D f .x/ D 2x4 ! 7x2 C 5 is
rigin and the ine y D x hen sketch the graph symmetric about the y-axis, and then graph the function. Find all
5x C 2y2 D x ! 1 D y4 C y2 real zeros of f. Round your answers to two decimal places.
S etch the graph of f .x/ D 2x C 3 and its inverse in the same
y D f .x/ D x3 ! 4x 2y D 5 ! x2
plane.

Objective T R
o eco e fa ar t t e s a es of Up to now, our approach to graphing has been based on plotting points and ma ing use
t e ra s of s as c f nct ons an
to cons er trans at on re ect on an of any symmetry that exists. ut this technique is usually not the preferred one. ater in
ert ca stretc n or s r n n of t e this text, we will analyze graphs by using other techniques. However, some functions
ra of a f nct on and their associated graphs occur so frequently that we find it worthwhile to memorize
them. Figure 2.32 shows six such basic functions.

y y y

f(x) = x2
2 f(x) = x3
f(x) = x 2 2

x x x
2 2 2

(a) (b) (c)

y y y

f(x) = x
f(x ) = 1x
2 2 f(x) = x 2

x x x
2 2 2

y
(d) (e) (f)

y = x2 + 2 FIGURE Six basic functions.

f(x) = x2 At times, by altering a function through an a gebraic manipulation, the graph of

2 the new function can be obtained from the graph of the original function by performing
a ge metric manipulation. For example, we can use the graph of f.x/ D x2 to graph
x y D x2 C 2. Note that y D f.x/ C 2. Thus, for each x, the corresponding ordinate for the
graph of y D x2 C 2 is 2 more than the ordinate for the graph of f.x/ D x2 . This means
that the graph of y D x2 C 2 is simply the graph of f.x/ D x2 shifted, or trans ated, 2
FIGURE raph of units upward. (See Figure 2.33.) We say that the graph of y D x2 C2 is a transf rmati n
y D x2 C 2. of the graph of f.x/ D x2 . Table 2.2 gives a list of basic types of transformations.
 Section 2.7 rans at ons an e ect ons 119

Table . T
How to Transform raph of y D f.x/
Equation to btain raph of Equation

y D f .x/ C c shift c units upward

y D f .x/ ! c shift c units downward
y D f .x ! c/ shift c units to right
y D f .x C c/ shift c units to left
y D !f .x/ re ect about x-axis
y D f .!x/ re ect about y-axis
y D cf .x/ c > 1 vertically stretch away from x-axis by a
factor of c
y D cf .x/ c < 1 vertically shrin toward x-axis by a
factor of c

E AM LE H T

S etch the graph of y D .x ! 1/3 .

S We observe that .x ! 1/3 is x3 with x replaced by x ! 1. Thus, if f.x/ D x3 ,
then y D .x ! 1/3 D f.x ! 1/, which has the form f.x ! c/, where c D 1. From
Table 2.2, the graph of y D .x ! 1/3 is the graph of f.x/ D x3 shifted 1 unit to the right.
(See Figure 2.34.)

f (x) = x3
y = (x - 1)3
1
x
-1 1
-1

FIGURE raph of y D .x ! 1/3 .

Now ork Problem 3 G

E AM LE S R
p
S etch the graph of y D ! 12 x.
p p
S We can do this problem in two steps. First, observe that 12 x x is x
p p
multiplied by 12 . Thus, if f.x/ D x, then 12 x D 12 f.x/, which has the form cf.x/,
p
where c D 12 . So the graph of y D 12 x is the graph of f shrun vertically toward
the x-axis by a factor of 12 (Transformation , Table 2.2 see Figure 2.35). Second, the
p p
minus sign in y D ! 12 x causes a re ection in the graph of y D 12 x about the x-axis
(Transformation 5, Table 2.2 see Figure 2.35).
120 C nct ons an ra s

f(x) = x
2
1
y= x
1 2

x
1 4
1
y=-2 x

p p
FIGURE To graph y D ! 12 x, shrin y D x and re ect
result about x-axis.

Now ork Problem 5 G

R BLEMS
p
In Pr b ems use the graphs f the functi ns in Figure raph the function y D 3 x C k for k D 0; 1; 2; 3; !1; !2,
and transf rmati n techniques t p t the gi en functi ns and !3. bserve the vertical translations compared to the first
3 graph.
y D x3 ! 1 y D !x2 yD
xC2 1
raph the function y D for k D 0; 1; 2; 3; !1; !2, and
p 2 xCk
yD! x!2 yD y D jxj ! 2 !3. bserve the horizontal translations compared to the first
3x
graph.
1p
y D jx C 1j ! 2 yD! x!2 y D 2 C .x C 3/3 raph the function y D kx3 for k D 1; 2; 12 , and 3. bserve the
3
vertical stretching and shrin ing compared to the first graph.
p 5 raph the function for k D !2. bserve that the graph is the same
y D .x ! 1/2 C 1 y D !x yD
2!x as that obtained by stretching the re ection of y D x3 about the
x-axis by a factor of 2.
In Pr b ems describe hat must be d ne t the graph f
y D f .x/ t btain the graph f the gi en equati n
y D !1=2. f .x ! 5/ C 1/ y D 2. f .x ! 1/ ! 4/
y D f .!x/ ! 5 y D f .3x/

Objective F S
o sc ss f nct ons of se era When we defined a functi n f W X % from X to in Section 2.1, we did so for sets
ar a es an to co te f nct on
a es o sc ss t ree ens ona X and without requiring that they be sets of numbers. We have not often used that
coor nates an s etc s e generality yet. ost of our examples have been functions from .!1; 1/ to .!1; 1/.
s rfaces We also saw in Section 2.1 that for sets X and we can construct the new set X #
whose elements are ordered pairs .x; y/ with x in X and y in . It follows that for any
three sets X, , and the notion of a function f W X # % is already covered by the
basic definition. Such an f is simply a rule that assigns to each element .x; y/ in X #
at most one element of , denoted by f..x; y//. There is general agreement that in this
situation one should drop a layer of parentheses and write simply f.x; y/ for f..x; y//.
o note here that even if each of X and are sets of numbers, say X D .!1; 1/ D ,
then X# is definitely n t a set of numbers. In other words, an rdered pair f numbers
is n t a number.
The graph of a function f W X % is the subset of X # consisting of all ordered
pairs of the form .x; f.x//, where x is in the domain of f. It follows that the graph of
a function f W X # % is the subset of .X # / # consisting of all ordered
pairs of the form ..x; y/; f.x; y//, where .x; y/ is in the domain of f. The ordered pair
..x; y/; f.x; y// has its first coordinate given by .x; y/, itself an ordered pair, while its
second coordinate is the element f.x; y/ in . ost people prefer to replace .X # / #
with X # # , an element of which is an ordered triple .x; y; z/, with x in X, y in ,
and z in . These elements are easier to read than the ..x; y/; z/, which are the o cial
 Section 2.8 nct ons of e era ar a es 121

elements of .X# /# . In fact, we can de ne an ordered triple .x; y; z/ to be a shorthand

for ..x; y/; z/ if we wish.
efore going further, it is important to point out that these very general consider-
ations have been motivated by a desire to ma e mathematics applicable. any people
when confronted with mathematical models built around functions, and equations relat-
ing them, express both an appreciation for the elegance of the ideas and a s epticism
about their practical value. A common complaint is that in practice there are factors
unaccounted for in a particular mathematical model. Translated into the context we are
developing, this complaint frequently means that the functions in a mathematical model
should involve more variables than the modeler originally contemplated. eing able to
add new variables, to account for phenomena that were earlier thought to be insignifi-
cant, is an important aspect of robustness that a mathematical model should possess. If
we now how to go from one variable to two variables, where the two variables can
be construed as an ordered pair and hence a single variable of a new ind, then we can
iterate the procedure and deal with functions of as many variables as we li e.
For sets X1 , X2 , : : : , Xn and , a function f W X1 # X2 # & & & # Xn % in our
general sense provides the notion of a -valued function of n-variables. In this case,
an element of the domain of f is an ordered n tuple .x1 ; x2 ; & & & ; xn /, with xi in Xi for
i D 1; 2; & & & ; n, for which f.x1 ; x2 ; & & & ; xn / is defined. The graph of f is the set of all
ordered n C 1-tuples of the form .x1 ; x2 ; & & & ; xn ; f.x1 ; x2 ; & & & ; xn //, where .x1 ; x2 ; & & & ; xn /
is in the domain of f.
Suppose a manufacturer produces two products, and . Then the total cost
depends on the levels of production of b th and . Table 2.3 is a schedule that
indicates total cost at various levels. For example, when 5 units of and 6 units of
are produced, the total cost c is 17. In this situation, it seems natural to associate the
number 17 with the rdered pair .5; 6/:
.5; 6/ ‘ 17
The first element of the ordered pair, 5, represents the number of units of produced,
while the second element, 6, represents the number of units of produced. Corre-
sponding to the other production situations shown, we have
.5; 7/ ‘ 1
.6; 6/ ‘ 1
and
.6; 7/ ‘ 20

Table . C F T
No. of Units No. of Units Total Cost
of Produced, x of Produced, y of Production, c

5 6 17
5 7 1
6 6 1
6 7 20

This listing can be considered to be the definition of a function

cWX# % .!1; 1/
where X D f5; 6g and D f6; 7g with
c.5; 7/ D 1 c.6; 7/ D 20
c.5; 6/ D 17 c.6; 6/ D 1
We say that the total-cost schedule can be described by c D c.x; y/, a function of the
two independent variables x and y. The letter c is used here for both the dependent
122 C nct ons an ra s

variable and the name of the rule that defines the function. f course, the range of c is
the subset f17; 1 ; 1 ; 20g of .!1; 1/. ecause negative costs are unli ely to ma e
sense, we might want to refine c and construe it to be a function c W X # % Œ0; 1/.
Realize that a manufacturer might very well produce 17, say, different products, in
which case cost would be a function
c W X1 # X2 # : : : # X17 % Œ0; 1/
In this situation, c.x1 ; x2 ; : : : x17 / would be the cost of producing x1 units of Product 1,
x2 units of Product 2, x3 units of Product 3, , and x17 units of Product 17. The study
of functions of two variables allows us to see general patterns li e this.
ost people were acquainted with certain functions of two variables long before
they ever heard of functions, as the following example illustrates.

E AM LE F T

a a.x; y/ D x C y is a function of two variables. Some function values are

a.1; 1/ D 1 C 1 D 2
a.2; 3/ D 2 C 3 D 5
We have a W .!1; 1/ # .!1; 1/ % .!1; 1/.
b m.x; y/ D xy is a function of two variables. Some function values are
m.2; 2/ D 2 & 2 D 4
m.3; 2/ D 3 & 2 D 6
The domain of both a and m is all of .!1; 1/ # .!1; 1/. bserve that if we were
to define division as a function d W .!1; 1/ # .!1; 1/ % .!1; 1/
with d.x; y/ D x ' y then the domain of d is .!1; 1/ # ..!1; 1/ ! f0g/, where
.!1; 1/ ! f0g is the set all real numbers except 0.
G
Turning to another function of two variables, we see that the equation
2
zD
x2 C y2
defines z as a function of x and y:
2
z D f.x; y/ D
x2 C y2
The domain of f is all ordered pairs of real numbers .x; y/ for which the equation
has meaning when the first and second elements of .x; y/ are substituted for x and
y, respectively, in the equation. This requires that x2 C y2 ¤ 0. However, the only
pair .x; y/ of real numbers for which x2 C y2 D 0 is .0:0/. Thus, the domain of f is
.!1; 1/ # .!1; 1/ ! f.0; 0/g. To find f.2; 3/, for example, we substitute x D 2 and
A L IT I y D 3 into the expression 2=.x2 C y2 / and obtain z D f.2; 3/ D 2=.22 C 32 / D 2=13.
The cost per day for manufacturing
both 12-ounce and 20-ounce coffee E AM LE F T
mugs is given by c D 160 C 2x C 3y,
where x is the number of 12-ounce xC3
a f.x; y/ D is a function of two variables. ecause the denominator is zero
mugs and y is the number of y!2
20-ounce mugs. What is the cost when y D 2, the domain of f is all .x; y/ such that y ¤ 2. Some function values are
per day of manufacturing
a 500 12-ounce and 700 20-ounce
0C3
f.0; 3/ D D3
mugs 3!2
b 1000 12-ounce and 750 20-ounce 3C3
f.3; 0/ D D !3
mugs 0!2
Note that f.0; 3/ ¤ f.3; 0/.
 Section 2.8 nct ons of e era ar a es 123

b h.x; y/ D 4x defines h as a function of x and y. The domain is all ordered pairs of

real numbers. Some function values are
h.2; 5/ D 4.2/ D
h.2; 6/ D 4.2/ D
Note that the function values are independent of the choice of y.
c If z2 D x2 C y2 and x D 3 and y D 4, then z2 D 32 C 42 D 25. Consequently,
z D ˙5. Thus, with the ordered pair .3; 4/, we cann t associate exactly one output
number. Hence z2 D x2 C y2 does not define z as a function of x and y.
Now ork Problem 1 G
E AM LE T H I

n hot and humid days, many people tend to feel uncomfortable. In the United States,
the degree of discomfort is numerically given by the temperature humidity index, THI,
which is a function of two variables, td and t :
THI D f.td ; t / D 15 C 0:4.td C t /
where td is the dry-bulb temperature (in degrees Fahrenheit) and t is the wet-bulb
temperature (in degrees Fahrenheit) of the air. Evaluate the THI when td D 0 and
t D 0.
S We want to find f. 0; 0/:
f. 0; 0/ D 15 C 0:4. 0 C 0/ D 15 C 6 D 3
When the THI is greater than 75, most people are uncomfortable. In fact, the THI was
once called the discomfort index. any electric utilities closely follow this index
so that they can anticipate the demand that air conditioning places on their systems.
z After our study of Chapter 4, we will be able to describe the similar Humidex, used in
Canada.
Now ork Problem 3 G
1 From the second paragraph in this section it follows that a function f W .!1; 1/ #
-1
.!1; 1/ % .!1; 1/, where we write z D f.x; y/, will have a graph consisting of
-1
1
ordered triples of real numbers. The set of all ordered triples of real numbers can be
1
y pictured as providing a dimensional rectangular coordinate system. Such a system
x
-1 is formed when three mutually perpendicular real-number lines in space intersect at the
origin of each line, as in Figure 2.36. The three number lines are called the x-, y-, and
FIGURE 3-dimensional z-axes, and their point of intersection is called the origin of the system. The arrows
rectangular coordinate system.
indicate the positive directions of the axes, and the negative portions of the axes are
shown as dashed lines.
To each point P in space, we can assign a unique ordered triple of numbers, called
the c rdinates of P. To do this see Figure 2.37(a) , from P, we construct a line per-
pendicular to the x, y-plane that is, the plane determined by the x- and y-axes. et
z

z
4

z0 (2, 3, 4)

P (x0, y0, z0) (0, 0, 0)

(2, 0, 0) 3
x0
y0 y y
x x (2, 3, 0)
Q

(a) (b)

FIGURE Points in space.

124 C nct ons an ra s

f(x, y) (x, y, f(x, y))

Graph of z = f(x, y)

y y
x

FIGURE raph of a function of two variables.

be the point where the line intersects this plane. From , we construct perpendiculars
to the x- and y-axes. These lines intersect the x- and y-axes at x0 and y0 , respectively.
From P, a perpendicular to the z-axis is constructed that intersects the axis at z0 . Thus,
we assign to P the ordered triple .x0 ; y0 ; z0 /. It should also be evident that with each
ordered triple of numbers we can assign a unique point in space. ue to this one-to-
one correspondence between points in space and ordered triples, an ordered triple can
be called a point. In Figure 2.37(b), points .2; 0; 0/, .2; 3; 0/, and .2; 3; 4/ are shown.
Note that the origin corresponds to .0; 0; 0/. Typically, the negative portions of the axes
are not shown unless required.
We represent a function of two variables, z D f.x; y/, geometrically as follows:
To each ordered pair .x; y/ in the domain of f, we assign the point .x; y; f.x; y//. The
set of all such points is the graph of f. Such a graph appears in Figure 2.3 . We can
consider z D f.x; y/ as representing a surface in space in the same way that we have
considered y D f.x/ as representing a cur e in the p ane. Not all functions y D f.x/
describe aesthetically pleasing curves in fact most do not and in the same way we
stretch the meaning of the word surface.
We now give a brief discussion of s etching surfaces in space. We begin with planes
that are parallel to a coordinate plane. y a coordinate plane we mean a plane con-
taining two coordinate axes. For example, the plane determined by the x- and y-axes is
the x; y-plane. Similarly, we spea of the x; z-plane and the y; z-plane. The coordinate
planes divide space into eight parts, called ctants. In particular, the part containing all
Names are not usually assigned to the points .x; y; z/ such that x, y, and z are positive is called the first octant.
remaining seven octants. Suppose S is a plane that is parallel to the x; y-plane and passes through the point
.0; 0; 5/. See Figure 2.3 (a). Then the point .x; y; z/ will lie on S if and only if z D 5
that is, x and y can be any real numbers, but z must equal 5. For this reason, we say
that z D 5 is an equation of S. Similarly, an equation of the plane parallel to the
x; z-plane and passing through the point .0; 2; 0/ is y D 2 Figure 2.3 (b) . The equation
x D 3 is an equation of the plane passing through .3; 0; 0/ and parallel to the y; z-plane
Figure 2.3 (c) . Next, we loo at planes in general.

z z z
Plane z = 5
Plane y = 2
(0, 0, 5) Plane x = 3

y y y
x x x (3, 0, 0)
(0, 2, 0)
(a) (b) (c)

FIGURE Planes parallel to coordinate planes.

 Section 2.8 nct ons of e era ar a es 125

In space, the graph of an equation of the form

Ax C By C Cz C D0
where is a constant and A, B, and C are constants that are not all zero, is a plane.
Since three distinct points (not lying on the same line) determine a plane, a convenient
way to s etch a plane is to first determine the points, if any, where the plane intersects
the x-, y-, and z-axes. These points are called intercepts.

E AM LE G

S etch the plane 2x C 3y C z D 6.

S The plane intersects the x-axis when y D 0 and z D 0. Thus, 2x D 6,
which gives x D 3. Similarly, if x D z D 0, then y D 2 if x D y D 0, then z D 6.
Therefore, the intercepts are .3; 0; 0/, .0; 2; 0/, and .0; 0; 6/. After these points are plot-
ted, a plane is passed through them. The portion of the plane in the first octant is shown
in Figure 2.40(a) note, however, that the plane extends indefinitely into space.

z z

6 6

2x + 3y + z = 6
2x + z = 6 3y + z = 6
x, z-trace y, z-trace

2 y 2 y
3 3 2x + 3y = 6
x, y-trace
x x

(a) (b)

FIGURE The plane 2x C 3y C z D 6 and its traces.

Now ork Problem 19 G

A surface can be s etched with the aid of its traces. These are the intersections
of the surface with the coordinate planes. To illustrate, for the plane 2x C 3y C z D 6
in Example 4, the trace in the x y-plane is obtained by setting z D 0. This
gives 2x C 3y D 6, which is an equation of a ine in the x; y-plane. Similarly, set-
z ting x D 0 gives the trace in the y; z-plane: the line 3y C z D 6. The x; z-trace is the line
2x C z D 6. See Figure 2.40(b).
4
E AM LE S S

S etch the surface 2x C z D 4.

S This equation has the form of a plane. The x- and z-intercepts are .2; 0; 0/
2
k y and .0; 0; 4/, and there is no y-intercept, since x and z cannot both be zero. Setting y D 0
x 2x + z = 4 gives the x; z-trace 2x C z D 4, which is a line in the x; z-plane. In fact, the intersection
of the surface with any plane y D k is also 2x C z D 4. Hence, the plane appears as in
FIGURE The plane Figure 2.41.
2x C z D 4. Now ork Problem 21 G
ur final examples deal with surfaces that are not planes but whose graphs can be
easily obtained.
126 C nct ons an ra s

E AM LE S S

Note that this equation places no S etch the surface z D x2 .

restriction on y.
S The x; z-trace is the curve z D x2 , which is a parabola. In fact, for any fixed
value of y, we get z D x2 . Thus, the graph appears as in Figure 2.42.
z
Now ork Problem 25 G
z = x2
E AM LE S S

S etch the surface x2 C y2 C z2 D 25.

S Setting z D 0 gives the x; y-trace x2 C y2 D 25, which is a circle of radius 5.
x y Similarly, the y; z-, x; z-traces are the circles y2 Cz2 D 25 and x2 Cz2 D 25, respectively.
Note also that since x2 C y2 D 25 ! z2 , the intersection of the surface with the plane
FIGURE The surface
z D k, where !5 " k " 5, is a circle. For example, if z D 3, the intersection is the
z D x2 .
circle x2 C y2 D 16. If z D 4, the intersection is x2 C y2 D . That is, cross sec-
tions of the surface that are parallel to the x y-plane are circles. The surface appears in
z
Figure 2.43 it is a sphere.

5 x2 + y2 + z2 = 25 Now ork Problem 27 G

For a function f W X # % , we have seen that the graph of f, being a subset
of X # # , is three dimensional for numerical examples. Admittedly, constructing
such a graph on paper can be challenging. There is another pictorial presentation of a
5 y function z D f.x; y/ for f W .!1; 1/ # .!1; 1/ % .!1; 1/, which is entirely
5
two dimensional. et be a number in the range of f. The equati n f.x; y/ D has a
x graph in the x; y-plane that, in principle, can be constructed and labeled. If we repeat
this construction in the same plane for several other values, i say, in the range of f
then we have a set of curves, called level curves, which may provide us with a useful
FIGURE The surface visualization of f.
x2 C y2 C z2 D 25.
There are are at least two examples of this technique that are within everyday
experience for many people. For the first, consider a geographic region that is small
enough to be considered planar, and coordinatize it. (A city with a rectangular grid of
numbered avenues and streets can be considered to be coordinatized by these.) At any
given time, temperature in degrees Fahrenheit is a function of place .x; y/. We might
write D .x; y/. n a map of the region we might connect all places that currently
have a temperature of 70ı F with a curve. This is the curve .x; y/ D 70. If we put
several other curves, such as .x; y/ D 6 and .x; y/ D 72, on the same map, then
we have the ind of map that appears on televised weather reports. The curves in this
case are called is therms the prefix is comes from the ree is s meaning equal .
For the next, again referring to geography, observe that each place .x; y/ has a definite
altitude A D A.x; y/. A map of a mountainous region with points of equal altitude con-
nected by what are called c nt ur ines is called a t p graphic map, and the generic
term e e cur es is particularly apt in this case.
In Chapter 7 we will encounter a number of inear functions of several variables.
If we have P D ax C by, expressing profit P as a function of production x of a product
and production y of a product , then the level curves ax C by D are called is pr t
ines.

E AM LE L C

S etch a family of at least four level curves for the function z D x2 C y2 .

S For any pair .x; y/, x2 C y2 $ 0, so the range of z D 2 2
p x 2 C y 2is contained in
Œ0; 1/. n the other hand, for any $ 0 we can write D . / C 0 , which shows
that the range of z D x2 C y2 is all p of Œ0; 1/. For $ 0 we recognize the graph of
x2 C y2 D as a circle of radius centered at the origin .0; 0/. If we ta e to be
4, , 16, and 25, then our level curves are concentric circles of radii 2, 3, 4, and 5,
 Section 2.8 nct ons of e era ar a es 127

y respectively. See Figure 2.44. Note that the level curve x2 C y2 D 0 consists of the
point .0; 0/ and no others.
5 Now ork Problem 29 G
An example of a function of three variables is D .x; y; z/ D xyz. It provides the
volume of a bric with side lengths x, y, and z if x, y, and z are all positive.
-5 1 2 3 4 5
x An e ips id is a surface which in standard position is given by an equation of
x2 y2 z2
the form 2 C 2 C 2 D 1, for a, b, and c positive numbers called the radii. No one of
a b c
the variables is a function of the other two. If two of the numbers a, b, and c are equal
-5
and the third is larger, then the special ind of ellipsoid that results is called a pr ate
spher id, examples of which are provided by both a football and a rugby ball. In any
event, the volume of space enclosed by an ellipsoid with radii a, b, and c is given by
FIGURE evel curves for 4
z D x2 C y2 . D .a; b; c/ D "abc, and this is another example of a function of three (positive)
3
variables.
In the context of functions of several variables, it is also interesting to consider
functions whose a ues are ordered pairs. For any set X, one of the simplest is the diag
na function & W X % X # X given by &.x/ D .x; x/. We mentioned in Example 1(b)
that ordinary multiplication is a function m W .!1; 1/ # .!1; 1/ % .!1; 1/.
If we let & denote the diagonal function for .!1; 1/, then we see that the composite
m ı & of the rather simple-minded functions & and m is the more interesting function
y D x2 .

R BLEMS
In Pr b ems determine the indicated functi n a ues f r the Genetics Under certain conditions, if two brown-eyed
gi en functi ns parents have exactly k children, the probability that there will be
f .x; y/ D 4x ! y2 C 3I f .1; 2/ exactly r blue-eyed children is given by
$ %r $ %k!r
2
f .x; y/ D 3x y ! 4yI f .2; !1/ kŠ 1 3
P.r; k/ D r D 0; 1; 2; : : : ; k
rŠ.k ! r/Š 4 4
g.x; y; z/ D 2x.3y C z/I g.3; 0; !1/
Find the probability that, out of a total of seven children, exactly
g.x; y/ D x3 C xy C y3 I g.1; b/, where b is the unique two will be blue eyed.
solution of t3 C t C 1 D 0
rs In Pr b ems nd equati ns f the p anes that satisfy the
h.r; s; t; u/ D 2 I h.!3; 3; 5; 4/ gi en c nditi ns
t ! u2
h.r; s; t; u/ D ruI h.1; 5; 3; 1/ Parallel to the x z-plane and also passes through the point
g.pA ; p / D 2pA .p2A ! 5/I g.4; / .0; 2; 0/
p Parallel to the y z-plane and also passes through the point
g.pA ; p / D p2A p C I g.4; /
.!2; 0; 0/
F.x; y; z/ D 17I F.6; 0; !5/
Parallel to the x y-plane and also passes through the point
2x
F.x; y; z/ D I F.1; 0; 3/ .2; 7; 6/
.y C 1/z
Parallel to the y z-plane and also passes through the point
f .x; y/ D .x C y/2 I f .a C h; b/ . 6; !2; 2/
f .x; y/ D x2 y ! 3y3 I f .r C t; r/ In Pr b ems sketch the gi en surfaces
Ecology A method of ecological sampling to determine x C 2y C 3z D 1 2x C y C 2z D 6
animal populations in a given area involves first mar ing all the
animals obtained in a sample of R animals from the area and then 3x C 6y C 2z D 12 2x C 3y C 5z D 1
releasing them so that they can mix with unmar ed animals. At a 3x C y D 6 zD1!y
later date a second sample is ta en of animals, and the number z D 4 ! x2 y D z2
of these, that are mar ed S, is noted. ased on R, , and S, an
x2 C y2 C z2 D x2 C 4y2 D 1
estimate of the total population of animals in the sample area is
given by In Pr b ems sketch at east three e e cur es f r the gi en
functi n
R
D f .R; ; S/ D zDxCy z D x2 ! y2
S
Find f .200; 200; 50/. This method is called the
mark and recapture pr cedure.4 4
E. P. dum, Ec gy (New or : Holt, Rinehart and Winston, 1 66).
128 C nct ons an ra s

Chapter 2 Review
I T S E
S Functions
ordered pair .x; y/ relation function domain range
independent variable Ex. 2, p.
dependent variable function value, f.x/ Ex. 3, p.
f.x C h/ ! f.x/
difference quotient, Ex. 4, p.
h
demand function supply function Ex. 5, Ex. 6, p. , 0
S Special Functions
constant function polynomial function (linear and quadratic) Ex. 1, Ex. 2, p. 1, 2
rational function case-defined function Ex. 3, Ex. 4, p. 2, 2
absolute value, jxj factorial, rŠ Ex. 5, Ex. 6, p. 3
S Combinations of Functions
f C g f ! g fg f=g composite function, f ı g Ex. 1, Ex. 2, p. 7,
S Inverse Functions
inverse function, f !1 one-to-one function Ex. 1, p. 102
S Graphs in Rectangular Coordinates
rectangular coordinate system coordinate axes origin x; y-plane
coordinates of a point quadrant graph of equation
x-intercept y-intercept Ex. 1, p. 106
graph of function function-value axis zeros of function Ex. 4, p. 10
vertical-line test horizontal-line test Ex. , p. 110
S Symmetry
x-axis symmetry y-axis symmetry Ex. 1, p. 113
symmetry about origin symmetry about y D x Ex. 6, p. 117
S ranslations and Re ections
horizontal and vertical translations Ex. 1, p. 11
stretching and re ection Ex. 2, p. 11
S Functions of Several Variables
z D f.x; y/ Ex. 2, p. 122
graph of z D f.x; y/ Ex. 4, p. 125
level curves Ex. , p. 126

S
A function f is a rule that assigns at most one output f.x/ to Two functions f and g can be combined to form a sum,
each possible input x. A function is often specified by a for- difference, product, quotient, or composite as follows:
mula that prescribes what must be done to an input x to obtain
f.x/. To obtain a particular function value f.a/, we replace
each x in the formula by a.
The domain of a function f W X % consists of all . f C g/.x/ D f.x/ C g.x/
inputs x for which the rule defines f.x/ as an element of . f ! g/.x/ D f.x/ ! g.x/
the range consists of all elements of of the form f.x/. . fg/.x/ D f.x/g.x/
Some special types of functions are constant functions, $ %
polynomial functions, and rational functions. A function that f f.x/
.x/ D
is defined by more than one expression depending on the ind g g.x/
of input is called a case-defined function. . f ı g/.x/ D f.g.x//
A function has an inverse if and only if it is one-to-one.
In economics, supply (demand) functions give a corre-
spondence between the price p of a product and the number
of units q of the product that producers (consumers) will sup- A rectangular coordinate system allows us to represent equa-
ply (buy) at that price. tions in two variables (in particular, those arising from
 Chapter 2 e e 129

functions) geometrically. The graph of an equation in x and y Symmetry about Replace y by !y in given equation.
consists of all points .x; y/ that correspond to the solutions of x-axis Symmetric if equivalent equation
the equation. We plot a su cient number of points and con- is obtained.
nect them (where appropriate) so that the basic shape of the Symmetry about Replace x by !x in given equation.
graph is apparent. Points where the graph intersects the x- and y-axis Symmetric if equivalent equation
y-axes are called x-intercepts and y-intercepts, respectively. is obtained.
An x-intercept is found by letting y be 0 and solving for x a
Symmetry about Replace x by !x and y by !y in
y-intercept is found by letting x be 0 and solving for y.
origin given equation.
The graph of a function f is the graph of the equation
Symmetric if equivalent equation
y D f.x/ and consists of all points .x; f.x// such that x is in
is obtained.
the domain of f. From the graph of a function, it is easy to
determine the domain and the range. Symmetry about Interchange x and y in given
The fact that a graph represents a function can be deter- yDx equation.
mined by using the vertical-line test. A vertical line cannot Symmetric if equivalent equation is
cut the graph of a function at more than one point. obtained.
The fact that a function is one-to-one can be determined Sometimes the graph of a function can be obtained from
by using the horizontal-line test on its graph. A horizontal that of a familiar function by means of a vertical shift upward
line cannot cut the graph of a one-to-one function at more or downward, a horizontal shift to the right or left, a re ection
than one point. When a function passes the horizontal-line about the x-axis or y-axis, or a vertical stretching or shrin -
test, the graph of the inverse can be obtained by re ecting ing away from or toward the x-axis. Such transformations are
the original graph in the line y D x. indicated in Table 2.2 in Section 2.7.
When the graph of an equation has symmetry, the mirror- A function of two variables is a function whose domain
image effect allows us to s etch the graph by plotting fewer consists of ordered pairs. A function of n variables is a
points than would otherwise be needed. The tests for sym- function whose domain consists of ordered n-tuples. The
metry are as follows: graph of a real-valued function of two variables requires a
three-dimensional coordinate system. evel curves provide
another technique to visualize functions of two variables.

R
8̂
In Pr b ems gi e the d main f each functi n < !q C 1 if ! 1 " q < 0
x g.x/ D x2 C 3jx C 2j
f .x/ D 2 f .q/ D q2 C 1 if 0 " q < 5
x ! 6x C 5 :̂ 3
q ! if 5 " q " 7
F.t/ D 7t C 4t2 G.x/ D 1 $ % $ %
p p 1 1
x!2 s!5 f ! , f .0/, f , f .5/, f .6/
h.x/ D .s/ D 2 2
x!3 4
In Pr b ems nd the functi n a ues f r the gi en functi n In Pr b ems nd and simp ify a f .x C h/ and b
f .x C h/ ! f .x/
f .x/ D 2x2 ! 3x C 5 f .0/, f .!2/, f .5/, f ."/ h
$ %
1 f .x/ D 1 ! 3x f .x/ D 11x2 C 4
h.x/ D 7I h.4/; h ; h.!156/; h.x C 4/
100 7
p f .x/ D 3x2 C x ! 2 f .x/ D
G.x/ D 4 x ! 3I G.3/; G.1 /; G.t C 1/; G.x3 / xC1
3x C 2 If f .x/ D 3x ! 1 and g.x/ D 2x C 3, find the following.
F.x/ D F.!1/, F.0/, F.4/, F.x C 2/ a . f C g/.x/ b . f C g/.4/ c . f ! g/.x/
x!5
p f
uC4 d . fg/.x/ e . fg/.1/ f .x/
h.u/ D I h.5/; h.!4/; h.x/; h.u ! 4/ g
u
$ % g . f ı g/.x/ h . f ı g/.5/ i .g ı f /.x/
.t ! 2/3 1
.t/ D .!1/, .0/, , .x2 / If f .x/ D x3 and g.x/ D 2x C 1, find the following.
5 3
! a . f C g/.x/ b . f ! g/.x/ c . f ! g/.!6/
!3 if x < 1 f f
f .x/ D d . fg/.x/ e .x/ f .1/
4 C x2 if x > 1 g g
f .4/, f .!2/, f .0/, f .1/ g . f ı g/.x/ h .g ı f /.x/ i .g ı f /.2/
130 C nct ons an ra s

In Pr b ems nd . f ı g/.x/ and .g ı f /.x/ In Figure 2.45, which graphs represent functions of x
1
f .x/ D 2 ; g.x/ D x C 1 y y y
x
x!2 1
f .x/ D , g.x/ D p
3 x
p
f .x/ D x C 2; g.x/ D x3 x x x

f .x/ D 2; g.x/ D 3

(a) (b) (c)

In Pr b ems and nd the intercepts f the graph f each
equati n and test f r symmetry ab ut the x axis the y axis the FIGURE iagram for Problem 37.
rigin and y D x n t sketch the graph 2 3
If f .x/ D .x ! x C 7/ , find a f .2/ and b f .1:1/ rounded
x2 y2 to two decimal places.
y D 2x C x3 D4
x2 C y2 C 1
Find all real roots of the equation
5x3 ! 7x2 D 4x ! 2
In Pr b ems and nd the x and y intercepts f the graphs
rounded to two decimal places.
f the equati ns A s test f r symmetry ab ut the x axis the
y axis and the rigin hen sketch the graphs Find all real roots of the equation
y D 4 C x2 y D 3x ! 7 x3 C x C 1 D 0
rounded to two decimal places.
Find all real zeros of
In Pr b ems graph each functi n and gi e its d main and
range A s determine the intercepts f .x/ D x.2:1x2 ! 3/2 ! x3 C 1
p rounded to two decimal places.
G.u/ D u C 4 f .x/ D j2xj ! 2
2 p etermine the range of
y D g.t/ D D !.u/ D !u !
jt ! 4j !2:5x ! 4 if x < 0
f .x/ D
raph the following case-defined function, and give its 6 C 4:1x ! x2 if x $ 0
domain and range: From the graph of f .x/ D !x3 C 0:04x C 7, find a the range
! and b the intercepts rounded to two decimal places.
2 if x " 0 p
y D f .x/ D
2 ! x if x > 0 From the graph of f .x/ D x C 5.x2 ! 4/, find a the
p minimum value of f .x/ b the range of f, and c all real zeros
Use the graph of f .x/ D x to s etch the graph of of f rounded to two decimal places.
p
y D x ! 2 ! 1. raph y D f .x/ D x4 C xk for k D 0, 1, 2, 3, and 4. For which
Use the graph of f .x/ D x2 to s etch the graph of values of k does the graph have a symmetry about the
1 y-axis b symmetry about the origin
y D ! x2 C 2.
2 S etch the graph of x C 2y C 3z D 6.
rend Equation The pro ected annual sales (in dollars)
of a new product are given by the equation S etch the graph of 3x C y C 5z D 10.
S D 150;000 C 3000t, where t is the time in years from 2005. Construct three level curves for P D 5x C 7y.
Such an equation is called a trend equati n. Find the
pro ected annual sales for 2010. Is S a function of t Construct three level curves for C D 2x C 10y.
 nes Para o as
an ste s

F
or the problem of industrial pollution, some people advocate a mar et-based
3.1 nes
solution: et manufacturers pollute, but ma e them pay for the privilege. The
3.2 cat ons an near more pollution, the greater the fee, or levy. The idea is to give manufacturers
nct ons an incentive not to pollute more than necessary.
oes this approach wor In the figure below, curve 1 represents the cost per ton1
3.3 a rat c nct ons
of cutting pollution. A company polluting indiscriminately can normally do some pol-
3.4 ste s of near lution reduction at a small cost. As the amount of pollution is reduced, however, the cost
at ons per ton of further reduction rises and ultimately increases without bound. This is illus-
trated by curve 1 rising indefinitely as the total tons of pollution produced approaches 0.
3.5 on near ste s
( ou should try to understand why this m de is reasonably accurate.)
3.6 cat ons of ste s of ine 2 is a levy scheme that goes easy on clean-running operations but charges
at ons an increasing per-ton fee as the total pollution amount goes up. ine 3, by contrast, is
a scheme in which low-pollution manufacturers pay a high per-ton levy while gross
C er 3 e e
polluters pay less per ton (but more overall). uestions of fairness aside, how well will
each scheme wor as a pollution control measure
Faced with a pollution levy, a company tends to cut pollution s ng as it sa es
m re in e y c sts than it incurs in reducti n c sts. The reduction efforts continue until
the reduction costs exceed the levy savings.
The latter half of this chapter deals with systems of equations. Here, curve 1 and
line 2 represent one system of equations, and curve 1 and line 3 represent another. nce
you have learned how to solve systems of equations, you can return to this page and
verify that the line 2 scheme leads to a pollution reduction from amount A to amount
B, while the line 3 scheme fails as a pollution control measure, leaving the pollution
level at A.
cleanup or levy1
Cost per ton of

2
B A
Total tons of pollution

1 Technically, this is the margina cost per ton (see Section 11.3).

131
132 C nes Para o as an ste s

Objective L
o e e o t e not on of s o e an
erent for s of e at ons of nes S L
any relationships between quantities can be represented conveniently by straight
lines. ne feature of a straight line is its steepness. For example, in Figure 3.1, line
1 rises faster as it goes from left to right than does line 2 . In this sense, 1 is steeper.
To measure the steepness of a line, we use the notion of s pe. In Figure 3.2, as we
move along line from (1, 3) to (3, 7), the x-coordinate increases from 1 to 3, and the
y-coordinate increases from 3 to 7. The average rate of change of y with respect to x is
the ratio
change in y vertical change 7!3 4
D D D D2
change in x horizontal change 3!1 2
The ratio of 2 means that for each 1-unit increase in x, there is a 2-unit increase in y.
ue to the increase, the line rises from left to right. It can be shown that, regardless of
which two points on are chosen to compute the ratio of the change in y to the change
in x, the result is always 2, which we call the s pe of the line.

et .x1 ; y1 / and .x2 ; y2 / be two different points on a nonvertical line. The slope of
the line is
! "
y2 ! y1 vertical change
mD D
x2 ! x1 horizontal change

Having no slope does not mean having A vertical line does not have a slope, because any two points on it must have
a slope of zero. x1 D x2 see Figure 3.3(a) , which gives a denominator of zero in Equation (1). For
a horizontal line, any two points must have y1 D y2 . See Figure 3.3(b). This gives a
numerator of zero in Equation (1), and hence the slope of the line is zero.
y
L
7 (3, 7)
6
y
L1 5 Vertical change = 4
4
(1, 3)
3 Slope = 4
=2
2

L2 2
Horizontal
1 change = 2
x x
1 2 3

FIGURE ine 1 is steeper than 2. FIGURE Slope of a line.

y y

(x1, y1)
(x2, y2)
y1 = y2 (x2, y2)
(x1, y1)
x x
x1 = x2

(a) No slope (b) Slope of zero

FIGURE Vertical and horizontal lines.

 Section 3. nes 133

p (price)

(2, 4)
Increase of 1 unit
1
Decrease of 2 unit

(8, 1)

q (quantity)

FIGURE Price quantity line.

This example shows how the slope can be E AM LE R

interpreted.
The line in Figure 3.4 shows the relationship between the price p of a widget (in dollars)
A L IT I and the quantity q of widgets (in thousands) that consumers will buy at that price. Find
A doctor purchased a new car in and interpret the slope.
2012 for 62,000. In 2015, he sold it to S In the slope formula (1), we replace the x s by q s and the y s by p s.
a friend for 50,000. raw a line show-
Either point in Figure 3.4 may be chosen as .q1 ; p1 /. etting .2; 4/ D .q1 ; p1 / and
ing the relationship between the selling
price of the car and the year in which it
. ; 1/ D .q2 ; p2 /, we have
was sold. Find and interpret the slope.
p2 ! p1 1!4 !3 1
mD D D D!
q2 ! q1 !2 6 2

The slope is negative, ! 12 . This means that, for each 1-unit increase in quantity (one
thousand widgets), there corresponds a decrease in price of 12 (dollar per widget). ue
to this decrease, the line fa s from left to right.
Now ork Problem 3 G

In summary, we can characterize the orientation of a line by its slope:

m=2

m= 1
2
ero slope: horizontal line
Undefined slope: vertical line
m=0 Positive slope: line rises from left to right
m = - 12 Negative slope: line falls from left to right

m = -2
ines with different slopes are shown in Figure 3.5. Notice that the c ser the s pe
FIGURE Slopes of lines. is t 0, the m re near y h riz nta is the ine he greater the abs ute a ue f the s pe
the m re near y ertica is the ine. We remar that two lines are parallel if and only if
they have the same slope or are both vertical.
y

(x, y) E L
If we now a point on a line and the slope of the line, we can find an equation whose
Slope = m
graph is that line. Suppose that line has slope m and passes through the point .x1 ; y1 /.
(x1, y1) If (x, y) is any other point on (see Figure 3.6), we can find an algebraic relationship
x between x and y. Using the slope formula on the points .x1 ; y1 / and (x, y) gives
y ! y1
FIGURE ine through Dm
x ! x1
.x1 ; y1 / with slope m.
y ! y1 D m.x ! x1 /
134 C nes Para o as an ste s

E ery point on satisfies Equation (2). It is also true that e ery point satisfying Equa-
tion (2) must lie on . Thus, Equation (2) is an equation for and is given a spe-
cial name:

y ! y1 D m.x ! x1 /
is a point slope form f an equati n f the ine thr ugh .x1 ; y1 / ith s pe m.

E AM LE S F
A L IT I
Find an equation of the line that has slope 2 and passes through (1, !3).
A new applied mathematics pro-
gram at a university has grown in enroll- S Using a point-slope form with m D 2 and .x1 ; y1 / D .1; !3/ gives
ment by 14 students per year for the past
y ! y1 D m.x ! x1 /
5 years. If the program enrolled 50 stu-
dents in its third year, what is an equa- y ! .!3/ D 2.x ! 1/
tion for the number of students S in the
program as a function of the number of y C 3 D 2x ! 2
years since its inception
which can be rewritten as
2x ! y ! 5 D 0
Now ork Problem 9 G
An equation of the line passing through two given points can be found easily, as
Example 3 shows.

E AM LE L T
A L IT I
Find an equation of the line passing Find an equation of the line passing through (!3, ) and (4, !2).
through the given points. A temperature
of 41ı F is equivalent to 5ı C, and a tem- S
perature of 77ı F is equivalent to 25ı C.
S First we find the slope of the line from the given points. Then we sub-
stitute the slope and one of the points into a point-slope form.

The line has slope

!2 ! 10
mD D!
4 ! .!3/ 7
Choosing .4; !2/ as .x1 ; y1 / would give
the same result. Using a point-slope form with (!3, ) as .x1 ; y1 / gives
10
y! D! Œx ! .!3/!
7
10
y ! D ! .x C 3/
7
7y ! 56 D !10x ! 30
y
10x C 7y ! 26 D 0
y = mx + b Now ork Problem 13 G
(0, b) y-intercept Recall that a point (0, b) where a graph intersects the y-axis is called a y-intercept
Slope m (Figure 3.7). If the slope m and y-intercept b of a line are nown, an equation for the
x line is by using a point-slope form with .x1 ; y1 / D .0; b/
y ! b D m.x ! 0/
FIGURE ine with slope m and Solving for y gives y D mx C b, called the s pe intercept f rm of an equation of
y-intercept b. the line:
 Section 3. nes 135

y D mx C b
is the slope intercept form f an equati n f the ine ith s pe m and y intercept b.

E AM LE S I F

Find an equation of the line with slope 3 and y-intercept !4.

S Using the slope-intercept form y D mx C b with m D 3 and b D !4 gives
y D 3x C .!4/
y D 3x ! 4

Now ork Problem 17 G

E AM LE F S I L
A L IT I
ne formula for the recommended Find the slope and y-intercept of the line with equation y D 5.3 ! 2x/.
dosage (in milligrams) of medication
S
for a child t years old is
1 S We will rewrite the equation so it has the slope-intercept form
yD .t C 1/a
24 y D mx C b. Then the slope is the coe cient of x and the y-intercept is the constant
where a is the adult dosage. For an over- term.
the-counter pain reliever, a D 1000.
Find the slope and y-intercept of this
equation. We have

y D 5.3 ! 2x/
y D 15 ! 10x
y D !10x C 15
y
Thus, m D !10 and b D 15, so the slope is !10 and the y-intercept is 15.
Now ork Problem 25 G
(a, b)
x=a If a ertica line passes through .a; b/ (see Figure 3. ), then any other point .x; y/
(x, y) lies on the line if and only if x D a. The y-coordinate can have any value. Hence,
a
x an equation of the line is x D a. Similarly, an equation of the h riz nta line passing
through .a; b/ is y D b. (See Figure 3. .) Here the x-coordinate can have any value.
FIGURE Vertical line
through .a; b/.
E AM LE E H L
y
a An equation of the vertical line through .!2; 3/ is x D !2. An equation of the
horizontal line through .!2; 3/ is y D 3.
y=b (x, y)
b The x-axis and y-axis are horizontal and vertical lines, respectively. ecause .0; 0/
b (a, b) lies on both axes, an equation of the x-axis is y D 0, and an equation of the y-axis
is x D 0.
x
Now ork Problems 21 and 23 G
From our discussions, we can show that every straight line is the graph of an equa-
FIGURE Horizontal line
tion of the form Ax C By C C D 0, where A, B, and C are constants and A and B are not
through .a; b/.
both zero. We call this a general linear equation (or an equati n f the rst degree) in
the variables x and y, and x and y are said to be linearly related. For example, a general
linear equation for y D 7x ! 2 is .!7/x C .1/y C .2/ D 0. Conversely, the graph of a
general linear equation is a straight line. Table 3.1 gives the various forms of equations
of straight lines.
136 C nes Para o as an ste s

Table .1 F E S L
Point-slope form y ! y1 D m.x ! x1 /
Slope-intercept form y D mx C b
eneral linear form Ax C By C C D 0
o not confuse the forms of equations of
horizontal and vertical lines. Remember Vertical line xDa
which one has the form x D constant Horizontal line yDb
and which one has the form y D constant.

Example 3 suggests that we could add another entry to the table. For if we now that
y2 ! y1
points .x1 ; y1 / and .x2 ; y2 / are points on a line, then the slope of that line is m D
x2 ! x1
y2 ! y1
and we could say that y ! y1 D .x ! x1 / is a t p int f rm f r an equati n f
x2 ! x1
a ine passing thr ugh p ints .x1 ; y1 / and .x2 ; y2 /. Whether one chooses to remember
many formulas or a few problem-solving principles is very much a matter of individual
taste.
A L IT I
E AM LE C F E L
Find a general linear form of the
Fahrenheit Celsius conversion equa- a Find a general linear form of the line whose slope-intercept form is
tion whose slope-intercept form is 2
FD C C 32.
yD! xC4
5 3
S etting one side to be 0, we obtain
2
xCy!4D0
3
This illustrates that a general linear form which is a general linear form with A D 23 , B D 1, and C D !4. An alternative general
of a line is not unique. form can be obtained by clearing fractions:
2x C 3y ! 12 D 0
b Find the slope-intercept form of the line having a general linear form
A L IT I 3x C 4y ! 2 D 0
S etch the graph of the Fahrenheit S We want the form y D mxCb, so we solve the given equation for y. We have
Celsius conversion equation that you
found in the preceding Apply It. How 3x C 4y ! 2 D 0
could you use this graph to convert a
Celsius temperature to Fahrenheit 4y D !3x C 2
3 1
yD! xC
4 2
which is the slope-intercept form. Note that the line has slope ! 34 and y-intercept 12 .
Now ork Problem 37 G
E AM LE G G L E
y
2x - 3y + 6 = 0 S etch the graph of 2x ! 3y C 6 D 0.
S
(0, 2)
S Since this is a general linear equation, its graph is a straight line. Thus,
we need only determine two different points on the graph in order to s etch it. We
x will find the intercepts.
(-3, 0)

If x D 0, then !3y C 6 D 0, so the y-intercept is 2. If y D 0, then 2x C 6 D 0,

so the x-intercept is !3. We now draw the line passing through .0; 2/ and .!3; 0/. (See
FIGURE raph of
Figure 3.10.)
2x ! 3y C 6 D 0.
Now ork Problem 27 G
 Section 3. nes 137
L
As stated previously, there is a rule for parallel lines:

L ines are para e if and n y if they ha e the same s pe r

are b th ertica .
It follows that any line is parallel to itself.
There is also a rule for perpendicular lines. oo bac to Figure 3.5 and observe
that the line with slope ! 12 is perpendicular to the line with slope 2. The fact that the
slope of either of these lines is the negative reciprocal of the slope of the other line is
not a coincidence, as the following rule states.

L ines ith s pes m1 and m2 , respecti e y are perpen

dicu ar t each ther if and n y if
1
m1 D !
m2
re er any h riz nta ine and any ertica ine are perpendicu ar t each ther.
Rather than simply remembering this equation for the perpendicularity condition,
observe why it ma es sense. For if two lines are perpendicular, with neither vertical,
then one will necessarily rise from left to right while the other will fall from left to
right. Thus, the slopes must have different signs. Also, if one is steep, then the other is
relatively at, which suggests a relationship such as is provided by reciprocals.

E AM LE L
A L IT I
Show that a triangle with vertices Figure 3.11 shows two lines passing through .3; !2/. ne is parallel to the line
at A.0; 0/, B.6; 0/, and C.7; 7/ is not a y D 3x C 1, and the other is perpendicular to it. Find equations of these lines.
right triangle.
y
y = 3x + 1

(a) parallel

x
(3, -2)
(b) perpendicular

FIGURE ines parallel and perpendicular to y D 3x C 1 (Example ).

S The slope of y D 3x C 1 is 3. Thus, the line through .3; !2/ that is para e
to y D 3x C 1 also has slope 3. Using a point-slope form, we get
y ! .!2/ D 3.x ! 3/
y C 2 D 3x !
y D 3x ! 11
The slope of a line perpendicu ar to y D 3x C 1 must be ! 13 (the negative reciprocal
of 3). Using a point-slope form, we get
1
y ! .!2/ D ! .x ! 3/
3
1
yC2D! xC1
3
1
yD! x!1
3
Now ork Problem 55 G
138 C nes Para o as an ste s

R BLEMS
In Pr b ems nd the s pe f the straight ine that passes x C 3y D 0, !5x C 3y ! 17 D 0
thr ugh the gi en p ints y D 3; x D ! 13
.1; 3/, .4; 7/ .!2; 10/, .5; 3/ x D 3; x D !3
.6; !2/, . ; !3/ (2, !4), (3, !4)
3x C y D 4; x ! 3y C 1 D 0
(5, 3), (5, ! ) .1; !2/, .3; 5/
x ! 2 D 3; y D 2
(5, !2), (4, !2) (1, !7), ( , 0)
In Pr b ems nd an equati n f the ine satisfying the gi en
In Pr b ems nd a genera inear equati n c nditi ns Gi e the ans er in s pe intercept f rm if p ssib e
.Ax C By C C D 0/ f the straight ine that has the indicated
pr perties and sketch each ine Passing through .0; 7/ and parallel to 2y D 4x C 7
Passes through .!1; 7/ and has slope !5 Passing through (2, ! ) and parallel to x D !4
Passes through the origin and has slope 75 Passing through (2, 1) and parallel to y D 2
Passes through .!2; 2/ and has slope 2 Passing through (3, !4) and parallel to y D 3 C 2x
3
Perpendicular to y D 3x ! 5 and passing through (3, 4)
Passes through .! 52 ; 5/ and has slope 1
3
Perpendicular to 6x C 2y C 43 D 0 and passing through .1; 5/
Passes through (!6, 1) and (1, 4)
Passing through .5; 2/ and perpendicular to y D !3
Passes through .5; 2/ and .6; !4/
Passing through .4; !5/ and perpendicular to the line
Passes through .!3; !4/ and .!2; ! / 2x
Passes through .1; 1/ and .!2; !3/ 3y D ! C 3
5
Has slope 2 and y-intercept 4 Passing through (!7, !5) and parallel to the line
2x C 3y C 6 D 0
Has slope 5 and y-intercept !7
Passing through (!4, 10) and parallel to the y-axis
Has slope ! 12 and y-intercept !3
A straight line passes through .!1; !3/ and .!1; 17/. Find
Has slope 0 and y-intercept ! 12 the point on it that has an x-coordinate of 5.
Is horizontal and passes through .3; !2/ A straight line has slope 3 and y-intercept .0; 1/. oes the
Is vertical and passes through .!1; !1/ point .!1; !2/ lie on the line
Passes through (2, !3) and is vertical Stoc In 1 6, the stoc in a computer hardware company
Passes through the origin and is horizontal traded for 37 per share. However, the company was in trouble
and the stoc price dropped steadily, to per share in 2006.
In Pr b ems nd if p ssib e the s pe and y intercept f raw a line showing the relationship between the price per share
the straight ine determined by the equati n and sketch the and the year in which it traded for the time interval 1 6, 2006 ,
graph with years on the x-axis and price on the y-axis. Find and interpret
the slope.
y D 4x ! 6 x C 7 D 14
In Pr b ems nd an equati n f the ine describing the
3x C 5y ! D 0 yC4D7
f ing inf rmati n
x D !5 x ! D 5y C 3
ome Runs In one season, a ma or league baseball player
y D 7x y ! 7 D 3.x ! 4/ has hit 14 home runs by the end of the third month and 20 home
yD3 6y ! 24 D 0 runs by the end of the fifth month.
Business A delicatessen owner starts her business with
In Pr b ems nd a genera inear f rm and the
debts of 100,000. After operating for five years, she has
s pe intercept f rm f the gi en equati n
accumulated a profit of 40,000.
2x D 5 ! 3y 3x C 7y D 13
Due Date The length, , of a human fetus at least 12 wee s
4x C y ! 5 D 0 3.x ! 4/ ! 7.y C 1/ D 2 old can be estimated by the formula D 1:53t ! 6:7, where is in
x 2y 3 1 centimeters and t is in wee s from conception. An obstetrician
! C D !4 yD xC
2 3 4 300 uses the length of a fetus, measured by ultrasound, to determine
In Pr b ems determine hether the ines are para e the approximate age of the fetus and establish a due date for the
perpendicu ar r neither mother. The formula must be rewritten to result in an age, t, given
a fetal length, . Find the slope and -intercept of the equation. Is
y D 3x ! 7, y D 3x C 3 there any physical significance to either of the intercepts
y D 4x C 3; y D 5 C 4x Discus hrow A mathematical model can approximate the
y D 5x C 2; !5x C y ! 3 D 0 winning distance for the lympic discus throw by the
formula d D 1 4 C t, where d is in feet and t D 0 corresponds to
y D x; y D !x
the year 1 4 . Find a general linear form of this equation.
x C 3y C 5 D 0; y D !3x
 Section 3.2 cat ons an near nct ons 139

Campus Map A coordinate map of a college campus Revenue Equation A small business predicts its revenue
gives the coordinates .x; y/ of three ma or buildings as follows: growth by a straight-line method with a slope of 50,000 per year.
computer center, .3:5; !1:5/ engineering lab, .0:5; 0:5/ and In its fifth year, it had revenues of 330,000. Find an equation that
library .!1; !2:5/. Find the equations (in slope-intercept form) of describes the relationship between revenues, R, and the number of
the straight-line paths connecting a the engineering lab with the years, , since it opened for business.
computer center and b the engineering lab with the library. Are raph y D !0: x C 7:3 and verify that the y-intercept is 7.3.
these two paths perpendicular to each other
raph the lines whose equations are
Geometry Show that the points A(0, 0), B(0, 4), C(2, 3),
and (2, 7) are the vertices of a parallelogram. ( pposite sides of y D 1:5x C 1
a parallelogram are parallel.)
y D 1:5x ! 1
Approach Angle A small plane is landing at an airport
with an approach angle of 45 degrees, or slope of !1. The plane and
begins its descent when it has an elevation of 3600 feet. Find the y D 1:5x C 2:5
equation that describes the relationship between the craft s
What do you observe about the orientation of these lines Why
altitude and distance traveled, assuming that at distance 0 it
would you expect this result from the equations of the lines
starts the approach angle. raph your equation on a graphing
themselves
calculator. What does the graph tell you about the approach
if the airport is 3 00 feet from where the plane starts its raph the line y D 7:1x C 5:4. Find the coordinates
landing of any two points on the line, and use them to estimate the
slope. What is the actual slope of the line
Cost Equation The average daily cost, C, for a room at a
city hospital has risen by 61.34 per year for the years 2006 Show that if a line has x-intercept x0 and y-intercept y0 , both
through 2016. If the average cost in 2010 was 112 .50, what is different from 0, then xx0 C yy0 D 1, equivalently y0 x C x0 y D x0 y0 ,
an equation that describes the average cost during this decade, as a
is an equation of the line. For 0 " x " x0 and 0 " y " y0 interpret
function of the number of years, , since 2006
the last equation geometrically.

Objective A L F
o e e o t e not on of e an an any situations in economics can be described by using straight lines, as evidenced by
s c r es an to ntro ce near
f nct ons Example 1.

A L IT I E AM LE L
A sporting-goods manufacturer Suppose that a manufacturer uses 100 lb of material to produce products A and , which
allocates 1000 units of time per day to require 4 lb and 2 lb of material per unit, respectively. If x and y denote the number of
ma e s is and s i boots. If it ta es
units produced of A and , respectively, then all levels of production are given by the
units of time to ma e a s i and 14 units
of time to ma e a boot, find an equation
combinations of x and y that satisfy the equation
to describe all possible production 4x C 2y D 100 where x; y # 0
levels of the two products.
Thus, the levels of production of A and are linearly related. Solving for y gives
y (units of B) y D !2x C 50 slope-intercept form
4x + 2y = 100
(y = -2x + 50) so the slope is !2. The slope re ects the rate of change of the level of production of
with respect to the level of production of A. For example, if 1 more unit of A is to be
(0, 50)
50
produced, it will require 4 more pounds of material, resulting in 42 D 2 fe er units of
40
(10, 30)
. Accordingly, as x increases by 1 unit, the corresponding value of y decreases by 2
30 units. To s etch the graph of y D !2x C 50, we can use the y-intercept .0; 50/ and the
20 fact that when x D 10, y D 30. (See Figure 3.12.)
10
x Now ork Problem 21 G
10 20 (units of A)

FIGURE inearly related

S C
production levels. For each price level of a product, there is a corresponding quantity of that product
that consumers will demand (that is, purchase) during some time period. Usually, the
higher the price, the smaller is the quantity demanded as the price falls, the quantity
demanded increases. If the price per unit of the product is given by p and the corre-
sponding quantity (in units) is given by q, then an equation relating p and q is called a
140 C nes Para o as an ste s

p p

Demand curve Supply curve

(Price/unit)

(Price/unit)
d (c, d)

b (a, b)

q q
a c
(Quantity per unit of time) (Quantity per unit of time)
(a) (b)

FIGURE emand and supply curves.

demand equation. Its graph is called a demand curve. Figure 3.13(a) shows a demand
curve. In eeping with the practice of most economists, the horizontal axis is the q-axis
and the vertical axis is the p-axis. We will assume that the price per unit is given in
dollars and the period is one wee . Thus, the point .a; b/ in Figure 3.14(a) indicates
that, at a price of b dollars per unit, consumers will demand a units per wee . Since
negative prices or quantities are not meaningful, both a and b must be nonnegative.
For most products, an increase in the quantity demanded corresponds to a decrease in
price. Thus, a demand curve typically falls from left to right, as in Figure 3.13(a).
In response to various prices, there is a corresponding quantity of product that
Typically, a demand curve falls from left pr ducers are willing to supply to the mar et during some time period. Usually, the
to right and a supply curve rises from left higher the price per unit, the larger is the quantity that producers are willing to supply
to right. However, there are exceptions. as the price falls, so will the quantity supplied. If p denotes the price per unit and q
For example, the demand for insulin denotes the corresponding quantity, then an equation relating p and q is called a supply
could be represented by a vertical line, equation, and its graph is called a supply curve. Figure 3.13(b) shows a supply curve.
since this demand can remain constant
regardless of price. If p is in dollars and the period is one wee , then the point .c; d/ indicates that, at a
price of d dollars each, producers will supply c units per wee . As before, c and d are
nonnegative. A supply curve usually rises from left to right, as in Figure 3.13(b). This
indicates that a producer will supply more of a product at higher prices.
bserve that a function whose graph either falls from left to right or rises from left
to right thr ugh ut its entire d main will pass the horizontal line test of Section 2.5.
Certainly, the demand curve and the supply curve in Figure 3.14 are each cut at most
once by any horizontal line. Thus, if the demand curve is the graph of a function
p D .q/, then will have an inverse and we can solve for q uniquely to get
q D !1 .p/. Similarly, if the supply curve is the graph of a function p D S.q/, then S
is also one-to-one, has an inverse S!1 , and we can write q D S!1 .p/.
We will now focus on demand and supply curves that are straight lines (Figure 3.14).
They are called inear demand and inear supply curves. Such curves have equations
in which p and q are linearly related. ecause a demand curve typically falls from left
to right, a linear demand curve has a negative slope. See Figure 3.14(a). However,
the slope of a linear supply curve is positive, because the curve rises from left to right.
See Figure 3.14(b).
p p

Linear demand Linear supply

curve curve

Negative Positive
slope slope
q q

(a) (b)

FIGURE inear demand and supply curves.

 Section 3.2 cat ons an near nct ons 141

A L IT I E AM LE F E
The demand per wee for 50-inch
television sets is 1200 units when the Suppose the demand per wee for a product is 100 units when the price is 5 per unit
price is 575 each and 00 units when and 200 units at 51 each. etermine the demand equation, assuming that it is linear.
the price is 725 each. Find the demand S
equation for the sets, assuming that it is
linear.
S Since the demand equation is linear, the demand curve must be a straight
line. We are given that quantity q and price p are linearly related such that p D 5
when q D 100 and p D 51 when q D 200. Thus, the given data can be rep-
resented in a q; p-coordinate plane see Figure 3.14(a) by points .100; 5 / and
.200; 51/. With these points, we can find an equation of the line that is, the demand
equation.

The slope of the line passing through .100; 5 / and .200; 51/ is

51 ! 5 7
mD D!
200 ! 100 100

An equation of the line (point-slope form) is

p ! p1 D m.q ! q1 /
7
p!5 D! .q ! 100/
100
80
Simplifying gives the demand equation

7
pD! q C 65
100
0 1000
0
Customarily, a demand equation (as well as a supply equation) expresses p, in terms of
FIGURE raph of demand q and actually defines a function of q. For example, Equation (1) defines p as a function
7
function p D ! 100 q C 65. of q and is called the demand functi n for the product. (See Figure 3.15.)
Now ork Problem 15 G
L F
A inear functi n was defined in Section 2.2 to be a polynomial function of degree 1.
Somewhat more explicitly,

A function f is a inear functi n if and only if f.x/ can be written in the form
f.x/ D ax C b, where a and b are constants and a ¤ 0.

Suppose that f.x/ D ax C b is a linear function, and let y D f.x/. Then y D ax C b,

which is an equation of a straight line with slope a and y-intercept b. Thus, the graph
f a inear functi n is a straight ine that is neither ertica n r h riz nta We say that
the function f.x/ D ax C b has slope a.
142 C nes Para o as an ste s

A L IT I E AM LE G L F
A computer repair company
charges a fixed amount plus an hourly f(x) g(t)
rate for a service call. If x is the number
3
of hours needed for a service call, f(x) = 2x - 1 5
the total cost of a call is described 2 g(t) = 15 - 2t
3 3
by the function f .x/ D 40x C 60.
x f (x) t g(t)
raph the function by finding and
0 -1 x 0 5 t
plotting two points. -1 2 6
2 3 6 1

(a) (b)

FIGURE raphs of linear functions.

a raph f.x/ D 2x ! 1.
S Here f is a linear function (with slope 2), so its graph is a straight line. Since
two points determine a straight line, we need only plot two points and then draw a line
through them. See Figure 3.16(a). Note that one of the points plotted is the vertical-
axis intercept, !1, which occurs when x D 0.
15 ! 2t
b raph g.t/ D .
3
S Notice that g is a linear function, because we can express it in the form
g.t/ D at C b.
15 ! 2t 15 2t 2
g.t/ D D ! D! tC5
3 3 3 3
The graph of g is shown in Figure 3.16(b). Since the slope is ! 23 , observe that as t
increases by 3 units, g.t/ decreases by 2.
Now ork Problem 3 G
E AM LE L F
A L IT I
The height of children between the Suppose f is a linear function with slope 2 and f.4/ D . Find f.x/.
ages of 6 years and 10 years can be S Since f is linear, it has the form f.x/ D ax C b. The slope is 2, so a D 2,
modeled by a linear function of age t in and we have
years. The height of one child changes
by 2.3 inches per year, and she is 50.6 f.x/ D 2x C b
inches tall at age . Find a function that
describes the height of this child at age t.
Now we determine b. Since f.4/ D , we replace x by 4 in Equation (2) and solve
for b:
f.4/ D 2.4/ C b
D Cb
0Db
Hence, f.x/ D 2x.
Now ork Problem 7 G
E AM LE L F
A L IT I
An antique nec lace is expected If y D f.x/ is a linear function such that f.!2/ D 6 and f.1/ D !3, find f.x/.
to be worth 360 after 3 years and S
640 after 7 years. Find a function that
describes the value of the nec lace after S The function values correspond to points on the graph of f. With these
x years.
points we can determine an equation of the line and hence the linear function.
 Section 3.2 cat ons an near nct ons 143

The condition that f.!2/ D 6 means that when x D !2, then y D 6. Thus, .!2; 6/
lies on the graph of f, which is a straight line. Similarly, f.1/ D !3 implies that .1; !3/
also lies on the line. If we set .x1 ; y1 / D .!2; 6/ and .x2 ; y2 / D .1; !3/, the slope of
the line is given by
y2 ! y1 !3 ! 6 !
mD D D D !3
x2 ! x1 1 ! .!2/ 3
We can find an equation of the line by using a point-slope form:
y ! y1 D m.x ! x1 /
y ! 6 D !3Œx ! .!2/!
y ! 6 D !3x ! 6
y D !3x
ecause y D f.x/, f.x/ D !3x. f course, the same result is obtained if we set
.x1 ; y1 / D .1; !3/.
Now ork Problem 9 G
In many studies, data are collected and plotted on a coordinate system. An analysis
of the results may indicate a functional relationship between the variables involved. For
example, the data points may be approximated by points on a straight line. This would
indicate a linear functional relationship, such as the one in the next example.

w(weight) E AM LE H

675 (25, 675) In testing an experimental diet for hens, it was determined that the average live weight
(in grams) of a hen was statistically a linear function of the number of days d after
the diet began, where 0 " d " 50. Suppose the average weight of a hen beginning the
diet was 40 grams and 25 days later it was 675 grams.
a etermine as a linear function of d.
40 d(days) S Since is a linear function of d, its graph is a straight line. When d D 0
25 50
(the beginning of the diet), D 40. Thus, .0; 40/ lies on the graph. (See Figure 3.17.)
Similarly, .25; 675/ lies on the graph. If we set .d1 ; 1 / D .0; 40/ and
FIGURE inear function
.d2 ; 2 / D .25; 675/, the slope of the line is
describing diet for hens.
2! 1 675 ! 40 635 127
mD D D D
d2 ! d1 25 ! 0 25 5
Using a point-slope form, we have
! D m.d ! d1 /
1
127
! 40 D .d ! 0/
5
127
! 40 D d
5
127
D d C 40
5
which expresses as a linear function of d.
b Find the average weight of a hen when d D 10.
S When d D 10, D 127 5
.10/ C 40 D 254 C 40 D 2 4. Thus, the average
weight of a hen 10 days after the beginning of the diet is 2 4 grams.
Now ork Problem 19 G
144 C nes Para o as an ste s

R BLEMS
In Pr b ems nd the s pe and ertica axis intercept f the Depreciation Suppose the value of a mountain bi e
inear functi n and sketch the graph decreases each year by 10 of its original value. If the original
y D f .x/ D !4x y D f .x/ D 6x C 3 value is 1 00, find an equation that expresses the value of the
bi e t years after purchase, where 0 " t " 10. S etch the equation,
h.t/ D 5t ! 7 f .s/ D 3.5 ! s/ choosing t as the horizontal axis and as the vertical axis. What is
5!q the slope of the resulting line This method of considering the
p.q/ D h.q/ D 0:5q C 0:25
3 value of equipment is called straight ine depreciati n.
In Pr b ems nd f x if f is a inear functi n that has the
Depreciation A new television depreciates 120 per year,
gi en pr perties
and it is worth 340 after four years. Find a function that describes
slope D 4, f .1/ D 7 f .0/ D 3; f .4/ D !5 the value of this television, if x is the age of the television in years.
f .1/ D 2; f .!2/ D slope D !5, f . 14 / D Appreciation A new house was sold for 1,1 3,000 six
slope D ! 23 , f .! 23 / D ! 23 f .2/ D 7, f .3/ D 14 years after it was built and purchased. The original owners
calculated that the house appreciated 53,000 per year while they
f .!2/ D !1; f .!4/ D !3 owned it. Find a linear function that describes the appreciation of
slope D 0:01; f .0:1/ D 0:01 the building, in thousands of dollars, if x is the number of years
since the original purchase.
Demand Equation Suppose consumers will demand
60 units of a product when the price is 15.30 per unit and Appreciation A house purchased for 245,000 is expected
35 units when the price is 1 .30 each. Find the demand equation, to double in value in 15 years. Find a linear equation that
assuming that it is linear. Find the price per unit when 40 units are describes the house s value after t years.
demanded.
otal Cost A company s yearly total production cost C is
Demand Equation The demand per wee for a C is typically given by C D C.n/ D F C cn, where F is fixed cost and
26,000 copies when the price is 12 each, and 10,000 copies when cn, the variable cost, is cost per item, c, times the number of items
the price is 1 each. Find the demand equation for the C , produced, n. If, in 2010, 1000 items were produced at a total cost
assuming that it is linear. of 3500 and, in 2015, 1500 items were produced at a total cost of
Supply Equation A laptop manufacturer will produce 5000, determine the linear function C of n. What are the
3,000,000 units when the price is 00, and 2,000,000 units when numerical values of F and c
the price is 700. Assume that price, p, and quantity, q, produced Sheep’s ool Length For sheep maintained at high
are linearly related and find the supply equation. environmental temperatures, respiratory rate, r (per minute),
Supply Equation Suppose a manufacturer of shoes will increases as wool length, (in centimeters), decreases.2 Suppose
place on the mar et 50 (thousand pairs) when the price is 35 sheep with a wool length of 2 cm have an (average) respiratory
(dollars per pair) and 35 when the price is 30. Find the supply rate of 160, and those with a wool length of 4 cm have a
equation, assuming that price p and quantity q are linearly related. respiratory rate of 125. Assume that r and are linearly related. a
Find an equation that gives r in terms of . b Find the respiratory
Cost Equation Suppose the cost to produce 10 units of a rate of sheep with a wool length of 1 cm.
product is 40 and the cost of 20 units is 70. If cost, c, is linearly
related to output, q, find a linear equation relating c and q. Find Isocost Line In production analysis, an is c st ine is a line
the cost to produce 35 units. whose points represent all combinations of two factors of
production that can be purchased for the same amount. Suppose a
Cost Equation An advertiser goes to a printer and is farmer has allocated 20,000 for the purchase of x tons of
charged for 100 copies of one yer and 3 for 200 copies of fertilizer (costing 200 per ton) and y acres of land (costing 2000
another yer. This printer charges a fixed setup cost plus a charge per acre). Find an equation of the isocost line that describes the
for every copy of single-page yers. Find a function that describes various combinations that can be purchased for 20,000. bserve
the cost of a printing ob, if x is the number of copies made. that neither x nor y can be negative.
Electricity Rates An electric utility company charges Isoprofit Line A manufacturer produces products X and
residential customers 12.5 cents per ilowatt-hour plus a base for which the profits per unit are 7 and , respectively. If x units
charge each month. ne customer s monthly bill comes to 51.65 of X and y units of are sold, then the total profit P is given by
for 3 0 ilowatt-hours. Find a linear function that describes the P D P.x; y/ D 7x C y, where x; y # 0. a S etch the graph of
total monthly charges for electricity if x is the number of this equation for P D 260. The result is called an is pr t ine
ilowatt-hours used in a month. and its points represent all combinations of sales that produce a
Demand Equation If the price of a product and demand profit of 260. It is an example of a e e cur e for the function
for it are nown to be linearly related, with a demand for 120 units P.x; y/ D 7x C y of two variables as introduced in Section 2. .
when the price is 0 and a demand for 0 units when the price is b etermine the slope for P D 260. c For P D 60, determine
150, determine the equation. int Recall the form for a linear the slope. d Are isoprofit lines always parallel
equation developed in Problem 76 of Section 3.1.

2
Adapted from . E. Fol , r., extb k f En ir nmenta Physi gy 2nd ed.
(Philadelphia: ea Febiger, 1 74).
 Section 3.3 a rat c nct ons 145

Grade Scaling For reasons of comparison, a professor after the diet was initiated, where 0 " d " 100. If the weight of
wants to rescale the scores on a set of test papers so that the a pig beginning the diet was 21 g, and thereafter the pig gained
maximum score is still 100 but the average is 65 instead of 56. 6.3 g every 10 days, determine as a function of d, and find the
a Find a linear equation that will do this. int ou want 56 to weight of a pig 55 days after the beginning of the diet.
become 65 and 100 to remain 100. Consider the points (56, 65)
and (100, 100) and, more generally, (x, y), where x is the old score
and y is the new score. Find the slope and use a point-slope form.
Express y in terms of x. b If 62 on the new scale is the lowest
passing score, what was the lowest passing score on the
original scale
Profit Coe cients A company ma es two products, and
. If the company ma es a profit from selling 1 unit of and b
profit from selling 1 unit of , then it is clear that its total profit P Cric et Chirps iologists have found that the number of
from selling x units of and y units of is given by P D ax C by. chirps made per minute by cric ets of a certain species is related
If, moreover, it is nown that a profit of P can be made by selling to the temperature. The relationship is very close to being linear.
40 units of and 0 units of r by selling 0 units of and 30 At 6 ı F, the cric ets chirp about 124 times a minute. At 0ı F,
units of , determine the profit coe cients a and b in terms of P. they chirp about 172 times a minute. a Find an equation that
Psychology In a certain learning experiment involving gives Fahrenheit temperature, t, in terms of the number of chirps,
repetition and memory,3 the proportion, p, of items recalled c, per minute. b If you count chirps for only 15 seconds, how
was estimated to be linearly related to the effective study time, can you quic ly estimate the temperature
t (in seconds), where t is between 5 and . For an effective study
time of 5 seconds, the proportion of items recalled was 0.32.
For each 1-second increase in study time, the proportion recalled
increased by 0.05 . a Find an equation that gives p in terms of t.
b What proportion of items was recalled with seconds of
effective study time
Diet for Pigs In testing an experimental diet for pigs, it
was determined that the (average) live weight, (in ilograms), of
a pig was statistically a linear function of the number of days, d,

Objective F
o s etc ara o as ar s n fro In Section 3.3, a quadratic functi n was defined as a polynomial function of degree 2.
a rat c f nct ons
In other words,

A function f is a quadratic functi n if and only if f.x/ can be written in the form
f.x/ D ax2 C bx C c, where a, b, and c are constants and a ¤ 0.

For example, the functions f.x/ D x2 ! 3x C 2 and F.t/ D !3t2 are quadratic.
1
However, g.x/ D 2 is n t quadratic, because it cannot be written in the form
x
g.x/ D ax2 C bx C c.
The graph of the quadratic function y D f.x/ D ax2 C bx C c is called a parabola
and has a shape li e the curves in Figure 3.1 . If a > 0, the graph extends upward
indefinitely, and we say that the parabola pens up ard Figure 3.1 (a) . If a < 0, the
parabola pens d n ard Figure 3.1 (b) .
Each parabola in Figure 3.1 is symmetric about a vertical line, called the axis of
symmetry of the parabola. That is, if the page were folded on one of these lines, then
the two halves of the corresponding parabola would coincide. The axis (of symmetry)
is n t part of the parabola, but is a useful aid in s etching the parabola.
Each part of Figure 3.1 shows a point labeled vertex, where the axis cuts the
parabola. If a > 0, the vertex is the lowest point on the parabola. This means that

3
. . Hintzman, Repetition and earning, in he Psych gy f earning
Vol. 10, ed. . H. ower (New or : Academic Press, Inc., 1 76), p. 77.
146 C nes Para o as an ste s

Parabola: y = f(x) = ax2 + bx + c

y y

Vertex
Axis of
symmetry
Axis of
symmetry

x x

Vertex
a 7 0, opens upward a 6 0, opens downward
(a) (b)

FIGURE Parabolas.

f.x/ has a minimum value at this point. y performing algebraic manipulations on

ax2 C bx C c (referred to as c mp eting the square), we can determine not only this
minimum value, but also where it occurs. We have
f.x/ D ax2 C bx C c D .ax2 C bx/ C c
b2
Adding and subtracting gives
4a
! "
2 b2 b2
f.x/ D ax C bx C Cc!
4a 4a
! 2 "
b b b2
D a x2 C x C 2 C c !
a 4a 4a
so that
! "
b 2 b2
f.x/ D a x C Cc!
2a 4a
! "
b 2
Since x C # 0 and a > 0, it follows that f.x/ has a minimum value when
2a
b b
xC D 0 that is, when x D ! . The y-coordinate corresponding to this value of
2a
! " 2a
b
x is f ! . Thus, the vertex is given by
2a
! ! ""
b b
vertex D ! ; f !
2a 2a
This
! is " also the vertex of a parabola that opens downward .a < 0/, but in this case
b
f ! is the maximum value of f.x/. See Figure 3.1 (b).
2a
bserve that a function whose graph is a parabola is not one-to-one, in either
the opening upward or opening downward case, since many horizontal lines will cut
the graph twice. However, if we restrict the domain of a quadratic function to either
# " ! $
b b
! ; 1 or !1; ! , then the restricted function will pass the horizontal line
2a 2a
test and therefore be one-to-one. (There are many other restrictions of a quadratic func-
tion that are one-to-one however, their domains consist of more than one interval.) It
follows that such restricted quadratic functions have inverse functions.
The point where the parabola y D ax2 C bx C c intersects the y-axis (that is, the
y-intercept) occurs when x D 0. The y-coordinate of this point is c, so the y-intercept
is c. In summary, we have the following.
 Section 3.3 a rat c nct ons 147

G F
The graph of the quadratic function y D f.x/ D ax2 C bx C c is a parabola.
If a > 0, the parabola opens upward. If a < 0, it opens downward.
! ! ""
b b
The vertex is ! ; f ! .
2a 2a
The y-intercept is c.

We can quic ly s etch the graph of a quadratic function by first locating the vertex,
the y-intercept, and a few other points, such as those where the parabola intersects
the x-axis. These x intercepts are found by setting y D 0 and solving for x. nce the
intercepts and vertex are found, it is then relatively easy to pass the appropriate parabola
through these points. In the event that the x-intercepts are very close to the vertex or
that no x-intercepts exist, we find a point on each side of the vertex, so that we can give
a reasonable s etch of the parabola. eep in mind that passing a (dashed) vertical line
through the vertex gives the axis of symmetry. y plotting points to one side of the
axis, we can use symmetry and obtain corresponding points on the other side.

E AM LE G F
A L IT I
A car dealership believes that the
raph the quadratic function y D f.x/ D !x2 ! 4x C 12.
daily profit from the sale of minivans is S Here a D !1, b D !4, and c D 12. Since a < 0, the parabola opens
given by P.x/ D !x2 C2xC3 , where downward and, thus, has a highest point. The x-coordinate of the vertex is
x is the number of minivans sold. Find
b !4
the function s vertex and intercepts, and ! D! D !2
graph the function. If their model is cor- 2a 2.!1/
rect, comment on the viability of deal- The y-coordinate is f.!2/ D !.!2/2 ! 4.!2/ C 12 D 16. Thus, the vertex is .!2; 16/,
ing in minivans. so the maximum value of f.x/ is 16. Since c D 12, the y-intercept is 12. To find the
x-intercepts, we let y be 0 in y D !x2 ! 4x C 12 and solve for x:
0 D !x2 ! 4x C 12
0 D !.x2 C 4x ! 12/
0 D !.x C 6/.x ! 2/
Hence, x D !6 or x D 2, so the x-intercepts are !6 and 2. Now we plot the vertex,
axis of symmetry, and intercepts. See Figure 3.1 (a). Since .0; 12/ is t units to the
right of the axis of symmetry, there is a corresponding point t units to the eft of the
axis with the same y-coordinate. Thus, we get the point .!4; 12/. Through all points,
we draw a parabola opening downward. See Figure 3.1 (b).

y y

Vertex 16 16

12 12

8 8
y = f(x) = -x2 -4x + 12
Axis of
symmetry 4 4
-6
x x
-6 -4 -2 2 -4 -2 2
-4 -4

-8 -8

(a) (b)

FIGURE raph of parabola y D f .x/ D !x2 ! 4x C 12.

Now ork Problem 15 G

148 C nes Para o as an ste s

E AM LE G F

raph p D 2q2 .
S Here p is a quadratic function of q, where a D 2, b D 0, and c D 0. Since
p a > 0, the parabola opens upward and, thus, has a lowest point. The q-coordinate of
the vertex is
8 b 0
! D! D0
q p 2a 2.2/
p = 2q2
2 8
-2 8 q
and the p-coordinate is 2.0/2 D 0. Consequently, the minimum value of p is 0 and the
-2 2 vertex is .0; 0/. In this case, the p-axis is the axis of symmetry. A parabola opening
upward with vertex at .0; 0/ cannot have any other intercepts. Hence, to draw a reason-
FIGURE raph of parabola
able graph, we plot a point on each side of the vertex. If q D 2, then p D . This gives
p D 2q2 . the point .2; / and, by symmetry, the point .!2; /. (See Figure 3.20.)
Now ork Problem 13 G
Example 3 illustrates that finding
intercepts may require use of the
quadratic formula. E AM LE G F

raph g.x/ D x2 ! 6x C 7.
A L IT I S Here g is a quadratic function, where a D 1, b D !6, and c D 7. The
A man standing on a pitcher s parabola opens upward, because a > 0. The x-coordinate of the vertex (lowest point) is
mound throws a ball straight up with
an initial velocity of 32 feet per sec- b !6
! D! D3
ond. The height, h, of the ball in feet t 2a 2.1/
seconds after it was thrown is described
by the function h.t/ D !16t2 C 32t C , and g.3/ D 32 ! 6.3/ C 7 D !2, which is the minimum value of g.x/. Thus, the vertex
for t # 0. Find the function s vertex and is .3; !2/. Since c D 7, the vertical-axis intercept is 7. To find x-intercepts, we set
intercepts, and graph the function. g.x/ D 0.

0 D x2 ! 6x C 7

The right side does not factor easily, so we will use the quadratic formula to solve for
x:
g(x) p p
g(x) = x2 - 6x + 7 !b ˙ b2 ! 4ac !.!6/ ˙ .!6/2 ! 4.1/.7/
xD D
2a 2.1/
7 p p p
6˙ 6˙ 4$2 6˙2 2
D D D
2 2 2
p
6 2 2 p
D ˙ D3˙ 2
3- 2 3+ 2 2 2
3 p p
6
x Therefore, the x-intercepts are 3C 2 and 3 ! 2. After plotting the vertex, intercepts,
and (by symmetry) the point .6; 7/, we draw a parabola opening upward in Figure 3.21.
-2
Now ork Problem 17 G
FIGURE raph of parabola
g.x/ D x2 ! 6x C 7.
E AM LE G F

raph y D f.x/ D 2x2 C 2x C 3 and find the range of f.

S This function is quadratic with a D 2, b D 2, and c D 3. Since a > 0, the
graph is a parabola opening upward. The x-coordinate of the vertex is
b 2 1
! D! D!
2a 2.2/ 2
 Section 3.3 a rat c nct ons 149
y

5
Range: y 9 2

3
5
x y 2

-2 7 y = f(x) = 2x2 + 2x + 3
1 7
x
-2 - 12 1

FIGURE raph of y D f .x/ D 2x2 C 2x C 3.

and the y-coordinate is 2.! 12 /2 C 2.! 12 / C 3 D 52 . Thus, the vertex is .! 12 ; 52 /. Since

c D 3, the y-intercept is 3. A parabola opening upward with its vertex above the
x-axis has no x-intercepts. In Figure 3.22 we plotted the y-intercept, the vertex, and
an additional point .!2; 7/ to the left of the vertex. y symmetry, we also get the point
.1; 7/. Passing a parabola through these points gives the desired graph. From the figure,
we see that the range of f is all y # 52 , that is, the interval Π52 ; 1/.

Now ork Problem 21 G

E AM LE F G I

For the parabola given by the function

y D f.x/ D ax2 C bx C c

determine the inverse of the restricted function given by g.x/ D ax2 C bx C c, for
b
x # ! . (We now that this restricted function passes the horizontal line test, so g
2a
does have an inverse.) raph g and g!1 in the same plane, in the case where a D 2,
b D 2, and c D 3.
! "
b
S We begin by observing that, for a > 0, the range of g is Œg ! ; 1/,
! " 2a
b
while, for a < 0, the the range of g is .!1; g ! !. (It follows that, for
! " 2a
b
a > 0, the domain of g!1 is Œg ! ; 1/, while for a < 0, the domain of g!1 is
! " 2a
b
.!1; g ! !. We now follow the procedure described in Example 5 of Section 2.4.
2a
b
For x # ! , we solve y D ax2 C bx C c for x in terms of y. We apply the
2a p
2 !b ˙ b2 ! 4a.c ! y/
quadratic formula to ax C bx C .c ! y/ D 0, giving x D .
p 2a
2
!b C b ! 4a.c ! x/
Now in case a > 0, g!1 .x/ D , while, in case a < 0, g!1 .x/ D
p 2a
!b ! b2 ! 4a.c ! x/
. The signs are chosen to ensure that the values of g!1 lie in the
2a
b
domain of g, which, for either a > 0 or a < 0 is Œ! ; 1/.
2a
To complete the exercise, observe that in Figure 3.22 we have provided the graph
of y D 2x2 C 2x C 3. For the tas at hand, we redraw that part of the curve that lies
to the right of the axis of symmetry. This provides the graph of g. Next we provide a
150 C nes Para o as an ste s

6
5
4
3
2
1
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6

FIGURE raph of g and g!1 .

dotted copy of the line y D x. Finally, we draw the mirror image of g in the line y D x
to obtain the graph of g!1 as in Figure 3.23.
Now ork Problem 27 G
E AM LE M R
A L IT I
The demand function for a pub- The demand function for a manufacturer s product is p D 1000 ! 2q, where p is the
lisher s line of coo boo s is price (in dollars) per unit when q units are demanded (per wee ) by consumers. Find the
level of production that will maximize the manufacturer s total revenue, and determine
p D 6 ! 0:003q
this revenue.
where p is the price (in dollars) per unit
when q units are demanded (per day) by S
consumers. Find the level of production
S To maximize revenue, we determine the revenue function, r D f.q/.
that will maximize the manufacturer s
total revenue, and determine this rev- Using the relation
enue. total revenue D .price/.quantity/
we have
r D pq
The formula for total revenue is part of
the repertoire of relationships in business Using the demand equation, we can express p in terms of q, so r will be a function
and economics. of q.

We have
r D pq
D .1000 ! 2q/q
r D 1000q ! 2q2
Note that r is a quadratic function of q, with a D !2, b D 1000, and c D 0. Since
a < 0 (the parabola opens downward) and r attains a maximum at the vertex .q; r/,
where
b 1000
qD! D! D 250
2a 2.!2/
The maximum value of r is given by
r.250/ D 1000.250/ ! 2.250/2
D 250;000 ! 125;000 D 125;000
 Section 3.3 a rat c nct ons 151

r Thus, the maximum revenue that the manufacturer can receive is 125,000, which
r = 1000q - 2q2 occurs at a production level of 250 units. Figure 3.24 shows the graph of the revenue
125,000 function. nly that portion for which q # 0 and r # 0 is drawn, since quantity and
revenue cannot be negative.

q
Now ork Problem 29 G
250 500

FIGURE raph of revenue

function.

R BLEMS
In Pr b ems state hether the functi n is quadratic per unit when q units are demanded (per day). Find the level of
1 production that maximizes the manufacturer s total revenue, and
f .x/ D 5x2 g.x/ D 2 determine this revenue.
2x ! 4
g.x2 / D !5 ! 11x2 k. / D 2 2 .2 2 C 2/ Revenue The demand function for an o ce supply
company s line of plastic rulers is p D 0: 5 ! 0:00045q, where p is
h.q/ D .3 ! q/2 f .t/ D 2t.3 ! t/ C 4t
the price (in dollars) per unit when q units are demanded (per day)
2
s ! by consumers. Find the level of production that will maximize the
f .s/ D g.x/ D .x2 C 2/2 manufacturer s total revenue, and determine this revenue.
2
In Pr b ems d n t inc ude a graph Revenue The demand function for an electronics
company s laptop computer line is p D 2400 ! 6q, where p is the
a For the parabola y D f .x/ D 3x2 C 5x C 1, find the vertex. price (in dollars) per unit when q units are demanded (per wee )
b oes the vertex correspond to the highest point or the lowest by consumers. Find the level of production that will maximize the
point on the graph manufacturer s total revenue, and determine this revenue.
Repeat Problem if y D f .x/ D x2 C 4x ! 1. Mar eting A mar eting firm estimates that n months after
2 the introduction of a client s new product, f.n/ thousand
For the parabola y D f .x/ D x C x ! 6, find a the households will use it, where
y-intercept, b the x-intercepts, and c the vertex.
10
Repeat Problem 11 if y D f .x/ D 5 ! x ! 3x2 . f .n/ D n.12 ! n/; 0 " n " 12

In Pr b ems graph each functi n Gi e the ertex and Estimate the maximum number of households that will use the
intercepts and state the range product.
y D f .x/ D x2 ! 6x ! 7 y D f .x/ D x2 Profit A manufacturer s profit P from producing and
y D g.x/ D !2x ! 6x 2 2
y D f .x/ D x ! 4 selling q items is given by P.q/ D !2q2 C 00q ! 50; 000.
2
etermine the quantity that maximizes profit and the maximum
s D h.t/ D t C 6t C s D h.t/ D 2t2 ! 3t ! 5 profit.
y D f .x/ D !5 C 3x ! 3x2 Psychology A prediction made by early psychology
y D .x/ D 1 ! x ! x2 relating the magnitude of a stimulus, x, to the magnitude of a
response, y, is expressed by the equation y D kx2 , where k is
t D f .s/ D s2 ! s C 14
a constant of the experiment. In an experiment on pattern
t D f .s/ D s2 C 6s C 11 recognition, k D 3. Find the function s vertex and graph the
In Pr b ems state hether f x has a maximum a ue r a equation. (Assume no restriction on x.)
minimum a ue and nd that a ue Biology iologists studied the nutritional effects on rats
f .x/ D 23x2 ! 12x C 10 f .x/ D !7x2 ! 2x C 6 that were fed a diet containing 10 protein.4 The protein
consisted of yeast and corn our. y varying the percentage, P, of
f .x/ D 4x ! 50 ! 0:1x2 f .x/ D x.x C 3/ ! 12
yeast in the protein mix, the group estimated that the average
In Pr b ems and restrict the quadratic functi n t th se x weight gain (in grams) of a rat over a period of time was
satisfying x # here is the x c rdinate f the ertex f the
1 2
parab a etermine the in erse f the restricted functi n Graph f .P/ D !P C 2P C 20; 0 " P " 100
the restricted functi n and its in erse in the same p ane 50
f .x/ D x2 ! 2x C 4 f .x/ D !x2 ! 1 Find the maximum weight gain.
eight of Ball Suppose that the height, s, of a ball thrown
Revenue The demand function for a manufacturer s vertically upward is given by
product is p D f .q/ D 100 ! 10q, where p is the price (in dollars)
s D !4: t2 C 62:3t C 1:

4
Adapted from R. ressani, The Use of east in Human Foods, in Sing e Ce
Pr tein ed. R. I. ateles and S. R. Tannenbaum (Cambridge, A: IT Press,
1 6 ).
152 C nes Para o as an ste s

where s is in meters and t is elapsed time in seconds. (See Roc et Launch A toy roc et is launched straight up from
Figure 3.25.) After how many seconds will the ball reach its the roof of a garage with an initial velocity of 0 feet per second.
maximum height What is the maximum height The height, h, of the roc et in feet, t seconds after it was released,
is described by the function h.t/ D !16t2 C 0t C 14. Find the
function s vertex and intercepts, and graph the function.
Max
Area Express the area of the rectangle shown in
Figure 3.26 as a quadratic function of x. For what value of x will
the area be a maximum

11 - x
x

FIGURE iagram for Problem 40.

s=0 Enclosing Plot A building contractor wants to fence
in a rectangular plot ad acent to a straight highway, using the
FIGURE all thrown upward (Problem 36). highway for one side, which will be left unfenced. (See
Archery A boy standing on a hill shoots an arrow straight Figure 3.27.) If the contractor has 500 feet of fence, find the
up with an initial velocity of 5 feet per second. The height, h, of dimensions of the maximum enclosed area.
the arrow in feet, t seconds after it was released, is described by
the function h.t/ D !16t2 C 5t C 22. What is the maximum
height reached by the arrow How many seconds after release
does it ta e to reach this height
x x
Long Fall At 2 meters, the ur halifa in ubai has
been the world s tallest building since 200 . If an ob ect were to
FIGURE iagram for Problem 41.
fall from the top of it, then after t seconds the ob ect would be at a
height above the ground of . 2 ! 4: t2 / meters and traveling at Find two numbers whose sum is 7 and whose product is a
! : t meters per second. How fast would it be traveling at the maximum.
moment of impact when it hit the ground (Ignore air resistance.)

Objective S L E
o so e s ste s of near e at ons
n ot t o an t ree ar a es T S
s n t e tec n e of e nat on
a t on or s st t t on When a situation must be described mathematically, it is not unusual for a set of equa-
n C a ter ot er et o s
are s o n tions to arise. For example, suppose that the manager of a factory is setting up a pro-
duction schedule for two models of a new product. odel A requires 4 resistors and
transistors. odel requires 5 resistors and 14 transistors. From its suppliers, the fac-
tory gets 335 resistors and 50 transistors each day. How many of each model should
the manager plan to ma e each day so that all the resistors and transistors are used
It s a good idea to construct a table that summarizes the important information.
Table 3.2 shows the number of resistors and transistors required for each model, as
well as the total number available.

Table .
odel A odel Total Available

Resistors 4 5 335
Transistors 14 50

Suppose we let x be the number of model A made each day and y be the number
of model . Then these require a total of 4x C 5y resistors and x C 14y transistors.
Since 335 resistors and 50 transistors are available, we have
%
4x C 5y D 335
x C 14y D 50
 Section 3.4 ste s of near at ons 153

y One intersection y y
L1
L1 point (x0, y0) L2
L2 L1, L2
No intersection Infinitely many
point intersection points
x x x

FIGURE inear system FIGURE inear system FIGURE inear system

(one solution). (no solution). (infinitely many solutions).

We call this set of equations a system of two linear equations in the variables x and
y. The problem is to find values of x and y for which b th equations are true simu ta
ne us y. A pair .x; y/ of such values is called a s uti n of the system.
Note that a sing e solution is given by an
rdered pair of values. Since Equations (1) and (2) are linear, their graphs are straight lines call these lines
1 and 2 . Now, the coordinates of any point on a line satisfy the equation of that line
that is, they ma e the equation true. Thus, the coordinates of any point of intersection
of 1 and 2 will satisfy both equations. This means that a point of intersection gives a
solution of the system.
If 1 and 2 are drawn on the same plane, there are three situations that could occur:

1 and 2 may intersect at exactly one point, say, .a; b/. (See Figure 3.2 .) Thus,
the system has the solution x D a and y D b.
1 and 2 may be parallel and have no points in common. (See Figure 3.2 .) In this
case, there is no solution.
1 and 2 may be the same line. (See Figure 3.30.) Here the coordinates of any
point on the line are a solution of the system. Consequently, there are infinitely
many solutions.

ur main concern in this section is algebraic methods of solving a system of linear

equations. We will successively replace the system by other systems that have the same
solutions. eneralizing the terminology of Section 0.7, in the subsection titled Equiv-
alent Equations, we say that two systems are equivalent if their sets of solutions are
equal. The replacement systems have progressively more desirable forms for determin-
ing the solution. ore precisely, we see an equivalent system containing an equation
in which one of the variables does not appear. (In this case we say that the variable
has been e iminated.) In dealing with systems of inear equations, our passage from a
system to an equivalent system will always be accomplished by one of the following
procedures:

Interchanging two equations

ultiplying one equation by a nonzero constant
Replacing an equation by itself plus a multiple of another equation

We will return to these procedures in more detail in Chapter 6. For the moment, since
we will also consider nonlinear systems in this chapter, it is convenient to express our
solutions in terms of the very general principles of Section 0.7 that guarantee equiva-
lence of equations.
We will illustrate the elimination procedure for the system in the problem originally
posed:
%
4x C 5y D 335
x C 14y D 50

To begin, we will obtain an equivalent system in which x does not appear in one equa-
tion. First we find an equivalent system in which the coe cients of the x-terms in each
154 C nes Para o as an ste s

equation are the same except for their sign. ultiplying Equation (3) by that is,
multiplying both sides of Equation (3) by and multiplying Equation (4) by !4 gives
%
36x C 45y D 3015
!36x ! 56y D !3400
The left and right sides of Equation (5) are equal, so each side can be added to the
corresponding side of Equation (6). This results in
!11y D !3 5
which has only one variable, as planned. Solving gives
y D 35
so we obtain the equivalent system
%
36x C 45y D 3015
y D 35
Replacing y in Equation (7) by 35, we get
36x C 45.35/ D 3015
36x C 1575 D 3015
36x D 1440
x D 40
Thus, the original system is equivalent to
%
x D 40
y D 35
We can chec our answer by substituting x D 40 and y D 35 into b th of the rigina
equations. In Equation (3), we get 4.40/ C 5.35/ D 335 equivalently, 335 D 335.
In Equation (4), we get .40/ C 14.35/ D 50 equivalently, 50 D 50. Hence, the
solution is
x D 40 and y D 35
Each day the manager should plan to ma e 40 of model A and 35 of model . ur
procedure is referred to as elimination by addition. Although we chose to eliminate x
first, we could have done the same for y by a similar procedure.

E AM LE E A M
A L IT I
A computer consultant has Use elimination by addition to solve the system.
200,000 invested for retirement, part %
3x ! 4y D 13
at and part at . If the total
yearly income from the investments is 3y C 2x D 3
17,200, how much is invested at each S Aligning the x- and y-terms for convenience gives
rate
%
3x ! 4y D 13
2x C 3y D 3
To eliminate y, we multiply Equation ( ) by 3 and Equation (10) by 4:
%
x ! 12y D 3
x C 12y D 12
Adding Equation (11) to Equation (12) gives 17x D 51, from which x D 3. We have
the equivalent system
%
x ! 12y D 3
xD3
 Section 3.4 ste s of near at ons 155

Replacing x by 3 in Equation (13) results in

.3/ ! 12y D 3
y !12y D 12
2x + 3y = 3
3x - 4y = 13
y D !1

x
so the original system is equivalent to
(3, - 1)
%
y D !1
xD3

FIGURE inear system of The solution is x D 3 and y D !1. Figure 3.31 shows a graph of the system.
Example 1: one solution.
Now ork Problem 1 G
The system in Example 1,

%
3x ! 4y D 13
2x C 3y D 3

can be solved another way. We first choose one of the equations for example, Equa-
tion (15) and solve it for one variable in terms of the other, say x in terms of y. Hence,
Equation (15) is equivalent to 3x D 4y C 13, which is equivalent to

4 13
xD yC
3 3

and we obtain
( 4 13
xD yC
3 3
2x C 3y D 3

Substituting the right side of Equation (17) for x in Equation (1 ) gives

! "
4 13
2 yC C 3y D 3
3 3

Thus, x has been eliminated. Solving Equation (1 ), we have

26
yC C 3y D 3
3 3
y C 26 C y D clearing fractions
17y D !17
y D !1

Replacing y in Equation (17) by !1 gives x D 3, and the original system is equivalent to

%
xD3
y D !1

as before. This method is called elimination by substitution.

156 C nes Para o as an ste s

E AM LE M E S
A L IT I
Two species of deer, A and , living Use elimination by substitution to solve the system
in a wildlife refuge are given extra food %
x C 2y ! D 0
in the winter. Each wee , they receive
2 tons of food pellets and 4.75 tons of 2x C 4y C 4 D 0
hay. Each deer of species A requires S It is easy to solve the first equation for x. oing so gives the equivalent
4 pounds of the pellets and 5 pounds system
of hay. Each deer of species requires %
2 pounds of the pellets and 7 pounds of x D !2y C
hay. How many of each species of deer 2x C 4y C 4 D 0
will the food support so that all of the
food is consumed each wee Substituting !2y C for x in Equation (21) yields
2.!2y C / C 4y C 4 D 0
!4y C 16 C 4y C 4 D 0
The latter equation simplifies to 20 D 0. Thus, we have the system
%
x D !2y C
20 D 0
Since Equation (23) is ne er true, there is no solution of the original system. The
reason is clear if we observe that the original equations can be written in slope-intercept
form as
1
yD! xC4
2
and
1
yD! x!1
2
These equations represent straight lines having slopes of ! 12 but different y-intercepts
namely, 4 and !1. That is, they determine different parallel lines. (See Figure 3.32.)
y

4 x + 2y - 8 = 0

2x + 4y + 4 = 0 Distinct parallel lines

x
-1

A L IT I FIGURE inear system of Example 2: no solution.

Two species of fish, A and , are

raised in one pond at a fish farm where
Now ork Problem 9 G
they are fed two vitamin supplements.
Each day, they receive 100 grams of the E AM LE AL S I M S
first supplement and 200 grams of the
second supplement. Each fish of species Solve
8
A requires 15 mg of the first supple- < x C 5y D 2
ment and 30 mg of the second supple-
1 5
ment. Each fish of species requires : xC yD1
20 mg of the first supplement and 40 mg 2 2
of the second supplement. How many of S We begin by eliminating x from the second equation. ultiplying Equa-
each species of fish will the pond sup- tion (25) by !2, we have
port so that all of the supplements are %
x C 5y D 2
consumed each day
!x ! 5y D !2
 Section 3.4 ste s of near at ons 157

Adding Equation (26) to Equation (27) gives

%
x C 5y D 2
0D0

ecause Equation (2 ) is a ays true, any solution of Equation (2 ) is a solution of the

system. Now let us see how we can express our answer. From Equation (2 ), we have
x D 2 ! 5y, where y can be any real number, say, r. Thus, we can write x D 2 ! 5r.
The complete solution is
x D 2 ! 5r
yDr

where r is any real number. In this situation r is called a parameter, and we say that
we have a one-parameter family of solutions. Each value of r determines a particular
solution. For example, if r D 0, then x D 2 and y D 0 is a solution if r D 5, then
x D !23 and y D 5 is another solution. Clearly, the given system has infinitely many
solutions.
It is worthwhile to note that by writing Equations (24) and (25) in their slope-
intercept forms, we get the equivalent system
y 8̂
1 2
L1: x + 5y = 2 ˆ
<y D ! x C
L1, L2 5 5
L2: 12 x + 52 y = 1
ˆ
x :̂ y D ! 1 x C 2
5 5
in which both equations represent the same line. Hence, the lines coincide (Figure 3.33),
FIGURE inear system of and Equations (24) and (25) are equivalent. The solution of the system consists of the
Example 3: infinitely many solutions. coordinate pairs of all points on the line x C 5y D 2, and these points are given by our
parametric solution.
Now ork Problem 13 G
E AM LE M

A chemical manufacturer wishes to fill an order for 500 liters of a 25 acid solution.
(Twenty-five percent by volume is acid.) If solutions of 30 and 1 are available in
stoc , how many liters of each must be mixed to fill the order
S et x and y be the number of liters of the 30 and 1 solutions, respec-
tively, that should be mixed. Then,

x C y D 500

To help visualize the situation, we draw the diagram in Figure 3.34. In 500 liters of a
25 solution, there will be 0:25.500/ D 125 liters of acid. This acid comes from two
sources: 0.30x liters of it come from the 30 solution, and 0.1 y liters of it come from
the 1 solution. Hence,

0:30x C 0:1 y D 125

These two equations form a system of two linear equations in two un nowns. Solving
the first for x gives x D 500 ! y. Substituting in the second gives

0:30.500 ! y/ C 0:1 y D 125

1 1
Solving this equation for y, we find that y D 20 3
liters. Thus, x D 500!20 3
D 2 1 23
liters. (See Figure 3.35.)
158 C nes Para o as an ste s

500 Liters
500 0.30x + 0.18y = 125

x Liters y Liters
+ =
x + y = 500
0.30x 0.18y 0.25(500)
is acid. is acid. is acid.
0 500
30% Solution 18% Solution 25% Solution 0

FIGURE ixture problem. FIGURE raph for Example 4.

Now ork Problem 25 G

T S
The methods used in solving a two-variable system of linear equations can be used to
solve a three-variable system of linear equations. A general linear equation in the
three variables x y and is an equation having the form

Ax C By C Cz D

where A, B, C, and are constants and A, B, and C are not all zero. For example,
2x ! 4y C z D 2 is such an equation. eometrically, a general linear equation in three
variables represents a p ane in space, and a solution to a system of such equations is the
intersection of planes. Example 5 shows how to solve a system of three linear equations
in three variables.

E AM LE S T L S
A L IT I
A coffee shop specializes in blend- Solve
ing gourmet coffees. From type A, 8̂
2x C y C z D 3
type , and type C coffees, the owner <
wants to prepare a blend that will sell for !x C 2y C 2z D 1
.50 for a 1-pound bag. The cost per :̂
pound of these coffees is 12, , and x ! y ! 3z D !6
7, respectively. The amount of type
is to be twice the amount of type A. How S This system consists of three linear equations in three variables. From Equa-
much of each type of coffee will be in tion (32), x D y C 3z ! 6. y substituting for x in Equations (30) and (31), we obtain
the final blend 8̂
< 2.y C 3z ! 6/ C y C z D 3
!.y C 3z ! 6/ C 2y C 2z D 1
:̂ x D y C 3z ! 6

Simplifying gives
8̂
< 3y C 7z D 15
y ! z D !5
:̂
x D y C 3z ! 6

Note that x does not appear in Equations (33) and (34). Since any solution of the original
system must satisfy Equations (33) and (34), we will consider their solution first:
%
3y C 7z D 15
y ! z D !5

From Equation (34), y D z ! 5. This means that we can replace Equation (33) by

3.z ! 5/ C 7z D 15 that is; zD3

 Section 3.4 ste s of near at ons 159

Since z is 3, we can replace Equation (34) with y D !2. Hence, the previous system is
equivalent to
%
zD3
y D !2
The original system becomes
8
<z D 3
y D !2
:
x D y C 3z ! 6
from which x D 1. The solution is x D 1, y D !2, and z D 3, which you should verify.
Now ork Problem 15 G
ust as a two-variable system may have a one-parameter family of solutions, a
three-variable system may have a one-parameter or a two-parameter family of solu-
tions. The next two examples illustrate.

E AM LE F S

Solve 8̂
< x ! 2y D 4
2x ! 3y C 2z D !2
:̂
4x ! 7y C 2z D 6
S Note that since Equation (35) can be written x ! 2y C 0z D 4, we can view
Equations (35) to (37) as a system of three linear equations in the variables x, y, and z.
From Equation (35), we have x D 2y C 4. Using this equation and substitution, we can
eliminate x from Equations (36) and (37):
8̂
< x D 2y C 4
2.2y C 4/ ! 3y C 2z D !2
:̂ 4.2y C 4/ ! 7y C 2z D 6

which simplifies to give

8̂
< x D 2y C 4
y C 2z D !10
:̂
y C 2z D !10
ultiplying Equation (40) by !1 gives
8̂
< x D 2y C 4
y C 2z D !10
:̂ !y ! 2z D 10

Adding the second equation to the third yields

8
< x D 2y C 4
y C 2z D !10
:
0D0
Since the equation 0 D 0 is always true, the system is equivalent to
%
x D 2y C 4
y C 2z D !10

Solving Equation (42) for y, we have

y D !10 ! 2z
160 C nes Para o as an ste s

which expresses y in terms of z. We can also express x in terms of z. From

Equation (41),

x D 2y C 4
D 2.!10 ! 2z/ C 4
D !16 ! 4z

Thus, we have
%
x D !16 ! 4z
y D !10 ! 2z
Since no restriction is placed on z, this suggests a parametric family of solutions. Setting
z D r, we have the following family of solutions of the given system:
x D !16 ! 4r
y D !10 ! 2r
zDr
ther parametric representations of the where r can be any real number. We see, then, that the given system has infinitely
solution are possible. many solutions. For example, setting r D 1 gives the particular solution x D !20,
y D !12, and z D 1. There is nothing special about the name of the parameter. In fact,
since z D r, we could consider z to be the parameter.
Now ork Problem 19 G
E AM LE T F S

Solve the system

%
x C 2y C z D 4
2x C 4y C 2z D
S This is a system of two linear equations in three variables. We will eliminate
x from the second equation by first multiplying that equation by ! 12 :
%
x C 2y C z D 4
!x ! 2y ! z D !4
Adding the first equation to the second gives
%
x C 2y C z D 4
0D0
From the first equation, we obtain
x D 4 ! 2y ! z
Since no restriction is placed on either y or z, they can be arbitrary real numbers, giving
us a two-parameter family of solutions. Setting y D r and z D s, we find that the
solution of the given system is
x D 4 ! 2r ! s
yDr
zDs
where r and s can be any real numbers. Each assignment of values to r and s results
in a solution of the given system, so there are infinitely many solutions. For example,
letting r D 1 and s D 2 gives the particular solution x D 0; y D 1, and z D 2. As in the
last example, there is nothing special about the names of the parameters. In particular,
since y D r and z D s, we could consider y and z to be the two parameters.
Now ork Problem 23 G
 Section 3.4 ste s of near at ons 161

R BLEMS
In Pr b ems s e the systems a gebraica y
% %
x C 4y D 3 4x C 2y D
3x ! 2y D !5 5y ! 4x D 5
% %
2x C 3y D 1 3x C y D 13
x C 2y D 0 !x C 7y D 3
% %
uC D5 2p C q D 16
u! D7 3p C 3q D 33
% %
x ! 2y D !7 4x C 12y D 12
5x C 3y D ! 2x C 4y D 12
%
x C 3y C 2 D !x C 2y C 3
3x C y C 1 D x C
(
5x C 7y C 2 D y ! 4x C 6
21
2
x ! 43 y ! 11
4
D 32 x C 23 y C 5
4
(2 (1
3
x C 12 y D 2 2
z! 1
4
D 1
6
3 5 1 1 1
xC 6
y D ! 11
2 2
z C 4
D 6 Speed of Raft n a trip on a raft, it too 12 hour to travel
% %
2p C 3q D 5 3x ! 2y D 5 10 miles downstream. The return trip too 34 hour. Find the speed
10p C 15q D 25 !6x C 4y D 10 of the raft in still water and the speed of the current.
8 8
< 2x C y C 6z D 3 < x C y C z D !1 Furniture Sales A manufacturer of dining-room sets
x ! y C 4z D 1 3x C y C z D 1 produces two styles: early American and contemporary. From
: 3x C 2y ! 2z D 2 : 4x ! 2y C 2z D 0 past experience, management has determined that 20 more of
8 8 the early American styles can be sold than the contemporary
< x C 4y C 3z D 10 < x C 2y C z D 4 styles. A profit of 250 is made on each early American set sold,
4x C 2y ! 2z D !2 2x ! 4y ! 5z D 26 whereas a profit of 350 is made on each contemporary set. If, in
: 3x ! y C z D 11 : 2x C 3y C z D 10
the forthcoming year, management desires a total profit of
% % 130,000, how many units of each style must be sold
2x C 4z D 0 2y C 3z D 1
y!z D 3 3x ! 4z D 0 Survey National Surveys was awarded a contract to
8 8 perform a product-rating survey for Crispy Crac ers. A total of
< x ! y C 2z D 0 < x ! 2y ! z D 0 250 people were interviewed. National Surveys reported that
2x C y ! z D 0 2x ! 4y ! 2z D 0 62.5 more people li ed Crispy Crac ers than disli ed them.
: x C 2y ! 3z D 0 : !x C 2y C z D 0
However, the report did not indicate that 16 of those
% %
x ! 3y C z D 5 7x C y C z D 5 interviewed had no comment. How many of those surveyed li ed
!2x C 6y ! 2z D !10 6x C y C z D 3 Crispy Crac ers How many disli ed them How many had no
comment
Mixture A chemical manufacturer wishes to fill an order
for 00 gallons of a 25 acid solution. Solutions of 20 and 35 Equali ing Cost United Products Co. manufactures
are in stoc . How many gallons of each solution must be mixed to calculators and has plants in the cities of Exton and Whyton. At
fill the order the Exton plant, fixed costs are 5000 per month, and the cost of
producing each calculator is 5.50. At the Whyton plant, fixed
Mixture A gardener has two fertilizers that contain costs are 6000 per month, and each calculator costs 4.50 to
different concentrations of nitrogen. ne is 3 nitrogen and the produce. Next month, United Products must produce 1000
other is 11 nitrogen. How many pounds of each should she mix calculators. How many must be made at each plant if the total cost
to obtain 20 pounds of a concentration at each plant is to be the same
Fabric A textile mill produces fabric made from different Coffee Blending The oonloon coffee chain used to retail
fibers. From cotton, polyester, and nylon, the owners want to three types of coffee that sold for 12.00, 13.00 and 15.00 per
produce a fabric blend that will cost 3.25 per pound to ma e. pound. To simplify operations they decide to ma e a blend that
The cost per pound of these fibers is 4.00, 3.00, and 2.00, they can sell for 14.00 per pound and that uses the same amounts
respectively. The amount of nylon is to be the same as the amount of the two cheaper coffees. How much of each type are needed to
of polyester. How much of each fiber will be in the final fabric ma e a 100 pound batch of the blend
axes A company has taxable income of 75 ,000. The Commissions A company pays its salespeople on a basis
federal tax is 35 of that portion left after the state tax has been of a certain percentage of the first 100,000 in sales, plus a
paid. The state tax is 15 of that portion left after the federal tax certain percentage of any amount over 100,000 in sales. If
has been paid. Find the federal and state taxes. one salesperson earned 500 on sales of 175,000 and another
Airplane Speed An airplane travels 1500 m in 2 h with salesperson earned 14, 00 on sales of 2 0,000, find the two
the aid of a tailwind. It ta es 2 h, 30 min, for the return trip, ying percentages.
against the same wind. Find the speed of the airplane in still air
and the speed of the wind.
162 C nes Para o as an ste s

early Profits In news reports, profits of a company this Investments A total of 35,000 was invested at three
year . / are often compared with those of last year . /, but actual interest rates: 7, , and . The interest for the first year was
values of and are not always given. This year, a company had 2 30, which was not reinvested. The second year the amount
profits of 25 million more than last year. The profits were up originally invested at earned 10 instead, and the other rates
30 . etermine and from these data. remained the same. The total interest the second year was 2 60.
Fruit Pac aging The lo t .com rganic Produce How much was invested at each rate
Company has 3600 lb of onut Peaches that it is going to pac age iring or ers A company pays s illed wor ers in its
in boxes. Half of the boxes will be loose filled, each containing assembly department 16 per hour. Semis illed wor ers in that
20 lb of peaches, and the others will be pac ed with -lb department are paid .50 per hour. Shipping cler s are paid
clamshells ( ip-top plastic containers), each containing 2.2 lb of 10 per hour. ecause of an increase in orders, the company
peaches. etermine the number of boxes and the number of needs to hire a total of 70 wor ers in the assembly and shipping
clamshells that are required. departments. It will pay a total of 725 per hour to these
Investments A person made two investments, and the employees. ecause of a union contract, twice as many
percentage return per year on each was the same. f the total semis illed wor ers as s illed wor ers must be employed. How
amount invested, 40 of it minus 1000 was invested in one many semis illed wor ers, s illed wor ers, and shipping cler s
venture, and at the end of 1 year the person received a return of should the company hire
400 from that venture. If the total return after 1 year was 1200, Solvent Storage A 10,000-gallon railroad tan car is to be
find the total amount invested. filled with solvent from two storage tan s, A and B. Solvent from
Production Run The Roc ywood arden Furniture A is pumped at the rate of 25 gal min. Solvent from B is pumped
company ma es three products: chairs, side tables, and coffee at 35 gal min. Usually, both pumps operate at the same time.
tables. A chair requires 10 units of wood, 3 units of bolts, and However, because of a blown fuse, the pump on A is delayed 5
3 units of washers. A side table requires 4 units of wood, 1 unit of minutes. oth pumps finish operating at the same time. How
bolts, and 1 unit of washers. A coffee table requires units of many gallons from each storage tan will be used to fill the car
wood, 2 units of bolts, and 3 units of washers. The company has
in stoc 1 40 units of wood, 510 units of bolts, and 5 0 units of
washers. Roc ywood is going out of business and wants to
use up all its stoc . To do this how many chairs, side tables,
and coffee tables should Roc ywood ma e in its final
production run

Objective N S
o se s st t t on to so e non near A system of equations in which at least one equation is not linear is called a nonlinear
s ste s
system. We can often solve a nonlinear system by substitution, as was done with linear
systems. The following examples illustrate.

E AM LE S N S

Solve
%
x2 ! 2x C y ! 7 D 0
3x ! y C 1 D 0
S

S If a nonlinear system contains a linear equation, we usually solve the

linear equation for one variable and substitute for that variable in the other equation.

Solving Equation (2) for y gives

y D 3x C 1
Substituting into Equation (1) and simplifying, we have
x2 ! 2x C .3x C 1/ ! 7 D 0
x2 C x ! 6 D 0
.x C 3/.x ! 2/ D 0
x D !3 or x D 2
 Section 3.5 on near ste s 163

y If x D !3, then Equation (3) implies that y D ! if x D 2, then y D 7. The logic

3x - y + 1 = 0 behind our calculations shows that if a pair .a; b/ satisfies the nonlinear system of this
(2, 7)
example then either .a; b/ D .!3; ! / or .a; b/ D .2; 7/. We also need to show that if
either .a; b/ D .!3; ! / or .a; b/ D .2; 7/ then .a; b/ satisfies the nonlinear system.
This last step amounts to so-called eri cati n that each of .!3; ! / and .2; 7/ do
satisfy the system. It is n t pti na (although verification is often left to the reader).
x2 - 2x + y - 7 = 0 Here is a sample. The calculations .!3/2 ! 2.!3/ C .! / ! 7 D C 6 ! ! 7 D 0
x and 3.!3/ ! .! / C 1 D ! C C 1 D 0 show that .!3; ! / is a solution. Similarly,
.2/2 ! 2.2/ C .7/ ! 7 D 4 ! 4 C 7 ! 7 D 0 and 3.2/ ! .7/ C 1 D 6 ! 7 C 1 D 0 show
that .2; 7/ is a solution.
The solution pairs .!3; ! / and .2; 7/ can be seen geometrically on the graph of
the system in Figure 3.36. Notice that the graph of Equation (1) is a parabola and the
(-3, -8) graph of Equation (2) is a line. The solutions are the intersection points .!3; ! / and
.2; 7/.

FIGURE Nonlinear system of

Now ork Problem 1 G
equations. E AM LE S N S
This example illustrates the need for
verification of all possible solutions. Solve ( p
yD xC2
xCyD4
S Solving the second equation, which is linear, for y gives
yD4!x
Substituting into the first equation yields
p
6 4!xD xC2
16 ! x C x2 D x C 2 squaring both sides
2
x ! x C 14 D 0

-6 10
.x ! 2/.x ! 7/ D 0
Intersection
X=2 Y=2 Thus, x D 2 or x D 7. From Equation (4), if x D 2, then y D 2 if x D 7, then y D !3.
At this point we now that .2; 2/ and .7; !3/ are the only possible solution pairs.
-2 p p
The calculations .2/ C 2 D 4 D 2 D .2/ and .2/ C .2/ D 2 C 2 D 4 show that
FIGURE Nonlinear system of
Example 2. the pair .2; 2/ is a solution.
p p
The calculation .7/ C 2 D D 3 ¤ .!3/ shows that the pair .7; !3/ does
not satisfy the first equation of the system and this is enough to declare that .7; !3/ is
not a solution of the system. (The fact that .7; !3/ does satisfy the second equation is
now irrelevant. A solution must satisfy a the equations of a system.
The graph of the system bears out the fact that there is only one point of intersection
of the curves defined by the equations in the system. (See Figure 3.37.)
Now ork Problem 13 G

R BLEMS
In Pr b ems s e the gi en n n inear system
% % % %
yDx !2
y D x3 y D 4 C 2x ! x2 x2 C 4x ! y D !4
2x C y D 3 x ! 2y D 0 y D x2 C 1 y ! x2 ! 4x C 3 D 0
% 2 % 2 % p %
p D 5!q y ! x2 D 2 pD q y D 1=x
p D qC1 x ! y D 14 p D q2 x ! y D !1
% % 2 % 2 %
y D x2 p !qC1 D 0 x D y2 C 13 x2 C y2 C 2xy D 1
x!y D 1 5q ! 3p ! 2 D 0 y D x2 ! 15 2x ! y D 2
164 C nes Para o as an ste s

8̂
x2 raphically solve the system
% ˆ
<y D C1 (
x D yp
C1 x!1
2y D x3
yD2 xC2 ˆ 1
:̂ y D y D ! x2
x!1
angents In calculus, the notion of a tangent to a curve to one-decimal-place accuracy.
y D f .x/ at a point .a; f .a// on the curve is of great importance. raphically solve the system
(Roughly spea ing, such a tangent is a line incident with the point
.a; f .a// but with no other points on the curve.) Find the tangent %
y D x2 ! 2x C 1
line to the curve y D x2 at the point .2; 4/ (which is clearly on
y D x3 C x2 ! 2x C 3
y D x2 ).
Awning The shape of a decorative awning over a storefront to one-decimal-place accuracy.
can be described by the function y D 0:06x2 C 0:012x C , where
y is the height of the edge of the awning (in feet) above the raphically solve the system
sidewal and x is the distance (in feet) from the center of the %
store s doorway. A vandal po es a stic through the awning, y D x3 C 6x C 2
piercing it in two places. The position of the stic can be y D 2x C 3
described by the function y D 0: 12x C 5. Where are the holes in
the awning caused by the vandal to one-decimal-place accuracy.
raphically determine how many solutions there are to the In Pr b ems graphica y s e the equati n by treating it
system 8 as a system R und ans ers t t decima p aces
< 1
yD 2 0: x2 C 2x D 6 where x # 0
x p
: ! xC3D1!x x3 ! 3x2 D x !
y D 2 ! x2

Objective A S E
o so e s ste s escr n
e r an rea e en o nts E
p
Recall from Section 3.2 that an equation that relates price per unit and quantity demanded
12 (supplied) is called a demand equati n supp y equati n . Suppose that, for product ,
the demand equation is
(Dollars)

(540, 9)
8
1
(1080, 6) pD! q C 12
1 0
4
and the supply equation is
q 1
500 1000 1500 pD qC
(Units/week) 300
Demand equation: p = - 180 q + 12
1
where q, p # 0. The corresponding demand and supply curves are the lines in
Figures 3.3 and 3.3 , respectively. In analyzing Figure 3.3 , we see that consumers
FIGURE emand curve. will purchase 540 units per wee when the price is per unit, 10 0 units when the
p price is 6, and so on. Figure 3.3 shows that when the price is per unit producers
will place 300 units per wee on the mar et, at 10 they will supply 600 units, and
12
(600, 10)
so on.
When the demand and supply curves of a product are represented on the same
(Dollars)

8 (300, 9)
coordinate plane, the point .m; n/ where the curves intersect is called the point of equi
librium. (See Figure 3.40.) The price n, called the equilibrium price, is the price at
4
which consumers will purchase the same quantity of a product that producers wish to
sell at that price. In short, n is the price at which stability in the producer consumer
relationship occurs. The quantity m is called the equilibrium quantity.
500 1000 1500
q To determine precisely the equilibrium point, we solve the system formed by the
(Units/week) supply and demand equations. et us do this for our previous data, namely, the system
1 8̂
Supply equation: p = 300 q+8 1
ˆ
<p D ! q C 12 demand equation
FIGURE Supply curve.
1 0
ˆ
:̂ p D 1 q C supply equation
300
 Section 3.6 cat ons of ste s of at ons 165

p 1
p= 300 q+8

p 12
Supply curve Equilibrium
price 9.50 (450, 9.50) Equilibrium point

Equilibrium price = n
8
1
p = - 180 q + 12
n (m, n) Equilibrium point 4

Demand curve
q
450 1000
q
m
Equilibrium quantity = m Equilibrium quantity

FIGURE Equilibrium. FIGURE Equilibrium.

1
y substituting q C for p in the demand equation, we get
300
1 1
qC D! q C 12
300 1 0
! "
1 1
C qD4
300 1 0
q D 450 equilibrium quantity
Thus,
1
pD .450/ C
300
D :50 equilibrium price
and the equilibrium point is (450, .50). Therefore, at the price of .50 per unit, manu-
facturers will produce exactly the quantity (450) of units per wee that consumers will
purchase at that price. (See Figure 3.41.)

E AM LE T E E

et p D q C 50 be the supply equation for a manufacturer s product, and suppose

100
7
the demand equation is p D ! q C 65.
100
a If a tax of 1.50 per unit is to be imposed on the manufacturer, how will the original
equilibrium price be affected if the demand remains the same
S efore the tax, the equilibrium price is obtained by solving the system
8̂
ˆ
<p D q C 50
100
ˆ
:̂ p D ! 7 q C 65
100
y substitution,
7
! q C 65 D q C 50
100 100
15
15 D q
100
100 D q
166 C nes Para o as an ste s

and

pD .100/ C 50 D 5
100
Thus, 5 is the original equilibrium price. efore the tax, the manufacturer supplies
q units at a price of p D q C 50 per unit. After the tax, he will sell the same q units
100 ! "
for an additional 1.50 per unit. The price per unit will be q C 50 C 1:50, so
100
the new supply equation is

pD q C 51:50
100
Solving the system
8̂
ˆ
<p D q C 51:50
100
ˆ
:̂ p D ! 7 q C 65
100
will give the new equilibrium price:
7
q C 51:50 D ! q C 65
100 100
15
q D 13:50
100
qD 0

pD . 0/ C 51:50 D 5 :70
100
The tax of 1.50 per unit increases the equilibrium price by 0.70. (See Figure 3.42.)
Note that there is also a decrease in the equilibrium quantity from q D 100 to q D 0,
because of the change in the equilibrium price. (In the problems, you are as ed to find
the effect of a subsidy given to the manufacturer, which will reduce the price of the
product.)
p

70
Supply curve after tax
Supply curve before tax
(90, 58.70)

60
(100, 58)

51.5
50
Demand curve

q
100 200

FIGURE Equilibrium before and after tax.

b etermine the total revenue obtained by the manufacturer at the equilibrium point
both before and after the tax.
S If q units of a product are sold at a price of p dollars each, then the total
revenue is given by

yTR D pq
 Section 3.6 cat ons of ste s of at ons 167

efore the tax, the revenue at (100,5 ) is (in dollars)

yTR D .5 /.100/ D 5 00
After the tax, it is
yTR D .5 :70/. 0/ D 52 3
which is a decrease.
Now ork Problem 15 G
E AM LE E N

Find the equilibrium point if the supply and demand equations of a product are
q 000
pD C 10 and p D , respectively.
40 q
S Here the demand equation is not linear. Solving the system
8̂ q
ˆ
< p D 40 C 10
ˆ 000
:̂ p D
q
by substitution gives
000 q
D C 10
q 40
320;000 D q2 C 400q multiplying both sides by 40q
q2 C 400q ! 320;000 D 0
.q C 00/.q ! 400/ D 0
q D ! 00 or q D 400
We disregard q D ! 00, since q represents quantity. Choosing q D 400, we have
p D . 000=400/ D 20, so the equilibrium point is (400,20). (See Figure 3.43.)

p
Demand
8000
p= q

Supply
q
p= + 10
40
(400, 20)
20
10
q
80 160 240 320 400

FIGURE Equilibrium with nonlinear demand.

G
B E
Suppose a manufacturer produces product A and sells it at per unit. Then the total
revenue yTR received (in dollars) from selling q units is
yTR D q total revenue
168 C nes Para o as an ste s

y (Revenue, costs
in dollars) Break-even point

22
yTC = 9
q + 5000
(m, n)
5000

yTR = 8q

q
500 1000

FIGURE rea -even chart.

The difference between the total revenue received for q units and the total cost of
q units is the manufacturer s profit:

profit D total revenue ! total cost

(If profit is negative, then we have a loss.) Total cost, yTC , is the sum of total variable
costs yVC and total fixed costs yFC :

yTC D yVC C yFC

Fixed costs are those costs that, under normal conditions, do not depend on the level
of production that is, over some period of time they remain constant at all levels of
output. (Examples are rent, o cers salaries, and normal maintenance.) Variable costs
are those costs that vary with the level of production (such as the cost of materials,
labor, maintenance due to wear and tear, etc.). For q units of product A, suppose that

yFC D 5000 fixed cost

22
and yVC D q variable cost

Then
22
yTC D q C 5000 total cost

The graphs of total cost and total revenue appear in Figure 3.44. The horizontal
axis represents the level of production, q, and the vertical axis represents the total dollar
value, be it revenue or cost. The brea even point is the point at which total revenue
equals total cost .TR D TC/. It occurs when the levels of production and sales result
in neither a profit nor a loss to the manufacturer. In the diagram, called a break e en
chart the brea -even point is the point .m; n/ at which the graphs of yTR D q and
yTC D 22 q C 5000 intersect. We call m the brea even quantity and n the brea even
revenue. When total cost and revenue are linearly related to output, as in this case, for
any production level greater than m, total revenue is greater than total cost, resulting
in a profit. However, at any level less than m units, total revenue is less than total cost,
resulting in a loss. At an output of m units, the profit is zero. In the following example,
we will examine our data in more detail.

E AM LE B E L

A manufacturer sells a product at per unit, selling all that is produced. Fixed cost is
5000 and variable cost per unit is 22 (dollars).

a Find the total output and revenue at the brea -even point.
 Section 3.6 cat ons of ste s of at ons 169
22
S At an output level of q units, the variable cost is yVC D q and the total
revenue is yTR D q. Hence,
yTR D q
22
yTC D yVC C yFC D q C 5000

At the brea -even point, total revenue equals total cost. Thus, we solve the system
formed by the foregoing equations. Since
yTR D yTC
8000 we have
22
Total cost qD q C 5000
50
Total revenue
q D 5000
q D 00
0 1000
0 Hence, the desired output is 00 units, resulting in a total revenue (in dollars) of
FIGURE Equilibrium point yTR D . 00/ D 7200
( 00, 7200).
(See Figure 3.45.)
b Find the profit when 1 00 units are produced.
S Since profit D total revenue ! total cost, when q D 1 00 we have
# $
22
yTR ! yTC D .1 00/ ! .1 00/ C 5000

D 5000
The profit when 1 00 units are produced and sold is 5000.
c Find the loss when 450 units are produced.
S When q D 450,
# $
22
yTR ! yTC D .450/ ! .450/ C 5000 D !2500

A loss of 2500 occurs when the level of production is 450 units.

d Find the output required to obtain a profit of 10,000.
S In order to obtain a profit of 10,000, we have
profit D total revenue ! total cost
! "
22
10;000 D q ! q C 5000

50
15;000 D q
q D 2700
Thus, 2700 units must be produced.
Now ork Problem 9 G

E AM LE B E

etermine the brea -even quantity of anufacturing Co., given the following
data: total fixed cost, 1200 variable cost per unit, 2 total revenue for selling q units,
p
yTR D 100 q.
170 C nes Para o as an ste s

S For q units of output,

p
yTR D 100 q
yTC D 2q C 1200
Equating total revenue to total cost gives
p
100 q D 2q C 1200
p
50 q D q C 600 dividing both sides by 2
Squaring both sides, we have
2500q D q2 C 1200q C .600/2
y
yTC = 2q + 1200 0 D q2 ! 1300q C 360; 000
3000
y the quadratic formula,
p
1300 ˙ 250;000
2000 Break-even qD
points 2
1300 ˙ 500
yTR = 100 q qD
2
q D 400 or q D 00
Although both q D 400 and q D 00 are brea -even quantities, observe in Figure 3.46
q
400 900 that when q > 00, total cost is greater than total revenue, so there will always be
a loss. This occurs because here total revenue is not linearly related to output. Thus,
FIGURE Two brea -even points. producing more than the brea -even quantity does not necessarily guarantee a profit.
Now ork Problem 21 G

R BLEMS
In Pr b ems y u are gi en a supp y equati n and a demand yTR D 2q yTR D 0:25q
equati n f r a pr duct If p represents price per unit in d ars yTC D 5q C 300 yTC D 0:16q C 360
and q represents the number f units per unit f time nd the 00
equi ibrium p int In Pr b ems and sketch the system yTR D 0 ! yTR D 0:1q2 C q
qC3 yTC D 3q C 400
3 5
Supply: p D 100
q C 5, emand: p D ! 100 q C 11 yTC D 1:1q C 37:3
Supply: p D 1
1500
q 1
C 4, emand: p D ! 2000 qC Business Supply and demand equations for a certain
product are
Supply: 35q ! 2p C 250 D 0,
emand: 65q C p ! 537:5 D 0 3q ! 200p C 1 00 D 0
Supply: 246p ! 3:25q ! 2460 D 0,
emand: 410p C 3q ! 14;452:5 D 0 and
2
Supply: p D 2q C 20, emand: p D 200 ! 2q 3q C 100p ! 1 00 D 0
Supply: p D q2 C 5q C 100,
emand: p D 700 ! 5q ! q2 respectively, where p represents the price per unit in dollars and
p q represents the number of units sold per time period.
Supply: p D q C 10, emand: p D 20 ! q
a Find the equilibrium price algebraically, and derive it
2240
Supply: p D 14 q C 6, emand: p D graphically.
q C 12 b Find the equilibrium price when a tax of 27 cents per unit is
imposed on the supplier.
In Pr b ems y R represents t ta re enue in d ars and y C
represents t ta c st in d ars f r a manufacturer If q represents Business A manufacturer of a product sells all that is
b th the number f units pr duced and the number f units s d produced. The total revenue is given by yTR D :5q, and the total
nd the break e en quantity Sketch a break e en chart in cost is given by yTC D q C 500, where q represents the number
Pr b ems and of units produced and sold.
a Find the level of production at the brea -even point, and draw
yTR D 4q yTR D 14q the brea -even chart.
yTC D 2q C 5000 yTC D 40
3
q C 1200 b Find the level of production at the brea -even point if the
xed cost increases by 10 .
 Section 3.6 cat ons of ste s of at ons 171

Business A manufacturer sells a product at .35 per unit, onroe Forging Co. ou are specifically requested to determine
selling all produced. The fixed cost is 2116, and the variable cost the following:
is 7.20 per unit. At what level of production will there be a profit a Net profit or loss based on the pricing proposal
of 4600 At what level of production will there be a loss of b Unit sales volume under the proposed price that is required to
1150 At what level of production will the brea -even point ma e the same 40,000 profit that is now earned at the current
occur price and unit sales volume
Business The mar et equilibrium point for a product Use the following data in your analysis:
occurs when 13,500 units are produced at a price of 4.50 per
unit. The producer will supply no units at 1, and the consumers Current Proposal of Vice
will demand no units at 20. Find the supply and demand perations President of Sales
equations if they are both linear.
Business A manufacturer of a children s toy will brea Unit price 2.50 2.00
even at a sales volume of 200,000. Fixed costs are 40,000, and Unit sales volume 200,000 units 2 0,000 units
each unit of output sells for 5. etermine the variable cost Variable cost
per unit.
Total 350,000 4 0,000
Business The igfoot Sandal Co. manufactures sandals for Per unit 1.75 1.75
which the material cost is 0. 5 per pair and the labor cost is
0. 6 per pair. Additional variable costs amount to 0.32 per pair. Fixed cost 110,000 110,000
Fixed costs are 70,500. If each pair sells for 2.63, how many Profit 40,000
pairs must be sold for the company to brea even

Business Suppose products A and have demand and

supply equations that are related to each other. If qA and q are the
quantities produced and sold of A and , respectively, and pA and
p are their respective prices, the demand equations are

qA D 7 ! pA C p

and

q D 24 C pA ! p

and the supply equations are

Business a Find the brea -even points for Pear-shaped
Corp, which sells all it produces, if the variable cost per unit is qA D !3 C 4pA ! 2p
p
1 3, fixed costs are 2 3 and yTR D q, where q is the number of
thousands of units of output produced. and
b raph the total revenue curve and the total cost curve in the
same plane. q D !5 ! 2pA C 4p
c Use your answer in a and examination of the curves in b to Eliminate qA and q to get the equilibrium prices.
report the quantity interval in which maximum profit occurs.
Business The supply equation for a product is
Business A company has determined that the demand
equation for its product is p D 1000=q, where p is the price per p D q2 ! 4
unit for q units produced and sold in some period. etermine the
quantity demanded when the price per unit is a 4, b 2, and and the demand equation is
c 0.50. For each of these prices, determine the total revenue
4
that the company will receive. What will be the revenue pD
regardless of the price int Find the revenue when the price q!2
is p dollars. Here p represents price per unit in dollars and q > 2 represents
Business Using the data in Example 1, determine how the number of units (in thousands) per unit time. raph both
original equilibrium price will be affected if the company is given equations and use the graphs to determine the equilibrium
a government subsidy of 1.50 per unit. quantity to one decimal place.
Business The onroe Forging Company sells a Business For a manufacturer, the total-revenue equation is
corrugated steel product to the Standard anufacturing Company p
and is in competition on such sales with other suppliers of the yTR D 20:5 q C 4 ! 41
Standard anufacturing Co. The vice president of sales of
and the total-cost equation is
onroe Forging Co. believes that by reducing the price of the
product, a 40 increase in the volume of units sold to the yTC D 0:02q3 C 10:4;
Standard anufacturing Co. could be secured. As the manager of
the cost and analysis department, you have been as ed to analyze where q represents (in thousands) both the number of units
the proposal of the vice president and submit your produced and the number of units sold. raph a brea -even chart
recommendations as to whether it is financially beneficial to the and find the brea -even quantity.
172 C nes Para o as an ste s

Chapter 3 Review
I T S E
S Lines
slope of a line point-slope form slope-intercept form Ex. 1, p. 133
general linear equation in x and y linearly related Ex. 7, p. 136
S Applications and Linear Functions
demand equation demand curve supply equation supply curve Ex. 2, p. 141
Ex. 3, p. 142
S Quadratic Functions
parabola axis of symmetry vertex Ex. 1, p. 147
S Systems of Linear Equations
system of equations equivalent systems elimination by addition Ex. 1, p. 154
elimination by substitution parameter Ex. 3, p. 156
general linear equation in x, y, and z Ex. 5, p. 15
S onlinear Systems
nonlinear system Ex. 1, p. 162
S Applications of Systems of Equations
point of equilibrium equilibrium price equilibrium quantity Ex. 1, p. 165
brea -even point brea -even quantity brea -even revenue Ex. 3, p. 16

S
The orientation of a nonvertical line is characterized by the In economics, supply functions and demand functions
slope of the line given by have the form p D f.q/ and play an important role. Each
y2 ! y1 gives a correspondence between the price p of a product and
mD the number of units q of the product that manufacturers (or
x2 ! x1
consumers) will supply (or purchase) at that price during
where .x1 ; y1 / and .x2 ; y2 / are two different points on the line. some time period.
The slope of a vertical line is not defined, and the slope of a A quadratic function has the form
horizontal line is zero. ines rising from left to right have
f.x/ D ax2 C bx C c .a ¤ 0/
positive slopes lines falling from left to right have nega-
tive slopes. Two lines are parallel if and only if they have The graph of f is a parabola that opens upward if a > 0 and
the same slope or both are vertical. Two lines with nonzero downward if a < 0. The vertex is
slopes m1 and m2 are perpendicular to each other if and only ! ! ""
1 b b
if m1 D ! . Any horizontal line and any vertical line are ! ;f !
m2 2a 2a
perpendicular to each other.
and the y-intercept is c. The axis of symmetry and the x- and
asic forms of equations of lines are as follows:
y-intercepts, are useful in s etching the graph.
y ! y1 D m.x ! x1 / point-slope form A system of linear equations can be solved with the
method of elimination by addition or elimination by substi-
y D mx C b slope-intercept form tution. A solution may involve one or more parameters. Sub-
xDa vertical line stitution is also useful in solving nonlinear systems.
yDb horizontal line Solving a system formed by the supply and demand
equations for a product gives the equilibrium point, which
Ax C By C C D 0 general indicates the price at which consumers will purchase the
same quantity of a product that producers wish to sell at that
The linear function
price.
f.x/ D ax C b .a ¤ 0/ Profit is total revenue minus total cost, where total cost
is the sum of fixed costs and variable costs. The brea -even
has a straight line for its graph. points are the points where total revenue equals total cost.
 Chapter 3 e e 173

R
8 8̂
The slope of the line through (2,5) and .3; k/ is 4. Find k. 3y C x
< xC yCzD6 < 2x C D
The slope of the line through .4; 2/ and .7; k/ is 0. Find k. 2x C y C z D 7 3
: x C 3y ! z D 4 :̂ y C 5x C 2y D 7
8 4
In Pr b ems determine the s pe intercept f rm and a % < 3
genera inear f rm f an equati n f the straight ine that has x2 ! y C 5x D 2 yD
xC2
the indicated pr perties x2 C y D 3 :
xCy!2D0
Passes through .!2; 3/ and has y-intercept !1 8
% < x!y!zD0
Passes through (!1, !1) and is parallel to the line y D 3x ! 4 x C 2z D !2
!x C y ! z D 0
xCyCzD5 : !x ! y C z D 0
Passes through . ; 3/ and has slope 3 % %
x!y!zD0 2x ! 5y C 6z D 1
Passes through (3, 5) and is vertical
2x ! 2y C 3z D 0 4x ! 10y C 12z D 2
Passes through .5; 7/ and is horizontal
Suppose a and b are linearly related so that a D 0 when
Passes through (1, 2) and is perpendicular to the line b D !3 and a D 3 when b D !5. Find a general linear form of an
!3y C 5x D 7 equation that relates a and b. Also, find a when b D 3.
Has y-intercept !3 and is perpendicular to 2y C 5x D 2 emperature and eart Rate When the temperature,
etermine whether the point .3; 11/ lies on the line through (in degrees Celsius), of a cat is reduced, the cat s heart rate,
.2; 7/ and .4; 13/. r (in beats per minute), decreases. Under laboratory conditions,
a cat at a temperature of 36ı C had a heart rate of 206, and at a
In Pr b ems determine hether the ines are para e temperature of 30ı C its heart rate was 122.
perpendicu ar r neither If r is linearly related to , where is between 26 and 3 ,
a determine an equation for r in terms of , and b determine
x C 4y C 2 D 0; x ! 2y ! 2 D 0 the cat s heart rate at a temperature of 27ı C.
y ! 5 D 3.x ! 1/; 4x C 12y ! 7 D 0
x ! 3 D 2.y C 4/; y D 4x C 2
2x C 7y ! 4 D 0; 6x C 21y D 0
y D 5x C 2; 10x ! 2y D 3
y D 7x; yD7
Suppose f is a linear function such that f .!1/ D !3 and f .x/
In Pr b ems rite each ine in s pe intercept f rm and decreases by three units for every two-unit increase in x. Find f .x/.
sketch hat is the s pe f the ine If f is a linear function such that f .!1/ D and f .2/ D 5, find
5x ! 3y D 7 x D !3y C 4 f .x/.
4 ! 3y D 0 3x ! 5y D 0 Maximum Revenue The demand function for a
manufacturer s product is p D f .q/ D 200 ! 2q, where p is the
In Pr b ems graph each functi n F r th se that are inear price (in dollars) per unit when q units are demanded. Find the
gi e the s pe and the ertica axis intercept F r th se that are level of production that maximizes the manufacturer s total
quadratic gi e a intercepts and the ertex revenue, and determine this revenue.
y D f .x/ D 17 ! 5x s D g.t/ D 3 ! 5t ! 2t2 Sales ax The difference in price of two items before a
7 sales tax is imposed is 2.00. The difference in price after
y D f .x/ D ! x2 y D f .x/ D 3x ! 7 the sales tax is imposed is allegedly 3.10. Show that this scenario
y D h.t/ D 3 C 2t C t 2 y D k.t/ D !3 ! 3t is not possible.
s D g.t/ D !5t y D F.x/ D .2x ! 1/2 Equilibrium Price If the supply and demand equations
of a certain product are 120p ! q ! 240 D 0 and
y D F.x/ D !.x2 C 2x C 3/ y D f .x/ D 5x C 2 100p C q ! 1200 D 0, respectively, find the equilibrium price.
Demand A company is aware that members of its industry
In Pr b ems s e the gi en system
% % invariably have linear demand functions. The company has data
2x ! y D 6 12x ! 4y D 7 showing that when 5030 units of their product were demanded
3x C 2y D 5 y D 3x ! 5 their price was 2 per unit and when 6075 units were
% % demanded their price was 2 per unit. Write the company s
7x C 5y D 5 2x C 4y D
demand equation.
6x C 5y D 3 3x C 6y D 12
8̂ 8̂ Brea Even Point A manufacturer of a certain product
1 1 1 1 1 sells all that is produced. etermine the brea -even point if the
< x! yD2 < x! yD
2 3 3 4 12 product is sold at 16 per unit, fixed cost is 10,000, and variable
:̂ 3 x C 1 y D 3 :̂ 4 x C 3y D 5 cost is given by y VC D q, where q is the number of units
4 2 3 3 produced (y VC expressed in dollars).
174 C nes Para o as an ste s

emperature Conversion Celsius temperature, C, is a raphically solve the linear system

linear function of Fahrenheit temperature, F. Use the facts that %
3x C 4y D 20
32ı F is the same as 0ı C and 212ı F is the same as 100ı C to find 7x C 5y D 64
this function. Also, find C when F D 50.
raphically solve the linear system
%
0:2x ! 0:3y D 2:6
0:3x C 0:7y D 4:1
Round x and y to two decimal places.
raphically solve the nonlinear system
8
< 3
yD where x > 0
7x
: 2
Pollution In one province of a developing nation, water yDx !
pollution is analyzed using a supply-and-demand model. The Round x and y to two decimal places.
0:0042
en ir nmenta supp y equati n D 0:01 3 ! describes raphically solve the nonlinear system
p %
the levy per ton, (in dollars), as a function of total pollution, p y D x3 C 1
(in tonnes per square ilometer), for p # 0:22 5. The y D 2 ! x2
0:037 Round x and y to two decimal places.
en ir nmenta demand equati n D 0:0005 C , describes
p raphically solve the equation
the per-tonne abatement cost as a function of total pollution for
x2 C 4 D x3 ! 3x
p > 0. Find the expected equilibrium level of total pollution to
two decimal places.5 by treating it as a system. Round x to two decimal places.

5
See Hua Wang and avid Wheeler, Pricing Industrial Pollution in China: An
Economic Analysis of the evy System, World an Policy Research
Wor ing Paper #1644, September 1 6.
 onent a
an o ar t c
nct ons
ust as biological viruses spread through contact between organisms, so computer
4.1 onent a nct ons viruses spread when computers interact via the Internet. Computer scientists
4.2 o ar t c nct ons study how to fight computer viruses, which cause a lot of damage in the form
of deleted and corrupted files. ne thing computer scientists do is devise math-
4.3 Pro ert es of o ar t s ematical models of how quic ly viruses spread. This spread can be remar ably rapid.
Within three days of an identifiable virus being reported, over 100,000 new cases have
4.4 o ar t c an
onent a at ons subsequently been reported
Exponential functions, which this chapter discusses in detail, provide one plausi-
C er 4 e e ble model. Consider a computer virus that hides in an email attachment and, once the
attachment is downloaded, automatically sends a message with a similar attachment to
every address in the host computer s email address boo . If the typical address boo
contains 20 addresses, and if the typical computer user retrieves his or her email once
a day, then a virus on a single machine will have infected 20 machines after one day,
202 D 400 machines after two days, 203 D 000 after three days, and, in general, after
t days, the number of infected computers will be given by the exponential function
.t/ D 20t .
This model assumes that all the computers involved are lin ed, via their address
boo lists, into a single, well-connected group. Exponential models are most accu-
rate for small values of t this model, in particular, ignores the slowdown that occurs
when most emails start going to computers already infected, which happens as several
days pass. For example, our model tells us that after eight days, the virus will infect
20 D 25:6 billion computers more computers than actually exist ut despite its
limitations, the exponential model does explain why new viruses often infect many
thousands of machines before antivirus experts have had time to react.

175
176 C onent a an o ar t c nct ons

Objective E F
o st e onent a f nct ons an The functions of the form f.x/ D bx , for constant b, are important in mathematics,
t er a cat ons to s c areas as
co o n nterest o at on business, economics, science, and other areas of study. An excellent example is
ro t an ra oact e eca f.x/ D 2x . Such functions are called exp nentia functi ns. ore precisely,

It is important to note that an exp nentia

functi n such as 2x is entirely different
from a p er functi n such as x2 . The The function f defined by
exponential function has a variable f.x/ D bx
exponent the power function has a
variable base. where b > 0; b ¤ 1, and the exponent x is any real number, is called an exponential
function with base b.

The reason we require b ¤ 1 in the definition is that if f.x/ D 1x , we have

f.x/ D 1x D 1 for all x and this is a particular constant function. Constant functions are
already nown to us. If we were to consider b D 0, then f.x/ D 0x would be undefined
for negative values of x. If we too b < 0, then f.x/ D bx would be undefined for such
x-values as x D 1=2. We repeat that each b in .0; 1/ [ .1; 1/ gives us an example of a
function fb .x/ D bx that is defined for a real x.
For the moment, consider the exponential function 3x . Since the x in bx can p be any
real number, it is not at first clear what value is meant by something li e 3 2 , where
the exponent is an irrational number. Stated simply, we use approximations. ecause
p p p5
2 D 1:41421 : : :, 3 is approximately 31:4 D 37=5 D 37 , which is defined. etter
2
p
100
approximations
p
are 31:41 D 3141=100 D 3p141 , and so on. In this way, the meaning of
3 becomes clear. A calculator value of 3 2 is (approximately) 4:72 0.
2

When we wor with exponential functions, it is often necessary to apply rules for
To review exponents, refer to Section 0.3. exponents. These rules are as follows, where x and y are real numbers and b and c are
positive.

R E ! "x
b bx
b b D bxCy
x y
D x
c c
bx
D bx!y b1 D b
by
.bx /y D bxy b0 D 1
1
.bc/x D bx cx b!x D
bx

Some functions that do not appear to have the exponential form bx can be put in
A L IT I that form by applying the preceding rules. For example, 2!x D 1=.2x / D . 12 /x and
32x D .32 /x D x .
The number of bacteria in a cul-
ture that doubles every hour is given by
.t/ D A ! 2t , where A is the number
originally present and t is the number of E AM LE B G
hours the bacteria have been doubling.
Use a graphing calculator to plot this The number of bacteria present in a culture after t minutes is given by
function for various values of A > 1. ! "t
How are the graphs similar How does 4
the value of A alter the graph .t/ D 300
3
 Section 4. onent a nct ons 177
! "t
4
Note that .t/ is a constant multiple of the exponential function
3
a How many bacteria are present initially
S Here we want to find .t/ when t D 0. We have
! "0
4
.0/ D 300 D 300.1/ D 300
3
Thus, 300 bacteria are initially present.
b Approximately how many bacteria are present after 3 minutes
S
! "3 ! "
4 64 6400
.3/ D 300 D 300 D " 711
3 27
Hence, approximately 711 bacteria are present after 3 minutes.
Now ork Problem 31 G

G E F
A L IT I
Suppose an investment increases E AM LE G E F b>1
by 10 every year. a e a table of
the factor by which the investment raph the exponential functions f.x/ D 2x and f.x/ D 5x .
increases from the original amount for
S y plotting points and connecting them, we obtain the graphs in Figure 4.1.
0 to 4 years. For each year, write an
expression for the increase as a power of For the graph of f.x/ D 5x , because of the unit distance chosen on the y-axis, the points
1
some base. What base did you use How .#2; 25 /, (2,25), and (3,125) are not shown.
does that base relate to the problem We can ma e some observations about these graphs. The domain of each function
Use your table to graph the multiplica- consists of all real numbers, and the range consists of all positive real numbers. Each
tive increase as a function of the number graph has y-intercept (0,1). oreover, the graphs have the same general shape.
of years. Use your graph to determine
Each rises from left to right. As x increases, f.x/ also increases. In fact, f.x/ increases
when the investment will double.
without bound. However, in quadrant I, the graph of f.x/ D 5x rises more quic ly than
that of f.x/ D 2x because the base in 5x is greater than the base in 2x (that is, 5 > 2).
oo ing at quadrant II, we see that as x becomes very negative, the graphs of both
functions approach the x-axis. We say that the x-axis is an asympt te for each graph.
This implies that the function values get very close to 0.

y
x 2x y x 5x

1 8 1
-2 4 -2 25

1 5 1
-1 2 -1 5
4
0 1 0 1
f(x) = 2x f(x) = 5x
2
1 2 1 1 5
x x
2 4 -2 -1 1 2 3 -1 1 2 25

3 8 3 125
(a) (b)

FIGURE raphs of f .x/ D 2x and

f .x/ D 5x .

Now ork Problem 1 G

178 C onent a an o ar t c nct ons

The observations made in Example 2 are true for all exponential functions whose
base b is greater than 1. Example 3 will examine the case for a base between 0 and 1,
that is 0 < b < 1.

E AM LE G E F 0<b<1
A L IT I
# 1 $x
Suppose the value of a car depre- raph the exponential function f.x/ D 2
.
ciates by 15 every year. a e a
table of the factor by which the value S y plotting points and connecting them, we obtain the graph in Figure 4.2.
decreases from the original amount for Notice that the domain consists of all real numbers, and the range consists of all positive
0 to 3 years. For each year, write an real numbers. The graph has y-intercept (0,1). Compared to the graphs in Example 2,
expression for the decrease as a power the graph here fa s from left to right. That is, as x increases, f.x/ decreases. Notice that
of some base. What base did you use as x becomes very positive, f.x/ ta es on values close to 0 and the graph approaches
How does that base relate to the prob- the x-axis. However, as x becomes very negative, the function values are unbounded.
lem Use your table to graph the mul-
tiplicative decrease as a function of the
number of years. Use your graph to
y
determine when the car will be worth
half as much as its original price.
8
1 x
x 2

-3 8
There are two basic shapes for the graphs
of exponential functions, and they depend -2 4
on the base involved: b in .0; 1/ or b in 4
.1; 1/. -1 2
1 x
f(x) = 2
0 1
2
1
1 2
1
y
x
1 -3 -2 -1 1
2 4

% &x
1
FIGURE raph of f .x/ D 2 .

f(x) = bx Now ork Problem 3 G

b71
1
x
Graph rises In general, the graph of an exponential function has one of two shapes, depending
from left to
right on the value of the base, b. This is illustrated in Figure 4.3. It is important to observe
(a) that in either case the graph passes the horizontal line test. Thus, all exponential func-
tions are one-to-one. The basic properties of an exponential function and its graph are
y summarized in Table 4.1.
Recall from Section 2.7 that the graph of one function may be related to that of
another by means of a certain transformation. ur next example pertains to this concept.

Table .1 E F f .x/ D bx
The domain of any exponential function is .#1; 1/.
f(x) = bx
06b61 The range of any exponential function is .0; 1/.
1
x The graph of f .x/ D bx has y-intercept (0,1).
Graph falls There is no x-intercept.
from left to
right If b > 1, the graph rises from left to right.
(b)
If 0 < b < 1, the graph fa s from left to right.
FIGURE eneral shapes
If b > 1, the graph approaches the x-axis as x becomes more and more negative.
of f .x/ D bx .
If 0 < b < 1, the graph approaches the x-axis as x becomes more and more positive.
 Section 4. onent a nct ons 179
y
f(x) = 2x

1 x-4
1 x y=
f(x) = 2
y = 2x - 3 2

1
x
1
x
-2 4

% &x!4
1
FIGURE raph of y D 2x # 3. FIGURE raph of y D 2 .

Example 4 ma es use of transformations E AM LE T E F

from Table 2.2 of Section 2.7.
a Use the graph of y D 2x to plot y D 2x # 3.
A L IT I S The function has the form f.x/ # c, where f.x/ D 2x and c D 3. Thus,
its graph is obtained by shifting the graph of f.x/ D 2x three units downward. (See
After watching his sister s money
grow for three years in a plan with Figure 4.4.)
# $x # $x!4
an yearly return, eorge started a b Use the graph of y D 12 to graph y D 12 .
savings account with the same plan. # $x
If y D 1:0 t represents the multi- S The function has the form f.x # c/, where f.x/ D 12 and c D 4. Hence,
# $x
plicative increase in his sister s account, its graph is obtained by shifting the graph of f.x/ D 12 four units to the right. (See
write an equation that will represent Figure 4.5.)
the multiplicative increase in eorge s
account, using the same time reference. Now ork Problem 7 G
If eorge has a graph of the multipli-
cative increase in his sister s money at
time t years since she started saving,
E AM LE G F C B
how could he use the graph to pro ect 2
the increase in his money raph y D 3x .
S Although this is not an exponential function, it does have a constant base.
We see that replacing x by #x results in the same equation. Thus, the graph is symmetric
about the y-axis. Plotting some points and using symmetry gives the graph in Figure 4.6.

2
y = 3x

x
-1 1

x 0 1 2

y 1 3 81

2
FIGURE raph of y D 3x .

Now ork Problem 5 G

180 C onent a an o ar t c nct ons

C I
Exponential functions are involved in compound interest, whereby the interest earned
by an invested amount of money (or principal) is reinvested so that it, too, earns inter-
est. That is, the interest is converted (or c mp unded) into principal, and hence, there
is interest on interest .
For example, suppose that 100 is invested at the rate of 5 compounded annually.
At the end of the first year, the value of the investment is the original principal ( 100),
plus the interest on the principal ( 100(0.05)):
100 C 100.0:05/ D 105
This is the amount on which interest is earned for the second year. At the end of the
second year, the value of the investment is the principal at the end of the first year
( 105), plus the interest on that sum ( 105(0.05)):
105 C 105.0:05/ D 110:25
Thus, each year the principal increases by 5 . The 110.25 represents the original
principal, plus all accrued interest it is called the accumu ated am unt or compound
amount. The difference between the compound amount and the original principal is
called the c mp und interest Here the compound interest is 110:25! 100 D 10:25.
ore generally, if a principal of P dollars is invested at a rate of 100r percent
compounded annually (for example, at 5 , r is 0.05), the compound amount after 1 year
is P C Pr, or, by factoring, P.1 C r/. At the end of the second year, the compound
amount is
P.1 C r/ C .P.1 C r//r D P.1 C r/.1 C r/ factoring
D P.1 C r/2
Actually, the calculation above using factoring is not necessary to show that the com-
pounded amount after two years is P.1Cr/2 . Since any amount P is worth P.1Cr/ a year
later, it follows that the amount of P.1 C r/ is worth P.1 C r/.1 C r/ D P.1 C r/2 a year
later, and one year later still the amount of P.1Cr/2 is worth P.1Cr/2 .1Cr/ D P.1Cr/3 .
This pattern continues. After four years, the compound amount is P.1 C r/4 . In
general, the c mp und am unt S f the principa P at the end f n years at the rate f
r c mp unded annua y is given by

S D P.1 C r/n
A L IT I Notice from Equation (1) that, for a given principal and rate, S is a function of n. In fact,
Suppose 2000 is invested at 13 S is a constant multiple of the exponential function with base 1 C r.
compounded annually. Find the value of
the investment after five years. Find the
E AM LE C A C I
interest earned over the first five years.
Suppose 1000 is invested for 10 years at 6 compounded annually.
S a Find the compound amount.
S We use Equation (1) with P D 1000, r D 0:06, and n D 10:

5000
S D 1000.1 C 0:06/10 D 1000.1:06/10 " 17 0: 5
4000 Figure 4.7 shows the graph of S D 1000.1:06/n . Notice that as time goes on, the
3000 compound amount grows dramatically.
2000 S = 1000 (1.06)n b Find the compound interest.
1000 S Using the results from part (a), we have
n
10 20 30 compound interest D S ! P
FIGURE raph of D 17 0: 5 ! 1000 D 7 0: 5
S D 1000.1:06/n .
Now ork Problem 19 G
 Section 4. onent a nct ons 181

Suppose the principal of 1000 in Example 6 is invested for 10 years as before,

but this time the compounding ta es place every three months (that is, quarter y) at the
rate of 1 12 per quarter. Then there are four interest periods per year, and in 10 years
there are 10.4/ D 40 interest periods. Thus, the compound amount with r D 0:015 is
now
1000.1:015/40 ! 1 14:02
and the compound interest is 14.02. Usually, the interest rate per interest period is
stated as an annual rate. Here we would spea of an annual rate of 6 compounded
quarterly, so that the rate per interest period, or the periodic rate, is 6 =4 D 1:5 .
This qu ted annual rate of 6 is called the nominal rate or the annual percentage
rate APR . Unless otherwise stated, all interest rates will be assumed to be annual
The abbreviation APR is a common one
and is found on credit card statements (nominal) rates. Thus, a rate of 15 compounded monthly corresponds to a periodic
and in advertising. rate of 15 =12 D 1:25 .
n the basis of our discussion, we can generalize Equation (1). The formula

S D P.1 C r/n
A nominal rate of 6 does not
necessarily mean that an investment gives the c mp und am unt S f a principa P at the end f n interest peri ds at the
increases in value by 6 in a year s time. peri dic rate f r.
The increase depends on the frequency of We have seen that for a principal of 1000 at a nominal rate of 6 over a period
compounding. of 10 years, annual compounding results in a compound interest of 7 0. 5, and with
quarterly compounding the compound interest is 14.02. It is typical that for a given
nominal rate, the more frequent the compounding, the greater is the compound interest.
However, while increasing the compounding frequency always increases the amount of
interest earned, the effect is not unbounded. For example, with wee ly compounding
the compound interest is
! "
0:06 10.52/
1000 1 C " 1000 ! 21:4
52
and with daily compounding it is
! "
0:06 10.365/
1000 1 C " 1000 ! 22:03
365
Sometimes the phrase money is worth is used to express an annual interest rate.
Thus, saying that money is worth 6 compounded quarterly refers to an annual (nom-
inal) rate of 6 compounded quarterly.

G
Equation (2) can be applied not only to the growth of money, but also to other types of
growth, such as that of population. For example, suppose the population P of a town of
10,000 is increasing at the rate of 2 per year. Then P is a function of time t, in years.
It is common to indicate this functional dependence by writing
P D P.t/
Here, the letter P is used in two ways: n the right side, P represents the function on
the left side, P represents the dependent variable. From Equation (2), we have
P.t/ D 10;000.1 C 0:02/t D 10;000.1:02/t

A L IT I E AM LE G
A new company with five employ-
ees expects the number of employees to The population of a town of 10,000 grows at the rate of 2 per year. Find the population
grow at the rate of 120 per year. Find three years from now.
the number of employees in four years. S From the preceding discussion,
P.t/ D 10;000.1:02/t
182 C onent a an o ar t c nct ons

27,000 For t D 3, we have

P(t) = 10,000(1.02)t P.3/ D 10;000.1:02/3 ! 10;612

Thus, the population three years from now will be 10,612. (See Figure 4. .)
Now ork Problem 15 G
0 50
9000 T N e
FIGURE raph of population It is useful to conduct a thought experiment, based on the discussion following
function P.t/ D 10;000.1:02/t . Example 6, to introduce an important number. Suppose that a single dollar is invested
for one year with an APR of 100 (remember, this is a thought experiment ) com-
pounded annually. Then the compound amount S at the end of the year is given by

S D 1.1 C 1/1 D 21 D 2

Without changing any of the other data, we now consider the effect of increasing the
number of interest periods per year. If there are n interest periods per year, then the
compound amount is given by
! " ! "
1 n nC1 n
SD1 1C D
n n
! "
nC1 n
In Table 4.2 we give approximate values for for some values of n.
n

Table . A e
! "n
nC1
n
n
# $1
2
1 1 D 2:00000
# $2
3
2 2 D 2:25000
# $3
4
3 3 ! 2:37037
# $4
5
4 4 ! 2:44141
# $5
6
5 5 D 2:4 32
# $10
11
10 10 ! 2:5 374
# $100
101
100 100 ! 2:704 1
# $1000
1001
1000 1000 ! 2:716 2
# $10;000
10;001
10,000 10;000 ! 2:71 15
# $100;000
100;001
100,000 100;000 ! 2:71 27
# $1;000;000
1;000;001
1,000,000 1;000;000 ! 2:71 2

!"
nC1 n
Apparently, the numbers increase as n does. However, they do not
n
increase without bound. For example, it is possible to show that for any positive integer
! "
nC1 n
n, < 3. In terms of our thought experiment, this means that if you start with
n
1.00 invested at 100 , then, no matter how many interest periods there are per year,
 Section 4. onent a nct ons 183

you will always have less than 3.00 at!the end "
of a year. There is a smallest real number
nC1 n
that is greater than all of the numbers . It is denoted by the letter e, in honor
n
of the Swiss mathematician eonhard Euler (1707 17 3). The p number e is irrational,
so its decimal expansion is nonrepeating, li e those of ! and 2 that we mentioned in
! "
nC1 n
Section 0.1. However, each of the numerical values for can be considered to
n
be a decimal approximation of e. The approximate value . 1;000;001
1;000;000
/1;000;000 ! 2:71 2
gives an approximation of e that is correct to 5 decimal places. The approximation of
e correct to 12 decimal places is e ! 2:71 2 1 2 45 .

E F B e
The number e provides the most important base for an exponential function. In fact, the
exponential function with base e is called the natural exponential function and even
the exp nentia functi n to stress its importance. Although e may seem to be a strange
base, the natural exponential function has a remar able property in calculus (which we
will see in a later chapter) that ustifies the name. It also occurs in economic analysis and
problems involving growth or decay, such as population studies, compound interest,
and radioactive decay. Approximate values of ex can be found with a single ey on
most calculators. The graph of y D ex is shown in Figure 4. . The accompanying table
indicates y-values to two decimal places. f course, the graph has the general shape of
an exponential function with base greater than 1.

9
8
7
x y 6 y = ex
0.14 5
-2
4
-1 0.37 3
0 1 2
1
1 2.72 x
-2 -1 1 2
2 7.39

The graph of the natural exponential FIGURE raph of the natural exponential
function in Figure 4. is important.
function.

E AM LE G F I e
A L IT I a raph y D e!x . ! "x
The multiplicative decrease in pur- !x 1 1
S Since e D and 0 < < 1, the graph is that of an exponential
chasing power P after t years of in ation e e
at 6 can be modeled by P D e!0:06t . function falling from left to right. (See Figure 4.10.) Alternatively, we can consider the
raph the decrease in purchasing power
graph of y D e!x as a transformation of the graph of f.x/ D ex . ecause e!x D f."x/,
as a function of t years.
the graph of y D e!x is simply the re ection of the graph of f about the y-axis. (Compare
the graphs in Figures 4. and 4.10.)
b raph y D exC2 .
S The graph of y D exC2 is related to that of f.x/ D ex . Since exC2 is f.x C 2/,
we can obtain the graph of y D exC2 by horizontally shifting the graph of f.x/ D ex
two units to the left. (See Figure 4.11.)
184 C onent a an o ar t c nct ons

y
y
9
8
x y 7
y = e-x
6 y = e x +2
-2 7.39
5
-1 2.72 f(x) = e x
4
0 1 3
2 1
1 0.37 x
1 -2 -1 1
2 0.14 x
-2 -1 1 2

FIGURE raph of y D e!x . FIGURE raph of

y D exC2 .

E AM LE G

The pro ected population P of a city is given by

P D 100;000e0:05t

where t is the number of years after 2000. Predict the population for the year 2020.
S The number of years from 1 0 to 2010 is 20, so let t D 20. Then

P D 100;000e0:05.20/ D 100;000e1 D 100;000e ! 271; 2

Now ork Problem 35 G

In statistics, an important function used to model certain events occurring in nature
is the P iss n distributi n functi n:
e!! "n
f.n/ D n D 0; 1; 2; : : :
nŠ
The symbol " (read mu ) is a ree letter. In certain situations, f.n/ gives the proba-
bility that exactly n events will occur in an interval of time or space. The constant " is
the average, also called mean number of occurrences in the interval. The next example
illustrates the Poisson distribution.

E AM LE H C

A hemocytometer is a counting chamber divided into squares and is used in studying

the number of microscopic structures in a liquid. In a well- nown experiment,1 yeast
cells were diluted and thoroughly mixed in a liquid, and the mixture was placed in a
hemocytometer. With a microscope, the yeast cells on each square were counted. The
probability that there were exactly n yeast cells on a hemocytometer square was found
to fit a Poisson distribution with " D 1: . Find the probability that there were exactly
four cells on a particular square.

1 R. R. So al and F. . Rohlf, Intr ducti n t Bi statistics (San Francisco: W. H. Freeman and Company, 1 73).
 Section 4. onent a nct ons 185

S We use the Poisson distribution function with " D 1: and n D 4:

e!! "n
f.n/ D
nŠ
e!1: .1: /4
f.4/ D ! 0:072
4Š
For example, this means that in 400 squares we would expect 400.0:072/ ! 2 squares
to contain exactly 4 cells. (In the experiment, in 400 squares the actual number observed
was 30.)
G
R
Radioactive elements are such that the amount of the element decreases with respect to
time. We say that the element decays. It can be shown that, if is the amount at time
t, then
!"t
D 0e

where 0 and $ (a ree letter read lambda ) are positive constants. Notice that
involves an exponential function of t. We say that follows an exponential law of
decay. If t D 0, then D 0 e0 D 0 # 1 D 0 . Thus, the constant 0 represents
the amount of the element present at time t D 0 and is called the initial amount. The
constant $ depends on the particular element involved and is called the decay constant.
ecause decreases as time increases, suppose we let be the length of time it
ta es for the element to decrease to half of the initial amount. Then at time t D , we
have D 0 =2. Equation (3) implies that
0
D 0 e!"
2
We will now use this fact to show that over any time interval of length , half of the
amount of the element decays. Consider the interval from time t to t C , which has
length . At time t, the amount of the element is 0 e!"t , and at time t C it is
!".tC / !"t !"
0e D 0e e D . 0 e!" /e!"t
0 !"t 1
D e D . 0 e!"t /
2 2
which is half of the amount at time t. This means that if the initial amount present, 0 ,
were 1 gram, then at time , 12 gram would remain at time 2 , 14 gram would remain
and so on. The value of is called the half life of the radioactive element. Figure 4.12
shows a graph of radioactive decay.

N0

N = N 0e-nt
N 0 /2

Half-life = T
N 0 /4
N 0 /8
t
T 2T 3T

FIGURE Radioactive decay.

186 C onent a an o ar t c nct ons

E AM LE R

A radioactive element decays such that after t days the number of milligrams present
is given by
D 100e!0:062t
100 a How many milligrams are initially present
S This equation has the form of Equation (3), D 0 e!"t , where 0 D 100
N = 100e-0.062t
and $ D 0:062. 0 is the initial amount and corresponds to t D 0. Thus, 100 milligrams
are initially present. (See Figure 4.13.)
b How many milligrams are present after 10 days
0 50 S When t D 10,
0 D 100e!0:062.10/ D 100e!0:62 ! 53:
FIGURE raph of Therefore, approximately 53. milligrams are present after 10 days.
radioactive decay function
D 100e!0:062t . Now ork Problem 47 G
R BLEMS
In Pr b ems graph each functi n probability P that a sub ect ma es a correct response on the nth
y D f .x/ D 4 x
y D f .x/ D 3 x trial is given by
% &x % &x 1
y D f .x/ D 15 y D f .x/ D 14 P D 1 " .1 " c/n!1 ; n $ 1; 0 < c < 1
2
2
y D f .x/ D 2.x!1/ y D f .x/ D 3.2/x 1
where c is a constant. Ta e c D and find P when n D 1, n D 2,
y D f .x/ D 3xC2 y D f .x/ D 3xC2 2
and n D 3.
y D f .x/ D 3x " 2 y D f .x/ D 3x!1 " 1
Express y D 34x as an exponential function in base 1.
y D f .x/ D 3!x y D f .x/ D 12 .2x=2 /
In Pr b ems nd a the c mp und am unt and b the
Pr b ems and refer t Figure hich sh s the graphs
c mp und interest f r the gi en in estment and annua rate
f y D 0:4x y D 2x and y D 5x
2000 for 5 years at 3 compounded annually
f the curves A, B, and C, which is the graph of y D 0:4x
5000 for 20 years at 5 compounded annually
f the curves A, B, and C, which is the graph of y D 2x
700 for 15 years at 7 compounded semiannually
4000 for 12 years at 7 12 compounded semiannually
A B C 1
3000 for 22 years at 4
compounded monthly
6000 for 2 years at compounded quarterly
5000 for 2 12 years at compounded monthly
500 for 5 years at 11 compounded semiannually
FIGURE 000 for 3 years at 6 14 compounded daily. (Assume that
there are 365 days in a year.)
Population The pro ected population of a city is given by Investment Suppose 1300 is placed in a savings account
P D 125; 000.1:11/t=20 , where t is the number of years after 1 5. that earns interest at the rate of 3.25 compounded monthly.
What is the pro ected population in 2015 a What is the value of the account at the end of three years
b If the account had earned interest at the rate of 3.5
Population For a certain city, the population P grows at the
compounded annually, what would be the value after three years
rate of 1.5 per year. The formula P D 1;527;000.1:015/t gives
the population t years after 1 . Find the population in a 1
and b 2000.
Paired Associate Learning In a psychological experiment
involving learning,2 sub ects were as ed to give particular
responses after being shown certain stimuli. Each stimulus was a
pair of letters, and each response was either the digit 1 or 2. After Investment A certificate of deposit is purchased for
each response, the sub ect was told the correct answer. In this 6500 and is held for three years. If the certificate earns 2
so-called paired ass ciate learning experiment, the theoretical compounded quarterly, what is it worth at the end of three years

2
. aming, athematica Psych gy (New or : Academic Press, Inc., 1 73).
 Section 4. onent a nct ons 187

Population Growth The population of a town of 5000 Evaluate f .0/, f .1/, and f .2/. Round your answers to three
grows at the rate of 3 per year. a etermine an equation that decimal places.
gives the population t years from now. b Find the population 1
Express ekt in the form bt . Express x in the form bx .
three years from now. ive your answer to (b) to the nearest e
integer. Radioactive Decay A radioactive element is such that
grams remain after t hours, where
Bacteria Growth acteria are growing in a culture, and
their number is increasing at the rate of 5 an hour. Initially, 400 D 12e!0:031t
bacteria are present. a etermine an equation that gives the a How many grams are initially present To the nearest tenth of a
number, , of bacteria present after t hours. b How many gram, how many grams remain after b 10 hours c 44 hours
bacteria are present after one hour c After four hours ive d ased on your answer to part (c), what is your estimate
your answers to (b) and (c) to the nearest integer. of the half-life of this element
Bacteria Reduction A certain medicine reduces the Radioactive Decay At a certain time, there are 100
bacteria present in a person by 10 each hour. Currently, milligrams of a radioactive substance. The substance decays so
100,000 bacteria are present. a e a table of values for the that after t years the number of milligrams present, , is given by
number of bacteria present each hour for 0 to 4 hours. For each
D 100e!0:045t
hour, write an expression for the number of bacteria as a product
of 100,000 and a power of 10 . Use the expressions to ma e How many milligrams, rounded to the nearest tenth of a
an entry in your table for the number of bacteria after t hours. milligram, are present after 20 years
Write a function for the number of bacteria after t hours.
Radioactive Decay If a radioactive substance has a
Recycling Suppose the amount of plastic being recycled half-life of years, how long does it ta e for 1 gram of the
increases by 32 every year. a e a table of the factor by which substance to decay to 1 gram
recycling increases over the original amount for 0 to 5 years.
Mar eting A mail-order company advertises in a national
Population Growth Cities A and presently have magazine. The company finds that, of all small towns, the
populations of 270,000 and 360,000, respectively. City A grows at percentage (given as a decimal) in which exactly x people respond
the rate of 6 per year, and grows at the rate of 4 per year. to an ad fits a Poisson distribution with " D 0:5. From what
etermine the larger and by how much the populations differ at percentage of small towns can the company expect exactly two
the end of five years. ive your answer to the nearest integer. people to respond Round your answer to four decimal places.
Pr b ems and in e a dec ining p pu ati n If a Emergency Room Admissions Suppose the number of
p pu ati n dec ines at the rate f r per time peri d then patients admitted into a hospital emergency room during a certain
the p pu ati n after t time peri ds is gi en by hour of the day has a Poisson distribution with mean 4. Find the
P D P0 .1 " r/t probability that during that hour there will be exactly two
emergency patients. Round your answer to four decimal places.
here P0 is the initia p pu ati n .the p pu ati n hen t D 0/
Population ecause of an economic downturn, the
population of a certain urban area declines at the rate
of 1.5 per year. Initially, the population is 350,000. To the
nearest person, what is the population after three years
Enrollment After a careful demographic analysis, a
university forecasts that student enrollments will drop by 3 per
year for the the next 12 years. If the university currently has
14,000 students, how many students will it have 12 years from % 1 &x
now raph y D 17x and y D 17 on the same screen.
In Pr b ems use a ca cu at r t nd the a ue r unded t etermine the intersection point.
f ur decima p aces f each expressi n et a > 0 be a constant. raph y D 2!x and y D 2a # 2!x on
e1:5 e4:6 e!0: e!2=3 the same screen, for constant values a D 2 and a D 3. bserve
that the graph of y D 2a # 2x appears to be the graph of y D 2!x
In Pr b ems and graph the functi ns
shifted a units to the right. Prove algebraically that, in this case,
y D "e !.xC1/ y D 2ex merely two observations predict what is true.
elephone Calls The probability that a telephone operator For y D 5x , find x if y D 3. Round your answer to two
will receive exactly x calls during a certain period is given by decimal places.
e!3 3x For y D 2x , find x if y D . Round your answer to two
PD decimal places.
xŠ
Find the probability that the operator will receive exactly four Cell Growth Cells are growing in a culture, and their
calls. Round your answer to four decimal places. number is increasing at the rate of 7 per hour. Initially, 1000
ormal Distribution An important function used in cells are present. After how many full hours will there be at least
economic and business decisions is the n rma distributi n density 3000 cells
functi n which, in standard form, is Bacteria Growth Refer to Example 1. How long will it
1 1 2 ta e for 1000 bacteria to be present Round your answer to
f .x/ D p e!. 2 /x
2! the nearest tenth of a minute.
188 C onent a an o ar t c nct ons

Demand Equation The demand equation for a new toy is Investment If 2000 is invested in a savings account that
earns interest at the rate of . compounded annually, after how
q D 100; 000.0: 5012/p
many full years will the amount at least double
Evaluate q to the nearest integer when p D 15.

Objective L F
o ntro ce o ar t c f nct ons an Since all exponential functions pass the horizontal line test, they are all one-to-one
t e r ra s Pro ert es of o ar t s
e sc sse n ect on functions. It follows that each exponential function has an inverse. These functions,
inverse to the exponential functions, are called the logarithmic functions.
ore precisely, if f.x/ D bx , the exponential function base b (where 0 < b < 1
or 1 < b), then the inverse function f !1 .x/ is called the garithm functi n base b
and is denoted logb x. It follows from our general remar s about inverse functions in
Section 2.4 that
y D logb x if and only if by D x
To review inverse functions, refer to
Section 2.4. and we have the following fundamental equations:

logb bx D x
and
blogb x D x

where Equation (1) holds for all x in ."1; 1/ and Equation (2) holds for all x in
.0; 1/. We recall that ."1; 1/ is the domain of the exponential function base b and
.0; 1/ is the range of the exponential function base b. It follows that .0; 1/ is the
domain of the logarithm function base b and ."1; 1/ is the range of the logarithm
function base b.
Stated otherwise, given positive x, logb x is the unique number with the property
that blogb x D x. The generalities about inverse functions also enable us to see immedi-
ately what the graph of a logarithmic function loo s li e.
In Figure 4.15 we have shown the graph of the particular exponential function
y D f.x/ D 2x , whose general shape is typical of exponential functions y D bx for
which the base b satisfies 1 < b. We have added a (dashed) copy of the line y D x. The
graph of y D f !1 .x/ D log2 x is obtained as the mirror image of y D f.x/ D 2x in the
line y D x.

8
7
6
5
4
3
2
1
x
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8
-1
-2
-3
-4
-5
-6
-7
-8

FIGURE raphs of y D 2x and y D log2 x.

 Section 4.2 o ar t c nct ons 189

In Table 4.3 we have tabulated the function values that appear as y-coordinates of
Table . S F
the dots in Figure 4.15.
It is clear that the exponential function base 2 and the logarithm function base 2
x 2x x log2 x und the effects of each other. Thus, for all x in the domain of 2x (which is ."1; 1/),
1 1 we have
"2 4 4 "2
1 1
"1 2 2 "1 log2 2x D x
0 1 1 0
1 2 2 1 and, for all x in the domain of log2 x which is the range of 2x (which is .0; 1/) we have
2 4 4 2
3 3 2log2 x D x

It cannot be said too often that

y D logb x means by D x

and conversely

Logarithmic and by D x means y D logb x

exponential forms

Logarithm Exponent In this sense, a garithm f a number is an exp nent logb x is the power to which
we must raise b to get x. For example,
log2 8 = 3 23 = 8

base base log2 D 3 because 23 D

FIGURE A logarithm can
be considered an exponent. We say that log2 D 3 is the garithmic f rm of the exp nentia f rm 23 D . (See
Figure 4.16.)

A L IT I E AM LE C E L F
If bacteria have been doubling
every hour and the current amount is 16 Exp nentia F rm garithmic F rm
times the amount first measured, then a Since it follows that log5 25 D 2
52 D 25
the situation can be represented by 16 D
2t . Represent this equation in logarith- b Since 34 D 1 it follows that log3 1 D 4
mic form. What does t represent c Since 100 D 1 it follows that log10 1 D 0

Now ork Problem 1 G

A L IT I
An earthqua e measuring .3 on
the Richter scale# can
$ be represented E AM LE C L E F
by :3 D log10 II0 , where I is the
garithmic F rm Exp nentia F rm
intensity of the earthqua e and I0 is
the intensity of a zero-level earthqua e. a log10 1000 D 3 means 103 D 1000
Represent this equation in exponential 1
form. b log64 D means 641=2 D
2
1 1
A L IT I c log2 D "4 means 2!4 D
16 16
Suppose a recycling plant has found
that the amount of material being recy- Now ork Problem 3 G
cled has increased by 50 every year
since the plant s first year of opera-
tion. raph each year as a function of E AM LE G L F b>1
the multiplicative increase in recycling
since the first year. abel the graph with Examine again the graph of y D log2 x in Figure 4.15. This graph is typical for a
the name of the function.
logarithmic function logb x with b > 1.
Now ork Problem 9 G
190 C onent a an o ar t c nct ons

A L IT I E AM LE G L F 0<b<1
Suppose a boat depreciates 20 raph y D log1=2 x.
every year. raph the number of years % 1 &x
the boat is owned as a function of the S To plot points, we plot the inverse function y D 2
and re ect the graph
multiplicative decrease in its original in the line y D x. (See Figure 4.17.)
value. abel the graph with the name of
the function.
y

6
5
4
3
2

1
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6

# $x
1
FIGURE raph of y D 2 and y D log1=2 x.

From the graph, we can see that the domain of y D log1=2 x is the set of all positive
% &x
real numbers, for that is the range of y D 12 , and the range of y D log1=2 x is the set
% &x
of all real numbers, which is the domain of y D 12 . The graph falls from left to right.
1
Numbers between 0 and 1 have positive base 2
logarithms, and the closer a number
1
is to 0, the larger is its base 2
logarithm. Numbers greater than 1 have negative base
1
2
logarithms. The logarithm of 1 is 0, regard ess f the base b and corresponds to the
x-intercept .1; 0/. This graph is typical for a logarithmic function with 0 < b < 1.
Now ork Problem 11 G
y y = logb x, y
b71
y = logb x,
06b61

x x
1 1

(a) (b)

FIGURE eneral shapes of y D logb x.

Summarizing the results of Examples 3 and 4, we can say that the graph of a log-
arithmic function has one of two general shapes, depending on whether b > 1 or
0 < b < 1. (See Figure 4.1 .) For b > 1, the graph rises from left to right as x
gets closer and closer to 0, the function values decrease without bound, and the graph
gets closer and closer to the y-axis. For 0 < b < 1, the graph falls from left to right as
 Section 4.2 o ar t c nct ons 191

x gets closer and closer to 0, the function values increase without bound, and the graph
gets closer and closer to the y-axis. In each case, note that
The domain of a logarithmic function is the interval .0; 1/. Thus, the logarithm of
a nonpositive number does not exist.
The range is the interval ."1; 1/.
The logarithm of 1 is 0, which corresponds to the x-intercept (1, 0).
The graph of the natural logarithmic Important in calculus are logarithms to the base e, called natural logarithms. We
function in Figure 4.1 is important, too.
use the notation ln for such logarithms:
ln x means loge x
The symbol ln x can be read natural log of x.
ogarithms to the base 10 are called common logarithms. They were frequently
used for computational purposes before the calculator age. The subscript 10 is usually
y
omitted from the notation:
y = ln x log x means log10 x
1 ost good calculators give approximate values for both natural and common log-
x arithms. For example, verify that ln 2 ! 0:6 315. This means that e0:6 315 ! 2.
1 e
Figure 4.1 shows the graph of y D ln x. ecause e > 1, the graph has the general
shape of that of a logarithmic function with b > 1 see Figure 4.1 (a) and rises from
left to right.
While the conventions about log, with no subscript, and ln are well established in
FIGURE raph of natural
logarithmic function. elementary boo s, be careful when consulting an advanced boo . In advanced texts,
log x means loge x, ln is not used at all, and logarithms base 10 are written explicitly
as log10 x.

E AM LE F L
A L IT I
The number of years it ta es for
a Find log 100.
an amount invested at an annual rate S Here the base is 10. Thus, log 100 is the power to which we must raise 10
of r and compounded continuously to to get 100. Since 102 D 100, log 100 D 2.
quadruple is a function of the annual
ln 4 b Find ln 1.
rate r given by t.r/ D . Use a calcu-
r S Here the base is e. ecause e0 D 1, ln 1 D 0.
lator to find the rate needed to quadruple
an investment in 10 years. c Find log 0:1.
1
S Since 0:1 D 10
D 10!1 , log 0:1 D "1.
Remember the way in which a logarithm d Find ln e!1 .
is an exponent.
S Since ln e!1 is the power to which e must be raised to obtain e!1 , clearly
!1
ln e D "1.
e Find log36 6.
p
S ecause 361=2 (D 36) is 6, log36 6 D 12 .
Now ork Problem 17 G
any equations involving logarithmic or exponential forms can be solved for an
un nown quantity by first transforming from logarithmic form to exponential form, or
vice versa. Example 6 will illustrate.

A L IT I E AM LE S L E E
The multiplicative increase m of an a Solve log2 x D 4.
amount invested at an annual rate of r
compounded continuously for a time t is S We can get an explicit expression for x by writing the equation in exponen-
given by m D ert . What annual percent- tial form. This gives
age rate is needed to triple the invest- 24 D x
ment in 12 years
so x D 16.
192 C onent a an o ar t c nct ons

b Solve ln.x C 1/ D 7.
S The exponential form yields e7 D x C 1. Thus, x D e7 " 1.
c Solve logx 4 D 2.
S In exponential form, x2 D 4 , so x D 7. We re ect x D "7 because a
negative number cannot be a base of a logarithmic function.
d Solve e5x D 4.
S We can get an explicit expression for x by writing the equation in logarith-
mic form. We have
ln 4 D 5x
ln 4
xD
5
Now ork Problem 49 G
R H L
From our discussion of the decay of a radioactive element in Section 4.1, we now that
the amount of the element present at time t is given by
!"t
D 0e

where 0 is the initial amount (the amount at time t D 0) and $ is the decay constant.
et us now determine the half-life of the element. At time , half of the initial amount
is present. That is, when t D ; D 0 =2. Thus, from Equation (3), we have
0 !"
D 0e
2
Solving for gives
1
D e!"
2
2 D e" ta ing reciprocals of both sides
To get an explicit expression for , we convert to logarithmic form. This results in
$ D ln 2
ln 2
D
$
Summarizing, we have the following:

If a radi acti e e ement has decay c nstant $, then the ha f ife f the e ement is
gi en by
ln 2
D
$
10

N = 10e-0.00501t E AM LE F H L

A 10-milligram sample of radioactive polonium 210 (which is denoted 210 Po/ decays
according to the equation
0 300
0 D 10e!0:00501t
FIGURE Radioactive decay where is the number of milligrams present after t days. (See Figure 4.20.) etermine
function D 10e!0:00501t . the half-life of 210 Po.
 Section 4.2 o ar t c nct ons 193

S Here the decay constant $ is 0.00501. y Equation (4), the half-life is given by
ln 2 ln 2
D D ! 13 .4 days
$ 0:00501
Now ork Problem 63 G
R BLEMS
In Pr b ems express each garithmic f rm exp nentia y and multiplicative increase in its original value. abel the graph with
each exp nentia f rm garithmica y the name of the function.
104 D 10;000 2 D log12 144 Cost Equation The cost for a firm producing q units of a
log2 1024 D 10 323=5 D 3 product is given by the cost equation
e3 ! 20:0 55 e0:33647 ! 1:4 c D .5q ln q/ C 15
ln 3 ! 1:0 61 log 7 ! 0: 450 Evaluate the cost when q D 12. (Round your answer to two
decimal places.)
In Pr b ems graph the functi ns
Supply Equation A manufacturer s supply equation is
y D f .x/ D log5 x y D f .x/ D log4 2x ! "
y D f .x/ D log1=4 x y D f .x/ D log1=5 x 5q
p D log 15 C
y D f .x/ D log2 .x C 4/ y D f .x/ D ln."x/
y D f .x/ D "2 ln x y D f .x/ D ln.x C 2/ where q is the number of units supplied at a price p per unit. At
what price will the manufacturer supply 1576 units
In Pr b ems e a uate the expressi n
Earthqua e The magnitude, , of an earthqua e and its
log6 36 log2 512 log5 625 energy, E, are related by the equation3
log16 4 log7 7 log 10; 000 ! "
p E
log 0:0001 log2 5 log5 1 1:5 D log
p 2:5 % 1011
1 1
log5 25
log2 log3 7 3 where is given in terms of Richter s preferred scale of 1 5 and
In Pr b ems nd x E is in ergs. Solve the equation for E.
log7 x D 3 log2 x D Biology For a certain population of cells, the number of
cells at time t is given by D 0 .2t=k /, where 0 is the number of
log5 x D 3 log4 x D 0
cells at t D 0 and k is a positive constant. a Find when t D k.
log x D "3 log! x D 1 b What is the significance of k c Show that the time it ta es to
ln x D "3 logx 25 D 2 have population 1 can be written
logx D 3 logx 4 D 13 1
t D k log2
logx 14 D "1 logx y D 1 0

log3 x D "3 logx .2x " 3/ D 1 Inferior Good In a discussion of an inferior good, Pers y4
logx .12 " x/ D 2 log6 36 D x " 1 solves an equation of the form
2 C log2 4 D 3x " 1 log3 .x C 2/ D "2 x22
logx .2x C / D 2 u0 D A ln.x1 / C
logx .16 " 4x " x2 / D 2 2
In Pr b ems nd x and express y ur ans er in terms f for x1 , where x1 and x2 are quantities of two products, u0 is a
natura garithms measure of utility, and A is a positive constant. etermine x1 .
e5x D 7 0:1e0:1x D 0:5 Radioactive Decay A 1-gram sample of radioactive lead
211 .211 Pb/ decays according to the equation D e!0:01 20t ,
e2x!5 C 1 D 4 6e2x " 1 D 12
where is the number of grams present after t minutes. How long
In Pr b ems use y ur ca cu at r t nd the appr ximate will it ta e until only 0.25 grams remain Express the answer to
a ue f each expressi n c rrect t e decima p aces the nearest tenth of a minute.
ln 11 ln 3:1 Radioactive Decay The half-life of radioactive actinium
ln 7:3 ln : 227 .227 Ac/ is approximately 21.70514 years. If a lab currently
has a 100-milligram sample, how many milligrams will it have
Appreciation Suppose an antique gains 10 in value every
one year from now
year. raph the number of years it is owned as a function of the

3
. E. ullen, An Intr ducti n t the he ry f Seism gy (Cambridge, U. .:
Cambridge at the University Press, 1 63).
4 A. . Pers y, An Inferior ood and a Novel Indifference ap, he
American Ec n mist, I , no. 1 (Spring 1 5).
194 C onent a an o ar t c nct ons

If logy x D 3 and logz x D 2, find a formula for z as an explicit Find the x-intercept of y D x3 ln x.
function of y only.
Use the graph of y D ex to estimate ln 2 to two decimal places
Solve for y as an explicit function of x if
Use the graph of y D ln x to estimate e2 to two decimal places.
x C 3e2y " D 0 etermine the x-values of points of intersection of the graphs
of y D .x " 2/2 and y D ln x. Round your answers to two decimal
Suppose y D f .x/ D x ln x. a For what values of x is y < 0
places.
( int etermine when the graph is below the x-axis.)
b etermine the range of f.

Objective L
o st as c ro ert es of The logarithmic function has many important properties. For example,
o ar t c f nct ons

1 logb .mn/ D logb m C logb n

which says that the logarithm of a product of two numbers is the sum of the logarithms
of the numbers. We can prove this property by deriving the exponential form of the
equation:

blogb mClogb n D mn

Using first a familiar rule for exponents, we have

blogb mClogb n D blogb m blogb n

D mn

where the second equality uses two instances of the fundamental equation (2) of Section
4.2. We will not prove the next two properties, since their proofs are similar to that of
Property 1.

m
2 logb D logb m " logb n
n
That is, the logarithm of a quotient is the difference of the logarithm of the numerator
and the logarithm of the denominator.

Table . C L
3 logb mr D r logb m
x log x x log x
That is, the logarithm of a power of a number is the exponent times the logarithm
2 0.3010 7 0. 451 of the number.
3 0.4771 0. 031
4 0.6021 0. 542
Table 4.4 gives the values of a few common logarithms. ost entries are approxi-
5 0.6 0 10 1.0000 mate. For example, log 4 ! 0:6021, which means 100:6021 ! 4. To illustrate the use of
6 0.77 2 e 0.4343 properties of logarithms, we will use this table in some of the examples that follow.

E AM LE F L U T

a Find log 56.

S og 56 is not in the table. ut we can write 56 as the product # 7. Thus,
by Property 1,

log 56 D log. # 7/ D log C log 7 ! 0: 031 C 0: 451 D 1:74 2

 Section 4.3 Pro ert es of o ar t s 195

b Find log 2 .
S y Property 2,
Although the logarithms in Example 1
can be found with a calculator, we will log D log " log 2 ! 0: 542 " 0:3010 D 0:6532
ma e use of properties of logarithms. 2
c Find log 64.
2
S Since 64 D , by Property 3,
2
log 64 D log D 2 log ! 2.0: 031/ D 1: 062
p
d Find log 5.
S y Property 3, we have
p 1 1
log 5 D log 51=2 D log 5 ! .0:6 0/ D 0:34 5
2 2
16
e Find log .
21
16
S log D log 16 " log 21 D log.42 / " log.3 # 7/
21
D 2 log 4 " Œlog 3 C log 7%
! 2.0:6021/ " Œ0:4771 C 0: 451% D "0:11 0
Now ork Problem 3 G
E AM LE R L E

1
a Express log in terms of log x.
x2
1
S log 2 D log x!2 D "2 log x Property 3
x
Here we have assumed that x > 0. Although log.1=x2 / is defined for x ¤ 0, the
expression "2 log x is defined only if x > 0. Note that we do have
1
log D log x!2 D "2 log jxj
x2
for all x ¤ 0.
1
b Express log in terms of log x, for x > 0.
x
S y Property 3,
1
log D log x!1 D "1 log x D " log x
x
Now ork Problem 21 G
From Example 2(b), we see that log.1=x/ D " log x. eneralizing gives the fol-
lowing property:

1
4 logb D " logb m
m
That is, the logarithm of the reciprocal of a number is the negative of the logarithm
of the number.

2 3
For example, log D " log .
3 2
196 C onent a an o ar t c nct ons

E AM LE L T S L

anipulations such as those in x

Example 3 are frequently used in a Write ln in terms of ln x, ln z, and ln .
z
calculus.
x
S ln D ln x " ln.z / Property 2
z
D ln x " .ln z C ln / Property 1
D ln x " ln z " ln

s
3 x5 .x " 2/
b Write ln in terms of ln x, ln.x " 2/, and ln.x " 3/.
x"3
s
' 5 (1=3
S
5
3 x .x " 2/ x .x " 2/ 1 x5 .x " 2/
ln D ln D ln
x"3 x"3 3 x"3
1
D flnŒx5 .x " 2/ % " ln.x " 3/g
3
1
D Œln x5 C ln.x " 2/ " ln.x " 3/%
3
1
D Œ5 ln x C ln.x " 2/ " ln.x " 3/%
3

Now ork Problem 29 G

A L IT I E AM LE C L
The Richter scale measure !of an "
I a Write ln x " ln.x C 3/ as a single logarithm.
earthqua e is given by R D log ,
I0
x
where I is the intensity of the earth- S ln x " ln.x C 3/ D ln Property 2
qua e and I0 is the intensity of a xC3
zero-level earthqua e. How much more
on the Richter scale is an earthqua e b Write ln 3 C ln 7 " ln 2 " 2 ln 4 as a single logarithm.
with intensity 00,000 times the inten-
sity of a zero-level earthqua e than an S ln 3 C ln 7 " ln 2 " 2 ln 4
earthqua e with intensity 000 times
the intensity of a zero-level earthqua e D ln 3 C ln 7 " ln 2 " ln.42 / Property 3
Write the answer as an expres-
sion involving logarithms. Simplify D ln 3 C ln 7 " Œln 2 C ln.42 /%
the expression by combining loga- D ln.3 # 7/ " ln.2 # 42 / Property 1
rithms, and then evaluate the resulting
expression. D ln 21 " ln 32
21
D ln Property 2
32

Now ork Problem 37 G

Since b0 D 1 and b1 D b, by converting to logarithmic forms we have the following
properties:

5 logb 1 D 0

6 logb b D 1
 Section 4.3 Pro ert es of o ar t s 197

E AM LE S L E
A L IT I
If an earthqua e is 10,000 times a Find ln e3x .
as intense as a zero-level earthqua e,
S y the fundamental equation (1) of Section 4.2 with b D e, we have
what is its measurement on the Richter
ln e3x D 3x.
scale Write the answer as a logarithmic
expression and simplify it. (See the pre- b Find log 1 C log 1000.
ceding Apply It for the formula.)
S y Property 5, log 1 D 0. Thus,

log 1 C log 1000 D 0 C log 103

D0C3 Fundamental equation (1) of
D3 Section 4.2 with b D 10
p
c Find log7 7.
p
S log7 7 D log7 7 =
D

! "
27
d Find log3 .
1
! " ! "
27 33
S log3 D log3 D log3 .3!1 / D "1
1 34
1
e Find ln e C log .
10
1
S ln e C log D ln e C log 10!1
10
D 1 C ."1/ D 0

Now ork Problem 15 G

2 2
o not confuse ln x with .ln x/ . We have

ln x2 D ln.x # x/

but

.ln x/2 D .ln x/.ln x/

Sometimes .ln x/2 is written as ln2 x. This is not a new formula but merely a nota-
tion. ore generally, some people write f 2 .x/ for . f.x//2 . We recommend avoiding
the notation f 2 .x/.

E AM LE U E S
2
a Find eln x .
2
S y (2) with b D e, eln x D x2 .
2
b Solve 10log x D 25 for x.
2
S 10log x D 25
x2 D 25 y Equation (2) of Section 4.2
xD˙5

Now ork Problem 45 G

198 C onent a an o ar t c nct ons

E AM LE E L B

Use a calculator to find log5 2.

S Calculators typically have eys for logarithms in base 10 and base e, but not
for base 5. However, we can convert logarithms in one base to logarithms in another
base. et us convert from base 5 to base 10. First, let x D log5 2. Then 5x D 2. Ta ing
the common logarithms of both sides of 5x D 2 gives
log 5x D log 2
x log 5 D log 2
log 2
xD ! 0:4307
log 5
If we had ta en natural logarithms of both sides, the result would be
x D .ln 2/=.ln 5/ ! 0:4307, the same as before.
G
eneralizing the method used in Example 7, we obtain the so-called change of
base formula:

C B F
loga m
7 logb m D
loga b

Some students find the change-of-base formula more memorable when it is

expressed in the form
.loga b/.logb m/ D loga m

in which the two instances of b apparently cancel. et us see how to prove this iden-
tity, for the ability to see the truth of such statements greatly enhances one s ability to
use them in practical applications. Since loga m D y precisely if ay D m, our tas is
equivalently to show that

a.loga b/.logb m/ D m

and we have
% &log m
a.loga b/.logb m/ D aloga b b
D blogb m
Dm

using a rule for exponents and fundamental equation (2) twice.

The change-of-base formula allows logarithms to be converted from base a to
base b.

E AM LE C B F

Express log x in terms of natural logarithms.

S We must transform from base 10 to base e. Thus, we use the change-of-base
formula (Property 7) with b D 10, m D x, and a D e:
loge x ln x
log x D log10 x D D
loge 10 ln 10

Now ork Problem 49 G

 Section 4.3 Pro ert es of o ar t s 199

R BLEMS
In Pr b ems et log 2 D a log 3 D b and log 5 D c In Pr b ems nd x
Express the indicated garithm in terms f a b and c 2 !5x/
eln.x D "15 4log4 xClog4 2 D 3
2 5
log 30 log 1024 log log 10log.x
2 C2x/
D3 e3 ln x D
3 2
27 6
log log log 100 In Pr b ems rite each expressi n in terms f natura
5 25
garithms
log 0:00003 log2 3 log2 3
log2 .2x C 1/ log2 .3x2 C 3x C 3/
In Pr b ems determine the a ue f the expressi n ith ut log3 .x2 C 1/ log7 .x2 C 1/
the use f a ca cu at r
p If eln z D 7ey , solve for y in terms of z.
log7 74 log11 .11 3 11/7 log 0:0000001
1 Statistics In statistics, the sample regression equation
10log 3:4 ln e2:77 ln e ln p
e y D abx is reduced to a linear form by ta ing logarithms of both
log3 1 1
log 10 C ln e3 eln e sides. Express log y in terms of x, log a, and log b and explain
what is meant by saying that the resulting expression is
In Pr b ems rite the expressi n in terms f ln x linear.
ln.x C 1/ and ln.x C 2/ Logarithm of a Sum In a study of military enlistments,
p5
x rown5 considers total military compensation C as the sum of
ln.x.x C 1/2 / ln
.x C 1/3 basic military compensation B (which includes the value of
x2 allowances, tax advantages, and base pay) and educational
ln ln.x.x C 1//3 benefits E. Thus, C D B C E. rown states that
.x C 1/3
! "!3 ! "
xC1 p E
ln 2 ln x.x C 1/.x C 2/ ln C D ln.B C E/ D ln B C ln 1 C
x .x C 2/ B
x.x C 1/ x2 .x C 1/ Verify this but explain why it is not really a formula for the
ln ln
x C 2p xC2 logarithm of a sum.
x x5
ln ln Earthqua e According to Richter,6 the magnitude of an
2
.x C 1/ .x C 2/ 3 .x C 1/2 .x C 2/3 earthqua e occurring 100 m from a certain type of seismometer
0 s 1 s is given by D log.A/ C 3, where A is the recorded trace
1 x 2 x2 .x C 2/3
ln @
5
A ln 4 amplitude (in millimeters) of the qua e. a Find the magnitude
xC2 xC1 .x C 1/5 of an earthqua e that records a trace amplitude of 10 mm.
b If a particular earthqua e has amplitude A1 and magnitude
In Pr b ems express each f the gi en f rms as a sing e 1 , determine the magnitude of a qua e with amplitude 10A1 in
garithm terms of 1 .
log 6 C log 4 log3 10 " log3 5 isplay the graph of y D log4 x.
1 isplay the graph of y D log4 .x C 2/.
3 log2 .2x/ " 5 log2 .x C 2/ 2 log x " log.x " 2/
2 ln x
isplay the graphs of y D log x and y D
7 log3 5 C 4 log3 17 5.2 log x C 3 log y " 2 log z/ ln 10
on the same screen. The graphs appear to be identical. Why
2 C 10 log 1:05
1 n the same screen, display the graphs of y D ln x and
.2 log 13 C 7 log 5 " 3 log 2/ y D ln.2x/. It appears that the graph of y D ln.2x/ is the graph of
3
y D ln x shifted upward. etermine algebraically the value of this
In Pr b ems determine the a ues f the expressi ns shift.
ith ut using a ca cu at r n the same screen, display the graphs of y D ln.2x/
e4 ln 3!3 ln 4 and y D ln.6x/. It appears that the graph of y D ln.6x/ is the graph
p p p p of y D ln.2x/ shifted upward. etermine algebraically the value of
log3 .ln. 7 C e3 C 7/ C ln. 7 C e3 " 7// this shift.
log6 54 " log6
p p p
log3 3 " log2 3 2 " log5 4 5

5
C. rown, ilitary Enlistments: What Can We earn from eographic
Variation he American Ec n mic Re ie 75, no. 1 (1 5), 22 34.
6
C. F. Richter, E ementary Seism gy (San Francisco: W. H. Freeman and
Company, 1 5 ).
200 C onent a an o ar t c nct ons

Objective L E E
o e e o tec n es for so n Here we solve garithmic and exp nentia equati ns. A logarithmic equation is an
o ar t c an e onent a e at ons
equation that involves the logarithm of an expression containing an un nown. For
example, 2 ln.x C 4/ D 5 is a logarithmic equation. n the other hand, an exponential
equation has the un nown appearing in an exponent, as in 23x D 7.
To solve some logarithmic equations, it is convenient to use the fact that, for any
base b, the function y D logb x is one-to-one. This means, of course, that
if logb m D logb n; then m D n
This follows from the fact that the function y D logb x has an inverse and is visually
apparent by inspecting the two possible shapes of y D logb x given in Figure 4.1 . In
either event, the function passes the horizontal line test of Section 2.5. Also useful for
solving logarithmic and exponential equations are the fundamental equations (1) and
(2) in Section 4.2.

E AM LE C

An experiment was conducted with a particular type of small animal.7 The logarithm of
the amount of oxygen consumed per hour was determined for a number of the animals
and was plotted against the logarithms of the weights of the animals. It was found that
log y D log 5: 34 C 0: 5 log x
where y is the number of microliters of oxygen consumed per hour and x is the weight
of the animal (in grams). Solve for y.
S We first combine the terms on the right side into a single logarithm:
log y D log 5: 34 C 0: 5 log x
D log 5: 34 C log x0: 5
Property 3 of Section 4.3
0: 5
log y D log.5: 34x / Property 1 of Section. 4.3
Since log is one-to-one, we have
y D 5: 34x0: 5

Now ork Problem 1 G

A L IT I
reg too a number and multiplied E AM LE S E E
it by a power of 32. ean started with
the same number and got the same result Find x if .25/xC2 D 53x!4 .
when she multiplied it by 4 raised to
S Since 25 D 52 , we can express both sides of the equation as powers of 5:
a number that was nine less than three
times the exponent that reg used. What .25/xC2 D 53x!4
power of 32 did reg use .52 /xC2 D 53x!4
52xC4 D 53x!4
A L IT I Since 5x is a one-to-one function,
The sales manager at a fast-food 2x C 4 D 3x " 4
chain finds that brea fast sales begin to xD
fall after the end of a promotional cam-
paign. The sales in dollars as a function
Now ork Problem 7 G
of the number of days d after the cam-
Some exponential equations can be solved by ta ing the logarithm of both sides
paign s end are given by
after the equation is put in a desirable form. The following example illustrates.
! "!0:1d
4
S D 00 :
3 E AM LE U L S E E
If the manager does not want sales to
drop below 450 per day before starting Solve 5 C .3/4x!1 D 12.
a new campaign, when should he start
such a campaign 7
R. W. Poole, An Intr ducti n t uantitati e Ec gy (New or : c raw-Hill, 1 74).
 Section 4.4 o ar t c an onent a at ons 201

S We first isolate the exponential expression 4x!1 on one side of the equation:
5 C .3/4x!1 D 12
.3/4x!1 D 7
7
4x!1 D
3
Now we ta e the natural logarithm of both sides:
ln 4x!1 D ln 7 " ln 3
Simplifying gives
.x " 1/ ln 4 D ln 7 " ln 3
ln 7 " ln 3
x"1D
ln 4
ln 7 " ln 3
xD C 1 ! 1:61120
ln 4
Now ork Problem 13 G
In Example 3, we used natural logarithms to solve the given equation. However,
logarithms in any base can be used. If we use common logarithms, we would obtain
log 7 " log 3
xD C 1 ! 1:61120
log 4
p
E AM LE E

12 The demand equation for a product is p D 121!0:1q . Use common logarithms to express
q in terms of p.
p = 121-0.1q S Figure 4.21 shows the graph of this demand equation for q $ 0. As is typical
6 of a demand equation, the graph falls from left to right. We want to solve the equation
for q. Ta ing the common logarithms of both sides of p D 121!0:1q gives

4 8
q log p D log.121!0:1q /
log p D .1 " 0:1q/ log 12
FIGURE raph of the log p
demand equation p D 121!0:1q . D 1 " 0:1q
log 12
log p
0:1q D 1 "
log 12
! "
log p
q D 10 1 "
log 12
Now ork Problem 43 G
To solve some exponential equations involving base e or base 10, such as 102x D 3,
the process of ta ing logarithms of both sides can be combined with the identity
logb br D r fundamental equation (1) from Section 4.2 to transform the equation
into an equivalent logarithmic form. In this case, we have

102x D 3
2x D log 3 logarithmic form
log 3
xD ! 0:23 6
2
202 C onent a an o ar t c nct ons

E AM LE R

In an article concerning predators and prey, Holling refers to an equation of the form
y D .1 " e!ax /
where x is the prey density, y is the number of prey attac ed, and and a are constants.
Verify his claim that

lnD ax
"y
S To find ax, we first solve the given equation for e!ax :
y D .1 " e!ax /
y
D 1 " e!ax
y
e!ax D 1 "
"y
e!ax D
Now we convert to logarithmic form:
"y
ln D "ax
"y
" ln D ax

ln D ax Property 4 of Section 4.3

"y
as was to be shown.
Now ork Problem 9 G
Some logarithmic equations can be solved by rewriting them in exponential forms.

E AM LE S L E
A L IT I
The Richter scale measure !of an " Solve log2 x D 5 " log2 .x C 4/.
I
earthqua e is given by R D log , S Here we must assume that both x and x C 4 are positive, so that their log-
I0
where I is the intensity of the earth- arithms are defined. oth conditions are satisfied if x > 0. To solve the equation, we
qua e, and I0 is the intensity of a zero- first place all logarithms on one side so that we can combine them:
level earthqua e. An earthqua e that log2 x C log2 .x C 4/ D 5
is 675,000 times as intense as a zero-
level earthqua e has a magnitude on the log2 .x.x C 4// D 5
Richter scale that is 4 more than another
earthqua e. What is the intensity of the In exponential form, we have
other earthqua e x.x C 4/ D 25
x2 C 4x D 32
x2 C 4x " 32 D 0 quadratic equation
.x " 4/.x C / D 0
x D 4 or xD"
ecause we must have x > 0, the only solution is 4, as can be verified by substituting
into the original equation. Indeed, replacing x by 4 in log2 x gives log2 4 D log2 22 D 2

C. S. Holling, Some Characteristics of Simple Types of Predation and Parasitism, he Canadian

Ent m gist 1, no. 7 (1 5 ), 3 5 .
 Section 4.4 o ar t c an onent a at ons 203

while replacing x by 4 in 5 " log2 .x C 4/ gives 5 " log2 .4 C 4/ D 5 " log2 . / D

5 " log2 23 D 5 " 3 D 2. Since the results are the same, 4 is a solution of the equation.
In solving a logarithmic equation, it is a good idea to chec for extraneous solu-
tions.
Now ork Problem 5 G

R BLEMS
In Pr b ems nd x r unded t three decima p aces Here is the percentage of a country s gross national product
log.7x C 2/ D log.5x C 3/ log x " log 5 D log 7 ( NP) that corresponds to foreign trade (exports plus imports),
and P is the country s population (in units of 100,000). Verify the
log 7 " log.x " 1/ D log 4 log2 x C 3 log2 2 D log2 2x claim that
ln."x/ D ln.x2 " 6/ ln.x C 7:5/ C ln 2 D 2 ln x
!0:1334 log P/
D 50P.0:206
e2x # e5x D e14 .e3x!2 /3 D e3 . 1/4x D
1 3 ou may assume that log 50 D 1:7. Also verify that, for any base
.27/2xC1 D e3x D 11 e4x D 4
3 b, .logb x/2 D logb .xlogb x /.
2e5xC2 D 17 5e2x!1 " 2 D 23 104=x D 6
5 Radioactivity The number of milligrams of a radioactive
7.10/0:2x D7 substance present after t years is given by
D3
5 102x
2.10/x C .10/xC1 D 4 2x D 5 D 100e!0:035t
72xC3 D 35xC7 D 11
4 a How many milligrams are present after 0 years
4x=2 D 20 2!2x=3 D
5 b After how many years will there be 20 milligrams present
5.3x " 6/ D 10 .4/5 3!x
"7D2 ive your answer to the nearest year.

127
D 67 log.x " 3/ D 3 Blood Sample n the surface of a glass slide is a grid that
3x divides the surface into 225 equal squares. Suppose a blood
log2 .x C 1/ D 4 log4 . x " 4/ D 2 sample containing red cells is spread on the slide and the cells
log4 .2x C 4/ " 3 D log4 3 ln.x C 3/ C ln.x C 5/ D 1 are randomly distributed. Then the number of squares containing
no cells is (approximately) given by 225e! =225 . If 100 of the
log.x " 3/ C log.x " 5/ D 1 squares contain no cells, estimate the number of cells the blood
sample contained.
log2 .5x C 1/ D 4 " log2 .3x " 2/
log.x C 2/2 D 2; where x > 0
! "
2
log2 D 3 C log2 x log.x C 1/ D log.x C 2/ C 1
x
Rooted Plants In a study of rooted plants in a certain
geographic region, it was determined that on plots of size A (in
square meters), the average number of species that occurred was S.
When log S was graphed as a function of log A, the result was
Population In Springfield the population P grows at
a straight line given by
the rate of 2 per year. The equation P D 1; 500; 000.1:02/t gives
the population t years after 2015. Find the value of t for which the
log S D log 12:4 C 0:26 log A
population will be 1, 00,000. ive your answer to the nearest
Solve for S. tenth of a year.
Mar et Penetration In a discussion of mar et penetration
Gross ational Product In an article, Taagepera and by new products, Hurter and Rubenstein11 refer to the function
Hayes10 refer to an equation of the form
q " pe!.tCC/.pCq/
2 F.t/ D
log D 1:7 C 0:206 log P " 0:1334.log P/ qŒ1 C e.tCC/.pCq/"

11
R. W. Poole, An Intr ducti n t uantitati e Ec gy (New or : A. P. Hurter, r., A. H. Rubenstein et al., ar et Penetration by New
c raw-Hill, 1 74). Innovations: The Technological iterature, echn gica F recasting and
10 S cia Change 11 (1 7 ), 1 7 221.
R. Taagepera and . P. Hayes, How Trade NP Ratio ecreases with
Country Size, S cia Science Research 6 (1 77), 10 32.
204 C onent a an o ar t c nct ons

x
where p, q, and C are constants. They claim that if F.0/ D 0, then Also, for any A and suitable b and a, solve y D Aba for x and
explain how the previous solution is a special case.
1 q Learning Equation Suppose that the daily output of units
CD" ln
pCq p of a new product on the tth day of a production run is given by
q D q.t/ D 100.1 " e!0:1t /
Show that their claim is true.
Demand Equation The demand equation for a consumer Such an equation is called a earning equati n and indicates that
product is q D 0 " 2p . Solve for p and express your answer in as time increases, output per day will increase. This may be due to
terms of common logarithms, as in Example 4. Evaluate p to two a gain in a wor er s proficiency at his or her ob. etermine, to the
decimal places when q D 60. nearest complete unit, the output a initially (that is when t D 0),
b on the first day, and c on the second day. d After how many
Investment The equation A D P.1:105/t gives the value days, correct to the nearest whole day, will a daily production run
A at the end of t years of an investment of P dollars compounded of 0 units be reached e Will production increase indefinitely
annually at an annual interest rate of 10.5 . How many years will
it ta e for an investment to double ive your answer to the Verify that 4 is the only solution to the logarithmic equation in
nearest year. Example 6 by graphing the function
Sales After t years the number of units of a product sold y D 5 " log2 .x C 4/ " log2 x
% &0: t
per year is given by q D 1000 12 . Such an equation is called a and observing when y D 0.
G mpertz equati n and describes natural growth in many areas of
Solve 23xC0:5 D 17. Round your answer to two decimal
study. Solve this equation for t in the same manner as in
places.
Example 4, and show that
! " Solve ln.x C 2/ D 5 " x. Round your answer to two decimal
3 " log q places.
log
log 2 raph the equation .3/2y " 4x D 5. ( int Solve for y as a
tD
log 0: function of x.)

Chapter 4 Review
I T S E
S Exponential Functions
exponential function, bx , for b > 1 and for 0 < b < 1 Ex. 2,3, p. 177,17
compound interest principal compound amount Ex. 6, p. 1 0
interest period periodic rate nominal rate
e natural exponential function, ex Ex. , p. 1 3
exponential law of decay initial amount decay constant half-life Ex. 11, p. 1 6
S Logarithmic Functions
logarithmic function, logb x common logarithm, log x Ex. 5, p. 1 1
natural logarithm, ln x Ex. 5, p. 1 1
S Properties of Logarithms
change-of-base formula Ex. , p. 1
S Logarithmic and Exponential Equations
logarithmic equation exponential equation Ex. 1, p. 200

S
An exponential function has the form f.x/ D bx . The graph The irrational number e ! 2:71 2 provides the most
of y D bx has one of two general shapes, depending on the important base for an exponential function. This base occurs
value of the base b. (See Figure 4.3.) The compound interest in economic analysis and many situations involving growth
formula of populations or decay of radioactive elements. Radioactive
elements follow the exponential law of decay,

S D P.1 C r/n D 0e
!"t

where is the amount of an element present at time t, 0

expresses the compounded future amount S of a principal is the initial amount, and $ is the decay constant. The time
P at periodic rate r, as an exponential function of the number required for half of the amount of the element to decay is
of interest periods n. called the half-life and denoted by .
 Chapter 4 e e 205

The logarithmic function is the inverse function of the logb mr D r logb m

exponential function, and vice versa. The logarithmic func- 1
tion with base b is denoted logb , and y D logb x if and logb D " logb m
only if by D x. The graph of y D logb x has one of two m
general shapes, depending on the value of the base b. (See logb 1 D 0
Figure 4.1 .) ogarithms with base e are called natural log- logb b D 1
arithms and are denoted ln those with base 10 are called
common logarithms and are denoted log. The half-life of logb br D r
a radioactive element can be given in terms of a natural blogb m D m
logarithm and the decay constant: D .ln 2/=$. loga m
Some important properties of logarithms are the logb m D
loga b
following:
logb .mn/ D logb m C logb n oreover, if logb m D logb n, then m D n. Similarly, if
m bm D bn , then m D n. any of these properties are used
logb D logb m " logb n in solving logarithmic and exponential equations.
n

R
In Pr b ems rite each exp nentia f rm garithmica y and Use natural logarithms to determine the value of log4 5.
each garithmic f rm exp nentia y p !
5
2
35 D 243 log7 343 D 3 log 1 3 D 14 If ln 2 D x and ln 3 D y, express ln in terms of x
1
105 D 100;000 e7 ! 10 6:63 log D1 and y.
p
In Pr b ems nd the a ue f the expressi n ith ut using a x2 3 x C 1
Express log p in terms of log x; log.x C 1/, and
ca cu at r 5 2
x C2
2
1 log.x C 2/.
log5 3125 log4 16 log3 1
1 Simplify 10log x C log 10x C log 10.
log1=5 625
log1=3 log 1 3
1
Simplify log C log 1000.
In Pr b ems nd x ith ut using a ca cu at r 1000
log5 625 D x logx 11 D "4 log2 x D "10 If ln y D x2 C 2, find y.
ln 1e D x ln.5x C 7/ D 0 eln.xC4/ D 7 S etch the graphs of y D .1=3/x and y D log1=3 x.
S etch the graph of y D 2xC3 .
In Pr b ems and et log 2 D a and log 3 D b Express the
gi en garithm in terms f a and b S etch the graph of y D "2 log2 x.
1024
log 000 log p In Pr b ems nd x
5
3
log.6x " 2/ D log. x " 10/ log 3x C log 3 D 2
In Pr b ems rite each expressi n as a sing e garithm 1
3x
2 D 16 xC1
43!x D 16
3 log 7 " 2 log 5 3 log2 x C 5 log2 y C 7 log z % &
log x C log.10x/ D 3 ln x!5 D ln 6
x!1
2 ln x C ln y " 3 ln z log6 2 " log6 4 " log6 3
ln.logx 3/ D 2 log3 x C log x D 5
1
3
ln x C 3 ln.x2 / " 2 ln.x " 1/ " 3 ln.x " 2/
4 log x C 2 log y " 3.log z C log / In Pr b ems nd x c rrect t three decima p aces
e3x D 14 103x=2 D 5 5.exC2 " 6/ D 10
In Pr b ems rite the expressi n in terms f ln x ln y and
ln z 7e3x!1 " 2 D 1 5xC1 D 11 35=x D 2
p
x7 y5 x p Investment If 2600 is invested for 6 12 years at 6
ln !3 ln ln 3 xyz
z .yz/2 compounded quarterly, find (a) the compound amount and (b) the
! 4 3 "5 ' r ( ! "3 ! "5 ! compound interest.
x y 1 y x y
ln ln ln
z2 x z y z Investment Find the compound amount of an investment
of 2000 for five years and four months at the rate of 12
Write log3 .x C 5/ in terms of natural logarithms. compounded monthly.
Write log2 .7x3 C 5/ in terms of common logarithms. Find the nominal rate that corresponds to a periodic rate of
We have log2 37 ! 5:20 45 and log2 7 ! 2: 0735. Find 1 16 per month.
log7 37.
206 C onent a an o ar t c nct ons

Bacteria Growth acteria are growing in a culture, and Sediment in ater The water in a idwestern la e
their number is increasing at the rate of 7 per hour. Initially, 500 contains sediment, and the presence of the sediment reduces
bacteria are present. a etermine an equation that gives the the transmission of light through the water. Experiments
number, , of bacteria present after t hours. b How many indicate that the intensity of light is reduced by 10 by
bacteria are present after one hour c After five hours, correct to passage through 20 cm of water. Suppose that the la e is
the nearest integer. uniform with respect to the amount of sediment contained by
Population Growth The population of a small town gr s the water. A measuring instrument can detect light at the
at the rate of "0:5 per year because the out ow of people to intensity of 0.17 of full sunlight. This measuring instrument
nearby cities in search of obs exceeds the birth rate. In 2006 the is lowered into the la e. At what depth will it first cease to
population was 6000. a etermine an equation that gives the record the presence of light ive your answer to the
population, P, t years from 2006. b Find what the population nearest 10 cm.
will be in 2016 (be careful to express your answer as an integer).
Revenue ue to ineffective advertising, the leer- ut
Razor Company finds that its annual revenues have been cut
sharply. oreover, the annual revenue, R, at the end of t years of
business satisfies the equation R D 200;000e!0:2t . Find the annual
revenue at the end of two years and at the end of three years.

Body Cooling In a discussion of the rate of cooling of

isolated portions of the body when they are exposed to low
temperatures, there occurs the equation12

t " e D. t " e/ e!at

where t is the temperature of the portion at time t; e is the

environmental temperature, the subscript refers to the initial
temperature difference, and a is a constant. Show that
1 . t " e/
aD ln
t t" e
Radioactivity A radioactive substance decays according Depreciation An alternative to straight-line depreciation
to the formula is dec ining ba ance depreciation. This method assumes that an
D 10e!0:41t item loses value more steeply at the beginning of its life than later
on. A fixed percentage of the value is subtracted each month.
where is the number of milligrams present after t hours. Suppose an item s initial cost is C and its useful life is months.
a etermine the initial amount of the substance present. b To Then the value, (in dollars), of the item at the end of n months is
the nearest tenth of a milligram, determine the amount present given by
after 1 hour and c after 5 hours. d To the nearest tenth of an ! "
hour, determine the half-life of the substance, and e determine 1 n
DC 1"
the number of hours for 0.1 milligram to remain.
Radioactivity If a radioactive substance has a half-life of so that each month brings a depreciation of 100 percent.
10 days, in how many days will 1 of the initial amount be present (This is called sing e dec ining ba ance depreciati n if the
Mar eting A mar eting-research company needs to annual depreciation were 200 percent, then we would spea of
determine how people adapt to the taste of a new toothpaste. In d ub e dec ining ba ance depreciati n.) A noteboo computer is
one experiment, a person was given a pasted toothbrush and was purchased for 1500 and has a useful life of 36 months. It
as ed periodically to assign a number, on a scale from 0 to 10, to undergoes double-declining-balance depreciation. After how
the perceived taste. This number was called the resp nse many months, to the nearest integer, does its value drop below
magnitude. The number 10 was assigned to the initial taste. After 500
conducting the experiment several times, the company estimated ln x
that the response magnitude is given by If y D f .x/ D , determine the range of f. Round values to
x
two decimal places.
R D 10e!t=50
etermine the points of intersection of the graphs of y D ln x
where t is the number of seconds after the person is given the and y D x " 2 with coordinates rounded to two decimal places.
toothpaste. a Find the response magnitude after 20 seconds,
correct to the nearest integer. b After how many seconds, Solve ln x D 6 " 2x. Round your answer to two decimal
correct to the nearest second, does a person have a response places.
magnitude of 3 Solve 63!4x D 15. Round your answer to two decimal places.

12
R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or :
c raw-Hill, 1 55).
 Chapter 4 e e 207

Bases We have seen that there are two inds of bases, b, for either ind corresponds to exactly one of the other ind. Who
exponential and logarithmic functions: those b in .0; 1/ and those would have thought it .1; 1/ so many numbers .0; 1/ so
b in .1; 1/. It might be supposed that there are m re of the second little space.
ind but this is not the case. Consider the function f: (0,1) & (1, isplay the graph of the equation .6/5y C x D 2. ( int Solve
1) given by f .x/ D 1=x. for y as an explicit function of x.)
a Show that the domain of f can be ta en to be .0; 1/. 2x
b Show that with domain .0; 1/ the range of f is .1; 1/. raph y D 2x and y D on the same screen. It appears that
c Show that f has an inverse g and determine a formula for g.x/. 2x
the graph of y D is the graph of y D 2x shifted three units to
The exercise shows that the numbers in .0; 1/ are in one-to-one
the right. Prove algebraically that this is true.
correspondence with the numbers in .1; 1/ so that every base of
 at e at cs
of nance

F
or people who li e cars and can afford a good one, a trip to an auto dealership
5.1 Co o n nterest can be a lot of fun. However, buying a car also has a side that many people
5.2 Present a e find unpleasant: the negotiating. The verbal tug-of-war with the salesperson
is especially di cult if the buyer is planning to pay on an installment plan and
5.3 nterest Co o n e does not understand the numbers being quoted.
Cont n o s How, for instance, does the fact that the salesperson is offering the car for 12, 00
5.4 nn t es translate into a monthly payment of 2 1.54 The answer is amortization. The term
comes via French from the atin root m rt meaning dead , from which we also
5.5 ort at on of oans get m rta and m rti ed. A debt that is gradually paid off is eventually illed, and
the payment plan for doing this is called an amortization schedule. The schedule is
5.6 Per et t es
determined by a formula we give in Section 5.4 and apply in Section 5.5.
C er 5 e e Using the formula, we can calculate the monthly payment for the car. If the buyer
ma es a 00 down payment on a 12, 00 car and pays off the rest over four years
at 4. APR compounded monthly, the monthly payment for principal and interest
only should be 272. 7. If the payment is higher than that, it may contain additional
charges such as sales tax, registration fees, or insurance premiums, which the buyer
should as about because some of them may be optional. Understanding the mathe-
matics of finance can help consumers ma e more informed decisions about purchases
and investments.

208
 Section 5. Co o n nterest 209

Objective C I
o e ten t e not on of co o n In this chapter we model selected topics in finance that deal with the time value of
nterest to nc e e ect e rates an
to so e nterest ro e s ose money, such as investments, loans, and so on. In later chapters, when more mathematics
so t ons re re o ar t s is at our disposal, certain topics will be revisited and expanded.
et us first review some facts from Section 4.1, where the notion of compound
interest was introduced. Under compound interest, at the end of each interest period,
the interest earned for that period is added to the principa (the invested amount) so that
it too earns interest over the next interest period. The basic formula for the value (or
c mp und am unt) of an investment after n interest periods under compound interest
is as follows:

C I F
For an original principal of P, the formula
S D P.1 C r/n
gives the compound amount S at the end of n interest peri ds at the peri dic
rate of r.

The compound amount is also called the accumu ated am unt and the difference
between the compound amount and the original principal, S ! P, is called the
c mp und interest.
Recall that an interest rate is usually quoted as an annua rate, called the n mina
rate or the annua percentage rate (APR). The periodic rate (or rate per interest period)
is obtained by dividing the nominal rate by the number of interest periods per year.
For example, let us compute the compound amount when 1000 is invested for
five years at the nominal rate of compounded quarterly. The rate per peri d is
0:0 =4, and the number of interest periods is 5 " 4.
From Equation (1), we have
! "
0:0 5!4
A calculator is handy while reading this S D 1000 1 C
chapter.
4
D 1000.1 C 0:02/20 # 14 5: 5

A L IT I
E AM LE C I
Suppose you leave an initial amount Suppose that 500 amounted to 5 .3 in a savings account after three years. If
of 51 in a savings account for three interest was compounded semiannually, find the nominal rate of interest, compounded
years. If interest is compounded daily
semiannually, that was earned by the money.
(365 times per year), use a graphing cal-
culator to graph the compound amount S S et r be the semiannual rate. There are 2 " 3 D 6 interest periods. From
as a function of the nominal rate of inter- Equation (1),
est. From the graph, estimate the nomi-
nal rate of interest so that there is 600 500.1 C r/6 D 5 :3
after three years. 5 :3
.1 C r/6 D
500
r
6 5 :3
1CrD
500
r
6 5 :3
rD ! 1 # 0:0275
500
Thus, the semiannual rate was 2.75 , so the nominal rate was 5 12 compounded
semiannually.
G
210 C at e at cs of nance

E AM LE M

At what nominal rate of interest, compounded yearly, will money double in eight years
S et r be the rate at which a principal of P doubles in eight years. Then the
compound amount is 2P. From Equation (1),
P.1 C r/ D 2P
.1 C r/ D 2
p
1CrD 2
p
r D 2 ! 1 # 0:0 05
Note that the doubling rate is
independent of the principal P. Hence, the desired rate is .05 .
G
We can determine how long it ta es for a given principal to accumulate to a par-
ticular amount by using logarithms, as Example 3 shows.

E AM LE C I
A L IT I
Suppose you leave an initial amount How long will it ta e for 600 to amount to 00 at an annual rate of 6 compounded
of 520 in a savings account at an quarterly
annual rate of 5.2 compounded daily
(365 days per year). Use a graphing cal- S The periodic rate is r D 0:06=4 D 0:015. et n be the number of interest
culator to graph the compound amount periods it ta es for a principal of P D 600 to amount to S D 00. Then, from Equa-
S as a function of the interest periods. tion (1),
From the graph, estimate how long it 00 D 600.1:015/n
ta es for the amount to accumulate to
750. 00
.1:015/n D
600
.1:015/n D 1:5
To solve for n, we first ta e the natural logarithms of both sides:
ln.1:015/n D ln 1:5
n ln 1:015 D ln 1:5 since ln mr D r ln m
ln 1:5
nD # 27:233
ln 1:015
The number of years that corresponds to 27.233 quarterly interest periods is
27:233 4 # 6: 0 3, which is about 6 years, 12 months. However, because interest is
calculated quarterly, we must wait until the next fully completed quarter, the t enty
eighth, to realize a principal that is in excess of 00 equivalently, 7 years after the
money was invested.
Now ork Problem 20 G
E R
If P dollars are invested at a nominal rate of 10 compounded quarterly for one year,
the principal will earn more than 10 that year. In fact, the compound interest is
! "
0:10 4
S!PDP 1C ! P D Œ.1:025/4 ! 1!P
4
# 0:103 13P
which is about 10.3 of P. That is, 10.3 is the approximate rate of interest com-
pounded annua y that is actually earned, and that rate is called the effective rate of
interest. The effective rate is independent of P. In general, the effective interest rate is
ust the rate of simp e interest earned over a period of one year. Thus, we have shown
 Section 5. Co o n nterest 211

that the nominal rate of 10 compounded quarterly is equivalent to an effective rate of

10.3 . Following the preceding procedure, we can generalize our result:

E R
The effective rate re that is equivalent to a nominal rate of r compounded n times
a year is given by
# r $n
re D 1 C !1
n

A L IT I
E AM LE E R
An investment is compounded What effective rate is equivalent to a nominal rate of 6 compounded a semiannually
monthly. Use a graphing calculator to and b quarterly
graph the effective rate re as a function
of the nominal rate r. Then use the S
graph to find the nominal rate that is a From Equation (3), the effective rate is
equivalent to an effective rate of .
! "
0:06 2
re D 1 C ! 1 D .1:03/2 ! 1 D 0:060 D 6:0
2
b The effective rate is
! "
0:06 4
re D 1 C ! 1 D .1:015/4 ! 1 # 0:061364 D 6:14
4
Now ork Problem 9 G
Example 4 illustrates that, for a given nominal rate r, the effective rate increases
as the number of interest periods per year .n/ increases. However, in Section 5.3 it
is shown that, regardless of how large n is, the maximum effective rate that can be
obtained is er ! 1, where e is the irrational number introduced in Section 4.1. We recall
that e # 2:71 2 .

E AM LE E R

To what amount will 12,000 accumulate in 15 years if it is invested at an effective rate

of 5
S Since an effective rate is the rate that is compounded annually, we have
S D 12;000.1:05/15 # 24; 47:14

Now ork Problem 15 G

E AM LE M

How many years will it ta e for money to double at the effective rate of r
S et n be the number of years it ta es for a principal of P to double. Then
the compound amount is 2P. Thus,
2P D P.1 C r/n
2 D .1 C r/n
ln 2 D n ln.1 C r/ ta ing logarithms of both sides
212 C at e at cs of nance

Hence,
ln 2
nD
ln.1 C r/
For example, if r D 0:06, the number of years it ta es to double a principal is
ln 2
# 11. years
ln 1:06

Now ork Problem 11 G

We remar that when alternative interest rates are available to an investor, effective
rates are used to compare them that is, to determine which of them is the best.
Example 7 illustrates.

E AM LE C I R
A L IT I
Suppose you have two investment If an investor has a choice of investing money at 6 compounded daily or 6 1 com-
opportunities. ou can invest 10,000 at pounded quarterly, which is the better choice
11 compounded monthly, or you can
invest 700 at 11.25 compounded S
quarterly. Which has the better effec-
tive rate of interest Which is the better S We determine the equivalent effective rate of interest for each nominal
investment over 20 years rate and then compare our results.

The respective effective rates of interest are

! "
0:06 365
re D 1 C ! 1 # 6:1
365
and
! "
0:06125 4
re D 1 C ! 1 # 6:27
4

Since the second choice gives the higher effective rate, it is the better choice (in
spite of the fact that daily compounding may be psychologically more appealing).
Now ork Problem 21 G
N I R
sua y, it is tacitly assumed that, for any interest rate r, we have r $ 0. In 2016, neg
ati e interest rates, r < 0, were in the news (although the notion has been around
for a long time). First, observe that a formula such as that given by Equation (1),
S D P.1 C r/n , ma es perfectly good sense for r < 0. To give a concrete example, if
r D !5 D !0:05 then the base .1 C r/ of the exponential expression in (1) becomes
0: 5 < 1, and it follows that P.1Cr/n is a decreasing function of the number n of inter-
est periods. After 1 interest period an initial amount of 100 is worth 5, after 2 interest
periods it is worth 100.0: 5/2 D 0:25, after 3 periods 100.0: 5/3 D 5:7375,
and so on.
If we are applying the equation S D P.1Cr/n to a deposit of P made to a ban , then
the depositor is the ender and the ban is in the position of the b rr er. Typically,
we expect the borrower to pay the lender, and in the case of positive interest rates this
is indeed the case. If r > 0 is the rate for one interest period, then the borrower pays the
lender r for each dollar borrowed for one interest period. If r < 0 then we can still say
that the borrower pays the lender r for each dollar borrowed for one period, but in the
case that r is negative this amounts to saying that the ender pays the b rr er jrj.
 Section 5. Co o n nterest 213

Why might an individual be willing to deposit money in a ban and pay the ban
to eep it there Well, if the money is sub ect to a large enough tax bill then a ban
account charging less than the tax bill and providing confidentiality might actually
appear attractive. f course many urisdictions actively combat this sort of tax evasion.
The negative interest rates in the news in 2016 were those offered by central ban s
(of countries) to the ordinary commercial ban s with which individuals and companies
do their ban ing. Rates offered by central ban s are usually considered to be a tool to
implement economic policy. When commercial ban s end to the central ban they have
virtually no ris . In di cult economic times very conservative ban s will prefer the
lower rates offered by the ris -free central ban over the higher rates they themselves
offer to ordinary borrowers, often with considerable ris . f course, if individuals and
companies are unable to borrow money then the economy of their country stagnates.
In 2016, central ban s of several countries sought to put more money in the hands
of individuals and businesses to boost their sluggish national economies. y offering
negative interest rates to commercial ban s, the central ban s enc uraged commercial
ban s to put their money in circulation by lending to individuals and companies rather
than hoarding cash at the central ban .

R BLEMS
In Pr b ems and nd (a) the c mp und am unt and ( ) the A 4000 certificate of deposit is purchased for 4000 and is
c mp und interest f r the gi en in estment and rate held for eleven years. If the certificate earns an effective rate of
6000 for eight years at an effective rate of 7 , what is it worth at the end of that period
How many years will it ta e for money to triple at the
750 for 12 months at an effective rate of 7
effective rate of r
In Pr b ems nd the e ecti e rate t three decima p aces niversity Costs Suppose attending a certain university
that c rresp nds t the gi en n mina rate cost 25,500 in the 200 2010 school year. This price included
2.75 compounded monthly tuition, room, board, boo s, and other expenses. Assuming an
effective 3 in ation rate for these costs, determine what the
5 compounded quarterly university costs were in the 2015 2016 school year.
3.5 compounded daily niversity Costs Repeat Problem 15 for an in ation rate of
6 compounded daily 2 compounded quarterly.
Find the effective rate of interest (rounded to three Finance Charge A ma or credit-card company has a
decimal places) that is equivalent to a nominal rate of 10 finance charge of 1 12 per month on the outstanding
compounded indebtedness. a What is the nominal rate compounded monthly
a yearly b semiannually b What is the effective rate
c quarterly d monthly
e daily
Find i the compound interest and ii the effective rate, to
four decimal places, if 1000 is invested for one year at an annual
rate of 5 compounded
a quarterly b monthly
c wee ly d daily
ver a five-year period, an original principal of 2000
accumulated to 2 50 in an account in which interest was
compounded quarterly. etermine the effective rate of interest, How long would it ta e for a principal of P to double if it is
rounded to two decimal places. invested with an APR of 7 , compounded monthly.

Suppose that over a six-year period, 1000 accumulated To what sum will 2000 amount in eight years if invested
to 1 5 in an investment certificate in which interest was at a 6 effective rate for the first four years and at 6
compounded quarterly. Find the nominal rate of interest, compounded semiannually thereafter
compounded quarterly, that was earned. Round your answer How long will it ta e for 100 to amount to 1000 if invested
to two decimal places. at 6 compounded monthly Express the answer in years,
In Pr b ems and nd h many years it u d take t rounded to two decimal places.
d ub e a principa at the gi en e ecti e rate Gi e y ur ans er t An investor has a choice of investing a sum of money
ne decima p ace at compounded annually or at 7. compounded
5 semiannually. Which is the better of the two rates
214 C at e at cs of nance

What nominal rate of interest, compounded monthly, ero Coupon Bond A zer c up n b nd is a bond that is
corresponds to an effective rate of 4.5 sold for less than its face value (that is, it is disc unted) and has
Savings Account A ban advertises that it pays interest no periodic interest payments. Instead, the bond is redeemed for
on savings accounts at the rate of 3 14 compounded daily. Find its face value at maturity. Thus, in this sense, interest is paid at
the effective rate if the ban assumes that a year consists of a maturity. Suppose that a zero-coupon bond sells for 420 and can
360 days or b 365 days in determining the dai y rate. Assume be redeemed in 14 years for its face value of 1000. The bond
that compounding occurs 365 times a year. earns interest at what nominal rate, compounded semiannually

Savings Account Suppose that 700 amounted to 01.06 Misplaced Funds Suppose that 1000 is misplaced in a
in a savings account after two years. If interest was compounded non-interest-bearing chec ing account and forgotten. Each year,
quarterly, find the nominal rate of interest, compounded quarterly, the ban imposes a service charge of 1.5 . After 20 years, how
that was earned by the money. much remains of the 1000 int Recall the notion of negati e
interest rates.
In ation As a hedge against in ation, an investor
purchased a 1 72 ran Torino in 1 0 for 0,000. It was sold in General Solutions Equation (1) can be solved for each of
2000 for 250,000. At what effective rate did the car appreciate the variables in terms of the other three. Find each of P, r, and n in
in value Express the answer as a percentage rounded to three this way. (There is no need to memorize any of the three new
decimal places. formulas that result. The point here is that by showing the general
solutions exist, we gain confidence in our ability to handle any
In ation If the rate of in ation for certain goods is 7 14 particular cases.)
compounded daily, how many years will it ta e for the average
price of such goods to double

Objective
o st resent a e an to so e Suppose that 100 is deposited in a savings account that pays 6 compounded annu-
ro e s n o n t e t e a e of
one s n e at ons of a e o ally. Then at the end of two years, the account is worth
ntro ce t e net resent a e of cas
o s 100.1:06/2 D 112:36

To describe this relationship, we say that the compound amount of 112.36 is the
future value of the 100, and 100 is the present value of the 112.36. Sometimes we
now the future value of an investment and want to find the present value. To obtain
a formula for doing this, we solve the equation S D P.1 C r/n for P. The result is
P D S=.1 C r/n D S.1 C r/"n .

The principal P that must be invested at the periodic rate of r for n interest periods
so that the compound amount is S is given by
P D S.1 C r/"n
and is called the present value of S.

E AM LE

Find the present value of 1000 due after three years if the interest rate is com-
pounded monthly.
S We use Equation (1) with S D 1000, r D 0:0 =12 D 0:0075, and n D
3.12/ D 36:

P D 1000.1:0075/"36 # 764:15

This means that 764.15 must be invested at compounded monthly to have 1000
in three years.
Now ork Problem 1 G
 Section 5.2 Present a e 215

If the interest rate in Example 1 were 10 compounded monthly, the present value
would be
! "
0:1 "36
P D 1000 1 C # 741:74
12
which is less than before. It is typical that the present value for a given future value
decreases as the interest rate per interest period increases.

E AM LE S T F

A trust fund for a child s education is being set up by a single payment so that at the
end of 15 years there will be 50,000. If the fund earns interest at the rate of 7 com-
pounded semiannually, how much money should be paid into the fund
S We want the present value of 50,000, due in 15 years. From Equation (1),
with S D 50;000, r D 0:07=2 D 0:035, and n D 15.2/ D 30, we have
P D 50;000.1:035/"30 # 17; 13: 2
Thus, 17, 13. 2 should be paid into the fund.
Now ork Problem 13 G
E
Suppose that r. Smith owes r. ones two sums of money: 1000, due in two years,
and 600, due in five years. If r. Smith wishes to pay off the total debt now by a
single payment, how much should the payment be Assume an interest rate of
compounded quarterly.
The single payment x due now must be such that it would grow and eventually pay
off the debts when they are due. That is, it must equal the sum of the present values of
the future payments. As shown in the timeline of Figure 5.1, we have
x D 1000.1:02/" C 600.1:02/"20
This equation is called an equati n f a ue. We find that
x # 1257:27
Figure 5.1 is a useful tool for visualizing Thus, the single payment now due is 1257.27. et us analyze the situation in more
the time value of money. Always draw detail. There are two methods of payment of the debt: a single payment now or two
such a timeline to set up an equation of payments in the future. Notice that Equation (2) indicates that the value n of all
value.
payments under one method must equal the value n of all payments under the other
method. In general, this is true not ust n but at any time. For example, if we multiply
both sides of Equation (2) by .1:02/20 , we get the equation of value
x.1:02/20 D 1000.1:02/12 C 600

Year
Year 0 1 2 3 4 5
0 1 2 3 4 5
x 1000 600
Single Value of
x 1000 600 12 periods
payment debts at
8 periods Debt year 5
1000 (1.02)12
Present 1000 (1.02)-8
Debt
value 20 periods Value of
of single payment
20 periods
debts 600 (1.02)-20 x (1.02)20 at year 5

FIGURE Replacing two future payments by a single FIGURE iagram for equation of value.
payment now.

The left side of Equation (3) gives the value five years from now of the single pay-
ment (see Figure 5.2), while the right side gives the value five years from now of all
216 C at e at cs of nance

payments under the other method. Solving Equation (3) for x gives the same result,
x # 1257:27. In general, an equation of value illustrates that when one is considering
two methods of paying a debt (or of ma ing some other transaction), at any time, the
value of all payments under one method must equal the value of all payments under
the other method.
In certain situations, one equation of value may be more convenient to use than
another, as Example 3 illustrates.

E AM LE E

A debt of 3000 due six years from now is instead to be paid off by three payments:
500 now, 1500 in three years, and a final payment at the end of five years. What
would this payment be if an interest rate of 6 compounded annually is assumed
S et x be the final payment due in five years. For computational convenience,
we will set up an equation of value to represent the situation at the end of that time,
for in that way the coe cient of x will be 1, as seen in Figure 5.3. Notice that at year 5
we compute the future values of 500 and 1500, and the present value of 3000. The
equation of value is
500.1:06/5 C 1500.1:06/2 C x D 3000.1:06/"1
so
x D 3000.1:06/"1 ! 500.1:06/5 ! 1500.1:06/2
# 475:6
Thus, the final payment should be 475.6 .
Year
0 1 2 3 4 5 6
500 1500 x 3000

1500 (1.06)2

500 (1.06)5

3000 (1.06)-1

FIGURE Time values of payments for Example 3.

Now ork Problem 15 G

When one is considering a choice of two investments, a comparison should be
made of the value of each investment at a certain time, as Example 4 shows.

E AM LE C I

Suppose that you had the opportunity of investing 5000 in a business such that the
value of the investment after five years would be 6300. n the other hand, you could
instead put the 5000 in a savings account that pays 6 compounded semiannually.
Which investment is better
S et us consider the value of each investment at the end of five years. At that
time the business investment would have a value of 6300, while the savings account
would have a value of 5000.1:03/10 # 671 :5 . Clearly, the better choice is putting
the money in the savings account.
Now ork Problem 21 G
 Section 5.2 Present a e 217

N
If an initial investment will bring in payments at future times, the payments are called
cash ows. The net present value, denoted NPV, of the cash ows is defined to be the
sum of the present values of the cash ows, minus the initial investment. If NPV > 0,
then the investment is profitable if NPV < 0, the investment is not profitable.

E AM LE N

Suppose that you can invest 20,000 in a business that guarantees you cash ows at the
ear Cash Flow end of years 2, 3, and 5 as indicated in the table to the left. Assume an interest rate of
7 compounded annually, and find the net present value of the cash ows.
2 10,000
S Subtracting the initial investment from the sum of the present values of the
3 000
cash ows gives
5 6000
NPV D 10; 000.1:07/"2 C 000.1:07/"3 C 6000.1:07/"5 ! 20; 000
# !457:31

Since NPV < 0, the business venture is not profitable if one considers the time value of
money. It would be better to invest the 20,000 in a ban paying 7 , since the venture
is equivalent to investing only 20;000 ! 457:31 D 1 ;542:6 .
Now ork Problem 19 G

R BLEMS
In Pr b ems nd the present a ue f the gi en future interest rate is compounded semiannually, how much is the
payment at the speci ed interest rate payment
6000 due in 20 years at 5 compounded annually A debt of 7000 due in five years is to be repaid by a payment
3500 due in eight years at 6 effective of 3000 now and a second payment at the end of five years. How
much should the second payment be if the interest rate is
4000 due in 12 years at 7 compounded semiannually
compounded monthly
2020 due in three years at 6 compounded monthly
A debt of 5000 due five years from now and 5000 due ten
000 due in 5 12 years at compounded quarterly years from now is to be repaid by a payment of 2000 in
6000 due in 6 12 years at 10 compounded semiannually two years, a payment of 4000 in four years, and a final payment
at the end of six years. If the interest rate is 2.5 compounded
000 due in five years at 10 compounded monthly annually, how much is the final payment
3
500 due in three years at 4
compounded quarterly A debt of 3500 due in four years and 5000 due in six years
7500 due in two years at 3 14 compounded daily is to be repaid by a single payment of 1500 now and three equal
payments that are due each consecutive year from now. If the
1250 due in 1 12 years at 13 12 compounded wee ly
interest rate is 7 compounded annually, how much are each of
A ban account pays 5.3 annual interest, compounded the equal payments
monthly. How much must be deposited now so that the account
Cash Flows An initial investment of 100,000 in a
contains exactly 12,000 at the end of one year
business guarantees the following cash ows:
Repeat Problem 11 for the nominal rate of 7.1 compounded
semiannually. ear Cash Flow
rust Fund A trust fund for a 10-year-old child is being 25,000
set up by a single payment so that at age 21 the child will receive
27,000. Find how much the payment is if an interest rate of 6 10 25,000
compounded semiannually is assumed. 11 30,000
A debt of 7500 due in five years and 2500 due in seven 12 50,000
years is to be repaid by a single payment now. Find how much the
payment is if the interest rate is 4 compounded quarterly. Assume an interest rate of 4 compounded quarterly.
a Find the net present value of the cash ows.
A debt of 600 due in three years and 00 due in four years b Is the investment profitable
is to be repaid by a single payment two years from now. If the
218 C at e at cs of nance

Cash Flows Repeat Problem 1 for the interest rate of 6

compounded semiannually.
Decision Ma ing Suppose that a person has the following
choices of investing 10,000:
a placing the money in a savings account paying 6
compounded semiannually Find the present value of 50,000 due in 20 years at a ban
b investing in a business such that the value of the investment rate of 5 compounded daily. Assume that the ban uses 360
after years is 16,000. days in determining the daily rate and that there are 365 days in a
Which is the better choice year that is, compounding occurs 365 times in a year.
A owes two sums of money: 1000 plus interest at 7 Promissory ote A pr miss ry n te is a written statement
compounded annually, which is due in five years, and 2000 plus agreeing to pay a sum of money either on demand or at a definite
interest at compounded semiannually, which is due in seven future time. When a note is purchased for its present value at a
years. If both debts are to be paid off by a single payment at the given interest rate, the note is said to be disc unted and the
end of six years, find the amount of the payment if money is worth interest rate is called the disc unt rate. Suppose a 10,000 note
6 compounded quarterly. due eight years from now is sold to a financial institution for
4700. What is the nominal discount rate with quarterly
Purchase Incentive A ewelry store advertises that for compounding
every 1000 spent on diamond ewelry, the purchaser receives
a 1000 bond at absolutely no cost. In reality, the 1000 is the Promissory ote a Repeat Problem 25 with monthly
full maturity value of a zero-coupon bond (see Problem 27 of compounding. b et r be the nominal discount rate in
Problems 5.1), which the store purchases at a heavily reduced Problem 25 and let s be the nominal discount rate in part (a).
price. If the bond earns interest at the rate of 7.5 compounded Prove, without reference to the future value and to the present
quarterly and matures after 20 years, how much does the bond value of the note, that
!r "
cost the store r
s D 12 3 1 C ! 1
4

Objective I C C
o e ten t e not on of co o n We have seen that when money is invested, at a given annual rate, the interest earned
nterest to t e s t at on ere nterest
s co o n e cont n o s o each year depends on how frequently interest is compounded. For example, more inter-
e e o n t s case for as for est is earned if it is compounded monthly rather than semiannually. We can successively
co o n a o nt an resent a e get still more interest by compounding it wee ly, daily, per hour, and so on. However,
for a given annual rate, there is a maximum interest that can be earned by increasing
the compounding frequency, and we now examine it.
Suppose a principal of P dollars is invested for t years at an annual rate of r. If
interest is compounded k times a year, then the rate per interest period is r=k, and
there are kt periods. From Section 4.1, recalled in Section 5.1, the compound amount
is given by
# r $kt
SDP 1C
k
If k, the number of interest periods per year, is increased indefinitely, as we did in the
thought experiment of Section 4.1 to introduce the number e, then the length of each
period approaches 0 and we say that interest is compounded continuously. We can
ma e this precise. In fact, with a little algebra we can relate the compound amount to
the number e. et m D k=r, so that
# ! " !rt !! " "rt !! " "rt
r $kt 1 k=r 1 m mC1 m
P 1C DP 1C DP 1C DP
k k=r m m
! "
nC1 n
In Section 4.1 we noted that, for n a positive integer, the numbers increase
n
as n does but they are nevertheless bounded. (For example, it can be shown that all of
! "
nC1 n
the numbers are less than 3.) We de ned e to be the least real number which
n
! "
nC1 n
is greater than all the values , where n is a positive integer. It can be shown
n
that it is not necessary to require that n be an integer. For any positive m, the numbers
 Section 5.3 nterest Co o n e Cont n o s 219
! "m
mC1
increase as m does but they remain bounded and the number e, as defined
m ! "
mC1 m
in Section 4.1, is the least real number that is greater than all the values .
m
In the case at hand, for fixed r, the numbers m D k=r increase as k (an integer)
does, but the m D k=r are not necessarily integers. However, if one accepts the truth of
!! " "rt
mC1 m
the preceding paragraph, then it follows that the compound amount P
m
rt
approaches the value Pe as k, and hence, m is increased indefinitely and we have the
following:

C A C I
The formula
S D Pert
gives the compound amount S of a principal of P dollars after t years at an annual
interest rate r compounded continuously.

The interest of 5.13 is the maximum E AM LE C A

amount of compound interest that can be
earned at an annual rate of 5 .
If 100 is invested at an annual rate of 5 compounded continuously, find the com-
pound amount at the end of

a 1 year.
b 5 years.

S
a Here P D 100, r D 0:05, and t D 1, so

S D Pert D 100e.0:05/.1/ # 105:13

We can compare this value with the value after one year of a 100 investment at an
annual rate of 5 compounded semiannually namely, 100.1:025/2 # 105:06.
b Here P D 100, r D 0:05, and t D 5, so

S D 100e.0:05/.5/ D 100e0:25 # 12 :40

Now ork Problem 1 G

We can find an expression that gives the effective rate that corresponds to an
annual rate of r compounded continuously. (From Section 5.1, the effective rate is the
rate compounded annually that gives rise to the same interest in a year as does the rate
and compounding scheme under consideration.) If re is the corresponding effective rate,
then, after one year a principal P accumulates to P.1 C re /. This must equal the accu-
mulated amount under continuous interest, Per . Thus, P.1 C re / D Per , from which it
follows that 1 C re D er , so re D er ! 1.

E R C I
The effective rate corresponding to an annual rate of r compounded continuously is
re D e r ! 1
220 C at e at cs of nance

E AM LE E R

Find the effective rate that corresponds to an annual rate of 5 compounded continu-
ously.
S The effective rate is
er ! 1 D e0:05 ! 1 # 0:0513
which is 5.13 .
Now ork Problem 5 G
If we solve S D Pert for P, we get P D S=ert D Se"rt . In this formula, P is the
principal that must be invested now at an annual rate of r compounded continuously so
that at the end of t years the compound amount is S. We call P the present value of S.

C I
The formula
P D Se"rt
gives the present value P of S dollars due at the end of t years at an annual rate of r
compounded continuously.

E AM LE T F

A trust fund is being set up by a single payment so that at the end of 20 years there will
be 25,000 in the fund. If interest is compounded continuously at an annual rate of 7 ,
how much money (to the nearest dollar) should be paid into the fund initially
S We want the present value of 25,000 due in 20 years. Therefore,
P D Se"rt D 25;000e".0:07/.20/
D 25;000e"1:4 # 6165
Thus, 6165 should be paid initially.
Now ork Problem 13 G
C
Terminology concerning rates is confusing because there are many different conven-
tions that depend on financial urisdictions and financial industries. In North America
what we have called the nominal rate is often called the APR, and the effective (annual)
rate is often called the AP , for annua percentage yie d, or the EAR. Whenever one
enters into a financial negotiation, it is always a good idea to ma e sure that all termi-
nology is mutually understood.
When interest is compounded, the effective rate should be used to compare differ-
ent schemes. It should be noted, though, that if a savings account bears interest com-
pounded monthly and has the same effective rate as another which is compounded
quarterly, say, then the account with more frequent compounding is probably a better
choice because the quarterly account may literally pay interest only four times a year.
If depositors want their money three months after an o cial quarter s end, they will
probably get no interest for the last three months. This argument carried to the limit
suggests that one should always opt for continuous compounding.
However, nobody seems able to name a financial institution that actually uses con-
tinuous compounding. Why is this We now that the actual cost of an interest scheme
is determined by the effective rate and, for any effective rate re that is acceptable to a
ban , there corresponds a continuous compounding rate r with re D er ! 1, namely
r D ln.re C1/. Note too that the calculation of future value via ert with continuous com-
pounding is truly easier than via .1 C r=n/nt . While the second expression is (usually)
 Section 5.3 nterest Co o n e Cont n o s 221

merely a rational number raised to what is often a positive integer, it really can t be
calculated, if the exponent is large, without a decent calculator. A so-called xy ey
is needed, and any such calculator has an Exp ey that will calculate ert as Exp.rt/.
(There is never any need to enter a decimal approximation of the irrational number e.)
n a typical decent calculator, .1 C r=n/nt requires about twice as many ey stro es
as ert . This paragraph does nothing to answer its question. We suggest that readers as
their ban managers why their ban s do not compound interest continuously. -)

R BLEMS
In Pr b ems and nd the c mp und am unt and c mp und What annual rate compounded continuously is equivalent to
interest if 4000 is in ested f r six years and interest is an effective rate of 3
c mp unded c ntinu us y at the gi en annua rate What annual rate r compounded continuously is equivalent to
6 14 a nominal rate of 6 compounded semiannually
If interest is compounded continuously at an annual rate of
In Pr b ems and nd the present a ue f 2500 due eight 0.07, how many years would it ta e for a principal P to triple
years fr m n if interest is c mp unded c ntinu us y at the ive your answer to the nearest year.
gi en annua rate If interest is compounded continuously, at what annual rate
1 12 will a principal double in 20 years ive the answer as a
percentage correct to two decimal places.
In Pr b ems nd the e ecti e rate f interest that Savings Options n uly 1, 2001, r. reen had 1000 in
c rresp nds t the gi en annua rate c mp unded c ntinu us y a savings account at the First National an . This account earns
interest at an annual rate of 3.5 compounded continuously. A
2 3 11 competing ban was attempting to attract new customers by
Investment If 100 is deposited in a savings account that offering to add 20 immediately to any new account opened with
earns interest at an annual rate of 4 12 compounded continuously, a minimum 1000 deposit, and the new account would earn
what is the value of the account at the end of two years interest at the annual rate of 3.5 compounded semiannually.
r. reen decided to choose one of the following three options on
Investment If 1500 is invested at an annual rate of 4 uly 1, 2001:
compounded continuously, find the compound amount at the end a eave the money at the First National an
of ten years. b ove the money to the competing ban
Stoc Redemption The board of directors of a corporation c eave half the money at the First National an and move the
agrees to redeem some of its callable preferred stoc in five years. other half to the competing ban .
At that time, 1,000,000 will be required. If the corporation can For each of these three options, find r. reen s accumulated
invest money at an annual interest rate of 5 compounded amount on uly 1, 2003.
continuously, how much should it presently invest so that the
future value is su cient to redeem the shares Investment a n April 1, 2006, s. Cheung invested
75,000 in a 10-year certificate of deposit that paid interest at the
rust Fund A trust fund is being set up by a single annual rate of 3.5 compounded continuously. When the
payment so that at the end of 30 years there will be 50,000 in the certificate matured on April 1, 2016, she reinvested the entire
fund. If interest is compounded continuously at an annual rate of accumulated amount in corporate bonds, which earn interest at the
6 , how much money should be paid into the fund initially rate of 4.5 compounded annually. What will be s. Cheung s
rust Fund As a gift for their newly born daughter s 21st accumulated amount on April 1, 2021
birthday, the Smiths want to give her at that time a sum of money b If s. Cheung had made a single investment of 75,000 in
that has the same buying power as does 21,000 on the date of her 2006 that matures in 2021 and has an effective rate of interest of
birth. To accomplish this, they will ma e a single initial payment 4 , would her accumulated amount be more or less than that in
into a trust fund set up specifically for the purpose. part (a) and by how much
a Assume that the annual effective rate of in ation is 3.5 . In Investment Strategy Suppose that you have 000 to
21 years, what sum will have the same buying power as does invest.
21,000 at the date of the Smiths daughter s birth a If you invest it with the First National an at the nominal rate
b What should be the amount of the single initial payment into of 5 compounded quarterly, find the accumulated amount at the
the fund if interest is compounded continuously at an annual rate end of one year.
of 3.5 b The First National an also offers certificates on which it
Investment Currently, the Smiths have 50,000 to invest pays 5.5 compounded continuously. However, a minimum
for 1 months. They have two options open to them: investment of 10,000 is required. ecause you have only 000,
a Invest the money in a certificate paying interest at the nominal the ban is willing to give you a 1-year loan for the extra 1000
rate of 5 compounded quarterly that you need. Interest for this loan is at an effective rate of ,
b Invest the money in a savings account earning interest at the and both principal and interest are payable at the end of the year.
annual rate of 4.5 compounded continuously. etermine whether or not this strategy of investment is preferable
How much money will they have in 1 months with each option to the strategy in part (a).
222 C at e at cs of nance

If interest is compounded continuously at an annual rate of the other three. Carry out the same derivation for the continuously
3 , in how many years will a principal double ive the answer compounded amount formula, S D Pert . (Again, there is no need
correct to two decimal places. to memorize any of the three other formulas that result, although
General Solutions In Problem 2 of Section 5.1 it was we have met one of them already. y seeing that the general
pointed out that the discrete y compounded amount formula, solutions are easy, we are informed that all particular solutions are
S D P.1 C r/n , can be solved for each of the variables in terms of easy, too.)

Objective A
o ntro ce t e not ons of or nar
ann t es an ann t es e o se
eo etr c ser es to o e t e resent A
a e an t e f t re a e of an
ann t o eter ne a ents to e It is best to define an annuity as any finite sequence of payments made at fixed periods
ace n a s n n f n of time over a given interval. The fixed periods of time that we consider will always be
of equal length, and we refer to that length of time as the payment period. The given
interval is the term of the annuity. The payments we consider will always be of equal
value. An example of an annuity is the depositing of 100 in a savings account every
three months for a year.
The word annuity comes from the atin word annus, which means year, and it is
li ely that the first usage of the word was to describe a sequence of annual payments.
We emphasize that the payment period can be of any agreed-upon length. The informal
definitions of annuity provided by insurance companies in their advertising suggest
that an annuity is a sequence of payments in the nature of pension income. However, a
sequence of rent, car, or mortgage payments fits the mathematics we wish to describe,
so our definition is silent about the purpose of the payments.
When dealing with annuities, it is convenient to mar time in units of payment
periods on a line, with time n , in other words the present, ta en to be 0. ur generic
annuity will consist of n payments, each in the amount R. With reference to such a
timeline (see Figure 5.4), suppose that the n payments (each of amount R) occur at
times 1; 2; 3; : : : ; n. In this case we spea of an ordinary annuity. Unless otherwise
specified, an annuity is assumed to be an ordinary annuity. Again with reference to
our timeline (see Figure 5.5), suppose now that the n equal payments occur at times
0; 1; 2; : : : ; n ! 1. In this case we spea of an annuity due. bserve that in any event,
the n C 1 different times 0; 1; 2; : : : ; n ! 1; n define n consecutive time intervals (each
of payment period length). We can consider that an ordinary annuity s payments are
at the end of each payment period while those of an annuity due are at the beginning
of each payment period. A sequence of rent payments is li ely to form an annuity due
because most landlords demand the first month s rent when the tenant moves in. y
contrast, the sequence of wage payments that an employer ma es to a regular full-time
employee is li ely to form an ordinary annuity because usually wages are for wor
d ne rather than for wor c ntemp ated.
We henceforth assume that interest is at the rate of r per payment period. For
either ind of annuity, a payment of amount R made at time k, for k one of the times
0; 1; 2; : : : ; n ! 1; n, has a value at time 0 and a value at time n. The value at time 0
is the present a ue of the payment made at time k. From Section 5.2 we see that the

0 1 2 3 n-1 n Time
R R R R R Payments

FIGURE rdinary annuity.

0 1 2 3 n-1 n Time
R R R R R Payments

FIGURE Annuity due.

 Section 5.4 nn t es 223

present value of the payment made at time k is R.1 C r/"k . The value at time n is the
future a ue of the payment made at time k. From Section 5.1 we see that the future
value of the payment made at time k is R.1 C r/n"k .

A
The present value of an annuity is the sum of the present a ues of all n payments. It
represents the amount that must be invested n to purchase all n of them. We consider
the case of an ordinary annuity and let A be its present value. y the previous paragraph
and Figure 5.6, we see that the present value is given by
A D R.1 C r/"1 C R.1 C r/"2 C % % % C R.1 C r/"n
From our wor in Section 1.6, we recognize this sum as that of the first n terms of the
geometric sequence with first term R.1 C r/"1 and common ratio .1 C r/"1 . Hence,
from Equation (16) of Section 1.6 we obtain
R.1 C r/"1 .1 ! .1 C r/"n /
AD
1 ! .1 C r/"1
R.1 ! .1 C r/"n /
D
.1 C r/.1 ! .1 C r/"1 /
R.1 ! .1 C r/"n /
D
.1 C r/ ! 1
1 ! .1 C r/"n
DR%
r
where the main simplification follows by replacing the factor .1Cr/"1 in the numerator
of the first line by .1 C r/ in the denominator of the second line.

A
The formula
1 ! .1 C r/"n
ADR%
r
gives the present value A of an ordinary annuity of R per payment period for n
periods at the interest rate of r per period.

The expression .1 ! .1 C r/"n /=r in Equation (1) is given a somewhat bizarre

notation in the mathematics of finance, namely an r , so that we have, by de niti n,
1 ! .1 C r/"n
an r D
r

Period
0 1 2 3 n-1 n
R R R R R Payments

R(1 + r)-1
Present
value of
ordinary R(1 + r)-2
annuity

R(1 + r)-n

FIGURE Present value of ordinary annuity.

224 C at e at cs of nance

With this notation, Equation (1) can be written as

A D Ran r
If we let R D 1 in Equation (2), then we see that an r represents the present value of
an annuity of 1 per payment period for n payment periods at the interest rate of r per
payment period. The symbol an r is sometimes read a angle n at r .
If we write
1 ! .1 C r/"n
an r D a.n; r/ D
r
Whenever a desired value of an r is not in then we see that an r is ust a function of two variables as studied in Section 2. . Indeed,
Appendix A, we will use a calculator to if we were to write
compute it.
1 ! .1 C y/"x
a.x; y/ D
y
then we see that, for fixed y, the function in question is a constant minus a multiple of
an exp nentia functi n f x. For x, a fixed positive integer, the function in question is
a rati na functi n f y.
f course an r is not the first deviation
p from the standard f.x/ nomenclature for
functions. We have already seen that x, jxj, nŠ, and log2 x are other creative notations
for particular common functions.
Selected values of an r are given, approximately, in Appendix A.

E AM LE A
A L IT I
iven a payment of 500 per month Find the present value of an annuity of 100 per month for 3 12 years at an interest rate
for six years, use a graphing calcula- of 6 compounded monthly.
tor to graph the present value A as a
function of the interest rate per month, S % & Substituting in Equation (2), we set R D 100, r D 0:06=12 D 0:005, and
r. etermine the nominal rate if the n D 3 12 .12/ D 42. Thus,
present value of the annuity is 30,000.
A D 100a42 0:005
From Appendix A, a42 0:005 # 37:7 300. Hence,
A # 100.37:7 300/ D 377 : 3
Thus, the present value of the annuity is 377 . 3.
Now ork Problem 5 G

A L IT I E AM LE A
Suppose a man purchases a house
with an initial down payment of 20,000 iven an interest rate of 5 compounded annually, find the present value of a gener-
and then ma es quarterly payments: alized annuity of 2000, due at the end of each year for three years, and 5000, due
2000 at the end of each quarter for six thereafter at the end of each year for four years. (See Figure 5.7.)
years and 3500 at the end of each quar-
ter for eight more years. iven an inter- S The present value is obtained by summing the present values of all pay-
est rate of 6 compounded quarterly, ments:
find the present value of the payments
and the list price of the house.
2000.1:05/"1 C 2000.1:05/"2 C 2000.1:05/"3 C 5000.1:05/"4
C5000.1:05/"5 C 5000.1:05/"6 C 5000.1:05/"7

Period
0 1 2 3 4 5 6 7

Payments 2000 2000 2000 5000 5000 5000 5000

FIGURE Annuity of Example 2.

 Section 5.4 nn t es 225

Rather than evaluating this expression, we can simplify our wor by considering the
payments to be an annuity of 5000 for seven years, minus an annuity of 3000 for
three years, so that the first three payments are 2000 each. Thus, the present value is
5000a7 0:05 ! 3000a3 0:05
# 5000.5:7 6373/ ! 3000.2:72324 /
# 20;762:12
Now ork Problem 17 G
E AM LE A
A L IT I
iven an annuity with equal pay- If 10,000 is used to purchase an annuity consisting of equal payments at the end of
ments at the end of each quarter for six each year for the next four years and the interest rate is 6 compounded annually, find
years and an interest rate of 4. com- the amount of each payment.
pounded quarterly, use a graphing cal-
culator to graph the present value A as S Here A D 10;000, n D 4, r D 0:06, and we want to find R. From Equa-
a function of the monthly payment R. tion (2),
etermine the monthly payment if the
present value of the annuity is 15,000. 10;000 D Ra4 0:06
Solving for R gives
10;000 10;000
RD # #2 5: 1
a4 0:06 3:465106
In general, the formula
A
RD
an r
gives the periodic payment R of an ordinary annuity whose present value is A.
Now ork Problem 19 G
E AM LE A A
A L IT I
A man ma es house payments of The premiums on an insurance policy are 50 per quarter, payable at the beginning of
1200 at the beginning of every month. each quarter. If the policyholder wishes to pay one year s premiums in advance, how
If the man wishes to pay one year s much should be paid, provided that the interest rate is 4 compounded quarterly
worth of payments in advance, how
much should he pay, provided that S We want the present value of an annuity of 50 per period for four periods at
the interest rate is 6. compounded a rate of 1 per period. However, each payment is due at the beginning of the payment
monthly period so that we have an annuity due. The given annuity can be thought of as an
initial payment of 50, followed by an ordinary annuity of 50 for three periods. (See
Figure 5. .) Thus, the present value is
50 C 50a3 0:01 # 50 C 50.2: 40 5/ # 1 7:05

Quarter
0 1 2 3
50 50 50 50

50 + 50a3 0.01

FIGURE Annuity due (present value).

226 C at e at cs of nance

An example of a situation involving an We remar that the general formula for the present value of an annuity due is
annuity due is an apartment lease for A D R C Ran"1 r that is,
which the first payment is made
immediately. A D R.1 C an"1 r /

Now ork Problem 9 G

F A
The future value of an annuity is the sum of the future a ues of all n payments. We
consider the case of an ordinary annuity and let S be its future value. y our earlier
considerations and Figure 5. , we see that the future value is given by
S D R C R.1 C r/ C R.1 C r/2 C % % % C R.1 C r/n"1

Period
0 1 2 n-2 n-1 n
Payments R R R R R

R(1 + r) Future
value of
ordinary
annuity
R(1 + r)2

R(1 + r)n - 1

FIGURE Future value of ordinary annuity.

Again from Section 1.6, we recognize this as the sum of the first n terms of a geometric
sequence with first term R and common ratio 1 C r. Consequently, using Equation (16)
of Section 1.6, we obtain
R.1 ! .1 C r/n / 1 ! .1 C r/n .1 C r/n ! 1
SD DR% DR%
1 ! .1 C r/ !r r

F A
The formula
.1 C r/n ! 1
SDR%
r
gives the future value S of an ordinary annuity of R (dollars) per payment period
for n periods at the interest rate of r per period.

The expression ..1 C r/n ! 1/=r is written sn r so that we have, by de niti n,

.1 C r/n ! 1
sn r D
r
and some approximate values of sn r are given in Appendix A. Thus,
S D Rsn r
It follows that sn r is the future value of an ordinary annuity of 1 per payment
period for n periods at the interest rate of r per period. i e an r , sn r is also a function
of two variables.

E AM LE F A

Find the future value of an annuity consisting of payments of 50 at the end of every
three months for three years at the rate of 6 compounded quarterly. Also, find the
compound interest.
 Section 5.4 nn t es 227
A L IT I S To find the amount of the annuity, we use Equation (4) with R D 50,
Suppose you invest in an RRSP by n D 4.3/ D 12, and r D 0:06=4 D 0:015:
depositing 2000 at the end of every tax
S D 50s12 0:015 # 50.13:041211/ # 652:06
year for the next 15 years. If the inter-
est rate is 5.7 compounded annually, The compound interest is the difference between the amount of the annuity and the sum
how much will you have at the end of of the payments, namely,
15 years 652:06 ! 12.50/ D 652:06 ! 600 D 52:06
Now ork Problem 11 G
E AM LE F A
A L IT I
Suppose you invest in an RRSP by At the beginning of each quarter, 50 is deposited into a savings account that pays 6
depositing 2000 at the beginning of compounded quarterly. Find the balance in the account at the end of three years.
every tax year for the next 15 years. If
the interest rate is 5.7 compounded S Since the deposits are made at the beginning of a payment period, we want
annually, how much will you have at the the future value of an annuity due as considered in Example 4. (See Figure 5.10.) The
end of 15 years given annuity can be thought of as an ordinary annuity of 50 for 13 periods, minus
the final payment of 50. Thus, the future value is
50s13 0:015 ! 50 # 50.14:236 30/ ! 50 # 661: 4

Period
0 1 2 11 12
50 50 50 50 50

1 period 50s13 0.015 - 50

12 periods

FIGURE Future value of annuity due.

The formula for the future value of an annuity due is S D RsnC1 r ! R, which is
S D R.snC1 r ! 1/

Now ork Problem 15 G

S F
ur final examples involve the notion of a sinking fund.

E AM LE S F

A sin ing fund is a fund into which periodic payments are made in order to satisfy
a future obligation. Suppose a machine costing 7000 is to be replaced at the end of
eight years, at which time it will have a salvage value of 700. In order to provide
money at that time for a new machine costing the same amount, a sin ing fund is set
up. The amount in the fund at the end of eight years is to be the difference between the
replacement cost and the salvage value. If equal payments are placed in the fund at the
end of each quarter and the fund earns compounded quarterly, what should each
payment be
S The amount needed after eight years is .7000 ! 700/ D 6300. et R
be the quarterly payment. The payments into the sin ing fund form an annuity with
n D 4. / D 32, r D 0:0 =4 D 0:02, and S D 6300. Thus, from Equation (4), we have
6300 D Rs32 0:02
6300 6300
RD # # 142:45
s32 0:02 44:227030
228 C at e at cs of nance

In general, the formula

S
RD
sn r

gives the periodic payment R of an annuity that is to amount to S.

Now ork Problem 23 G
E AM LE S F

A rental firm estimates that, if purchased, a machine will yield an annual net return of
1000 for six years, after which the machine would be worthless. How much should the
firm pay for the machine if it wants to earn 7 annually on its investment and also set
up a sin ing fund to replace the purchase price For the fund, assume annual payments
and a rate of 5 compounded annually.
S et x be the purchase price. Each year, the return on the investment is 0.07x.
Since the machine gives a return of 1000 a year, the amount left to be placed into the
fund each year is 1000 ! 0:07x. These payments must accumulate to x. Hence,
.1000 ! 0:07x/s6 0:05 D x
1000s6 0:05 ! 0:07xs6 0:05 D x
1000s6 0:05 D x.1 C 0:07s6 0:05 /
1000s6 0:05
Dx
1 C 0:07s6 0:05
1000.6: 01 13/
x#
1 C 0:07.6: 01 13/
# 4607: 2
Another way to loo at the problem is as follows: Each year, the 1000 must
x
account for a return of 0.07x and also a payment of into the sin ing fund. Thus,
s6 0:05
x
we have 1000 D 0:07x C , which, when solved, gives the same result.
s6 0:05
Now ork Problem 25 G

R BLEMS
In Pr b ems use Appendix A and nd the a ue f the gi en In Pr b ems and nd the present a ue f the gi en
expressi n annuity due
a4 0:035 a15 0:07 00 paid at the beginning of each six-month period for seven
years at the rate of compounded semiannually
s 0:0075 s12 0:0125
150 paid at the beginning of each month for five years at the
rate of 7 compounded monthly
In Pr b ems nd the present a ue f the gi en rdinary
annuity In Pr b ems nd the future a ue f the gi en rdinary
annuity
600 per year for six years at the rate of 6 compounded
annually 3000 per month for four years at the rate of compounded
monthly
1000 every month for three years at the m nth y rate of 1
compounded monthly 600 per quarter for four years at the rate of compounded
quarterly
2000 per quarter for 4 12 years at the rate of compounded 5000 per year for 20 years at the rate of 7 compounded
quarterly annually
1500 per month for 15 months at the rate of compounded 2500 every month for 4 years at the rate of 6 compounded
monthly monthly
 Section 5.4 nn t es 229

In Pr b ems and nd the future a ue f the gi en A owes the sum of 10,000 and agrees to pay the sum of
annuity due 1000 at the end of each year for ten years and a final payment at
1200 each year for 12 years at the rate of compounded the end of the eleventh year. How much should the final payment
annually be if interest is at 4 compounded annually

500 every quarter for 12 14 years at the rate of 5 In Pr b ems rather than using tab es use direct y the
compounded quarterly f ing f rmu as
For an interest rate of 4 compounded monthly, find the
present value of an annuity of 150 at the end of each month for 1 ! .1 C r/!n
eight months and 175 thereafter at the end of each month for a an r D
r
further two years.
.1 C r/n ! 1
Leasing O ce Space A company wishes to lease sn r D
temporary o ce space for a period of six months. The rental fee is r
1500 a month, payable in advance. Suppose that the company A Ar
wants to ma e a lump-sum payment at the beginning of the rental RD D
an r 1 ! .1 C r/!n
period to cover all rental fees due over the six-month period. If
money is worth compounded monthly, how much should the S Sr
payment be RD D
sn r .1 C r/n ! 1
An annuity consisting of equal payments at the end of each
quarter for three years is to be purchased for 15,000. If the Find s60 0:017 to five decimal places.
interest rate is 4 compounded quarterly, how much is each
payment Find a 0:052 to five decimal places.
Find 250a1 0 0:0235 to two decimal places.
Equipment Purchase A machine is purchased for 3000
down and payments of 250 at the end of every six months for six Find 1000s120 0:01 to two decimal places.
years. If interest is at compounded semiannually, find the Equal payments are to be deposited in a savings account at
corresponding cash price of the machine. the end of each quarter for 15 years so that at the end of that time
Suppose 100 is placed in a savings account at the end of there will be 5000. If interest is at 3 compounded quarterly,
each month for 50 months. If no further deposits are made, find the quarterly payment.
a how much is in the account after seven years, and b how Insurance Proceeds Suppose that insurance proceeds of
much of this amount is compound interest Assume that 25,000 are used to purchase an annuity of equal payments at the
the savings account pays compounded monthly. end of each month for five years. If interest is at the rate of 10
Insurance Settlement Options The beneficiary of an compounded monthly, find the amount of each payment.
insurance policy has the option of receiving a lump-sum payment Lottery ary ones won a state 4,000,000 lottery and
of 275,000 or 10 equal yearly payments, where the first payment will receive a chec for 200,000 now and a similar one each year
is due at once. If interest is at 3.5 compounded annually, find for the next 1 years. To provide these 20 payments, the State
the yearly payment. ottery Commission purchased an annuity due at the interest rate
Sin ing Find In 10 years, a 40,000 machine will have a of 10 compounded annually. How much did the annuity cost the
salvage value of 4000. A new machine at that time is expected to Commission
sell for 52,000. In order to provide funds for the difference Pension Plan Options Suppose an employee of a company
between the replacement cost and the salvage value, a sin ing is retiring and has the choice of two benefit options under the
fund is set up into which equal payments are placed at the end of company pension plan. ption A consists of a guaranteed payment
each year. If the fund earns 7 compounded annually, how much of 2100 at the end of each month for 20 years. Alternatively,
should each payment be under option , the employee receives a lump-sum payment equal
Sin ing Fund A paper company is considering the to the present value of the payments described under option A.
purchase of a forest that is estimated to yield an annual return of a Find the sum of the payments under option A.
60,000 for years, after which the forest will have no value. The b Find the lump-sum payment under option if it is determined
company wants to earn 6 on its investment and also set up a by using an interest rate of 6 compounded monthly. Round the
sin ing fund to replace the purchase price. If money is placed in answer to the nearest dollar.
the fund at the end of each year and earns 4 compounded An Early Start to Investing An insurance agent offers
annually, find the price the company should pay for the forest. services to clients who are concerned about their personal
Round the answer to the nearest hundred dollars. financial planning for retirement. To emphasize the advantages of
Sin ing Fund In order to replace a machine in the future, an early start to investing, she points out that a 25-year-old person
a company is placing equal payments into a sin ing fund at the who saves 2000 a year for 10 years (and ma es no more
end of each year so that after 10 years the amount in the fund is contributions after age 34) will earn more than by waiting 10
25,000. The fund earns 6 compounded annually. After 6 years, years and then saving 2000 a year from age 35 until retirement at
the interest rate increases and the fund pays 7 compounded age 65 (a total of 30 contributions). Find the net earnings
annually. ecause of the higher interest rate, the company (compound amount minus total contributions) at age 65 for both
decreases the amount of the remaining payments. Find the situations. Assume an effective annual rate of 7 , and suppose
amount of the new payment. Round your answer to the nearest that deposits are made at the beginning of each year. Round
dollar. answers to the nearest dollar.
230 C at e at cs of nance

Continuous Annuity An annuity in which R dollars is where r is the annual rate of interest compounded continuously.
paid each year by uniform payments that are payable continuously Find the present value of a continuous annuity of 365 a year for
is called a c ntinu us annuity The present value of a continuous 30 years at 3 compounded continuously.
annuity for t years is Profit Suppose a business has an annual profit of 40,000
for the next five years and the profits are earned continuously
throughout each year. Then the profits can be thought of as a
1 ! e!rt continuous annuity. (See Problem 36.) If money is worth 4
R% compounded continuously, find the present value of the profits.
r

Objective A L
o earn o to a ort e a oan an Suppose that a ban lends a borrower 1500 and charges interest at the nominal rate of
set an a ort at on sc e e
12 compounded monthly. The 1500 plus interest is to be repaid by equal payments
of R dollars at the end of each month for three months. ne could say that by paying
the borrower 1500, the ban is purchasing an annuity of three payments of R each.
Using the formula from Example 3 of the preceding section, we find that the monthly
payment is given by
A 1500 1500
RD D # # 510:0332
an r a3 0:01 2: 40 5
We will round the payment to 510.03, which may result in a slightly higher final
payment. However, it is not unusual for a ban to round up to the nearest cent, in which
case the final payment may be less than the other payments.
The ban can consider each payment as consisting of two parts: (1) interest on the
outstanding loan and (2) repayment of part of the loan. This is called amorti ing. A
loan is amorti ed when part of each payment is used to pay interest and the remaining
part is used to reduce the outstanding principal. Since each payment reduces the out-
standing principal, the interest portion of a payment decreases as time goes on. et us
analyze the loan ust described.
At the end of the first month, the borrower pays 510.03. The interest on the out-
standing principal is 0:01. 1500/ D 15. The balance of the payment, 510:03! 15 D
4 5:03, is then applied to reduce the principal. Hence, the principal outstanding is
now 1500 ! 4 5:03 D 1004: 7. At the end of the second month, the interest is
0:01. 1004: 7/ # 10:05. Thus, the amount of the loan repaid is 510:03 ! 10:05 D
4 : , and the outstanding balance is 1004: 7 ! 4 : D 504: . The interest
due at the end of the third and final month is 0:01. 504: / # 5:05, so the amount of
the loan repaid is 510:03! 5:05 D 504: . This would leave an outstanding balance
of 504: ! 504: D 0:01, so we ta e the final payment to be 510.04, and the debt
is paid off. As we said earlier, the final payment is ad usted to offset rounding errors.
An analysis of how each payment in the loan is handled can be given in a table called
any end-of-year mortgage statements an amorti ation schedule. (See Table 5.1.) The total interest paid is 30.10, which is
are issued in the form of an amortization often called the finance charge.
schedule.
Table .1 A S
Principal Principal
utstanding Payment Repaid at
at eginning Interest at End End of
Period of Period for Period of Period Period

1 1500 15 510.03 4 5.03

2 1004. 7 10.05 510.03 4 .
3 504. 5.05 510.04 504.
Total 30.10 1530.10 1500.00

In general, suppose that a loan of A dollars is to be repaid by a sequence of n equal

payments of R dollars, each made at the end of an agreed-upon period. The loan was
 Section 5.5 ort at on of oans 231

made at the beginning of the first period and we assume further that the interest rate is r
per period and compounded each period. It follows that the sum of the present values of
the n payments must equal the amount of the loan and we have A D Ran r equivalently,
A
RD .
an r
et k be any of the numbers 1; 2; % % % n. At the beginning of the kth period, k ! 1
payments have been made. It follows that the principal outstanding is the present value
(at time k!1) of the remaining n!.k!1/ D n!kC1 payments. y a time-line drawing
of the ind in Figure 5.6, it is seen that the principal outstanding at the beginning of
the kth period is Ran"kC1 r . Thus, the kth payment (at the end of the kth period) must
contain interest in the amount .Ran"kC1 r /r D Rran"kC1 r so the kth payment reduces
the amount owing by R ! Rrn"kC1 D R.1 ! rn"kC1 . f course, the total interest paid is
nR ! A D R.n ! an r /. We summarize these formulas, which describe the amortization
of the general loan, in Table 5.2. They allow us to ma e an am rtizati n schedu e as
shown in Table 5.1 for any loan.

Table . A F
A r
Periodic payment: R D DA%
an r 1 ! .1 C r/"n
Principal outstanding at beginning of kth period:

1 ! .1 C r/"nCk"1
Ran"kC1 r D R %
r
Interest in kth payment: Rran"kC1 r
Principal contained in kth payment: R.1 ! ran"kC1 r /
Total interest paid: R.n ! an r / D nR ! A

E AM LE A L

A person amortizes a loan of 170,000 for a new home by obtaining a 20-year mortgage
at the rate of 7.5 compounded monthly. Find a the monthly payment, b the total
interest charges, and c the principal remaining after five years.
a The number of payment periods is n D 12.20/ D 240 the interest rate per period
is r D 0:075=12 D 0:00625 and A D 170; 000. From Formula 1 in Table 5.2, the
monthly payment R is 170; 000=a240 0:00625 . Since a240 0:00625 is not in Appendix A,
we use the following equivalent, expanded formula and a calculator:
! "
0:00625
R D 170;000
1 ! .1:00625/"240
# 136 :51
b From Formula 5, the total interest charges are
240.136 :51/ ! 170;000 D 32 ;6 2:40 ! 170;000
D 15 ;6 2:40
c After five years, we are at the beginning of the 61st period. Using Formula 2 with
n ! k C 1 D 240 ! 61 C 1 D 1 0, we find that the principal remaining is
! "
1 ! .1:00625/"1 0
136 :51 # 147;733:74
0:00625

Now ork Problem 1 G

232 C at e at cs of nance

At one time, a very common type of installment loan involved the add-on method
of determining the finance charge. With that method, the finance charge is found by
applying a quoted annual interest rate under simp e interest to the borrowed amount
of the loan. The charge is then added to the principal, and the total is divided by the
number of m nths of the loan to determine the monthly installment payment.
In loans of this type, the borrower may not immediately realize that the true annual
rate is significantly higher than the quoted rate. To give a simple numerical exam-
ple, suppose that a 1000 loan is ta en for one year at interest under the add-on
method, with payments made monthly. The finance charge for this scheme is simply
1000.0:0 / D 0. Adding this to the loan amount gives 100 C 0 D 10 0 and
the monthly installment payment is 10 0=12 # 0: 3. We can now analyze this sit-
uation using the principles of this section. From that point of view, we simply have a
loan amount A D 1000, with n D 12 monthly payments each with payment amount
R D 10 0=12. We can now use A D Ran r and attempt to solve for r, the interest rate
per period.
1 ! .1 C r/"12 A 1000
D a12 r D D # 11:00 174312
r R 10 0=12
We cannot hope to algebraically solve for r in
1 ! .1 C r/"12
D 11:00 174312
r
but there are lots of approximation techniques available, one of which is simply to graph

1 D .1 ! .1 C / ^ .!12/=
2 D 11:00 174

on a graphing calculator and as for the first coordinate of the point of intersection.
oing so returns

r # 0:01351374

which corresponds to an annual rate of 12.0:01351374/ # 0:1622 D 16:22 Clearly,

it is very misleading to label this loan as a loan n closer examination, we see that
the aw in calculating loan payments by the add-on method is that it ta es no
account of the amount that is paid off the principal each month. The lender is effec-
tively obliged to pay interest on the original amount each month. The add-on method
loo s much easier than having to deal with the complexity of an r . As Albert Einstein
once said, Everything should be made as simple as possible but not simpler . Fortu-
nately, regulations concerning truth-in-lending laws have made add-on loans virtually
obsolete.
The annuity formula
1 ! .1 C r/"n
ADR%
r
cannot be solved for r in a simple closed form, which is why the previous example
required an approximation technique. n the other hand, solving the annuity formula
for n, to give the number of periods of a loan, is a straightforward matter. We have
Ar
D 1 ! .1 C r/"n
R
Ar R ! Ar
.1 C r/"n D 1 ! D
R R
!n ln.1 C r/ D ln.R ! Ar/ ! ln.R/ ta ing logs of both sides
ln.R/ ! ln.R ! Ar/
nD
ln.1 C r/
 Section 5.5 ort at on of oans 233

E AM LE L

uhammar Smith recently purchased a computer for 1500 and agreed to pay it off
by ma ing monthly payments of 75. If the store charges interest at the annual rate of
12 compounded monthly, how many months will it ta e to pay off the debt
S From the last equation on the previous page,
ln.75/ ! ln.75 ! 1500.0:01//
nD # 22:4
ln.1:01/
Therefore, it will require 23 months to pay off the loan (with the final payment less
than 75).
Now ork Problem 11 G

R BLEMS
A person borrows 000 from a ban and agrees to pay it off c the interest in the 24th payment
by equal payments at the end of each month for two years. If d the principal in the 24th payment
interest is at 13.2 compounded monthly, how much is each e the total interest paid.
payment A debt of 1 ,000 is being repaid by 15 equal semiannual
ronwen wishes to ma e a two-year loan and can afford payments, with the first payment to be made six months from
payments of 100 at the end of each month. If annual interest is at now. Interest is at the rate of 7 compounded semiannually.
6 compounded monthly, how much can she afford to borrow However, after two years, the interest rate increases to
compounded semiannually. If the debt must be paid off on the
Finance Charge etermine the finance charge on a
original date agreed upon, find the new annual payment. ive
36-month 000 auto loan with monthly payments if interest is at
your answer to the nearest dollar.
the rate of 4 compounded monthly.
A person borrows 2000 and will pay off the loan by equal
For a one-year loan of 500 at the rate of 15 compounded
payments at the end of each month for five years. If interest is at
monthly, find a the monthly installment payment and b the
the rate of 16. compounded monthly, how much is each
finance charge.
payment
Car Loan A person is amortizing a 36-month car loan of Mortgage A 245,000 mortgage for 25 years for a new
7500 with interest at the rate of 4 compounded monthly. Find home is obtained at the rate of .2 compounded monthly. Find
a the monthly payment, b the interest in the first month, and a the monthly payment, b the interest in the first payment,
c the principal repaid in the first payment. c the principal repaid in the first payment, and d the finance
Real Estate Loan A person is amortizing a 4 -month loan charge.
of 65,000 for a house lot. If interest is at the rate of 7.2 Auto Loan An automobile loan of 23,500 is to be
compounded monthly, find a the monthly payment, b the amortized over 60 months at an interest rate of 7.2 compounded
interest in the first payment, and c the principal repaid in the monthly. Find a the monthly payment and b the finance
first payment. charge.
In Pr b ems c nstruct am rtizati n schedu es f r the Furniture Loan A person purchases furniture for 5000
indicated debts Adjust the na payments if necessary and agrees to pay off this amount by monthly payments of 120.
10,000 repaid by three equal yearly payments with interest at If interest is charged at the rate of 12 compounded monthly,
5 compounded annually. how many fu payments will there be
000 repaid by eight equal semiannual payments with Find the monthly payment of a five-year loan for 500 if
interest at .5 compounded semiannually interest is at .24 compounded monthly.
00 repaid by five equal quarterly payments with interest at Mortgage ob and ary Rodgers want to purchase a new
10 compounded quarterly house and feel that they can afford a mortgage payment of 600 a
month. They are able to obtain a 30-year 7.6 mortgage
10,000 repaid by five equal monthly payments with interest
(compounded monthly), but must put down 25 of the cost of the
at compounded monthly
house. Assuming that they have enough savings for the down
A loan of 1300 is being paid off by quarterly payments of payment, how expensive a house can they afford ive your
110. If interest is at the rate of 6 compounded quarterly, how answer to the nearest dollar.
many fu payments will be made
Mortgage Suppose you have the choice of ta ing out a
A loan of 5000 is being amortized over 36 months at an 240,000 mortgage at 6 compounded monthly for either
interest rate of compounded monthly. Find 15 years or 25 years. How much savings is there in the finance
a the monthly payment charge if you were to choose the 15-year mortgage
b the principal outstanding at the beginning of the 36th month
234 C at e at cs of nance

n a 45,000 four-year loan, how much less is the monthly annual rate of 3 compounded monthly. If the monthly payment
payment if the loan were at the rate of .4 compounded monthly at the 3 rate is x dollars (x dollars is the homeowner s monthly
rather than at .6 compounded monthly payment) and the monthly payment at the rate is y dollars
ome Loan The federal government has a program to aid (y dollars is the monthly payment the ban must receive), then the
low-income homeowners in urban areas. This program allows government ma es up the difference y ! x to the ban each month.
certain qualified homeowners to obtain low-interest home The government does not want to bother with m nth y payments.
improvement loans. Each loan is processed through a commercial Instead, at the beginning of the loan, the government pays the
ban . The ban ma es home improvement loans at an annual rate present value of all such monthly differences, at an annual rate of
of compounded monthly. However, the government compounded monthly. If a qualified homeowner ta es out a
subsidizes the ban so that the loan to the homeowner is at the loan for 10,000 for four years, determine the government s
payment to the ban at the beginning of the loan.

Objective
o ntro ce t e not on of er et t
an s e ts of se ences

In this section we consider brie y the possibility of an in nite sequence of payments.

As in Section 5.4, we will measure time in payment periods starting n that is, at
time 0 and consider payments, each of amount R, at times 1; 2; : : : ; k; : : :. The last
sequence of dots is to indicate that the payments are to continue indefinitely. We can
visualize this on a timeline as in Figure 5.11. We call such an infinite sequence of
payments a perpetuity.

0 1 2 3 k Time
R R R R Payments

FIGURE Perpetuity.

Since there is no last payment, it ma es no sense to consider the future value of

such an infinite sequence of payments. However, if the interest rate per payment period
is r, we do now that the present a ue of the payment made at time k is R.1 C r/"k . If
we want to ascribe a present value to the entire perpetuity, we are led by this observation
and Figure 5.12 to define it to be
A D R.1 C r/"1 C R.1 C r/"2 C R.1 C r/"3 C % % % C R.1 C r/"k C % % %

0 1 2 3 k Time
R R R R Payments

R(1 + r)-1

R(1 + r)-2

R(1 + r)-3

R(1 + r)-k

FIGURE Present value of perpetuity.

With the benefit of Section 1.6, we recognize this sum as that of an infinite geometric
sequence with first term R.1 C r/"1 and common ratio .1 C r/"1 . Equation (17) of
Section 1.6 gives
X
1
R.1 C r/"1 R
AD R.1 C r/"k D D
kD1
1 ! .1 C r/"1 r
 Section 5.6 Per et t es 235

provided that j.1 C r/"1 j < 1. If the rate r is positive, then 1 < 1 C r so that 0 <
1
.1 C r/"1 D < 1 and the proviso is satisfied.
1Cr
In practical terms, this means that if an amount R=r is invested at time 0 in an
account that bears interest at the rate of r per payment period, then R can be withdrawn
at times 1; 2; : : : ; k; : : : indefinitely. It is easy to see that this ma es sense because if
R=r is invested at time 0, then at time 1 it is worth .R=r/.1 C r/ D R=r C R. If, at time
1, R is withdrawn, then R=r C R ! R D R=r remains and this process can be continued
indefinitely so that at any time k, the amount after the kth withdrawal is still R=r. In
other words, the withdrawals R are such that they consume only the interest earned since
the last withdrawal and leave the principal intact. Well-managed endowment funds are
run this way. The amount withdrawn each year to fund a scholarship, say, should not
exceed the amount earned in interest during the previous year.

E AM LE

alhousie University would li e to establish a scholarship worth 15,000 to be awarded

to the first year usiness student who attains the highest grade in ATH 1115, Com-
merce athematics. The award is to be made annually, and the Vice President, Finance,
believes that, for the foreseeable future, the university will be able to earn at least 2 a
year on investments. What principal is needed to ensure the viability of the scholarship
S The university needs to fund a perpetuity with payments R D 15;000 and
annual interest rate r D 0:02. It follows that 15;000=0:02 D 750;000 is needed.
Now ork Problem 5 G

L
P
An infinite sum, such as 1 "k
kD1 R.1 C r/ , which has arisen here, derives
P its meaning
from the associated nite partial sums. Here the nth partial sum is nkD1 R.1 C r/"k ,
which we recognize as Ran r , the present value of the annuity consisting of n equal
payments of R at an interest rate of r per payment period.
et .ck /1
kD1 be an infinite sequence as in Section 1.6. We say that the sequence has
limit and write
lim ck D
k!1

if e can make the a ues ck as c se as e ike t by taking k su cient y arge The

equation can be read as the limit of ck as k goes to infinity is equal to . A sequence
can fail to have a limit, but it can have at most one limit, so we spea of the limit .
We have already met an important example of this concept. In Section 4.1 we
defined the number
! e as
" the smallest real number that is greater than all of the real
nC1 n
numbers en D , for n any positive integer. In fact, we have also
n
lim en D e
n!1

A general infinite sequence .ck /1 1

kD1 determines a new sequence .sn /nD1 , where
Xn
sn D ck . We define
kD1

X
1 X
n
ck D lim sn D lim ck
n!1 n!1
kD1 kD1

This agrees with what we said about the sum of an infinite geometric sequence in
Section 1.6, and it is important to realize that the sums that arise for the present values
of annuities and perpetuities are but special cases of sums of geometric sequences.
236 C at e at cs of nance

However, we wish to ma e a simple observation by combining some of the equal-

ities of this section:
R X1 Xn
D R.1 C r/"k D lim R.1 C r/"k D lim Ran r
r kD1
n!1
kD1
n!1

and, ta ing R D 1, we get

1
lim an r D
n!1 r
We can verify this observation directly. In the defining equation
1 ! .1 C r/"n
an r D
r
only .1 C r/"n D 1=.1 C r/n depends on n. ecause 1 C r > 1, we can ma e the values
.1Cr/n as large as we li e by ta ing n su ciently large. It follows that we can ma e the
values 1=.1 C r/n as close as we li e to 0 by ta ing n su ciently large. It follows that
in the definition of an r , we can ma e the numerator as close as we li e to 1 by ta ing n
su ciently large and hence that we can ma e the whole fraction as close as we li e to
1=r by ta ing n su ciently large. It would be reasonable to write
1
a1 r D lim an r D
n!1 r

E AM LE L S

2n2 C 1
Find lim .
n!1 3n2 ! 5

S We first rewrite the fraction 2n2 C 1=3n2 ! 5.

2n2 C 1
2
2n C 1 n2
lim 2
D lim 2
n!1 3n ! 5 n!1 3n ! 5

n2
2
2n 1
2
C 2
D lim n 2 n
n!1 3n 5
2
! 2
n n
1
2C 2
D lim n
n!1 5
3! 2
n
So far we have only carried along the limit notation. We now observe that because
we can ma e the values n2 as large as we li e by ta ing n su ciently large, we can
ma e 1=n2 and 5=n2 as close as we li e to 0 by ta ing n su ciently large. It follows
that we can ma e the numerator of the main fraction as close as we li e to 2 and the
denominator of the main fraction as close as we li e to 3 by ta ing n su ciently large.
In symbols,
1
2n2 C 1 2C 2
lim D lim n D2
2
n!1 3n ! 5 n!1 5 3
3! 2
n

Now ork Problem 7 G

 Chapter 5 e e 237

R BLEMS
In Pr b ems nd the present a ue f the gi en perpetuity can earn per year on his money for the next 10 years, but he is
60 per month at the rate of 1.5 monthly only assuming that he will be able to get 5 per year after that.
a How much does Pierre need to pay each year for the first 10
5000 per month at the rate of 0.5 monthly years in order to ma e the planned withdrawals b Pierre s will
0,000 per year at the rate of 5 yearly states that, upon his death, any money left in his retirement
account is to be donated to the Princeton athematics
4000 per year at the rate of 10 yearly
epartment. If he dies immediately after receiving his 17th
Funding a Pri e The Commerce Society would li e to payment, how much will the Princeton athematics epartment
endow an annual prize of 120 to the student who is deemed to inherit
have exhibited the most class spirit. The Society is confident that
In Pr b ems nd the imit
it can invest indefinitely at an interest rate of at least 2.5 a year.
2
How much does the Society need to endow its prize n C 3n ! 6 nC7
lim lim
Retirement Planning Starting a year from now and ma ing n!1 n2 C 4 n!1 5n ! 3
! " ! "n
10 yearly payments, Pierre would li e to put into a retirement k C 1 2k n
lim lim
account enough money so that, starting 11 years from now, he can k!1 k n!1 n C 1
withdraw 30,000 per year until he dies. Pierre is confident that he

Chapter 5 Review
I T S E
S Compound Interest
effective rate Ex. 4, p. 211
S Present Value
present value Ex. 1, p. 214
future value equation of value net present value Ex. 3, p. 216
S Interest Compounded Continuously
compounded continuously Ex. 1, p. 21
S Annuities
annuity ordinary annuity annuity due Ex. 1, p. 224
present value of annuity, an r amount of annuity, sn r Ex. 2, p. 224
S Amorti ation of Loans
amortizing amortization schedules finance charge Ex. 1, p. 231
S Perpetuities
perpetuity Ex. 1, p. 235
limit of a sequence Ex. 2, p. 236

S
The concept of compound interest lies at the heart of any dis- Interest rates are usually quoted as an annual rate called
cussion dealing with the time value of money that is, the the nominal rate. The periodic rate is obtained by divid-
present value of money due in the future or the future value of ing the nominal rate by the number of interest periods each
money currently invested. Under compound interest, interest year. The effective rate is the annual simple-interest rate,
is converted into principal and earns interest itself. The basic which is equivalent to the nominal rate of r compounded n
compound-interest formulas are times a year and is given by

# r $n
S D P.1 C r/n future value
re D 1 C !1 effective rate
P D S.1 C r/"n present value n

Effective rates are used to compare different interest rates.

where S D compound amount (future value) If interest is compounded continuously, then
P D principal (present value)
S D Pert future value
r D periodic rate
P D Se"rt present value
n D number of interest periods
238 C at e at cs of nance

where S D compound amount .future value/ remaining part is used to reduce the principal. A complete
analysis of each payment is given in an amortization sched-
P D principal (present value)
ule. The following formulas deal with amortizing a loan of
r D annual rate A dollars, at the periodic rate of r, by n equal payments of
t D number of years R dollars each and such that a payment is made at the end
of each period:
and the effective rate is given by

re D e r ! 1 effective rate Periodic payment:

A r
RD DA%
An annuity is a sequence of payments made at fixed peri- an r 1 ! .1 C r/"n
ods of time over some interval. The mathematical basis for Principal outstanding at beginning of kth period:
formulas dealing with annuities is the notion of the sum of a
1 ! .1 C r/"nCk"1
geometric sequence that is, Ran"kC1 r D R %
r
X
n"1
a.1 ! rn /
sD ari D sum of geometric sequence Interest in kth payment:
iD0
1!r
Rran"kC1 r

where s D sum Principal contained in kth payment:

a D first term R.1 ! ran"kC1 r /
r D common ratio Total interest paid:
n D number of terms R.n ! an r / D nR ! A
An ordinary annuity is an annuity in which each payment
is made at the end of a payment period, whereas an annu-
ity due is an annuity in which each payment is made at the A perpetuity is an infinite sequence of payments made
beginning of a payment period. The basic formulas dealing at fixed periods of time. The mathematical basis for the for-
with ordinary annuities are mula dealing with a perpetuity is the notion of the sum of an
infinite geometric sequence that is,
1 ! .1 C r/"n
ADR% D Ran r present value
r X
1
a
.1 C r/n ! 1 sD ari D sum of infinite geometric sequence
SDR% D Rsn r future value iD0
1!r
r

where A D present value of annuity where s D sum

S D amount .future value/ of annuity a D first term
R D amount of each payment r D common ratio with jrj < 1
n D number of payment periods The basic formula dealing with perpetuities is
r D periodic rate
R
For an annuity due, the corresponding formulas are AD present value
r

A D R.1 C an"1 r / present value

where A D present value of perpetuity
S D R.snC1 r ! 1/ future value R D amount of each payment
r D periodic rate
A loan, such as a mortgage, is amortized when part of An infinite sum is defined as the limit of the sequence of
each installment payment is used to pay interest and the partial sums.
 Chapter 5 e e 239

R
Find the number of interest periods that it ta es for a principal A savings account pays interest at the rate of 2
to double when the interest rate is r per period. compounded semiannually. What amount must be deposited now
Find the effective rate that corresponds to a nominal rate of 5 so that 350 can be withdrawn at the end of every six months for
compounded monthly. the next 15 years
An investor has a choice of investing a sum of money at either Sin ing Fund A company borrows 5000 on which it will
.5 compounded annually or .2 compounded semiannually. pay interest at the end of each year at the annual rate of 11 . In
Which is the better choice addition, a sin ing fund is set up so that the 5000 can be repaid
at the end of five years. Equal payments are placed in the fund at
Cash Flows Find the net present value of the following cash the end of each year, and the fund earns interest at the effective
ows, which can be purchased by an initial investment of ,000: rate of 6 . Find the annual payment in the sin ing fund.
ear Cash Flow Car Loan A debtor is to amortize a 7000 car loan by
ma ing equal payments at the end of each month for 36 months. If
2 3400 interest is at 4 compounded monthly, find a the amount of
each payment and b the finance charge.
4 3500
5 3600 A person has debts of 500 due in three years with interest at
5 compounded annually and 500 due in four years with interest
at 6 compounded semiannually. The debtor wants to pay off
these debts by ma ing two payments: the first payment now, and
Assume that interest is at 5 compounded semiannually.
the second, which is double the first payment, at the end of the
A debt of 1500 due in five years and 2000 due in seven third year. If money is worth 7 compounded annually, how
years is to be repaid by a payment of 2000 now and a second much is the first payment
payment at the end of three years. How much should the second
Construct an amortization schedule for a credit card bill of
payment be if interest is at 3 compounded annually
5000 repaid by three monthly payments with interest at 24
Find the present value of an annuity of 250 at the end of each compounded monthly.
month for four years if interest is at 6 compounded monthly.
Construct an amortization schedule for a loan of 15,000
For an annuity of 200 at the end of every six months for repaid by five monthly payments with interest at compounded
6 12 years, find a the present value and b the future value at an monthly.
interest rate of compounded semiannually.
Find the present value of an ordinary annuity of 460 every
Find the amount of an annuity due that consists of 13 yearly month for nine years at the rate of 6 compounded monthly.
payments of 150, provided that the interest rate is 4
Auto Loan etermine the finance charge for a 4 -month
compounded annually.
auto loan of 11,000 with monthly payments at the rate of 5.5
Suppose 500 is initially placed in a savings account and 500 compounded monthly.
is deposited at the end of every month for the next year. If interest
is at 6 compounded monthly, how much is in the account at the
end of the year
 atr e ra
atrices, the sub ect of this chapter, are simply arrays of numbers. atri-
6.1 atr ces ces and matrix algebra have potential application whenever numerical
6.2 atr t on an information can be meaningfully arranged into rectangular bloc s.
ca ar t cat on ne area of application for matrix algebra is computer graphics. An
ob ect in a coordinate system can be represented by a matrix that contains the coordi-
6.3 atr t cat on nates of each corner. For example, we might set up a connect-the-dots scheme in which
6.4 o n ste s the lightning bolt shown is represented by the matrix to its right.
e cn atr ces
y
6.5 o n ste s
e cn atr ces
(-2, 4) (0, 4)
Cont n e x y

6.6 n erses 0 0
(-3, 1)
-2 4
6.7 eont ef s n t t t (0, 0)
x 0 4
na s s (-1, -1) 2 -2
-1 -1
C er 6 e e (2, -2) 0 -5
-3 1

(0, -5)

Computer graphics often show ob ects rotating in space. Computationally, rotation

is effected by matrix multiplication. The lightning bolt is rotated cloc wise 52 degrees
about the origin by matrix multiplication, involving a matrix whose entries are functions
t11 , t12 , t21 , and t22 of the rotation angle (with t11 D t22 and t12 D !t21 ):

(1.92, 4.04)
(-1.06, 2.98)
(3.15, 2.46) 0 0 0 0
-2 4 1.92 4.04
(-1.40, 0.17) 0 4 3.15 2.46
x 2 -2 t11 (52º) t12 (52º)
= -0.34 -2.81
(0, 0) -1 -1 t21 (52º) t22 (52º) -1.40 0.17
0 -5 -3.94 -3.08
-3 1 -1.06 2.98
(-3.94, -3.08) (-0.34, -2.81)

240
 Section 6. atr ces 241

Objective M
o ntro ce t e conce t of a atr Finding ways to describe many situations in mathematics and economics leads to the
an to cons er s ec a t es of
atr ces study of rectangular arrays of numbers. Consider, for example, the system of linear
equations
8̂
<3x C 4y C 3z D 0
2x C y ! z D 0
:̂ x ! 6y C 2z D 0

If we are organized with our notation, eeping the x s in the first column, the y s in
the second column, and so on, then the features that characterize this system are the
numerical coe cients in the equations, together with their relative positions. For this
reason, the system can be described by the rectangular arrays
2 3 2 3
3 4 3 0
42 1 !15 and 4 0 5
!6 2 0

one for each side of the equations, each being called a matrix (plural: matrices, pro-
nounced may0 -tri-sees). We consider such rectangular arrays to be ob ects in them-
Vertical bars, j j, around a rectangular selves, and our custom, as ust shown, will be to enclose them by brac ets. Parentheses
array do n t mean the same thing as are also commonly used. In symbolically representing matrices, we use capital letters
brac ets or parentheses. such as A, B, C, and so on.
In economics it is often convenient to use matrices in formulating problems and
displaying data. For example, a manufacturer who produces products , , and could
Table .1 represent the units of labor and material involved in one wee s production of these
Product items as in Table 6.1. ore simply, the data can be represented by the matrix
! "
10 12 16
AD
5 7
abor 10 12 16
The horizontal rows of a matrix are numbered consecutively from top to bottom, and the
aterial 5 7
vertical columns are numbered from left to right. For the foregoing matrix A, we have
column 1 column 2 column 3
" #
row 1 10 12 16
DA
row 2 5 7
Since A has two rows and three columns, we say that A has size 2 " 3 (read 2 by 3 )
or that A is 2 " 3, where the number of rows is specified first. Similarly, the matrices
2 3
2 3 1 2
1 6 !2 6!3 47
B D 4 5 1 !45 and C D 6 4 5
7
65
!3 5 0
7 !
have sizes 3 " 3 and 4 " 2, respectively.
The numbers in a matrix are called its entries. To denote the entries in a matrix
A of size 2 " 3, say, we use the name of the matrix, with d ub e subscripts to indicate
p siti n, consistent with the conventions above:
! "
A11 A12 A13
A21 A22 A23
The row subscript appears to the left of
the column subscript. In general, Aij and
Aji are different. For the entry A12 (read A sub one-two or ust A one-two ), the first subscript, 1,
specifies the row and the second subscript, 2, the column in which the entry appears.
Similarly, the entry A23 (read A two-three ) is the entry in the second row and the
third column. eneralizing, we say that the symbol Aij denotes the entry in the ith row
and jth column. In fact, a matrix A is a function of two variables with A.i; j/ D Aij .
242 C atr e ra

If A is m " n, write mN for the set f1; 2; : : : mg. Then, the domain of A is m N " nN , the set
of all ordered pairs .i; j/ with i in m
N and j in nN , while the range is a subset of the set of
real numbers, .!1; 1/.
ur concern in this chapter is the manipulation and application of various types of
matrices. For completeness, we now give a formal definition of a matrix.

A rectangular array of numbers A consisting of m horizontal rows and n vertical

columns,
2 3
A11 A12 # # # A1n
6 A21 A22 # # # A2n 7
6 7
6 # # ### # 7
6 7
6 # # ### # 7
6 7
4 # # ### # 5
Am1 Am2 # # # Amn

is called an m"n matrix and m"n is the si e of A. For the entry Aij , the row subscript
is i and the column subscript is j.

The number of entries in an m " n matrix is mn. For brevity, an m " n matrix can
The matrix ŒAij ! has Aij as its genera be denoted by the symbol ŒAij !m!n or, more simply, ŒAij !, when the size is understood
entry. from the context.
A matrix that has exactly one row, such as the 1 " 4 matrix

A D Œ1 7 12 3!

is called a row vector. A matrix consisting of a single column, such as the 5 " 1 matrix
2 3
1
6!27
6 7
6 157
6 7
4 5
16

is called a column vector. bserve that a matrix is 1 " 1 if and only if it is both a row
vector and a column vector. It is safe to treat 1 " 1 matrices as mere numbers. In other
words, we can write Œ7! D 7, and, more generally, Œa! D a, for any real number a.

E AM LE S M
A L IT I
A manufacturer who uses raw mate- a The matrix Œ1 2 0! has size 1 " 3.
rials A and is interested in trac ing the 2 3
costs of these materials from three dif-
1 !6
ferent sources. What is the size of the b The matrix 45 15 has size 3 " 2.
matrix she would use 4
c The matrix 7 has size 1 " 1.
2 3
1 3 7 !2 4
d The matrix 4 11 5 6 5 has size 3 " 5 and .3/.5/ D 15 entries.
6 !2 !1 1 1

Now ork Problem 1a G

 Section 6. atr ces 243

A L IT I E AM LE C M
An analysis of a wor place uses a a Construct a three-entry column matrix A such that A21 D 6 and Ai1 D 0 otherwise.
3 " 5 matrix to describe the time spent
on each of three phases of five differ- S Since A11 D A31 D 0, the matrix is
ent pro ects. Pro ect 1 requires 1 hour 2 3
for each phase, pro ect 2 requires twice 0
as much time as pro ect 1, pro ect 3 A D 465
requires twice as much time as pro ect 0
2; : : : ; and so on. Construct this time-
analysis matrix. b If ŒAij ! is 3 " 4 and Aij D i C j, find A.
S Here i D 1; 2; 3 and j D 1; 2; 3; 4, and A has .3/.4/ D 12 entries. Since
Aij D i C j, the entry in row i and column j is obtained by adding the numbers i and j.
Hence, A11 D 1 C 1 D 2, A12 D 1 C 2 D 3, A13 D 1 C 3 D 4, and so on. Thus,
2 3 2 3
1C1 1C2 1C3 1C4 2 3 4 5
A D 42 C 1 2 C 2 2 C 3 2 C 45 D 43 4 5 65
3C1 3C2 3C3 3C4 4 5 6 7

c Construct the 3 " 3 matrix I, given that I11 D I22 D I33 D 1 and Iij D 0 otherwise.
S The matrix is given by
2 3
1 0 0
I D 40 1 05
0 0 1

Now ork Problem 11 G

E M
We now define what is meant by saying that two matrices are equa .

atrices A and B are equal if and only if they have the same size and Aij D Bij for
each i and j (that is, corresponding entries are equal).

Thus,
" #
! "
2
1C1 2 2 1
D
2#3 0 6 0

but
! "
1
Œ1 1! ¤ and Œ1 1! ¤ Œ1 1 1! different sizes
1
A matrix equation can define a system of equations. For example, suppose that
! " ! "
x yC1 2 7
D
2z 5 4 2
y equating corresponding entries, we must have
8̂
ˆ xD2
<
yC1D7
ˆ 2z D 4
:̂ 5 D 2

Solving gives x D 2; y D 6; z D 2, and D 25 .

244 C atr e ra

T M
If A is a matrix, the matrix formed from A by interchanging its rows with its columns
is called the transp se of A.

The transpose of an m " n matrix A, denoted AT , is the n " m matrix whose ith row
is the ith column of A.

E AM LE T M
! "
1 2 3
If A D , find AT .
4 5 6
S atrix A is 2 " 3, so AT is 3 " 2. Column 1 of A becomes row 1 of AT ,
column 2 becomes row 2, and column 3 becomes row 3. Thus,
2 3
1 4
AT D 42 55
3 6

Now ork Problem 19 G

bserve that the columns of AT are the rows of A. Also, if we ta e the transpose of
our answer, the original matrix A is obtained. That is, the transpose operation has the
property that
.AT /T D A

S M
Certain types of matrices play important roles in matrix theory. We now consider some
of these special types.
An m " n matrix whose entries are all 0 is called the m " n ero matrix and is
denoted by 0m!n or, more simply, by 0 if its size is understood. Thus, the 2 " 3 zero
matrix is
! "
0 0 0
0D
0 0 0
and, in general, we have
2 3
0 0 ### 0
60 0 ### 07
6 7
6# # ### #7
0D6
6#
7
6 # ### #7
7
4# # ### #5
0 0 ### 0

A matrix having the same number of columns as rows for example, n rows and
n columns is called a square matrix of order n. That is, an m " n matrix is square if
and only if m D n. For example, matrices
2 3
2 7 4
46 2 05 and Œ3!
4 6 1

are square with orders 3 and 1, respectively.

 Section 6. atr ces 245

In a square matrix A of order n, the entries A11 , A22 , A33 ; : : : ; Ann lie on the diagonal
extending from the upper left corner to the lower right corner of the matrix and are said
to constitute the main diagonal. Thus, in the matrix
1 2 3
4 5 6
7 8 9

the main diagonal (see the shaded region) consists of A11 D 1, A22 D 5, and A33 D .
A square matrix A is called a diagonal matrix if all the entries that are off the main
diagonal are zero that is, if Aij D 0 for i ¤ j. Examples of diagonal matrices are
2 3
! " 3 0 0
1 0
and 40 6 05
0 1
0 0
A square matrix A is said to be an upper triangular matrix if all entries be
the main diagonal are zero that is, if Aij D 0 for i > j. Similarly, a matrix A is said to
It follows that a matrix is diagonal if and be a lower triangular matrix if all entries ab e the main diagonal are zero that is,
only if it is both upper triangular and
if Aij D 0 for i < j. When a matrix is either upper triangular or lower triangular, it is
lower triangular.
called a triangular matrix. Thus, the matrices
7 0 0 0
5 1 1
3 2 0 0
0 -3 7 and
6 5 -4 0
0 0 4
1 6 0 1

are upper and lower triangular matrices, respectively, and are therefore triangular
matrices.

R BLEMS
et Find the f ing entries
" # 2 3 2 3
1 2 3 1 1 A21 A42
1 !6 2
AD B D 44 5 65 C D 42 2 5 A24 A34
!4 2 1
7 3 3 A44 A55
2 3
" # 1 2 3 4 What are the third row entries
60 h i
1 0 6 1 6 077
D ED4 FD 6 2 Write the lower triangular matrix A, of order 3, for which all
2 3 0 0 2 05
0 0 6 1 entries n t required t be satisfy Aij D i ! j.
2 3 2 3 a Construct the matrix A D ŒAij ! if A is 2 " 3 and
5 1 6 2 Aij D !i C 2j.
G D 465 D 40 0 05 D Œ4! b Construct the 2 " 4 matrix C D Œ.i C j/2 !.
1 0 0 0
a Construct the matrix B D ŒBij ! if B is 2 " 2 and
Bij D .!1/i!j .i2 ! j2 /.
a State the size of each matrix.
b Construct the 2 " 3 matrix D Œ.!1/i . j3 /!.
b Which matrices are square
c Which matrices are upper triangular lower triangular If A D ŒAij ! is 12 " 10, how many entries does A have If
d Which are row vectors Aij D 1 for i D j and Aij D 0 for i ¤ j, find A33 , A52 , A10;10 , and
e Which are column vectors A12;10 .
ist the main diagonal of
In Pr b ems et 2 3 2 3
2 3 2 4 !2
6 7 5 x2 1 2y
7 !2 14 6 6 0 !17 7 6 p 7
66 a b 4 y 35
2 3 !27 4!4 6 !3 15
A D ŒAij ! D 6
45
7 y z 1
4 1 05 2 5 7 1
0 2 0
Write the zero matrix a of order 3 and b of size 2 " 4.
What is the order of A If A is a 7 " matrix, what is the size of AT
246 C atr e ra

In Pr b ems nd AT respectively. a How many white extreme models were sold in

" # anuary b How many blue deluxe models were sold in
h i
6 !3 February c In which month were more purple regular models
AD AD 2 4 6
2 4 sold d Which models and which colors sold the same number
2 3 2 3 of units in both months e In which month were more deluxe
2 5 !3 0 !2 3 0
models sold f In which month were more red widgets sold
A D 40 3 6 25 A D 4 3 4 55
7 !2 1 0 5 !6 g How many widgets were sold in anuary
2 3 Input Output Matrix Input output matrices, which were
et " #
1 0 0 developed by W. W. eontief, indicate the interrelationships that
7 0
AD B D 40 2 05 exist among the various sectors of an economy during some
0 6
0 10 !3 period of time. A hypothetical example for a simplified economy
2 3 2 3 is given by matrix at the end of this problem. The consuming
0 0 0 2 0 !1
C D 40 0 0 5 D 40 4 0 5 sectors are the same as the producing sectors and can be thought
0 0 0 0 0 6 of as manufacturers, government, steel industry, agriculture,
households, and so on. Each row shows how the output of a given
a Which are diagonal matrices sector is consumed by the four sectors. For example, of the total
b Which are triangular matrices output of industry A, 50 went to industry A itself, 70 to , 200 to
A matrix is symmetric if AT D A. Is the matrix of Problem 1 C, and 360 to all others. The sum of the entries in row 1 namely,
symmetric 6 0 gives the total output of A for a given period. Each column
If " # gives the output of each sector that is consumed by a given sector.
1 0 !1 For example, in producing 6 0 units, industry A consumed
AD 50 units of A, 0 of , 120 of C, and 420 from all other producers.
7 0
For each column, find the sum of the entries. o the same for each
verify the general property that .AT /T D A by finding AT and then row. What do we observe in comparing these totals Suppose
.AT /T . sector A increases its output by 10 namely, by 6 units.
In Pr b ems s e the matrix equati n Assuming that this results in a uniform 10 increase of all its
" # " # 2 3 2 3 inputs, by how many units will sector have to increase its
x 3 2 3 output Answer the same question for C and for all other
3x 2y ! 1 6 45 75 D 4 5
D y5 producers .
z 5 7 15
z 4 !5 4
2 3 2 3 C NSU ERS
4 2 1 4 2 1 ‚ …„ ƒ
43x y 3z5 D 46 7 5
Industry Industry Industry All ther
0 7 0 PR UCERS A C Consumers
" # " #
2 3
2x 7 y 7 Industry A 50 70 200 360
D
7 2y 7 y Industry 6 0 30 270 3207
D 6 7
Industry C 4120 240 100 10505
Inventory A grocer sold 125 cans of tomato soup, 275
cans of beans, and 400 cans of tuna. Write a row vector that gives All ther 420 370 40 4 60
the number of each item sold. If the items sell for 0. 5, 1.03, Producers
and 1.25 each, respectively, write this information as a column
Find all the values of x for which
vector.
" p # " #
Sales Analysis The Widget Company has its monthly sales x2 C 2000x x2 2001 !x
reports given by means of matrices whose rows, in order, D
x2 ln.ex / 2001 ! 2000x x
represent the number of regular, deluxe, and extreme models sold,
and the columns, in order, give the number of red, white, blue, and
In Pr b ems and nd AT
purple units sold. The matrices for anuary and February are
2 3 2 3 " # 2 3
1 4 5 0 2 5 7 7 3 1 4 2
3 !4 5
D 43 5 2 75 F D 42 4 4 65 AD A D 41 7 3 65
!2 1 6
4 1 3 2 0 0 1 2 1 4 1 2

Objective M A S M
o e ne atr a t on an sca ar
t cat on an to cons er
M A
ro ert es re ate to t ese o erat ons Consider a snowmobile dealer who sells two models, eluxe and Super. Each is avail-
able in one of two colors, red and blue. Suppose that the sales for anuary and February
are represented by the matrices
eluxe Super eluxe Super
! " ! "
red 1 2 red 3 1
D FD
blue 3 5 blue 4 2
 Section 6.2 atr t on an ca ar t cat on 247

respectively. Each row of and F gives the number of each model sold for a given
color. Each column gives the number of each color sold for a given model. A matrix
representing total sales for each model and color over the two months can be obtained
by adding the corresponding entries in and F:
! "
4 3
7 7
This situation provides some motivation for introducing the operation of matrix addi-
tion for two matrices of the same size.

If A and B are both m " n matrices, then the sum A C B is the m " n matrix obtained
by adding corresponding entries of A and B so that .A C B/ij D Aij C Bij . If the size
of A is different from the size of B, then A C B is not defined.
A L IT I
An o ce furniture company manu- For example, let
factures des s and tables at two plants, ! " ! "
A and . atrix represents the pro- 3 0 !2 5 !3 6
AD and BD
duction of the two plants in anuary, 2 !1 4 1 2 !5
and matrix F represents the production
Since A and B are the same size (2 " 3), their sum is defined. We have
of the two plants in February. Write a ! " ! "
matrix that represents the total produc- 3C5 0 C .!3/ !2 C 6 !3 4
ACBD D
tion at the two plants for the two months, 2 C 1 !1 C 2 4 C .!5/ 3 1 !1
where
A
" # E AM LE M A
D
des s 120 0 2 3 2 3 2 3 2 3
tables 105 130 1 2 7 !2 1C7 2!2 0
a 43 45 C 4!6 45 D 4 3 ! 6 4 C 45 D 4!3 5
" #
5 6 3 0 5C3 6C0 6
des s 110 140
FD ! " ! "
tables 5 125 1 2 2
b C is not defined, since the matrices are not the same size.
3 4 1

Now ork Problem 7 G

If A, B, C, and have the same size, then the following properties hold for matrix
addition:

M A
ACBDBCA commutative property
A C .B C C/ D .A C B/ C C associative property
AC DAD CA identity property

Property 1 states that matrices can be added in any order, and Property 2 allows
These properties of matrix addition matrices to be grouped for the addition operation. Property 3 states that the zero matrix
correspond to properties of addition of plays the same role in matrix addition as does the number 0 in the addition of real
real numbers. numbers. These properties are illustrated in Example 2.

E AM LE M A

et
! " ! "
1 2 1 0 1 2
AD BD
!2 0 1 1 !3 1
! " ! "
!2 1 !1 0 0 0
CD D
0 !2 1 0 0 0
248 C atr e ra

a Show that A C B D B C A.
S
! " "!
1 3 3 1 3 3
ACBD BCAD
!1 !3 2 !1 !3 2
Thus, A C B D B C A.
b Show that A C .B C C/ D .A C B/ C C.
S
! " ! "
!2 2 1 !1 4 2
A C .B C C/ D A C D
1 !5 2 !1 !5 3
! " ! "
1 3 3 !1 4 2
.A C B/ C C D CCD
!1 !3 2 !1 !5 3
c Show that A C D A.
S
! " ! " ! "
1 2 1 0 0 0 1 2 1
AC D C D DA
!2 0 1 0 0 0 !2 0 1

Now ork Problem 1 G

E AM LE E

In Section 6.7 we will discuss a way of modelling an economy that consists of well-
defined sectors. For example, each sector might correspond to an entire industry, such
as oil, or agriculture, or manufacturing. For a large complex country li e Canada, say,
there would be a huge number of such sectors, but we can illustrate the idea here by
confining ourselves to a three-sector economy. Suppose the sectors of our economy
are in fact oil ( ), agriculture (A), and manufacturing ( ) and that there are four con-
sumers 1, 2, 3, and 4. (A consumer might be a neighboring country.) The needs, for
each consumer, of each of the three sectors can be represented by 1- by 3-row matrices.
If we agree to list the industries consistently in the order , A, , then the needs of
Consumer 1 might be 1 D Π3 2 5 ! meaning that Consumer 1 needs, in suitable units,
3 units of oil, 2 units of agriculture, and 5 units of manufacturing. Such a matrix is
often called a demand ect r. For the other consumers we might have

2 D Œ0 1 6! 3 D Œ1 5 3! 4 D Œ2 1 4!

If we write C for total consumer demand, we have C D 1C 2C 3C 4, so that

C D Œ3 2 5! C Œ0 1 6! C Œ1 5 3! C Œ2 1 4! D Œ6 1 !

Now ork Problem 41 G

S M
Returning to the snowmobile dealer, recall that February sales were given by the matrix
! "
3 1
FD
4 2
If, in arch, the dealer doubles February s sales of each model and color of snowmo-
bile, the sales matrix for arch could be obtained by multiplying each entry in F by 2,
yielding
! "
2.3/ 2.1/
D
2.4/ 2.2/
 Section 6.2 atr t on an ca ar t cat on 249

It seems reasonable to write this operation as

! " ! " ! "
3 1 2#3 2#1 6 2
D 2F D 2 D D
4 2 2#4 2#2 4

which is thought of as multiplying a matrix by a real number. In the context of matrices,

real numbers are often called sca ars. Indeed, we have the following definition.

If A is an m " n matrix and k is a real number, then by kA we denote the m " n matrix
obtained by multiplying each entry in A by k so that .kA/ij D kAij . This operation is
called scalar multiplication, and kA is called a s alar multiple of A.

For example,
! " ! " ! "
1 0 !2 !3.1/ !3.0/ !3.!2/ !3 0 6
!3 D D
2 !1 4 !3.2/ !3.!1/ !3.4/ !6 3 !12

E AM LE S M

et
! " ! " ! "
1 2 3 !4 0 0
AD BD D
4 !2 7 1 0 0

Compute the following.

a 5A
S
! " ! " ! "
1 2 5.1/ 5.2/ 5 10
5A D 5 D D
4 !2 5.4/ 5.!2/ 20 !10

2
b ! B
3
S
" # " #
2 ! 23 .3/ ! 23 .!4/ !2 3
! BD 2 2
D
3 ! 3 .7/ ! 3 .1/ ! 14
3
! 23

1
c A C 3B
2
S
! " ! "
1 1 1 2 3 !4
A C 3B D C3
2 2 4 !2 7 1
" # ! " "1 #
1
1 !12 !11
D 2 C D 2
2 !1 21 3 23 2

d 0A
S
! " ! "
1 2 0 0
0A D 0 D D0
4 !2 0 0
250 C atr e ra

e k0
S ! " ! "
0 0 0 0
k0 D k D D0
0 0 0 0

Now ork Problem 5 G

For A and B of the same size, and for any scalars k and , we have the following
properties of scalar multiplication:

S M
k.A C B/ D kA C kB
.k C /A D kA C A
k. A/ D .k /A
0A D 0
k0 D 0

Properties 4 and 5 were illustrated in Examples 4(d) and (e) the others will be illus-
trated in the problems.
We also have the following properties of the transpose operation, where A and B
are of the same size and k is any scalar:

.A C B/T D AT C BT
.kA/T D kAT

The first property states that the transp se f a sum is the sum f the transp ses.

S M
If A is any matrix, then the scalar multiple .!1/A is simply written as !A and is
called the negative of A:
!A D .!1/A

Thus, if
! "
3 1
AD
!4 5

then
! " ! "
3 1 !3 !1
!A D .!1/ D
!4 5 4 !5

Note that !A is the matrix obtained by multiplying each i entry of A by !1.

Subtraction of matrices is defined in terms of matrix addition:

ore simply, to find A ! B, we can

subtract each entry in B from the
corresponding entry in A.
If A and B are the same size, then, by A ! B, we mean A C .!B/.
 Section 6.2 atr t on an ca ar t cat on 251

E AM LE M S

2 3 2 3 2 3 2 3
2 6 6 !2 2 6 6 !2
a 4!4 15 ! 44 15 D 4!4 15 C .!1/ 44 15
A L IT I 3 2 0 3 3 2 0 3
A manufacturer of doors, win- 2 3 2 3
dows, and cabinets writes her yearly 2 6 !6 2
profit (in thousands of dollars) for D 4!4 15 C 4!4 !15
each category in a column vector as 3 2 0 !3
2 3
6
24
7 2 3 2 3
P D 431 5. Her fixed costs of pro- 2!6 6C2 !4
532 D 4!4 ! 4 1 ! 15 D 4! 05
duction can be described by the vector 3C0 2!3 3 !1
2 3
40
6 7 ! " ! "
C D 4305. She calculates that, with 6 0 3 !3
60 b If A D and B D , then
2 !1 1 2
a new pricing structure that generates
an income that is 0 of her competi- ! " ! " ! "
tor s income, she can double her profit, T 6 2 6 !6 0
assuming that her fixed costs remain A ! 2B D ! D
0 !1 2 4 !2 !5
the same. This calculation can be rep-
resented by
2 3 2 3 2 3
x1
6 7 6 7
40
6
24
7
Now ork Problem 17 G
0: 4x2 5 ! 4305 D 2 431 5
x3 60 532
Solve for x1 ; x2 ; and x3 ; which repre-
sent her competitor s income from each
category. E AM LE M E
! " ! " ! "
x 3 5
Solve the equation 2 1 ! D5 :
x2 4 !4
S

S We first write each side of the equation as a single matrix. Then, by

equality of matrices, we equate corresponding entries.

We have

! " ! " ! "

x1 3 5
2 ! D5
x2 4 !4
! " ! " ! "
2x1 3 25
! D
2x2 4 !20
! " ! "
2x1 ! 3 25
D
2x2 ! 4 !20

y equality of matrices, we must have 2x1 ! 3 D 25, which gives x1 D 14 from

2x2 ! 4 D !20, we get x2 D ! .
Now ork Problem 35 G
252 C atr e ra

R BLEMS
In Pr b ems perf rm the indicated perati ns C mpute the indicated matrices if p ssib e
" # " #
T
7 !3 34 3A C .B ! C/T 3BT C 4CT
C
!2 1 52 2B C BT AC T
!B . ! 2AT /T
" # " # " #
2 !7 7 !4 2 7 Express the matrix equation
C C " # " # " #
!6 4 !2 1 7 2
3 !4 2
2 3 2 3 2 3 x !y D3
2 !3 5 1 4 !2 6 2 7 4
4 5 ! 5!4 1
05 42 10 !125
!4 !2 3 2 0 0 7 as a system of linear equations and solve.
In the reverse of the manner used in Problem 35, write the
2Œ 2 !1 3 ! C 4Œ!2 0 1! ! 0Œ2 3 1! system
" # " # (
1 2 7 x C 2y D 7
Π3 5 1 ! C 24 C
3 4 2 3x C 4y D 14
" # " # " #
5 3 0 0 2 !6 7 1 as a matrix equation.
C7 !6
!2 6 0 0 7 1 6 !2 In Pr b ems s e the matrix equati ns
" # " # " #
2 3 2 3 x !2 6
1 !1 !6 3 !3 D4
6 7 6 7 y 4 !2
62 07 6 2 67
6 7 ! 36 7 2 3 2 3 2 3
43 !65 4 1 !25 " # " # " #
2 x !10
4 4 5 x 2 !4x 445 C 2 4 y5 D 4!245
5 !6 D
2 3 2 3 3 !2y 3y
2 7 1 !1 3 4 6 4z 14
4 3 0 35 C 2 4 1 !2 35 2 3 2 3 2 3 2 3
2 !1 0 10
!1 0 5 1 3 !5 x 405 C 2 4 05 C y 4 25 D 4 6 5
2 3 02 3 2 31 2 6 !5 2x C 12 ! 5y
1 0 0 1 2 0 4 !2 2
B C
3 40 1 05 ! 3 @40 !2 15 ! 4!3 21 ! 5A Production An auto parts company manufactures
0 0 1 0 0 1 0 1 0 distributors, spar plugs, and magnetos at two plants, I and II.
atrix X represents the production of the two plants for retailer ,
In Pr b ems c mpute the required matrices if and matrix represents the production of the two plants for
" # " # " # " # retailer . Write a matrix that represents the total production at
2 1 !6 !5 !2 !1 0 0 the two plants for both retailers, where
AD BD CD 0D
3 !3 2 !3 !3 3 0 0
2 I II 3 2 I II 3
IS 35 60 IS 10 45
!2C !.A ! B/ X D SP 4 50 7005 D SP 4 00 7005
A 35 50 A 15 10
2.0/ ACB!C
3.2A ! 3B/ 0.2A C 3B ! 5C/ Sales et matrix A represent the sales (in thousands of
dollars) of a toy company in 2007 in three cities, and let B
3.A ! C/ C 6 A C .C C B/
represent the sales in the same cities in 200 , where
A ! 2B C 3C 3C ! 2B " #
1
A C 3.2B C 5C/ 1
A ! 5.B C C/ Action 400 350 150
3 2 AD
Educational 450 2 0 50
In Pr b ems erify the equati ns f r the preceding " #
matrices A B and C Action 3 0 330 220
BD
3.A C B/ D 3A C 3B .3 C 4/B D 3B C 4B Educational 460 320 750
k1 .k2 A/ D .k1 k2 /A If the company buys a competitor and doubles its 200 sales in
k.A ! 2B C C/ D kA ! 2kB C kC 2010, what is the change in sales between 2003 and 2010
Suppose the prices of products A, , C, and are given, in
In Pr b ems et
that order, by the price row vector
2 3 " # " #
1 2 P D ŒpA pB pC p !
1 3 1 0
A D 40 !15 BD CD
4 !1 1 2 If the prices are to be increased by 16 , the vector for the new
7 0
" # prices can be obtained by multiplying P by what scalar
1 2 !1 Prove that .A ! B/T D AT ! BT . ( int Use the definition of
D
1 0 2 subtracti n and properties of the transpose operation.)
 Section 6.3 atr t cat on 253

In Pr b ems c mpute the gi en matrices if 4A C 3B

" # " # " #
3 !4 5 1 4 2 !1 1 3 2.3A C 4B/ C 5C
AD BD CD
!2 1 6 4 1 2 2 6 !6 2.3C ! A/ C 2B

Objective M M
o e ne t cat on of atr ces an esides the operations of matrix addition and scalar multiplication, the product AB
to cons er assoc ate ro ert es o
e ress a s ste as a s n e atr of matrices A and B can be defined under a certain condition, namely, that the number
e at on sn atr t cat on f c umns f A is equa t the number f r s f B. Although the following definition of
matrix mu tip icati n might not appear to be a natural one, a thorough study of matrices
shows that the definition ma es sense and is extremely practical for applications.

et A be an m " n matrix and B be an n " p matrix. Then the product AB is the m " p
matrix with entry .AB/ik given by
X
n
.AB/ik D Aij Bjk D Ai1 B1k C Ai2 B2k C # # # C Ain bnk
jD1

In words, .AB/ik is obtained by summing the products formed by multiplying, in

order, each entry in row i of A by the corresponding entry in column k of B. If the
number of columns of A is not equal to the number of rows of B, then the product
AB is not defined.

bserve that the definition applies when A is a row vector with n entries and B is
a column vector with n entries. In this case, A is 1 " n, B is n " 1, and AB is 1 " 1. (We
noted in Section 6.1 that a 1 " 1 matrix is ust a number.) In fact,

32
B1
h i 6 B2 7
6 7
if A D A1 A2 ### An and B D 6 :: 7
4 : 5
Bn
X
n
then AB D Aj Bj D A1 B1 C A2 B2 C # # # C An Bn
jD1

Returning to our general definition, it now follows that the number .AB/ik is the product
of the ith row of A and the kth column of B. This is very helpful when real computations
are performed.
Three points must be completely understood concerning this definition of AB. First,
the number of columns of A must be equal to the number of rows of B. Second, the
product AB has as many rows as A and as many columns as B.

A B = C
m*n n*p m*p

must be
the same
size of product

Third, the definition refers to the product AB, in that rder A is the left factor and B is
the right factor. For AB, we say that B is premu tip ied by A or A is p stmu tip ied by B.
254 C atr e ra

To apply the definition, let us find the product

2 3
! " 1 0 !3
2 1 !6 4
AB D 0 4 25
1 !3 2
!2 1 1
atrix A has size 2 " 3, .m " n/ and matrix B has size 3 " 3, .n " p/. The number of
columns of A is equal to the number of rows of B, .n D 3/, so the product AB is defined
and will be a 2 " 3, .m " p/ matrix that is,
! "
.AB/11 .AB/12 .AB/13
AB D
.AB/21 .AB/22 .AB/23
The entry .AB/11 is obtained by summing the products of each entry in row 1 of A by
the corresponding entry in column 1 of B. Thus,

row 1 entries of A

c11 = (2)(1) + (1)(0) + (-6)(-2) = 14.

column 1 entries of B

At this stage, we have

2 3
! 1 " 0 !3 ! "
2 1 !6 4 5 14 .AB/12 .AB/13
AB D 0 4 2 D
1 !3 2 .AB/21 .AB/22 .AB/23
!2 1 1
Here we see that .AB/11 D 14 is the product of the first row of A and the first column
of B. Similarly, for .AB/12 , we use the entries in row 1 of A and those in column 2 of B:

row 1 entries of A

c12 = (2)(0) + (1)(4) + (-6)(1) = -2.

column 1 entries of B

We now have 2 3
! " 1 0 !3 ! "
2 1 !6 4 5 14 !2 .AB/13
AB D 0 4 2 D
1 !3 2 .AB/21 .AB/22 .AB/23
!2 1 1
For the remaining entries of AB, we obtain
.AB/13 D .2/.!3/ C .1/.2/ C .!6/.1/ D !10
.AB/21 D .1/.1/ C .!3/.0/ C .2/.!2/ D !3
.AB/22 D .1/.0/ C .!3/.4/ C .2/.1/ D !10
.AB/23 D .1/.!3/ C .!3/.2/ C .2/.1/ D !7
Thus,
2 3
! " 1 0 !3 ! "
2 1 !6 4 5 14 !2 !10
AB D 0 4 2 D
1 !3 2 !3 !10 !7
!2 1 1
Note that if we reverse the order of the factors, then the product
2 3
1 0 !3 ! "
4 5 2 1 !6
BA D 0 4 2
1 !3 2
!2 1 1
 Section 6.3 atr t cat on 255
atrix multiplication is n t c mmutati e. is n t defined, because the number of columns of B does n t equal the number of rows
of A. This shows that matrix multiplication is not commutative. In fact, for matrices
A and B, even when both products are defined, it is usually the case that AB and BA
are different. he rder in hich the matrices in a pr duct are ritten is extreme y
imp rtant.

E AM LE S M T

et A be a 3 " 5 matrix and B be a 5 " 3 matrix. Then AB is defined and is a 3 " 3

matrix. oreover, BA is also defined and is a 5 " 5 matrix.
If C is a 3 " 5 matrix and is a 7 " 3 matrix, then C is undefined, but C is
defined and is a 7 " 5 matrix.

Now ork Problem 7 G

E AM LE M

Compute the matrix product

2 3
! " 2 1
2 !4 2 4
AB D 0 45
0 1 !3
2 2
S Since A is 2 " 3 and B is 3 " 2, the product AB is defined and will have size
2 " 2. y simultaneously moving the index finger of the left hand from right to left
along the rows of A and the index finger of the right hand down the columns of B, it
should not be di cult to mentally find the entries of the product. We obtain
2 3
! " 2 1 ! "
2 !4 2 4 5 !10
0 4 D
0 1 !3 !6 !2
2 2
bserve that BA is also defined and has size 3 " 3.
Now ork Problem 19 G
A L IT I
A boo store has 100 dictionaries, E AM LE M
2 3
70 coo boo s, and 0 thesauruses in 4
stoc . If the value of each dictionary is a Compute Œ1 2 3! 455.
2 , each coo boo is 22, and each
6
thesaurus is 16, use a matrix product
to find the total value of the boo store s S The product has size 1 " 1:
inventory. 2 3
4
Œ1 2 3! 455 D Œ32!
6
2 3
1
b Compute 425 Œ1 6!.
3
S The product has size 3 " 2:
2 3 2 3
1 1 6
425 Œ1 6! D 42 125
3 3 1
2 32 3 2 3
1 3 0 1 0 2 16 !3 11
c 4!2 2 15 45 !1 35 D 4 10 !1 05
1 0 !4 2 1 !2 !7 !4 10
256 C atr e ra
! "! " ! "
a11 a12 b11 b12 a b C a12 b21 a11 b12 C a12 b22
d D 11 11
a21 a22 b21 b22 a21 b11 C a22 b21 a21 b12 C a22 b22

Now ork Problem 25 G

Example 4 shows that even when the E AM LE M

matrix products AB and BA are both
defined and the same size, they are not Compute AB and BA if
necessarily equal. ! " ! "
2 !1 !2 1
AD and B D :
3 1 1 4
S We have
! "! " ! "
2 !1 !2 1 !5 !2
AB D D
3 1 1 4 !5 7
! "! " ! "
!2 1 2 !1 !1 3
BA D D
1 4 3 1 14 3
Note that although both AB and BA are defined, and the same size, AB and BA are not
equal.
Now ork Problem 37 G

A L IT I E AM LE C
The prices (in dollars per unit) for
three textboo s are represented by the Suppose that the prices (in dollars per unit) for products A, , and C are represented
price vector P D Œ26:25 34:75 2 :50!. by the price vector
A university boo store orders these Price of
boo s in the quantities given by the
2 3 A C
250
6 7 P D Œ2 3 4!
column vector D 43255. Find the
175
If the quantities (in units) of A, , and C that are purchased are given by the column
total cost (in dollars) of the purchase. vector
2 3
7 units of A
D 4 55 units of
11 units of C
then the total cost (in dollars) of the purchases is given by the entry in the cost vector
2 3
7
P D Œ2 3 4! 4 55 D Œ.2 # 7/ C .3 # 5/ C .4 # 11/! D Œ73!
11

Now ork Problem 27 G

E AM LE E

In Example 3 of Section 6.2, suppose that in the hypothetical economy the price of coal
is 10,000 per unit, the price of electricity is 20,000 per unit, and the price of steel is
40,000 per unit. These prices can be represented by the (column) price vector
2 3
10;000
P D 420;0005
40;000
Consider the steel industry. It sells a total of 30 units of steel at 40,000 per unit, and
its total income is therefore 1,200,000. Its costs for the various goods are given by the
 Section 6.3 atr t cat on 257

matrix product
2 3
10;000
SP D Œ30 5 0! 420;0005 D Œ400;000!
40;000
Hence, the profit for the steel industry is 1;200;000 ! 400;000 D 00;000.
Now ork Problem 67 G
atrix multiplication satisfies the following properties, provided that all sums and
products are defined:

M M
A.BC/ D .AB/C associative property
A.B C C/ D AB C AC; distributive properties
.A C B/C D AC C BC

E AM LE A

If
2 3
! " ! " 1 0
1 !2 3 0 !1
AD BD C D 40 25
!3 4 1 1 2
1 1
compute ABC in two ways.
S rouping BC gives
0 2 31
! " ! " 1 0
1 !2 B 3 0 !1 4 C
0 2 5A
4 @ 1 1
A.BC/ D
!3 2
1 1
! "! " ! "
1 !2 2 !1 !4 !
D D
!3 4 3 4 6 1
Alternatively, grouping AB gives
2 3
! "! "! 1 0
1 !2 3 0 !1 4
.AB/C D 0 25
!3 4 1 1 2
1 1
2 3
! " 1 0
1 !2 !5 4
D 0 25
!5 4 11
1 1
! "
!4 !
D
6 1
Note that A.BC/ D .AB/C.
G
E AM LE

Verify that A.B C C/ D AB C AC if

! " ! " ! "
1 0 !2 0 !2 1
AD BD CD
2 3 1 3 0 2
258 C atr e ra

S n the left side, we have

! " ! " ! "!
1 0 !2 0 !2 1
A.B C C/ D C
2 3 1 3 0 2
! "! " ! "
1 0 !4 1 !4 1
D D
2 3 1 5 !5 17
n the right side,
! "! " ! "! "
1 0 !2 0 1 0 !2 1
AB C AC D C
2 3 1 3 2 3 0 2
! " ! " ! "
!2 0 !2 1 !4 1
D C D
!1 !4 !5 17
Thus, A.B C C/ D AB C AC.
Now ork Problem 69 G
E AM LE R M C

Suppose that a building contractor has accepted orders for five ranch-style houses,
seven Cape Cod style houses, and 12 colonial-style houses. Then his orders can be
represented by the row vector
D Œ5 7 12!
Furthermore, suppose that the raw materials that go into each type of house are steel,
wood, glass, paint, and labor. The entries in the following matrix, R, give the number of
units of each raw material going into each type of house (the entries are not necessarily
realistic, but are chosen for convenience):

Steel Wood lass Paint abor

2 3
Ranch 5 20 16 7 17
Cape Cod 47 1 12 215 D R
Colonial 6 25 5 13

Each row indicates the amount of each raw material needed for a given type of house
each column indicates the amount of a given raw material needed for each type of
house. Suppose now that the contractor wishes to compute the amount of each raw
material needed to fulfill his orders. Then such information is given by the matrix
2 3
5 20 16 7 17
R D Œ5 7 12! 47 1 12 215
6 25 5 13
D Œ146 526 260 15 3 !

Thus, the contractor should order 146 units of steel, 526 units of wood, 260 units of
glass, and so on.
The contractor is also interested in the costs he will have to pay for these materials.
Suppose steel costs 2500 per unit, wood costs 1200 per unit, and glass, paint, and
labor cost 00, 150, and 1500 per unit, respectively. These data can be written as
the column cost vector
2 3
2500
612007
6 7
CD6 6 007
7
4 1505
1500
 Section 6.3 atr t cat on 259

Then the cost of each type of house is given by the matrix

2 3
2 3 2500 2 3
5 20 16 7 17 6 612007
7 75; 50
RC D 47 1 12 215 6 7 4 1;5505
6 007 D
6 25 5 13 4 1505 71;650
1500

Consequently, the cost of materials for the ranch-style house is 75, 50, for the Cape
Cod house 1,550, and for the colonial house 71,650.
The total cost of raw materials for all the houses is given by
2 3
75; 50
RC D .RC/ D Œ5 7 12! 4 1;5505 D Œ1; 0 ; 00!
71;650
The total cost is 1, 0 , 00.
Now ork Problem 65 G
Another property of matrices involves scalar and matrix multiplications. If k is a
scalar and the product AB is defined, then
k.AB/ D .kA/B D A.kB/
The product k.AB/ can be written simply as kAB. Thus,
kAB D k.AB/ D .kA/B D A.kB/
For example,
! "! " ! "! ! "
2 1 1 3 2 1 1 3
3 D 3
0 !1 2 0 0 !1 2 0
! "! "
6 3 1 3
D
0 !3 2 0
! "
12 1
D
!6 0
There is an interesting property concerning the transpose of a matrix product:

.AB/T D BT AT
In words, the transpose of a product of matrices is equal to the product of their
transposes in the re erse order.

This property can be extended to the case of more than two factors. For
example,
Here, we used the fact that .AT /T D A. .AT BC/T D CT BT .AT /T D CT BT A

E AM LE T

et
! " ! "
1 0 1 2
AD and BD
1 2 1 0
Show that .AB/T D BT AT .
S We have
! " ! "
1 2 1 3
AB D so .AB/T D
3 2 2 2
260 C atr e ra

Now,
! " ! "
T1 1 1
T 1
A D and B D
0 2 2 0
Thus,
! "! " ! "
1 1 1 1 1 3
BT AT D D D .AB/T
2 0 0 2 2 2

so .AB/T D BT AT .
G
ust as the zero matrix plays an important role as the identity in matrix addition,
there is a special matrix, called the identity matrix that plays a corresponding role in
matrix multiplication:

The n " n identity matrix, denoted In , is the diagonal matrix whose main diagonal
entries are 1 s.

For example, the identity matrices I3 and I4 are

2 3
2 3 1 0 0 0
1 0 0 60 1 0 07
I 3 D 40 1 05 and I4 D 6
40
7
0 1 05
0 0 1
0 0 0 1
When the size of an identity matrix is understood, we omit the subscript and simply
denote the matrix by I. It should be clear that

IT D I

The identity matrix plays the same role in matrix multiplication as does the number
1 in the multiplication of real numbers. That is, ust as the product of a real number and 1
is the number itself, the product of a matrix and the identity matrix is the matrix itself.
For example,
! " ! "! " ! "
2 4 2 4 1 0 2 4
ID D
1 5 1 5 0 1 1 5
and
! " ! "! " ! "
2 4 1 0 2 4 2 4
I D D
1 5 0 1 1 5 1 5
In general, if I is n " n and A has n columns, then AI D A. If B has n rows, then IB D B.
oreover, if A is n " n, then
AI D A D IA

E AM LE M I I 0

If " #
! " 2
3 2 5
! 15
AD BD
1 4 1
! 10 3
10
! " ! "
1 0 0 0
ID 0D
0 1 0 0
compute each of the following.
 Section 6.3 atr t cat on 261

a I!A
S ! " ! " ! "
1 0 3 2 !2 !2
I!AD ! D
0 1 1 4 !1 !3
b 3.A ! 2I/
S
! " ! "!
3 2 1 0
3.A ! 2I/ D 3 !2
1 4 0 1
! " ! "!
3 2 2 0
D3 !
1 4 0 2
! " ! "
1 2 3 6
D3 D
1 2 3 6
c A0
S ! "! " ! "
3 2 0 0 0 0
A0 D D D0
1 4 0 0 0 0
In general, if A0 and 0A are defined, then

A0 D 0 D 0A
d AB
S
! "" 2
# ! "
3 2 5
! 15 1 0
AB D D DI
1 4 1
! 10 3 0 1
10

Now ork Problem 55 G

If A is a square matrix, we can spea of a p er of A:

If A is a square matrix and p is a positive integer, then the pth power of A, written
Ap , is the product of p factors of A:
Ap D A
„#Aƒ‚ A
###…
p factors

If A is n " n, we define A0 D In .

We remar that Ip D I.

E AM LE M
! "
1 0
If A D , compute A3 .
1 2
S Since A3 D .A2 /A and
! "! " ! "
1 0 1 0 1 0
A2 D D
1 2 1 2 3 4
we have
! "! " ! "
3 1
2 0 1 0 1 0
A DA AD D
3 4 1 2 7

Now ork Problem 45 G

262 C atr e ra

M E
Systems of linear equations can be represented by using matrix multiplication. For
example, consider the matrix equation
2 3
! " x1 ! "
1 4 !2 4 5 4
x2 D
2 !3 1 !3
x3
The product on the left side has order 2 " 1 and, hence, is a column matrix. Thus,
! " ! "
x1 C 4x2 ! 2x3 4
D
2x1 ! 3x2 C x3 !3
y equality of matrices, corresponding entries must be equal, so we obtain the system
#
x1 C 4x2 ! 2x3 D 4
2x1 ! 3x2 C x3 D !3
Hence, this system of linear equations can be defined by matrix Equation (1). We usu-
ally describe Equation (1) by saying that it has the form
AX D B
where A is the matrix obtained from the coe cients of the variables, X is a column
matrix obtained from the variables, and B is a column matrix obtained from the con-
stants. atrix A is called the c e cient matrix for the system.
Notice that the ariab e in the matrix equation AX D B is the column vector X.
In the example at hand, X is a 3 " 1 column vector. A single s uti n of AX D B is a
column vector C, f the same size as X, with the property that AC D B. In the present
example, a single solution being a 3 " 1 column vector is the same thing as an ordered
triple of numbers. Indeed, if C is an n " 1 column vector, then CT is a 1 " n row vector,
To review n-tuples, see Section 2. . which agrees with the notion of an n-tuple of numbers. For a system that consists of
m linear equations in n un nowns, its representation in the form AX D B will have A,
m " n, and B, m " 1. The variable X will then be an n " 1 column vector, and a sing e
solution C will be an n " 1 column vector, completely determined by an n-tuple of
numbers.

A L IT I
E AM LE M F S U M M
Write the following pair of Write the system
lines in matrix form, using matrix #
multiplication. 2x1 C 5x2 D 4
1 5 x1 C 3x2 D 7
y D ! x C ;y D ! x C
5 5 3 3
in matrix form by using matrix multiplication.
S If
! " ! " ! "
2 5 x 4
AD XD 1 BD
3 x2 7

then the given system is equivalent to the single matrix equation

AX D B
that is,
! "! " ! "
2 5 x1 4
D
3 x2 7

Now ork Problem 59 G

 Section 6.3 atr t cat on 263

R BLEMS
2 3 2 3 2 32 3 " #" #
1 3 !2 0 !2 3 0 0 1 x
If A D 4!2 1 !15 ; B D 4!2 !25 ; and 40 a11 a12 x1
4 1 05 4y5
0 4 3 3 1 !1 a21 a22 x2
1 0 0 z
AB D C D ŒCij ! nd each f the f ing " #2 3 2 3" #
x 2 !3
2 1 3 4 15 40 x
C11 C22 C32 x2 15 1
4 7 x2
x3 2 1
C33 C31 C12
If A is 2 " 3 B is 3 " 1 C is 2 " 5 is 4 " 3 E is 3 " 2 In Pr b ems c mpute the required matrices if
" # " # 2 3
and F is 2 " 3 nd the size and number f entries f each f the !1 1
1 !2 !2 3 0
f ing AD BD CD4 0 35
0 3 1 !4 1
AE E EC 2 4
2 3
B FB EB
2 3 2 3
EET B E.AE/ E.FB/ 1 0 0 3 0 0 6 13 0 07
6 7
.F C A/B D 40 1 1 5 E D 40 6 0 5 F D 6 60 6
1
077
1 2 1 0 0 3 4 1
5
rite the identity matrix that has the f ing rder 0 0 3

4 6 2 3
1 0 0
In Pr b ems perf rm the indicated perati ns
2 3" I D 40 1 05
" #" # # 0 0 1
!1 1
2 !4 4 0 4 0 45 1 !2
3 2 !1 3 3 4
2 1 6F C EI 3A ! 2BC
2 3
" #2 3 4 B. C E/ 3I ! 23 FE E. C I/
1
2 0 3 4 5 Œ1 2 3!455 . C/A A.BC/
4
!1 4 5 6
7 In Pr b ems c mpute the required matrix if it exists gi en
2 32 3 that
1 4 !1 2 1 0 " # 2 3 2 3
4 0 0 25 40 !1 15 0 0 !1 1 0
1 !1 0
!2 1 1 1 1 2 AD B D 42 !1 05 C D 42 !15
0 1 1
0 0 2 0 1
2 32 3 2 3 2 3
4 2 !2 3 1 1 0 1 0 0 0 0 0
43 10 0 5 40 0 0 0 5
I D 40 1 0 5 0 D 40 0 0 5
1 0 2 0 1 0 1 0 0 1 0 0 0
2 3
1 5 !2 !1
Œ1 !2 5! 4 0 0 2 15 A2 AT A B3
!1 0 1 !3 A.BT /2 C .AIC/T AT .2CT /
2 3 " #
!2 1 .BAT /T .2I/T .2I/2 ! 2I2
7
Œ1 !4! 4 0 15 Œ 10 ! .AT CT B/0 A.I ! 0/ IT 0
5 0 .AB/.AB/T B2 ! 3B C 2I
" # 0" # " #1
0 1 @ 1 0 1 0 1 0 A In Pr b ems represent the gi en system by using matrix
C mu tip icati n
2 3 1 1 0 0 0 1
( 8
0" <3x C y C z D 2
# " #1 2 3 3x C y D 6
x!yC zD 4
1 2
@ !2 0 2 !1 0 2 A4 2x ! y D 5 :5x ! y C 2z D 12
3 C2 3 45
3 !1 1 1 1 !2 8
5 6
<2r ! s C 3t D
" #" # 5r ! s C 2t D 5
1 !1 !1 0 !1 0 0 :3r ! 2s C 2t D 11
0 3 2 1 2 1 1
0 31 Secret Messages Secret messages can be encoded by using
" # " #2 a code and an encoding matrix. Suppose we have the
1 !2
1 2 B 20 1 4 C
1 5A
4 @ 1
2 following code:
3 0 !2
3 0
a b c d e f g h i j k m
" # 0" #" #1 1 2 3 4 5 6 7 10 11 12 13
34 2 3
2 C 7@ A n p q r s t u x y z
56 01 45
14 15 16 17 1 1 20 21 22 23 24 25 26
264 C atr e ra
" #
1 1 b Find the matrix RC whose first entry gives the total purchase
et the encoding matrix be E D : Then we can encode a price and whose second entry gives the total transportation cost.
2 3 " #
message by ta ing every two letters of the message, converting 1
c et D , and then compute RC , which gives the total
them to their corresponding numbers, creating a 1 " 2 matrix, and 1
then multiplying each matrix on the right by E. Use this code and cost of materials and transportation for all houses being built.
matrix to encode the message winter is coming , leaving the
Perform the following calculations for Example 6.
slashes to separate words.
a Compute the amount that each industry and each consumer
Inventory A pet store has 6 ittens, 10 puppies, and have to pay for the goods they receive.
7 parrots in stoc . If the value of each itten is 55, each puppy is b Compute the profit earned by each industry.
150, and each parrot is 35, find the total value of the pet store s c Find the total amount of money that is paid out by all the
inventory using matrix multiplication. industries and consumers.
Stoc s A stoc bro er sold a customer 200 shares of d Find the proportion of the total amount of money
stoc A, 300 shares of stoc , 500 shares of stoc C, and 250 found in part (c) paid out by the industries. Find the proportion of
shares of stoc . The prices per share of A, , C, and are the total amount of money found in part (c) that is paid out by the
100, 150, 200, and 300, respectively. Write a row vector consumers.
representing the number of shares of each stoc bought. Write a Prove that if AB D BA, then .A C B/.A ! B/ D A2 ! B2 .
column vector representing the price per share of each stoc .
Using matrix multiplication, find the total cost of the stoc s. Show that if
" # " #
Construction Cost In Example , assume that the 1 2 2 !3
AD and B D 3
contractor is to build five ranch-style, two Cape Cod style, and 1 2 !1 2
four colonial-style houses. Using matrix multiplication, compute
the total cost of raw materials. then AB D 0. bserve that since neither A nor B is the zero
matrix, the algebraic rule for real numbers, If ab D 0, then either
Costs In Example , assume that the contractor wishes to a D 0 or b D 0 , does not hold for matrices. It can also be shown
ta e into account the cost of transporting raw materials to the that the cancellation law is not true for matrices that is, if
building site as well as the purchasing cost. Suppose the costs are AB D AC, then it is not necessarily true that B D C.
given in the following matrix:
et 1 and 2 be two arbitrary 3 " 3 diagonal matrices. y
Purchase
2 Transport
3 computing 1 2 and 2 1 , show that
Steel a oth 1 2 and 2 1 are diagonal matrices.
63500 507
61500 507 Wood b 1 and 2 c mmute meaning that 1 2 D 2 1 .
6 7
C D61000 1007 lass
6 7 In Pr b ems c mpute the required matrices gi en that
4 250 105 Paint
" # 2 3 " #
3500 0 abor 1:1 4:
3:2 !4:1 5:1 !1:2 1:5
AD B D 4!2:3 3:25 C D
!2:6 1:2 6: 2:4 6:2
a y computing RC, find a matrix whose entries give the 4:6 !1:4
purchase and transportation costs of the materials for each type of
house. A.2B/ 2:6.BC/ 3CA.!B/ C3

Objective S S R M
os o o to re ce a atr an In this section we illustrate a method by which matrices can be used to solve a system
to se atr re ct on to so e a
near s ste of linear equations. It is important here to recall, from Section 3.4, that two systems
of equations are equi a ent if they have the same set of solutions. It follows that in
attempting to solve a linear system, call it S1 , we can do so by solving any system S2
that is equivalent to S1 . If the solutions of S2 are more easily found than those of S1 ,
then replacing S1 by S2 is a useful step in solving S1 . In fact, the method we illustrate
amounts to finding a sequence of equivalent systems, S1 , S2 , S3 , # # #, Sn for which the
solutions of Sn are b i us. In our development of this method, nown as the meth d
f reducti n, we will first solve a system by the usual method of elimination. Then we
will obtain the same solution by using matrices.
et us consider the system
#
3x ! y D 1
x C 2y D 5
consisting of two linear equations in two un nowns, x and y. Although this system can
be solved by various algebraic methods, we will solve it by a method that is readily
adapted to matrices.
 Section 6.4 o n ste s e cn atr ces 265

For reasons that will be obvious later, we begin by replacing Equation (1) by Equa-
tion (2), and Equation (2) by Equation (1), thus obtaining the equivalent system,
#
x C 2y D 5
3x ! y D 1
ultiplying both sides of Equation (3) by !3 gives !3x ! 6y D !15. Adding the left
and right sides of this equation to the corresponding sides of Equation (4) produces an
equivalent system in which x is eliminated from the second equation:
#
x C 2y D 5
0x ! 7y D !14
Now we will eliminate y from the first equation. ultiplying both sides of Equation (6)
by ! 17 gives the equivalent system,
#
x C 2y D 5
0x C y D 2
From Equation ( ), y D 2 and, hence, !2y D !4. Adding the sides of !2y D !4 to
the corresponding sides of Equation (7), we get the equivalent system,
#
x C 0y D 1
0x C y D 2
This is a linear system for which the solution is indeed b i us and, because all the
systems introduced along the way are equivalent systems, the obvious solution of the
last system is also the solution of the original system: x D 1 and y D 2.
Note that in solving the original linear system, we successively replaced it by an
equivalent system that was obtained by performing one of the following three opera-
tions (called e ementary perati ns), which leave the solution unchanged:
Interchanging two equations
ultiplying one equation by a nonzero constant
Adding a constant multiple of the sides of one equation to the corresponding sides
of another equation
efore showing a matrix method of solving the original system,
#
3x ! y D 1
x C 2y D 5
we first need to define some terms. Recall from Section 6.3 that the matrix
! "
3 !1
1 2
is the coe cient matrix of this system. The entries in the first column correspond to
the coe cients of the x s in the equations. For example, the entry in the first row and
first column corresponds to the coe cient of x in the first equation and the entry in the
second row and first column corresponds to the coe cient of x in the second equation.
Similarly, the entries in the second column correspond to the coe cients of the y s.
Another matrix associated with this system is called the augmented coe cient
matrix and is given by
! "
3 !1 1
1 2 5
The first and second columns are the first and second columns, respectively, of the
coe cient matrix. The entries in the third column correspond to the constant terms
in the system: The entry in the first row of this column is the constant term of the
first equation, whereas the entry in the second row is the constant term of the second
equation. Although it is not necessary to include the vertical line in the augmented
coe cient matrix, it serves to remind us that the 1 and the 5 are the constant terms
266 C atr e ra

that appear on the right sides of the equations. The augmented coe cient matrix itself
completely describes the system of equations.
The procedure that was used to solve the original system involved a number of
equivalent systems. With each of these systems, we can associate its augmented coef-
ficient matrix. Following are the systems that were involved, together with their corre-
sponding augmented coe cient matrices, which we have labeled A, B, C, , and E.
# ! "
3x ! y D 1 3 !1 1
DA
x C 2y D 5 1 2 5
# ! "
x C 2y D 5 1 2 5
DB
3x ! y D 1 3 !1 1
# ! "
x C 2y D 5 1 2 5
DC
0x ! 7y D !14 0 !7 !14
# ! "
x C 2y D 5 1 2 5
D
0x C y D 2 0 1 2
# ! "
x C 0y D 1 1 0 1
DE
0x C y D 2 0 1 2

et us see how these matrices are related.

atrix B can be obtained from A by interchanging the first and second rows of A.
This operation corresponds to interchanging the two equations in the original system.
atrix C can be obtained from B by adding, to each entry in the second row of B,
!3 times the corresponding entry in the first row of B:
" ˇ #
1 2 ˇ 5
CD ˇ
3 C .!3/.1/ !1 C .!3/.2/ ˇ 1 C .!3/.5/
" ˇ #
1 2 ˇ 5
D ˇ
0 !7 ˇ !14

This operation is described as follows: the addition of !3 times the first row of B to the
second row of B.
atrix can be obtained from C by multiplying each entry in the second row of
C by ! 17 . This operation is referred to as multiplying the second row of C by ! 17 .
atrix E can be obtained from by adding !2 times the second row of to the
first row of .
bserve that E, which gives the solution, was obtained from A by successively
performing one of three matrix operations called elementary row operations:

E R
Interchanging two rows of a matrix
ultiplying a row of a matrix by a nonzero number
Adding a multiple of one row of a matrix to a different row of that matrix

These elementary row operations correspond to the three elementary operations used in
the algebraic method of elimination. Whenever a matrix can be obtained from another
by one or more elementary row operations we say that the matrices are equivalent.
Thus, A and E are equivalent. (We could also obtain A from E by performing simi-
lar row operations in the reverse order, so the term equi a ent is appropriate.) When
describing particular elementary row operations we will use the following notation for
convenience:
 Section 6.4 o n ste s e cn atr ces 267

otation Corresponding Row Operation

Ri $ Rj Interchange rows Ri and Rj .
kRi ultiply row Ri by the nonzero constant k.
kRi C Rj Add k times row Ri to row Rj (but leave Ri unchanged).

For example, writing

2 3 2 3
1 0 !2 !4R C R 1 0 !2
44 !2 15 !!!!!!!!! 40 !2
1 2
5
5 0 3 5 0 3

means that the second matrix was obtained from the first by adding !4 times row 1 to
row 2. Note that we write .!k/Ri as !kRi .
We are now ready to describe a matrix procedure for solving a system of linear
equations. First, we form the augmented coe cient matrix of the system then, by
means of elementary row operations, we determine an equivalent matrix that clearly
indicates the solution. et us be specific as to what we mean by a matrix that c ear y
indicates the s uti n. This is a matrix, called a reduced matrix which will be defined
below. It is convenient to define first a ero row of a matrix to be a row that consists
entire y of zeros. A row that is not a zero-row, meaning that it contains at east ne
nonzero entry, will be called a non ero row. The first nonzero entry in a nonzero-row
is called the leading entry.

R M
A matrix is said to be a reduced matrix provided that all of the following are true:
All zero-rows are at the bottom of the matrix.
For each nonzero-row, the leading entry is 1, and all other entries in the c umn
of the leading entry are 0.
The leading entry in each row is to the right of the leading entry in any row
above it.

It can be shown that each matrix is equivalent to exact y ne reduced matrix. To solve
a system, we find the reduced matrix such that the augmented coe cient matrix is
equivalent to it. In our previous discussion of elementary row operations, the matrix
! "
1 0 1
ED
0 1 2
is a reduced matrix.

E AM LE R M

For each of the following matrices, determine whether it is reduced or not reduced.
! " ! " ! "
1 0 1 0 0 0 1
a b c
0 3 0 1 0 1 0
! " 2 3 2 3
0 0 0 1 0 0 0 1 0 3
d
0 0 0 e 40 0 0 5 f 40 0 1 2 5
0 1 0 0 0 0 0
S
a Not a reduced matrix, because the leading entry in the second row is not 1
b Reduced matrix
c Not a reduced matrix, because the leading entry in the second row is not to the right
of the leading entry in the first row
268 C atr e ra

d Reduced matrix
e Not a reduced matrix, because the second row, which is a zero-row, is not at the
bottom of the matrix
f Reduced matrix
Now ork Problem 1 G

E AM LE R M

Reduce the matrix

2 3
0 0 1 2
43 !6 !3 05
6 !12 2 11

S To reduce the matrix, we must get the leading entry to be a 1 in the first
row, the leading entry a 1 in the second row, and so on, until we arrive at a zero-row,
if there are any. oreover, we must wor from left to right, because the leading
entry in each row must be to the eft of all other leading entries in the rows be
it.

S Since there are no zero-rows to move to the bottom, we proceed to find the
first column that contains a nonzero entry this turns out to be column 1. Accordingly,
in the reduced matrix, the leading 1 in the first row must be in column 1. To accomplish
this, we begin by interchanging the first two rows so that a nonzero entry is in row 1 of
column 1:

0 0 1 2 3 -6 -3 0
R1 4 R2
3 -6 -3 0 0 0 1 2
6 -12 2 11 6 -12 2 11

1
Next, we multiply row 1 by 3
so that the leading entry is a 1:

1
R1 1 -2 -1 0
3
0 0 1 2
6 -12 2 11

Now, because we must have zeros below (and above) each leading entry, we add !6
times row 1 to row 3:

-6R1 + R3 1 -2 -1 0
0 0 1 2
0 0 8 11

Next, we move to the right of column 1 to find the first column that has a nonzero
entry in row 2 or below this is column 3. Consequently, in the reduced matrix, the
leading 1 in the second row must be in column 3. The foregoing matrix already does
have a leading 1 there. Thus, all we need do to get zeros below and above the leading
1 is add 1 times row 2 to row 1 and add ! times row 2 to row 3:

(1)R2 + R1 1 -2 0 2
0 0 1 2
-8R2 + R3 0 0 0 -5
 Section 6.4 o n ste s e cn atr ces 269

Again, we move to the right to find the first column that has a nonzero entry in row
3 namely, column 4. To ma e the leading entry a 1, we multiply row 3 by ! 15 :

- 15 R3 1 -2 0 2
0 0 1 2
0 0 0 1

Finally, to get all other entries in column 4 to be zeros, we add !2 times row 3 to both
row 1 and row 2:

-2R3 + R1 1 -2 0 0
0 0 1 0
The sequence of steps that is used to -2R3 + R2 0 0 0 1
reduce a matrix is not unique however,
the reduced matrix is unique.
The last matrix is in reduced form.
Now ork Problem 9 G
The method of reduction described for solving our original system can be gen-
eralized to systems consisting of m linear equations in n un nowns. To solve such a
system as
8̂
ˆ A11 x1 C A12 x2 C # # # C A1n xn D B1
ˆ
ˆ
ˆ
ˆ A x C A22 x2 C # # # C A2n xn D B2
< 21 1
# # # #
ˆ
ˆ # # # #
ˆ
ˆ
ˆ # # # #
:̂A x C A x C # # # C A x D B
m1 1 m2 2 mn n m

involves

determining the augmented coe cient matrix of the system, which is

2 3
A11 A12 ### A1n B1
6 A21 A22 ### A2n B2 7
6 7
6 # # # # 7
6 7
6 # # # # 7
6 7
4 # # # # 5
A L IT I Am1 Am2 ### Amn Bm
An investment firm offers three
stoc portfolios: A, , and C. The and
number of bloc s of each type of determining the reduced matrix to which the augmented coe cient matrix is
stoc in each of these portfolios is
equivalent.
summarized in the following table:

Portfolio Frequently, step 2 is called reducing the augmented c e cient matrix.

A C
High 6 1 3
Ris : oderate 3 2 3
ow 1 5 3 E AM LE S S R
A client wants 35 bloc s of high-
ris stoc , 22 bloc s of moderate-ris
y using matrix reduction, solve the system
stoc , and 1 bloc s of low-ris stoc . 8
How many of each portfolio should be <2x C 3y D !1
suggested 2x C y D 5
: xC yD 1
270 C atr e ra

S Reducing the augmented coe cient matrix of the system, we have

2 3 -1 1 1 1
R1 4R3
2 1 5 2 1 5
1 1 1 2 3 -1

-2R1 + R2 1 1 1
0 -1 3
2 3 -1

-2R1 + R3 1 1 1
0 -1 3
0 1 -3

(-1)R2 1 1 1
0 1 -3
0 1 -3

-R2 + R1 1 0 4
0 1 -3
0 1 -3

-R2 + R3 1 0 4
0 1 -3
0 0 0

The last matrix is reduced and corresponds to the system

8
< x C 0y D 4
0x C y D !3
:0x C 0y D 0
Recall from Section 3.4 that a sing e
solution of a system of equations in two
un nowns is an rdered pair of values. Since the original system is equivalent to this system, it has a unique solution, namely,
ore generally, a single solution of a
system of equations in n un nowns is an
rdered n tup e of values.
xD 4
A L IT I
y D !3
A health spa customizes the diet
and vitamin supplements of each of its
clients. The spa offers three different
Now ork Problem 13 G
vitamin supplements, each containing
different percentages of the recom-
mended daily allowance (R A) of E AM LE S S R
vitamins A, C, and . ne tablet of
supplement provides 40 of the R A Using matrix reduction, solve
of A, 20 of the R A of C, and 10
of the R A of . ne tablet of supple- 8
ment provides 10 of the R A of A, < x C 2y C 4z ! 6 D 0
10 of the R A of C, and 30 of the 2z C y ! 3 D 0
: x C y C 2z ! 1 D 0
R A of . ne tablet of supplement
provides 10 of the R A of A, 50
of the R A of C, and 20 of the R A
of . The spa staff determines that one S Rewriting the system so that the variables are aligned and the constant terms
client should ta e 1 0 of the R A of appear on the right sides of the equations, we have
vitamin A, 200 of the R A of vita-
min C, and 1 0 of the R A of vitamin 8
<x C 2y C 4z D 6
each day. How many tablets of each
y C 2z D 3
supplement should she ta e each day :x C y C 2z D 1
 Section 6.4 o n ste s e cn atr ces 271

Reducing the augmented coe cient matrix, we obtain

1 2 4 6 1 2 4 6
-R1 + R3
0 1 2 3 0 1 2 3
1 1 2 1 0 -1 -2 -5

1 0 0 0
-2R2 + R1
0 1 2 3
(1)R2 + R3
0 0 0 -2

1 0 0 0
- 12 R3
0 1 2 3
0 0 0 1

1 0 0 0
-3R3 + R2
0 1 2 0
0 0 0 1

The last matrix is reduced and corresponds to

8
< xD0
y C 2z D 0
: 0D1
Whenever we get a row with all 0 s to the Since 0 ¤ 1, there are no values of x, y, and z for which all equations are satisfied
left side of the vertical rule and a nonzero simultaneously. Thus, the original system has no solution.
entry to the right, no solution exists.
Now ork Problem 15 G
E AM LE F S
A L IT I
Using matrix reduction, solve
A zoo veterinarian can purchase 8
animal food of four different types: A, <2x1 C 3x2 C 2x3 C 6x4 D 10
, C, and . Each food comes in the x2 C 2x3 C x4 D 2
:3x ! 3x3 C 6x4 D
same size bag, and the number of grams 1
of each of three nutrients in each bag
S Reducing the augmented coe cient matrix, we have
is summarized in the following table:
3
Food 2 3 2 6 10 1 1 3 5
- 12 R1 2
A C 0 1 2 1 2 0 1 2 1 2
N1 5 5 10 5 3 0 -3 6 9 3 0 -3 6 9
Nutrient N2 10 5 30 10
3
N3 5 15 10 25 1 2
1 3 5
-3R1 + R3
0 1 2 1 2
For one animal, the veterinarian deter-
mines that she needs to combine the 0 - 92 -6 -3 -6
bags to get 10,000 g of N1 , 20,000 g of
N2 , and 20,000 g of N3 . How many bags 1 0 -2 3
2
of each type of food should she order - 32 R2 + R1 2

9
0 1 2 1 2
R +
2 2
R3 3
0 0 3 2
3

3
1 1 0 -2 2
R
3 3
2
0 1 2 1 2
1
0 0 1 2
1

5
1 0 0 2
4
2R3 + R1
0 1 0 0 0
-2R3 + R2 1
0 0 1 2
1
272 C atr e ra

This matrix is reduced and corresponds to the system

8̂
ˆ 5
<x1 C 2 x4 D 4
x2 D 0
ˆ
:̂x3 C 1 x4 D 1
2

Thus,
x1 D 4 ! 52 x4
x2 D 0
x3 D 1 ! 12 x4
The system imposes no restrictions on x4 so that x4 may ta e on any real value. If
we append
x4 D x4
to the preceding equations, then we have expressed all four of the un nowns in terms
of x4 and this is the genera solution of the original system.
For each particular value of x4 , Equations ( ) (12) determine a particu ar solution
of the original system. For example, if x4 D 0, then a particu ar solution is
x1 D 4 x2 D 0 x3 D 1 x4 D 0
If x4 D 2, then
x1 D !1 x2 D 0 x3 D 0 x4 D 2
is another particular solution. Since there are infinitely many possibilities for x4 , there
are infinitely many solutions of the original system.
Recall (see Examples 3 and 6 of Section 3.4) that, if we li e, we can write x4 D r
and refer to this new variable r as a parameter. (However, there is nothing special about
the name r, so we could consider x4 as the parameter on which a the original variables
depend. Note that we can write x2 D 0 C 0x4 and x4 D 0 C 1x4 .) Writing r for the
parameter, the solution of the original system is given by
x1 D 4 ! 52 r
x2 D 0 C 0r
x3 D 1 ! 12 r
x4 D 0 C 1r
where r is any real number, and we spea of having a ne parameter fami y of solutions.
Now, with matrix addition and scalar multiplication at hand, we can say a little more
about such families. bserve that
2 3
2 3 2 3 ! 5
x1 4 6 27
6 x2 7 6 0 7 6 7
6 7 D 6 7 C r6 07
4 x3 5 4 1 5 6!1 7
4 25
x4 0 1

Readers familiar with analytic geometry will 2 see

3 that the solutions form a ine in
4
607
x1 x2 x3 x4 -space, passing through the p int 6 7
4 1 5 and in the directi n of the line
0
2 3
2 3 5
0 !
6 27
607 6 07
segment oining 6 7 6 7
4 0 5 and 6 ! 1 7.
4 25
0 1
Now ork Problem 17 G
 Section 6.4 o n ste s e cn atr ces 273

A system of linear equations has zero, Examples 3 5 illustrate the fact that a system of linear equations may have a unique
one, or infinitely many solutions. solution, no solution, or infinitely many solutions. It can be shown that these are the
n y possibilities.

R BLEMS
In Pr b ems determine hether the matrix is reduced r n t S e Pr b ems by using matrix reducti n
reduced axes A company has taxable income of 312,000. The
" # " # 2 3
1 0 0 federal tax is 25 of that portion that is left after the state tax has
1 2 1 0 0 3 40 1 05 been paid. The state tax is 10 of that portion that is left after the
7 0 0 0 1 2
0 0 0 federal tax has been paid. Find the company s federal and state
2 3 2 3 2 3 taxes.
1 1 0 0 0 0 0 0 1
60 17 60 1 0 07 61 0 07 Decision Ma ing A manufacturer produces two products,
6 7 6 7 6 7
40 05 40 0 1 05 40 1 05 A and . For each unit of A sold, the profit is 10, and for each
0 0 0 0 0 0 0 0 0 unit of sold, the profit is 12. From experience, it has been
found that 50 more of A can be sold than of . Next year the
In Pr b ems reduce the gi en matrix manufacturer desires a total profit of 54,000. How many units of
" # " # 2 3
2 4 6 each product must be sold
1 3 0 !2 0 1 41 2 35
4 0 1 2 0 4 Production Scheduling A manufacturer produces three
1 2 3
2 3 2 3 2 3 products: A, , and C. The profits for each unit of A, , and C
2 3 2 3 4 1 0 0 2 sold are 1, 2, and 3, respectively. Fixed costs are 17,000 per
61 !67 6 1 7 2 37 62 0 37 year, and the costs of producing each unit of A, , and C are 4,
6 7 6 7 6 7
44 5 4!1 4 2 05 40 !1 05 5, and 7, respectively. Next year, a total of 11,000 units of all
1 7 0 1 1 0 0 4 1 three products is to be produced and sold, and a total profit of
25,000 is to be realized. If total cost is to be 0,000, how many
S e the systems in Pr b ems by the meth d f reducti n units of each of the products should be produced next year
( (
3x C 5y D 25 x ! 3y D !11 Production Allocation National es Co. has plants for
x ! 2y D 1 4x C 3y D producing des s on both the East Coast and West Coast. At the
( ( East Coast plant, fixed costs are 20,000 per year and the cost of
3x C y D 4 3x C 2y ! z D 1 producing each des is 0. At the West Coast plant, fixed costs
are 1 ,000 per year and the cost of producing each des is 5.
12x C 4y D 2 !x ! 2y ! 3z D 1
Next year the company wants to produce a total of 00 des s.
( (
x C 2y C z ! 4 D 0 x C y ! 5z ! D 0 etermine the production order for each plant for the forthcoming
year if the total cost for each plant is to be the same.
3x C 2z ! 5 D 0 2x ! y ! z ! 1 D 0
8̂ 8̂
ˆ
< x1 ! 3x2 D 0 ˆ
< x1 C 4x2 D
2x1 C 2x2 D 3 3x1 ! x2 D 6
ˆ ˆ
:̂5x1 ! x2 D 1 :̂ x1 ! x2 D 2
8̂ 8̂
ˆ
<x C 3y D 2 ˆ
< xC y! zD 7
2x C 7y D 4 2x ! 3y ! 2z D 4
ˆ ˆ
:̂x C 5y C z D 5 :̂ x ! y ! 5z D 23
8̂ 8̂
ˆ 3x ! y C z D 12 ˆ x C 3z D !1 Vitamins A person is ordered by a doctor to ta e 10 units
ˆ
ˆ ˆ
ˆ of vitamin A, units of vitamin , and 1 units of vitamin E each
< xC yC zD 2 <3x C 2y C 11z D 1
day. The person can choose from three brands of vitamin pills.
ˆ
ˆ x C 2y ! z D !2 ˆ
ˆ x C y C 4z D 1 rand contains 2 units of vitamin A, 3 units of vitamin , and
ˆ ˆ
:̂2x C y ! 3z D 1 :̂2x ! 3y C 3z D ! 5 units of vitamin E brand has 1, 3, and 4 units, respectively
and brand has 1 unit of vitamin A, none of vitamin , and 1 of
8̂
vitamin E.
ˆ
ˆ x1 ! x2 ! x3 ! x4 ! x5 D 0
ˆ
<x C x ! x ! x ! x D 0
1 2 3 4 5
ˆ
ˆ x C x C x ! x ! x5 D0
ˆ 1 2 3 4
:̂x1 C x2 C x3 C x4 ! x5 D 0
8̂
ˆ
ˆ x1 C x2 C x3 C x4 D 0
ˆ
<x C x C x ! x D 0
1 2 3 4
ˆ x
ˆ 1 C x ! x ! x4 D0
ˆ 2 3
:̂x1 ! x2 ! x3 C x4 D 0
274 C atr e ra

a Find all possible combinations of pills that will provide Investments An investment company sells three types of
exactly the required amounts of vitamins. pooled funds, Standard (S), eluxe ( ), and old Star ( ).
b If brand costs 1 cent a pill, brand 6 cents, and brand Each unit of S contains 12 shares of stoc A, 16 of stoc , and
3 cents, are there any combinations in part (a) costing exactly of stoc C.
15 cents a day Each unit of contains 20 shares of stoc A, 12 of stoc , and
c What is the least expensive combination in part (a) the most 2 of stoc C.
expensive
Each unit of contains 32 shares of stoc A, 2 of stoc , and
Production A firm produces three products, A, , and C, 36 of stoc C.
that require processing by three machines, I, II, and III. The time Suppose an investor wishes to purchase exactly 220 shares of
in hours required for processing one unit of each product by the stoc A, 176 shares of stoc , and 264 shares of stoc C by
three machines is given by the following table: buying units of the three funds.
a Set up equations in s, for units of S, d, for units of , and g, for
A C units of whose solution would provide the number of units of S,
, and that will meet the investor s requirements exactly.
I 3 1 2 b Solve the system set up in (a) and show that it has infinitely
II 1 2 1 many solutions, if we naively assume that s, d, and g can ta e on
III 2 4 1 arbitrary real values.
c Pooled funds can be bought only in units that are non-negative
integers. In the solution to (b) above, it follows that we must
achine I is available for 440 hours, machine II for 310 hours, require each of s, d, and g to be non-negative integers. Enumerate
and machine III for 560 hours. Find how many units of each the solutions in (b) that remain after we impose this new
product should be produced to ma e use of all the available time constraint.
on the machines. d Suppose the investor pays 300 for each unit of S, 400 for
each unit of , and 600 for each unit of . Which of the possible
solutions from part (c) will minimize the total cost to the investor

Objective S S R
o foc s o r attent on on M C
non o o eneo s s ste s t at
n o e ore t an one ara eter n As we saw in Section 6.4, a system of linear equations may have a unique solution, no
t e r enera so t on an to so e solution, or infinitely many solutions. When there are infinitely many, the general solu-
an cons er t e t eor of
o o eneo s s ste s tion is expressed in terms of at least one parameter. For example, the general solution
in Example 5 was given in terms of the parameter r:

x1 D 4 ! 52 r
x2 D 0
x3 D 1 ! 12 r
x4 D r

At times, more than one parameter is necessary. In fact, we saw a very simple example
in Example 7 of Section 3.4. Example 1 illustrates further.

E AM LE T F S

Using matrix reduction, solve

8
<x1 C 2x2 C 5x3 C 5x4 D !3
x1 C x2 C 3x3 C 4x4 D !1
:x ! x ! x C 2x D 3
1 2 3 4

S The augmented coe cient matrix is

2 3
1 2 5 5 !3
41 1 3 4 !15
1 !1 !1 2 3
 Section 6.5 o n ste s e cn atr ces Cont n e 275

whose reduced form is

2 3
1 0 1 3 1
40 1 2 1 !25
0 0 0 0 0

Hence,
#
x1 C x3 C 3x4 D 1
x2 C 2x3 C x4 D !2

from which it follows that

x1 D 1 ! x3 ! 3x4
x2 D !2 ! 2x3 ! x4

Since no restriction is placed on either x3 or x4 , they can be arbitrary real numbers,

giving us a parametric family of solutions. Setting x3 D r and x4 D s, we can give the
solution of the given system as
x1 D 1 ! r ! 3s
x2 D !2 ! 2r ! s
x3 Dr
x4 Ds

where the parameters r and s can be any real numbers. y assigning specific values
to r and s, we get particular solutions. For example, if r D 1 and s D 2, then the
corresponding particular solution is x1 D !6; x2 D !6; x3 D 1, and x4 D 2. As in the
one-parameter case, we can now go further and write
2 3 2 3 2 3 2 3
x1 1 !1 !3
6 x2 7 6 !2 7 6 7 6 7
6 7D6 7 C r 6 !2 7 C s 6 !1 7
4 x3 5 4 0 5 4 15 4 05
x4 0 0 1
2 3
1
6 !2 7
which can be shown to exhibit the family of solutions as a p ane through 6 7
4 0 5 in
0
x1 x2 x3 x4 -space.
Now ork Problem 1 G
It is customary to classify a system of linear equations as being either h m gene us
or n nh m gene us depending on whether the constant terms are all zero.

The system
8̂
ˆ A x C A12 x2 C # # # C A1n xn D B1
ˆ 11 1
ˆ
ˆ
ˆ A x C A22 x2 C # # # C A2n xn D B2
< 21 1
# # # #
ˆ
ˆ # # # #
ˆ
ˆ
ˆ # # # #
:̂A x C A x C # # # C A x D B
m1 1 m2 2 mn n m

is called a homogeneous system if B1 D B2 D # # # D Bm D 0. The system is a

nonhomogeneous system if at least one of the Bi is not equal to 0.
276 C atr e ra

E AM LE N H S

The system (
2x C 3y D 4
3x ! 4y D 0
is nonhomogeneous because of the 4( in the first equation. The system
2x C 3y D 0
3x ! 4y D 0
is homogeneous.
G
If the homogeneous system (
2x C 3y D 0
3x ! 4y D 0
were solved by the method of reduction, first the augmented coe cient matrix would
be written
" ˇ #
2 3 ˇˇ 0
3 !4 ˇ 0
bserve that the last column consists entirely of zeros. This is typical of the augmented
coe cient matrix of any homogeneous system. We would then reduce this matrix by
using elementary row operations:
" ˇ # " ˇ #
2 3 ˇˇ 0 1 0 ˇˇ 0
! ### !
3 !4 ˇ 0 0 1 ˇ 0
The last column of the reduced matrix also consists only of zeros. This does not occur
by chance. When any elementary row operation is performed on a matrix that has a col-
umn consisting entirely of zeros, the corresponding column of the resulting matrix will
also be all zeros. For convenience, it will be our custom when solving a homogeneous
system by matrix reduction to delete the last column of the matrices involved. That is,
we will reduce only the c e cient matrix of the system. For the preceding system, we
would have
! " ! "
2 3 1 0
! ### !
3 !4 0 1
Here the reduced matrix, called the reduced c e cient matrix corresponds to the
system
#
x C 0y D 0
0x C y D 0
so the solution is x D 0 and y D 0.
et us now consider the number of solutions of the homogeneous system
8̂
ˆ A11 x1 C A12 x2 C # # # C A1n xn D 0
ˆ
ˆ
ˆ
ˆ A x C A22 x2 C # # # C A2n xn D 0
< 21 1
# # # #
ˆ
ˆ # # # #
ˆ
ˆ
ˆ # # # #
:̂A x C A x C # # # C A x D 0
m1 1 m2 2 mn n

ne solution always occurs when x1 D 0; x2 D 0; : : : ; and xn D 0, since each equation

is satisfied for these values. This solution, called the trivial solution, is a solution of
e ery homogeneous system and follows from the matrix equation
A0n D 0m
where 0n is the n " 1 column vector (and 0m is the m " 1 column vector).
 Section 6.5 o n ste s e cn atr ces Cont n e 277

There is a theorem that allows us to determine whether a homogeneous system has

a unique solution (the trivial solution only) or infinitely many solutions. The theorem
is based on the number of nonzero-rows that appear in the reduced coe cient matrix
of the system. Recall that a n nzer r is a row that does not consist entirely of zeros.

T
et A be the reduced coe cient matrix of a homogeneous system of m linear equa-
tions in n un nowns. If A has exactly k nonzero-rows, then k $ n. oreover,
if k < n, the system has infinitely many solutions, and
if k D n, the system has a unique solution (the trivial solution).

If a homogeneous system consists of m equations in n un nowns, then the coef-

ficient matrix of the system has size m " n. Thus, if m < n and k is the number of
nonzero rows in the reduced coe cient matrix, then k $ m, and, hence, k < n. y the
foregoing theorem, the system must have infinitely many solutions. Consequently, we
have the following corollary.

C
A homogeneous system of linear equations with fewer equations than un nowns
has infinitely many solutions.
The preceding theorem and corollary
apply only to homogeneous systems of
linear equations. For example, consider
the system
# E AM LE N S H S
x C y ! 2z D 3
2x C 2y ! 4z D 4
etermine whether the system
which consists of two linear equations in #
three un nowns. We cannot conclude x C y ! 2z D 0
that this system has infinitely many 2x C 2y ! 4z D 0
solutions, since it is not homogeneous.
Indeed, it is easy to verify that it has no has a unique solution or infinitely many solutions.
solution.
S There are two equations in this homogeneous system, and this number is
less than the number of un nowns (three). Thus, by the previous corollary, the system
has infinitely many solutions.
Now ork Problem 9 G
A L IT I
E AM LE S H S
A plane in three-dimensional
space can be written as
etermine whether the following homogeneous systems have a unique solution or
ax C by C cz D d. We can find
the possible intersections of planes in infinitely many solutions then solve the systems.
this form by writing them as systems 8
of linear equations and using reduction
< x ! 2y C z D 0
a 2x ! y C 5z D 0
to solve them. If d D 0 in each : x C y C 4z D 0
equation, then we have a homogeneous
system with either a unique solution or
infinitely many solutions. etermine S Reducing the coe cient matrix, we have
whether the intersection of the planes 2 3 2 3
5x C 3y C 4z D 0
1 !2 1 1 0 3
42 !1 55 ! # # # ! 40 1 15
6x C y C 7z D 0 1 1 4 0 0 0
3x C 1y C 2z D 0
The number of nonzero-rows, 2, in the reduced coe cient matrix is less than the num-
has a unique solution or infinitely many
solutions then solve the system. ber of un nowns, 3, in the system. y the previous theorem, there are infinitely many
solutions.
278 C atr e ra

Since the reduced coe cient matrix corresponds to

#
x C 3z D 0
yC zD0
the solution may be given in parametric form by
x D !3r
y D !r
zDr
where r is any real number.
8̂
ˆ 3x C 4y D 0
<
x ! 2y D 0
b
ˆ2x C y D 0
:̂2x C 3y D 0

S Reducing the coe cient matrix, we have

2 3 2 3
3 4 1 0
61 !27 6 17
6 7 ! # # # ! 60 7
42 15 40 05
2 3 0 0
The number of nonzero-rows (2) in the reduced coe cient matrix equals the number
of un nowns in the system. y the theorem, the system must have a unique solution,
namely, the trivial solution x D 0; y D 0.
Now ork Problem 13 G

R BLEMS
In Pr b ems s e the systems by using matrix reducti n F r Pr b ems determine hether the system has in nite y
8̂
many s uti ns r n y the tri ia s uti n n t s e the
ˆ
< C x ! y ! z D !3 systems
2 C 3x C 2y C 15z D 12 (
ˆ 2:17x ! 5:3y C 0:27z D 0
:̂2 C x C 2y C 5z D
3:51x ! 1:4y C 0:01z D 0
8̂ 8̂
( ˆ
ˆ
<2 C x C 10y C 15z D ! 5 5 C 7x ! 2y ! 5z D 0 <3x ! 4y D 0
! 5x C 2y C 15z D !10 x C 5y D 0
ˆ 7 ! 6x C y ! 5z D 0 ˆ
:̂ C x C 6y C 12z D :̂4x ! y D 0
8̂ 8̂ 8̂ 8̂
ˆ
ˆ 3 ! x ! 3y ! z D !2 ˆ 3 C x ! 10y ! 2z D 5 ˆ
<2x C 3y C 12z D 0 ˆ
<x C y C z D 0
ˆ
<2 ! 2x ! 6y ! 6z D !4 ˆ
<
2 ! 6y ! 2z D 2 3x ! 2y C 5z D 0 x ! zD0
ˆ ˆ
ˆ
ˆ 2 ! x ! 3y ! 2z D !2 ˆ
ˆ ! x ! 2y ! 2z D !1 :̂4x C y C 14z D 0 :̂x ! 2y ! 5z D 0
ˆ :̂ C x ! 4y D 3
:̂3 C x C 3y C 7z D 2 8̂
8̂ ˆ
8̂ C x C y C 2z D 4 <3x C 2y C z D 0
ˆ
ˆ
ˆ
ˆ ! 3x C y ! z D 5 ˆ
ˆ2 C x C 2y C 2z D 7 2x C 2y C z D 0
ˆ
< ! 3x ! y C 3z D 1 ˆ
< ˆ
:̂4x C y C z D 0
C 2x C y C 4z D 5
ˆ
ˆ 3 ! x C y C z D 11 ˆ
ˆ
ˆ ˆ
ˆ S e each f the f ing systems
:̂2 ! 6x ! y C 4z D 4 ˆ ! 2x C 3y ! 4z D 7
3 ( (
:̂4 ! 3x C 4y ! 6z D 2x C 3y D 0 2x ! 5y D 0
( 5x ! 7y D 0 x ! 20y D 0
4x1 ! 3x2 C 5x3 ! 10x4 C 11x5 D !
( (
2x1 C x2 C 5x3 C 3x5 D 6 x C 6y ! 2z D 0 4x C 7y D 0
8̂ 2x ! 3y C 4z D 0 2x C 3y D 0
ˆ
ˆ x1 C 3x3 C x4 C 4x5 D 1
8̂ 8̂
ˆ
< x2 C x3 ! 2x4 D 0 ˆ ˆ
< xC yD0 <2x C y C z D 0
ˆ2x1 ! 2x2 C 3x3 C 10x4 C 15x5 D 10
ˆ 7x ! 5y D 0 x ! y C 2z D 0
ˆ ˆ ˆ
:̂ x1 C 2x2 C 3x3 ! 2x4 C 2x5 D !2 :̂ x ! 4y D 0 :̂ x C y C z D 0
 Section 6.6 n erses 279

8̂ 8̂ 8̂ 8̂
ˆ xC y C zD0 ˆ x C y C 7z D 0 ˆ C x C y C 4z D 0 ˆ C x ! 2y ! 2z D 0
ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
<
< ! 7y ! 14z D 0 < x! y! zD0 < C x C 5z D 0 ! x D0
ˆ ! 2y ! 4z D 0 ˆ 2x ! 3y ! 6z D 0 ˆ 2 x 3y 4z D 0 ˆ
ˆ2 C x ! 3y ! 3z D 0
ˆ
ˆ ˆ
ˆ ˆ
ˆ
C C C
:̂ C 2x ! 3y ! 3z D 0
:̂ ! 5y ! 10z D 0 :̂3x C y C 13z D 0 :̂ ! 3x C 2y ! z D 0

Objective I
o eter ne t e n erse of an We have seen how useful the method of reduction is for solving systems of linear equa-
n ert e atr an to se n erses to
so e s ste s tions. ut it is by no means the only method that uses matrices. In this section, we will
discuss a different method that applies to certain systems of n linear equations in n
un nowns.
In Section 6.3, we showed how a system of linear equations can be written in matrix
form as the single matrix equation AX D B, where A is the coe cient matrix. For
example, the system
#
x1 C 2x2 D 3
x1 ! x2 D 1
can be written in the matrix form AX D B, where
! " ! " ! "
1 2 x 3
AD XD 1 BD
1 !1 x2 1
otivation for what we now have in mind is provided by loo ing at the procedure
for solving the algebraic equation ax D b. The latter equation is solved by simply
multiplying both sides by the multiplicative inverse of a, if it exists. (Recall that the
multiplicative inverse of a nonzero number a is denoted a"1 (which is 1 a) and has the
property that a"1 a D 1.) For example, if 3x D 11, then
11
3"1 .3x/ D 3"1 .11/ so x D
3
If we are to apply a similar procedure to the matrix equation
AX D B
then we need a multiplicative inverse of A that is, a matrix C such that CA D I. If we
have such a C, then we can simply multiply both sides of Equation (1) by C and get
C.AX/ D CB
.CA/X D CB
IX D CB
X D CB
This shows us that i there is a solution of AX D B, then the only possible solution
is the matrix CB. Since we now that a matrix equation can have no solutions, a unique
solution, or infinitely many solutions, we see immediately that this strategy cannot
possibly wor unless the matrix equation has a unique solution. For CB to actually be a
solution, we require that A.CB/ D B, which is the same as requiring that .AC/B D B.
However, since matrix multiplication is not commutative, our assumption that CA D I
does not immediately give us AC D I. Consider the matrix products below, for example:
h i! " ! "h i ! "
1 1 10
10 D Œ1! D I1 but 10 D ¤ I2
0 0 00
However, if A and C are square matrices of the same order n, then it can be proved
that AC D In follows from CA D In , so in this case we can finish the argument above
and conclude that CB is a solution, necessarily the only one, of AX D B. For a square
matrix A, when a matrix C exists satisfying CA D I, necessarily C is also square of the
same size as A and we say that it is an inverse matrix (or simply an in erse) of A.
280 C atr e ra

If A is a square matrix and there exists a matrix C such that CA D I, then C is called
an in erse of A, and A is said to be invertible.

E AM LE I M
! " ! "
1 2 7 !2
et A D and C D . Since
3 7 !3 1
! "! " ! "
7 !2 1 2 1 0
CA D D DI
!3 1 3 7 0 1
matrix C is an inverse of A.
A L IT I G
Secret messages can be encoded by
using a code and an encoding matrix. It can be shown that an invertible matrix has one and only one inverse that is, an
Suppose we have the following code: inverse is unique. Thus, in Example 1, matrix C is the n y matrix such that CA D I.
a b c d e f g h i j k m For this reason, we can spea of the inverse of an invertible matrix A, which we denote
1 2 3 4 5 6 7 10 11 12 13
by the symbol A!1 . Accordingly, A!1 A D I. oreover, although matrix multiplication
n p q r
14 15 16 17 1 1
s t u x y z
20 21 22 23 24 25 26 is not generally commutative, it is a fact that A!1 c mmutes ith A:

et the encoding matrix be E. Then we A!1 A D I D AA!1

can encode a message by ta ing every
two letters of the message, converting Returning to the matrix equation AX D B, Equation (1), we can now state the
them to their corresponding numbers, following:
creating a 2 " 1 matrix, and then mul-
tiplying each matrix by E. The mes- If A is an invertible matrix, then the matrix equation AX D B has the unique solution
sage can be unscrambled with a decod- X D A!1 B.
ing matrix that is the inverse of the
coding matrix---that is, E!1 . etermine The idea of an inverse matrix reminds us of inverse functions, studied in Sec-
whether the encoding matrices tion 2.4. Inverse functions can be used to further understand inverse matrices. et Rn
" # " #
1 3 !2 1:5 denote the set of n " 1 column matrices (and let Rm denote the set of m " 1 column
and matrices). If A is an m " n matrix, then f.X/ D AX defines a function f W Rn # Rm .
2 4 1 !0:5
If m D n, it can be shown that the function given by f.X/ D AX has an inverse, in
are inverses of each other. the sense of Section 2.4, if and only if A has an inverse matrix, A!1 , in which case
f!1 .X/ D A!1 X.
There is a caution to be observed here. In general, for a function f to have an inverse,
For general functi ns, if g ı f D I, it does say g, then we require b th g ı f D I and f ı g D I, where I is the identity function. It
not follow that f ı g D I. is a special fact about square matrices that CA D I implies also AC D I.
A L IT I
If f is a function that has an inverse, then any equation of the form f.x/ D b has a
Suppose the encoding matrix unique solution, namely x D f!1 .b/.
" #
1 3
ED
2 4
E AM LE U I S S
was used to encode a message. Use the
code from Apply It 12 and the inverse Solve the system
" #
!2 1:5
#
E!1 D x1 C 2x2 D 5
1 !0:5
3x1 C 7x2 D 1
to decode the message, bro en into the
following pieces: S In matrix form, we have AX D B, where
! " ! " ! "
2 ; 46; 65; 0 1 2 x 5
AD XD 1 BD
3 7 x2 1
61; 2
5 ; ; 57; 6
In Example 1, we showed that
! "
7 !2
60; 4; 21; 34; 76; 102 A!1 D
!3 1
 Section 6.6 n erses 281

Therefore,
! "! " ! "
7 !2 5 !1
X D A!1 B D D
!3 1 1 3
so x1 D !1 and x2 D 3.
Now ork Problem 19 G
In order that the method of Example 2 be applicable to a system, two conditions
must be met:
The system must have the same number of equations as there are un nowns.
The coe cient matrix must be invertible.

As far as condition 2 is concerned, we caution that not all n nzer square matrices
are invertible. For example, if
! "
0 1
AD
0 1
then
! "! " ! " ! "
a b 0 1 0 aCb 1 0
D ¤
c d 0 1 0 cCd 0 1

for any values of a, b, c, and d. Hence, there is no matrix that, when postmultiplied by
A, yields the identity matrix. Thus, A is not invertible.
There is an interesting mechanical procedure that allows us, simultaneously, to
determine whether or not a matrix is invertible and find its inverse if it is so. The pro-
cedure is based on an observation whose proof would ta e us too far afield. First, recall
that for any matrix A there is a sequence E1 ; E2 ; : : : ; Ek of elementary row operations
that, when applied to A, produce a reduced matrix. In other words, we have
E1 E2 Ek
A # A1 # A2 # $ $ $ # Ak
where Ak is a reduced matrix. We recall, too, that Ak is unique and determined by A
Every identity matrix is a reduced matrix, alone (even though there can be many sequences, of variable lengths, of elementary
but not every (square) reduced matrix is row operations that accomplish this reduction). If A is square, say n " n, then we might
an identity matrix. For example, any zero
matrix 0 is reduced. have Ak D In , the n " n identity matrix.

T
For square A and Ak as previously, A is invertible if and only if Ak D I. oreover,
if E1 ; E2 ; : : : ; Ek is a sequence of elementary row operations that ta es A to I, then
the same sequence ta es I to A!1 .

E AM LE I M

Apply the theorem to determine if the matrix

! "
1 0
AD
2 2
is invertible.
S We will augment A with a copy of the (2 " 2) identity matrix ( ust as
we have often augmented a matrix by a column vector). The result will be 2 " 4.
We will apply elementary row operations to the entire 2 " 4 matrix until the first n
columns form a reduced matrix. If the result is I, then, by the theorem, A is invertible,
but because we have applied the operations to the entire 2 " 4 matrix, the last n
columns will, also by the theorem, be transformed from I to A!1 , if A is in fact
invertible.
282 C atr e ra

S We have
! " ! "
1 0 1 0 !2R1 C R2 1 0 1 0
ŒA j I! D !!!!!!!!
2 2 0 1 0 2 !2 1
1
" #
R
2 2 1 0 1 0
!!!!!!!! D ŒI j B!
0 1 !1 12

Since ŒA j I! transforms with I to the left of the augmentation bar, the matrix A is
invertible and the matrix B to the right of the augmentation bar is A!1 . Specifically, we
conclude that " #
1 0
A!1 D 1
!1 2
Now ork Problem 1 G
This procedure is indeed a general one.
For the interested reader, we remar that
the matrix B in the method described is in M F I M
any event invertible and we always have
BA D R. If A is an n " n matrix f rm the n " .2n/ matrix ŒA j I! and perf rm e ementary r
perati ns unti the rst n c umns f rm a reduced matrix Assume that the resu t
is ŒR j B! s that e ha e
ŒA j I! ! $ $ $ ! ŒR j B!
If R D I then A is in ertib e and A!1 D B If R ¤ I then A is n t in ertib e
meaning that A!1 d es n t exist and the matrix B is f n particu ar interest t ur
c ncerns here

E AM LE F I M

etermine A!1 if A is invertible.

A L IT I 2 3
1 0 !2
We could extend the encoding
scheme used in Apply It 12 to a 3 " 3
a A D 44 !2 15
matrix, encoding three letters of a mes- 1 2 !10
sage at a time. Find the inverses of the S Following the foregoing procedure, we have
following 3 " 3 encoding matrices:
2 3 2 3 1 0 -2 1 0 0
3 1 2 2 1 2 [A 0 I] = 4 1 0 1 0
6 7 6 7 -2
E D 42 2 2 5 F D 43 2 3 5 1 2 -10 0 0 1
2 1 3 4 3 4

-4R1 + R2 1 0 -2 1 0 0
0 -2 9 -4 1 0
-1R1 + R3 0 2 -8 -1 0 1

1 0 -2 1 0 0
- 12 R2
0 1 - 92 2 - 12 0
0 2 -8 -1 0 1

1 0 -2 1 0 0
-2R2 + R3
0 1 - 92 2 - 12 0
0 0 1 -5 1 1

1 0 0 -9 2 2
2R3 + R1
9
0 1 0 - 41
2
4 9
2
R +
2 3
R2
0 0 1 -5 1 1
 Section 6.6 n erses 283

The first three columns of the last matrix form I. Thus, A is invertible and
2 3
! 2 2
6 7
A!1 D 4! 41 2
4 25
!5 1 1
! "
3 2
b AD
6 4
S We have
! " ! "
3 2 1 0 !2R1 C R2 3 2 1 0
ŒA j I! D !!!!!!!!
6 4 0 1 0 0 !2 1

1 " #
R
3 1 1 2 1
0
!!!!!!!
! 3 3
0 0 !2 1
The first two columns of the last matrix form a reduced matrix different from I. Thus,
A is not invertible.
Now ork Problem 7 G
Now we will solve a system by using the inverse.

A L IT I E AM LE U I S S
A group of investors has 500,000 Solve the system
to invest in the stoc s of three compa- 8
nies. Company A sells for 50 a share < x1 ! 2x3 D 1
and has an expected growth of 13 per 4x1 ! 2x2 C x3 D 2
year. Company sells for 20 per share : x C 2x ! 10x D !1
1 2 3
and has an expected growth of 15 per
year. Company C sells for 0 a share by finding the inverse of the coe cient matrix.
and has an expected growth of 10
S In matrix form the system is AX D B, where
per year. The group plans to buy twice 2 3
as many shares of Company A as of 1 0 !2
Company C. If the group s goal is 12 A D 44 !2 15
growth per year, how many shares of 1 2 !10
each stoc should the investors buy
is the coe cient matrix. From Example 4(a),
2 3
! 2 2
!1 6 41 7
A D 4! 2 4 25
!5 1 1
The solution is given by X D A!1 B:
2 3 2 323 2 3
x1 ! 2 2 1 !7
4x2 5 D 6
4! 41
74 5 4
2 D !175
25
2
4
x3 !5 1 1 !1 !4

so x1 D !7; x2 D !17, and x3 D !4.

Now ork Problem 27 G
It can be shown that a system of n linear equations in n un nowns has a unique solu-
tion if and only if the coe cient matrix is invertible. Indeed, in the previous example
the coe cient matrix is invertible, and a unique solution does in fact exist. When the
coe cient matrix is not invertible, the system will have either no solution or infinitely
many solutions.
284 C atr e ra

While the solution of a system using a matrix inverse is very elegant, we must
provide a caution. iven AX D B, the computational wor required to find A!1 is
greater than that required to reduce the augmented matrix of the system, namely ŒA j
B!. If there are several equations to solve, all with the same matrix of coe cients but
variable right-hand sides, say AX D B1 , AX D B2 , : : : , AX D Bk , then for suitably
The method of reduction in Sections 6.4
and 6.5 is a faster computation than that large k it might be faster to compute A!1 than to do k reductions, but a numerical analyst
of finding a matrix inverse. will in most cases still advocate in favour of the reductions. For even with A!1 in hand,
one still has to compute A!1 B and, if the order of A is large, this too ta es considerable
time.

E AM LE AC M T I N I

Solve the system

8
< x ! 2y C z D 0
2x ! y C 5z D 0
: x C y C 4z D 0

S The coe cient matrix is

2 3
1 !2 1
42 !1 55
1 1 4
Since
2 3
2 3 1 0 3 ! 13 0 2
1 !2 1 1 0 0 6 7 3
42 !1 5 0 1 05 ! $ $ $ ! 6
40 1 1 ! 23 07
5
1
3
1 1 4 0 0 1 0 0 0 1 !1 1
the coe cient matrix is not invertible. Hence, the system cann t be solved by inverses.
Instead, another method must be used. In Example 4(a) of Section 6.5, the solution was
found to be x D !3r; y D !r, and z D r, where r is any real number (thus, providing
infinitely many solutions).
Now ork Problem 31 G

R BLEMS
2 3 2 3
In Pr b ems if the gi en matrix is in ertib e nd its in erse 7 0 !2 2 !3 1
" # " # 4 0
6 1 2 4 1 05 42 0 15
7 1 3 6 !3 0 1 4 !6 1
2 3 2 3
" # " # 1 2 0 !1 2 !3
2 2 1 a 4 0 1 35 4 2 1 05
a in .!1; 1/
2 2 0 1 !1 0 1 4 !2 5
2 3 2 3 2 3 2 3
1 0 0 2 0 1 2 3 2 !1 3
42 !4 05 4!1 4 05 41 3 55 40 2 05
0 1 2 2 1 0 1 5 12 2 1 1
2 3 2 3 Solve AX D B if
1 2 3 2 0 0
40 0 45 40 0 05 " # " #
2 3 3
0 0 5 0 0 !4 A!1 D and B D
1 5 7
2 3 2 3
1 0 0 0 0
40 05 40 1 0 5 Solve AX D B if
0 1 0 0 0 2 3 3 2
2 0 1 !2
2 3 2 3
1 2 3 1 2 !1 A!1 D 40 3 05 and B D 4 05
40 1 25 40 1 45 1 0 4 4
0 0 1 1 !1 2
 Section 6.6 n erses 285

F r Pr b ems if the c e cient matrix f the system is b Suppose each model A requires 10 widgets and 14 shims and
in ertib e s e the system by using the in erse If n t s e the each model requires 7 widgets and 10 shims. The factory can
system by the meth d f reducti n obtain 00 widgets and 1130 shims each hour. How many cars of
( ( each model can it produce while using all the parts available
2 3
6x C 5y D 2 2x C 4y D 5 a 0 0
x C y D !3 !x C 3y D !2 If A D 40 b 05, where a, b, c ¤ 0, show that
( ( 0 0 c
3x C y D 5 6x C y D 2 2 3
1=a 0 0
3x ! y D 7 7x C y D 7
A!1 D 4 0 1=b 0 5
( ( 0 0 1=c
x C 3y D 7 2x C 6y D
a If A and B are invertible matrices with the same order,
3x ! y D 1 3x C y D 7
show that .AB/!1 D B!1 A!1 . int Consider .B!1 A!1 /.AB/.
8 8 b If
< x C 2y C z D 4 <x C y C z D 6 " # " #
3x C zD2 x ! y C z D !1 !1 1 3 !1 1 1
: x! yCzD1 :x ! y ! z D 4 A D
2 4
and B D
1 5
8 8
<x C y C z D 3 <x C y C z D 6 find .AB/!1 .
xCy!zD4 xCy!zD0 If A is invertible, it can be shown that .AT /!1 D .A!1 /T .
:x ! y ! z D 5 :x ! y C z D 2
Verify this identity for
8 8 " #
< x C 3y C 3z D 7 < x C 3y C 3z D 7 4 1
AD
2x C y C z D 4 2x C y C z D 4 2 !3
: xC yC zD4 : xC yC zD3
8̂ A matrix P is said to be rth g na if P!1 D PT . Is the
" #
ˆ C 2y C z D 4 1 3 !4
< matrix P D orthogonal
! x C 2z D 12
5 4 3
ˆ2 C x C z D 12
:̂ C 2x C y C z D 12 Secret Message A friend has sent a friend a secret message
8̂ that consists of three row matrices of numbers as follows:
ˆ x ! 3y ! z D !1
< 7 70!
C yC D0 R1 D Œ33 R2 D Œ57 133 20!
ˆ! C 2x ! 2y ! z D 6
:̂ yCzD4
R3 D Œ3 0 33!
oth friends have committed the following matrix to memory (the
F r Pr b ems and nd .I ! A/!1
f r the gi en matrix A first friend used it to code the message):
" # " # 2 3
!3 !1 !3 2 1 2 !1
AD
!2 4
AD
4 3 AD4 2 5 25
!1 !2 2
Auto Production Solve the following problems by using ecipher the message by proceeding as follows:
the inverse of the matrix involved. a Calculate the three matrix products R1 A!1 , R2 A!1 , and R3 A!1 .
a An automobile factory produces two models, A and . odel b Assume that the letters of the alphabet correspond to the
A requires 1 labor hour to paint and 12 labor hour to polish numbers 1 through 26, replace the numbers in the preceding three
model requires 1 labor hour for each process. uring each hour matrices by letters, and determine the message.
that the assembly line is operating, there are 100 labor hours Investing A group of investors decides to invest 500,000
available for painting and 0 labor hours for polishing. How many in the stoc s of three companies. Company sells for 60 a share
of each model can be produced each hour if all the labor hours and has an expected growth of 16 per year. Company E sells for
available are to be utilized 0 per share and has an expected growth of 12 per year.
Company F sells for 30 a share and has an expected growth of
per year. The group plans to buy four times as many shares of
company F as of company E. If the group s goal is 13.6 growth
per year, how many shares of each stoc should the investors buy
Investing The investors in Problem 43 decide to try a new
investment strategy with the same companies. They wish to buy
twice as many shares of company F as of company E, and they
have a goal of 14.52 growth per year. How many shares of each
stoc should they buy
286 C atr e ra

Objective L I A
o se t e et o s of t s c a ter to Input output matrices, which were developed by Wassily W. eontief, codify the
ana e t e ro ct on of sectors of an
econo supply and demand interrelationships that exist among the various sectors of an econ-
omy during some time period. The phrase input utput is used because the matrices
show the values of outputs of each industry that are sold as inputs to each industry and
for final use by consumers. eontief won the 1 73 Nobel Prize in economic science
for the development of the input output method and its applications to economic
problems.

T B E
Suppose that a simple economy has three interrelated sectors, which we will label 1,
2, and 3. These might, for example, be agriculture, energy, and manufacturing. For
any sector j, production of one unit of output of j will typically require inputs from all
sectors of the economy, including j itself. If we write Aij for the number of units of input
from sector i required to produce one unit of output from sector j, then the numbers Aij
determine a 3 " 3 matrix A. For example, suppose that
2 3
2 1 3
6 51 2 10 7
AD6
45
1
10
1
10
7
5
1 1 1
5 5 10

It is important to avoid confusing A with Reading down the first column of A, we see that to produce one unit of output from
its transpose, a common mista e in this sector 1 requires 25 of a unit of input from sector 1, 15 of a unit of input from sector 2,
context. and 15 of a unit of input from sector 3. Similarly, the requirements for sectors 2 and 3
can be read from columns 2 and 3, respectively. There may well be externa demands
on the economy, which is to say, for each sector, a demand for a certain number of units
of output that will not be used as inputs for any of the sectors 1, 2, and 3. Such external
demands might be in the form of exports or consumer needs. From the point of view of
this model, the only attribute of external demands that concerns us is that they do not
overlap with the demands described by matrix A.
Suppose further that there is an external demand for 0 units output from sector 1,
160 units output from sector 2, and 240 units output from sector 3. We will write
2 3
0
D 4 160 5
240
for this external demand so that, as shown, the entry in row i is the external demand for
sector i. A ey question that arises now is that of determining the levels of production
for each of sectors 1, 2, and 3 so that the external demand can be satisfied. We must
bear in mind that production must satisfy not only the external demand but also the
requirements imposed by the data that ma e up matrix A. For each sector, some of its
output must be directed as input for the three sectors of the economy (including itself)
and some of it must be directed toward the corresponding component of . This leads
us to the important conceptual equation:
production D internal demand C external demand
et Xi , for i D 1; 2; 3, denote the production required of sector i to satisfy
Equation (1). Then production can be regarded as the matrix
2 3
X1
X D 4 X2 5
X3
and Equation (1) becomes
X D internal demand C
 Section 6.7 eont ef s n t t t na s s 287

To understand interna demand we should begin by realizing that it will have three
components, say C1 , C2 , and C3 , where Ci is the amount of production from sector
i consumed by production of X. Writing C for the 3 " 1 matrix whose ith row is Ci ,
Equation (2) now becomes

XDCC

bserve that C1 will have to account for the output of sector 1 used in producing
X1 units of sector 1 output, plus X2 units of sector 2 output, plus X3 units of sector
3 output. It ta es A11 units of 1 to produce one unit of 1, so production of X1 units
of 1 requires A11 X1 units of 1. It ta es A12 units of 1 to produce one unit of 2, so
production of X2 units of 2 requires A12 X2 units of 1. It ta es A13 units of 1 to produce
one unit of 3, so production of X3 units of 3 requires A13 X3 units of 1. It follows that we
must have

C1 D A11 X1 C A12 X2 C A13 X3

a ing similar arguments for C2 and C3 , we deduce

C2 D A21 X1 C A22 X2 C A23 X3

C3 D A31 X1 C A32 X2 C A33 X3
and the last three equalities are easily seen to combine to give

C D AX

Substituting C D AX in Equation (3), we obtain the following equation and its

equivalents:

X D AX C
X ! AX D
IX ! AX D
.I ! A/X D

The last equation displayed is of the form X D , so to solve for production X we

need only reduce the augmented matrix ŒI ! A j !.
Although it is not entirely standard, we will refer to the matrix A in our discussion
as the Leontief matrix. The matrix I ! A is the c e cient matrix of the system whose
solution provides the production X needed to satisfy the external demand .

E AM LE I A

For the eontief matrix A and the external demand of this section, complete the
numerical determination of the production needed to satisfy .
S We have only to write the augmented matrix of the equation .I ! A/X D ,
which is evidently
2 3
3 1 3
! ! 0
6 51 2 10 7
6! 1 7
4 5 10 ! 10 160 5
! 15 ! 15 10 240

and reduce it. Using the techniques of Section 6.4, we have

2 3 2 3 2 3
6 !5 !3 00 2 2 ! !2400 2 2 ! !2400
# 4 !2 !1 1600 5 # 4 0 11 !10 ! 00 5 # 4 0 11 !10 ! 0 5
!2 !2 2400 0 !11 24 000 0 0 14 7200
288 C atr e ra
2 3
71 : 4
from which we deduce X % 4 3 4: 1 5. We remar that, although we have rounded
514:2
our answer, it does follow from the last displayed augmented matrix that the system
has a unique solution.
Now ork Problem 1 G

A A I
In Example 1 the solution is unique, and this is typical of examples that involve a
realistic eontief matrix A. In other words, it is typical that the coe cient matrix I ! A
is invertible. Thus, the typically unique solution of

.I ! A/X D

is typically obtainable as

X D .I ! A/!1

We cautioned in Section 6.6 that finding the inverse of a coe cient matrix is usually
not a computationally e cient way of solving a system of linear equations. We also said
that if there are several systems to be solved, all with the same coe cient matrix, then
the calculation of its inverse might in this case be useful. Such a possibility is presented
by eontief s model.
For a given subdivision of an economy into n sectors, it is reasonable to expect
that the n " n eontief matrix A will remain constant for a reasonable interval of time.
It follows that the coe cient matrix I ! A will also remain constant during the same
interval. uring this time interval, planners may want to explore a variety of demands
1 , 2 , : : :, k and, for any one of these, , determine the production X required to
satisfy . With .I ! A/!1 in hand, the planner has only to ca cu ate

X D .I ! A/!1

(rather than to s e .I ! A/X D by reducing ŒI ! A j !).

F L M
The eontief matrix is often determined from data of the ind that we now present.
A hypothetical example for an oversimplified two-sector economy will be given. As
before, we note that the two sectors can be thought of as being from agriculture, energy,
manufacturing, steel, coal, or the li e. The ther pr ducti n fact rs row consists of
costs to the respective sectors, such as labor, profits, and so on. The externa demand
entry here could be consumption by exports and consumers. The matrix we first con-
sider is somewhat larger than the eontief matrix:

Consumers (input)
Sector Sector External
1 2 emand
Pr ducers utput : 2 3 ta s
Sect r 6 240 500 460 7 1200
Sect r 6 360 200 40 7 1500
4 5
ther Pr ducti n Fact rs 600 00
ta s 1200 1500

Each sector appears in a row and in a column. The row of a sector shows the pur-
chases of the sector s output by all the sectors and by the externa demand. The entries
 Section 6.7 eont ef s n t t t na s s 289

represent the value of the products and might be in units of millions of dollars of prod-
uct. For example, of the total output of sector 1, 240 went as input to sector 1 itself
(for internal use), 500 went to sector 2, and 460 went directly to the external demand.
The total output of sector 1 is the sum of the sector demands and the final demand:
.240 C 500 C 460 D 1200/.
Each sector column gives the values of the sector s purchases for input from each
sector (including itself) as well as what it spent for other costs. For example, in order
to produce its 1200 units, sector 1 purchased 240 units of output from itself, 360 units
of output from 2, and 600 units of other costs such as labor.
Note that for each sector the sum of the entries in its row is equal to the sum of the
entries in its column. For example, the value of the total output of sector 1 (1200) is
equal to the value of the total input to sector 1.
An important assumption of input output analysis is that the basic structure of the
economy remains the same over reasonable intervals of time. This basic structure is
found in the relative amounts of inputs that are used to produce a unit of output. These
are found from particular tables of the ind above, as we now describe. In produc-
ing 1200 units of product, sector 1 purchases 240 units from sector 1, 360 units from
sector 2, and spends 600 units on other costs. Thus, f r each unit f utput, sector 1
240
spends 1200 D 15 on sector 1, 1200
360
D 103 600
on sector 2, and 1200 D 12 on other costs.
Combining these fixed ratios of sector 1 with those of sector 2, we can give the input
requirements per unit of output for each sector:

2 1 2 3 2 1 2 3
240 500 1 1
1 6 1200 1500 7 65 3 7 1
6 360 7 63 2 7
6 200 7 D 6 7
2 6 1200 1500 7 6 10 15 7 2
6 7 6 7
4 600 00
5 4 1
5
ther 1200 1500 2 15 ther

The sum of each column is 1, and because of this we do not need the last row. Each
entry in the bottom row can be obtained by summing the entries above it and subtracting
the result from 1. If we delete the last row, then the ijth entry of the resulting matrix
is the number of units of sector i s product needed to produce ne unit of sector j s
product. It follows that this matrix,
" #
1 1
5 3
AD 3 2
10 15

is the eontief matrix for the economy.

Now, suppose the external demand for sector 1 changes from 460 to 500 and the
external demand for sector 2 changes from 40 to 1200. We would li e to now how
production will have to change to meet these new external demands. ut we have
already seen how to determine the production levels necessary to meet a given external
demand when we now the eontief matrix A. Now, with
! "
500
D
1200

we have only to solve .I ! A/X D , which in the present case will be effected by
reducing
2 3
4
! 13 500
AD 4 5 5
3 13
! 10 15
1200
290 C atr e ra

or, since A is only 2 " 2, calculating X D .I ! A/!1 on a graphing calculator. With a

TI- 3 Plus,
! "
!1 1404:4
X D .I ! A/ D
1 70:7

is easily obtained. Notice too that we can also update the ther Production Factors
row of the data with which we started. From the row we discarded of the relativized
data, we now that 12 of sector 1 s output and 15 of sector 2 s output must be directed
to other production factors, so the unrelativized data will now be

! "
1
.1404:4 /; .1 70:7 / % Œ702:25; 7:75!
2 15

The issue of computational e ciency can be a serious one. While we treat this
topic of input output analysis with examples of economies divided into two or three
sectors, a model with 20 sectors might be more realistic in which case the eontief
matrix would have 400 entries.

E AM LE I A

iven the input output matrix

Sect r Externa
21 2 3 emand 3
Sect r 1 240 1 0 144 36
6 7
2 6120 36 4 156 7
6 7
3 6120 72 4 240 7
4 5
ther 120 72 240

suppose external demand changes to 77 for 1, 154 for 2, and 231 for 3. Find the pro-
duction necessary to satisfy the new external demand. (The entries are in millions of
dollars.)

S y examining the data, we see that to produce 600 units of 1 required

240 units of 1, 120 units of 2, and 120 units of 3. It follows that to produce ne unit
of 1 required 240
600
D 25 units of 1, 120
600
D 15 units of 2, and 120
600
D 15 units of 3. The
numbers 25 , 15 , and 1
5
constitute, in that order, the first c umn of the eontief matrix.

S We separately add the entries in the first three rows. The total values of
output for sectors 1, 2, and 3 are 600, 360, and 4 0, respectively. To get the eontief
matrix A, we divide the sector entries in each sector column by the total value of output
for that sector:
2 3 2 3
240 1 0 144 2 1 3
6 600 360 4 07 6 5 2 10 7
6 120 36 4 7 6 1 1 1 7
AD 6 600 4 07 D6 7
4 360 5 4 5 10 10 5
120 72 4 1 1 1
600 360 4 0 5 5 10

The external demand matrix is

2 3
77
D 41545
231
 Section 6.7 eont ef s n t t t na s s 291

The result of evaluating .I ! A/!1 on a TI- 3 Plus is

2 3
6 2:5
43 0 5
4 5

Now ork Problem 7 G

R BLEMS
A very simple economy consists of two sectors: agriculture find the output matrix if final demand changes to a 200 for
and forestry. To produce one unit of agricultural products requires education and 300 for government b 64 for education and 64
1 1
4
of a unit of agricultural products and 12 of a unit of forestry for government.
products. To produce one unit of forestry products requires 23 of a
iven the input output matrix
unit of agricultural products and no units of forestry products.
etermine the production levels needed to satisfy an external
Industry
demand for 400 units of agriculture and 600 units of forestry Fina
products. 2Grain Ferti izer Catt e emand 3
An economy consists of three sectors: coal, steel, and Industry Grain
6 15 30 45 10 7
railroads. To produce one unit of coal requires 10 1
of a unit of coal, Ferti izer 6 25 30 60 5 7
6 7
1
of a unit of steel, and 1
of a unit of railroad services. To Catt e 6 50 40 60 30 7
10 10 4 5
produce one unit of steel requires 13 of a unit of coal, 10 1
of a unit ther 10 20 15
1
of steel, and 10 of a unit of railroad services. To produce one unit
of railroad services requires 14 of a unit of coal, 13 of a unit of steel,
1
and 10 of a unit of railroad services. etermine the production a find the eontief matrix. b If external demand changes to 15
levels needed to satisfy an external demand for 300 units of coal, for grain, 10 for fertilizer, and 35 for cattle, write the augmented
200 units of steel, and 500 units of railroad services. matrix whose reduction will give the necessary production levels
Suppose that a simple economy consists of three sectors: needed to meet these new demands.
agriculture (A), manufacturing ( ), and transportation (T).
Economists have determined that to produce one unit of A iven the input output matrix
requires 11 units of A, 1 units of , and 1 units of C, while
Industry
3
production of one unit of requires 16 units of A, 14 units of ,
and 163
units of T, and production of one unit of T requires 15 1
units E ectric Fina
1 1 2 ater P er Agricu ture emand 3
of A, 3 units of , and 6 units of T. There is an external demand
for 40 units of A, 30 units of , and no units of T. etermine Industry ater 100 400 240 260
6 7
the production levels necessary to meet the external demand. E ectric P er 6 100 0 4 0 140 7
6 7
Agricu ture 6 300 160 240 500 7
iven the input output matrix 4 5
ther 500 160 240

Industry
Fina
2 Stee C a emand 3
find the output matrix if final demand changes to 500 for water,
Industry Stee 200 500 500 7 150 for electric power, and 700 for agriculture. Round the entries
6
C a 6 400 200 00 7 to two decimal places.
4 5
ther 600 00
iven the input output matrix

find the output matrix if final demand changes to 600 for steel and Industry Fina
05 for coal. Find the total value of the other production costs that 2G ernment Agricu ture anufacturing emand 3
this involves. Industry G ernment 400 200 200 200
6 7
Agricu ture 6
6 200 400 100 300 77
iven the input output matrix anufacturing 6
6 200 100 300 400 77
4 5
ther 200 300 400
Industry
Fina
2Educati n G ernment emand 3 with entries in billions of dollars, find the output matrix for the
Industry Educati n 40 120 40 economy if the final demand changes to 300 for government, 350
G ernment 6 4 120 0 0 7
5 for agriculture, and 450 for manufacturing. Round the entries to
ther 40 0 the nearest billion dollars.
292 C atr e ra

iven the input output matrix in Problem , find the output iven the input output matrix in Problem , find the output
matrix for the economy if the final demand changes to 250 for matrix for the economy if the final demand changes to 300 for
government, 300 for agriculture, and 350 for manufacturing. government, 400 for agriculture, and 500 for manufacturing.
Round the entries to the nearest billion dollars. Round the entries to the nearest billion dollars.

Chapter 6 Review
I T S E
S Matrices
matrix size entry, Aij row vector column vector Ex. 1, p. 242
equality of matrices transpose of matrix, AT zero matrix, 0 Ex. 3, p. 244
S Matrix Addition and Scalar Multiplication
addition and subtraction of matrices scalar multiplication Ex. 4, p. 24
S Matrix Multiplication
matrix multiplication identity matrix, I power of a matrix Ex. 12, p. 261
matrix equation, AX D B Ex. 13, p. 262
S Solving Systems by Reducing Matrices
coe cient matrix augmented coe cient matrix Ex. 3, p. 26
elementary row operation equivalent matrices reduced matrix Ex. 4, p. 270
parameter Ex. 5, p. 271
S Solving Systems by Reducing Matrices Continued
homogeneous system nonhomogeneous system trivial solution Ex. 4, p. 277
S Inverses
inverse matrix invertible matrix Ex. 1, p. 2 0
S Leontief’s Input Output Analysis
input output matrix eontief matrix Ex. 1, p. 2 7

S
A matrix is a rectangular array of numbers enclosed within reduced matrix is obtained. The reduced matrix ma es any
brac ets. There are a number of special types of matrices, solutions to the system obvious and allows the detection of
such as zero matrices, identity matrices, square matrices, and nonexistence of solutions. If there are infinitely many solu-
diagonal matrices. esides the operation of scalar multiplica- tions, the general solution involves at least one parameter.
tion, there are the operations of matrix addition and subtrac- ccasionally, it is useful to find the inverse of a (square)
tion, which apply to matrices of the same size. The product matrix. The inverse (if it exists) of a square matrix A is found
AB is defined when the number of columns of A is equal to by augmenting A with I and applying elementary row opera-
the number of rows of B. Although matrix addition is com- tions to ŒA j I! until A is reduced resulting in ŒR j B! (with R
mutative, matrix multiplication is not. y using matrix mul- reduced). If R D I, then A is invertible and A!1 D B. If R ¤ I,
tiplication, we can express a system of linear equations as the then A is not invertible, meaning that A!1 does not exist. If
matrix equation AX D B. the inverse of an n " n matrix A exists, then the unique solu-
A system of linear equations may have a unique solution, tion to AX D B is given by X D A!1 B. If A is not invertible,
no solution, or infinitely many solutions. The main method the system has either no solution or infinitely many solutions.
of solving a system of linear equations using matrices is by ur final application of matrices dealt with the interrela-
applying the three elementary row operations to the aug- tionships that exist among the various sectors of an economy
mented coe cient matrix of the system until an equivalent and is nown as eontief s input output analysis.

R
2 3" #
In Pr b ems simp ify 1 7
" # " # 42 5 1 0 !2
3 4 1 0 !3
0 6 1
2 !3 1 0
!5 1 2 4
2 3
" # " # 2 3
2
!2 !3
!4
1 0 Œ2 3 7! 40 !15
6 0 2 5 2
 Chapter 6 e e 293
" # 0" # " #1 2 3
0 1 1
2 3 @ 2 3 1 A
! et A D 40 0 15. Find the matrices A2 , A3 , A1000 , and
!1 3 7 6 4 4
0 0 0
0" # " #1 A!1 (if the inverse exists).
2 0 0 !5 A " #
!@ C2 2 0
7 6 !4 AD . Show that .AT /!1 D .A!1 /T .
0 4
" #2
A consumer wishes to supplement his vitamin inta e by
0 3
2 Œ4 2!T exact y 13 units of vitamin A, 22 units of vitamin , and 31 units
1 1
of vitamin C per wee . There are three brands of vitamin capsules
" # 0" #T 12 available. rand I contains 1 unit each of vitamins A, , and C per
1 3 0 @ 1 0 A capsule brand II contains 1 unit of vitamin A, 2 of , and 3 of C
3 3 6 1 3 and brand III contains 4 units of A, 7 of , and 10 of C.
a What combinations of capsules of brands I, II, and III will
In Pr b ems c mpute the required matrix if produce exact y the desired amounts
#" " #
1 1 1 0 b If brand I capsules cost 5 cents each, brand II 7 cents each, and
AD BD brand III 20 cents each, what combination will minimize the
!1 2 0 2
consumer s wee ly cost
.2A/T ! 3I2 A.2I/ ! A0T B3 C I5 Suppose that A is an invertible n " n matrix.
T T T a Prove that Ak is invertible, for any integer k, where we employ
.AB/ ! B A
the convention that A0 D I.
In Pr b ems and s e f r x and y b Prove that if B and C are n " n matrices such that
" # " # " #" # " #
5 15 1 x 2 1 3 4 ABA!1 D ACA!1 , then B D C.
Œx! D D c If A2 D A, find A.
7 y 2 y x 3 3 y " # " #
10 !3 6
In Pr b ems reduce the gi en matrices If A D and B D , find 3AB ! 4B2 .
" # " # 4 7 !7 !3
1 4 0 0 7
5 0 5 Solve the system
2 3 2 3 8
2 1 4 0 0 0 1 <7: x ! 4:3y C 2:7z D 11:1
41 0 15 40 0 0 05 3:4x C 5: y ! 7:6z D 10:
: 4:5x ! 6:2y ! 7:4z D 15:
4 1 6 1 0 0 0

In Pr b ems s e each f the systems by the meth d f by using the inverse of the coe cient matrix. Round the answers
reducti n to two decimal places.
( ( iven the input output matrix
2x ! 5y D 0 x ! y C 2z D 3
4x C 3y D 0 3x C y C z D 5
8 8 Industry Fina
< x C y C 2z D 1 <x C 2y C 3z D 1
2A B emand 3
3x ! 2y ! 4z D !7 x C 4y C 6z D 2
: 2x ! y ! 2z D 2 :x C 6y C z D 3 Industry A 10
6
20 4
7
B6415 14 10 7
5
In Pr b ems nd the in erses f the matrices
" # " # ther 5
1 5 0 1
3 1 0
find the output matrix if final demand changes to 10 for A and 5
2 3 2 3
1 3 !2 5 0 0 for B. ( ata are in tens of billions of dollars.)
44 1 05 4!5 2 15
3 !2 2 !5 1 3
In Pr b ems and s e the gi en system by using the in erse
f the c e cient matrix
8 8
<x C y D 3 < 5x D3
yCzD4 !5x C 2y C z D 0
:x C z D 5 :!5x C y C 3z D 2
 near Pro ra n

inear programming sounds li e something involving the writing of computer

7.1 near ne a t es n o
code. ut while linear programming is often done on computers, the
ar a es
programming part of the name actually comes from World War II era
7.2 near Pro ra n military terminology, in which training, supply, and unit-deployment plans
were called programs. Each program was a solution to a problem in resource allocation.
7.3 e e et o For example, suppose that military units in a combat theater need diesel fuel. Each
7.4 rt c a ar a es unit has a certain number of tan s, truc s, and other vehicles each unit uses its vehi-
cles to accomplish an assigned mission and each unit s mission has some relation to
7.5 n at on the overall goal of winning the campaign. What fuel distribution program will best
contribute to overall victory
7.6 e a
Solving this problem requires quantifying its various elements. Counting gallons
C er 7 e e of fuel and numbers of each type of vehicle is easy, as is translating gallons of fuel into
miles a vehicle can travel. uantifying the relation between vehicle miles and unit mis-
sion accomplishment includes identifying constraints: the maximum gallons per load
a tan er truc can carry, the minimum number of miles each unit must travel to reach
its combat ob ective, and so on. Additional quantitative factors include probabilities,
such as a unit s chances of winning a ey engagement if it maneuvers along one route
of travel rather than another.
uantifying complicated real-world problems in this way is the province of a sub-
ect called operations research. inear programming, one of the oldest and still one
of the most important tools of operations research, is used when a problem can be
described using equations and inequalities that are all linear. In practice, many phe-
nomena that are not linear can be su ciently well approximated by linear functions,
over a restricted domain, to allow use of the techniques of this chapter.

294
 Section 7. near ne a t es n o ar a es 295

Objective L I T
o re resent eo etr ca t e so t on Suppose a consumer receives a fixed income of 60 per wee and uses it a to purchase
of a near ne a t n t o ar a es
an to e ten t s re resentat on to products A and . If A costs 2 per ilogram and costs 3 per ilogram, then if our
a s ste of near ne a t es consumer purchases x ilograms of A and y ilograms of , his cost will be 2x C 3y.
Since he uses all of his 60, x and y must satisfy
2x C 3y D 60; where x; y ! 0
y
The solutions of this equation, called a budget equati n give the possible ordered pairs
2x + 3y = 60 (x, y Ú 0) of amounts of A and that can be purchased for 60. The graph of the equation is the
20 budget ine in Figure 7.1. Note that .15; 10/ lies on the line. This means that if 15 g
(x, y)
of A are purchased, then 10 g of must be bought, for a total cost of 60.
(15, 10)
n the other hand, suppose the consumer does not necessarily wish to spend all of
the 60. In this case, the possible ordered pairs are described by the inequality
x
30

FIGURE udget line. 2x C 3y " 60; where x; y ! 0

When inequalities in one variable were discussed in Chapter 1, their solutions

were represented geometrically by inter a s on the real-number line. However, for an
inequality in two variables, li e Inequality (1), the solution is usually represented by
a regi n in the coordinate plane. We will find the region corresponding to (1) after
considering such inequalities in general.

A linear inequality in the variables x and y is an inequality that can be written in

one of the forms
ax C by C c < 0 ax C by C c " 0 ax C by C c > 0 ax C by C c ! 0

y where a, b, and c are constants, and not both a and b are zero.
y = mx + b

eometrically, the solution (or graph) of a linear inequality in x and y consists of

y 7 mx + b all points (x, y) in the plane whose coordinates satisfy the inequality. For example, a
solution of x C 3y < 20 is the point .#2; 4/, because substitution gives
x
#2 C 3.4/ < 20;
y 6 mx + b 10 < 20; which is true.
Clearly, there are infinitely many solutions, which is typical of every linear inequality.
FIGURE A nonvertical line To consider linear inequalities in general, we first note that the graph of a nonver-
determines two half-planes. tical line y D mx C b separates the plane into three distinct parts (see Figure 7.2):

the line itself, consisting of all points .x; y/ whose coordinates satisfy the equation
y D mx C b
y the region ab e the line, consisting of all points .x; y/ whose coordinates satisfy
x=a the inequality y > mx C b (this region is called an open half plane)
the open half-plane be the line, consisting of all points .x; y/ whose coordinates
x6a x7a satisfy the inequality y < mx C b.

In the situation where the strict inequality < is replaced by " , the solution of y "
0 a x mx C b consists of the line y D mx C b, as well as the half-plane below it. In this case,
we say that the solution is a closed half plane. A similar statement can be made when
> is replaced by !. For a vertical line x D a (see Figure 7.3), we spea of a half-
plane to the right .x > a/ of the line or to its left .x < a/. Since any linear inequality
FIGURE A vertical line (in two variables) can be put into one of the forms we have discussed, we can say that
determines two half-planes. the s uti n f a inear inequa ity must be a ha f p ane.
296 C near Pro ra n

y y

y
5 (x0, y0)
2x + y = 5 5 - 2x0
y 6 5 - 2x
y0 y … 5 - 2x
5
2
x x
x0
x

FIGURE raph of 2x C y < 5. FIGURE Analysis of point satisfying y < 5 # 2x. FIGURE raph of y " 5 # 2x.

To apply these facts, we will solve the linear inequality

2x C y < 5

From our previous discussion, we now that the solution is a half-plane. To find it, we
eometrically, the solution of a linear begin by replacing the inequality symbol by an equality sign and then graphing the
inequality in one variable is an interval resulting ine 2x C y D 5. This is easily done by choosing two points on the line for
on the line, but the solution of a linear instance, the intercepts . 52 ; 0/ and .0; 5/. (See Figure 7.4.) ecause points on the line do
inequality in t variables is a regi n in not satisfy the < inequality, we used a dashed line to indicate that the line is not part
the plane.
of the solution. We must now determine whether the solution is the half-plane ab e
the line or the one be it. This can be done by solving the inequality for y. nce y is
A L IT I
isolated, the appropriate half-plane will be apparent. We have
To earn some extra money, you
ma e and sell two types of refrigera- y < 5 # 2x
tor magnets, type A and type . ou
have an initial start-up expense of 50.
The production cost for type A is 0. 0 From the aforementioned statement 3, we conclude that the solution consists of
per magnet, and the production cost for the half-plane be the line. Part of the region that does n t satisfy this inequality is
type is 0.70 per magnet. The price shaded in Figure 7.4. It will be our custom, generally, when graphing inequalities to
for type A is 2.00 per magnet, and the shade the part of the whole plane that does n t satisfy the condition. Thus, if .x0 ; y0 / is
price for type is 1.50 per magnet. et any point in the unshaded region, then its ordinate y0 is less than the number 5 # 2x0 .
x be the number of type A and y be the (See Figure 7.5.) For example, .#2; #1/ is in the region, and
number of type produced and sold.
Write an inequality describing revenue
greater than cost. Solve the inequality #1 < 5 # 2.#2/
and describe the region. Also, describe #1 <
what this result means in terms of mag-
nets.
If, instead, the original inequality had been y " 5 # 2x, then the line y D 5 # 2x
would have been included in the solution. We would indicate its inclusion by using a
y solid line rather than a dashed line. This solution, which is a closed half-plane, is shown
in Figure 7.6. eep in mind that a solid line is included in the solution and a dashed
line is not

5 E AM LE S L I
y…5
Find the region defined by the inequality y " 5.
x
S Since x does not appear, the inequality is assumed to be true for all values
of x. The region consists of the line y D 5, together with the half-plane below it. (See
Figure 7.7, where the solution is the unshaded part together with the line itself.)
FIGURE raph of
y " 5. Now ork Problem 7 G
 Section 7. near ne a t es n o ar a es 297

E AM LE S L I

Solve the inequality 2.2x # y/ < 2.x C y/ # 4.

S We first solve the inequality for y, so that the appropriate half-plane is obvi-
ous. The inequality is equivalent to
4x # 2y < 2x C 2y # 4
y 4x # 4y < 2x # 4
x #4y < #2x # 4
y7 +1
2 x dividing both sides by #4 and reversing the sense of the
y> C1
2 inequality
1
x
-2 Using a dashed line, we now s etch y D .x=2/ C 1 by noting that its intercepts are
(0, 1) and .#2; 0/. ecause the inequality symbol is >, we shade the half-plane below
FIGURE raph of the line. Thin of the shading as stri ing out the points that you do not want. (See
y > 2x C 1. Figure 7. .) Each point in the unshaded region is a solution.
Now ork Problem 1 G
S I
The solution of a system of inequalities consists of all points whose coordinates simul-
taneously satisfy all of the given inequalities. eometrically, it is the region that is
common to all the regions determined by the given inequalities. For example, let us
solve the system
8̂
<2x C y > 3
x!y
:̂ 2y # 1 > 0

We first rewrite each inequality so that y is isolated. This gives the equivalent system
8̂
<y > #2x C 3
y"x
:̂y > 1
2

Next, we s etch the corresponding lines y D #2x C 3, y D x, and y D 12 , using dashed

The point where the graphs of y D x and
y D #2x C 3 intersect is not included in lines for the first and third, and a solid line for the second. We then shade the region
the solution. e sure to understand why. that is below the first line, the region that is above the second line, and the region that
is below the third line. The region that is unshaded (Figure 7. ) together with any solid
line boundaries are the points in the solution of the system of inequalities.

y=x

y 7 -2x + 3
y…x
1
y7 2
1
y= 2

x
y = -2x + 3

FIGURE Solution of a system of linear

inequalities.
298 C near Pro ra n

A L IT I
E AM LE S S L I
A store sells two types of cam- Solve the system
eras. In order to cover overhead, it must !
y ! #2x C 10
sell at least 50 cameras per wee , and
in order to satisfy distribution require- y!x#2
ments, it must sell at least twice as many
S The solution consists of all points that are simultaneously on or above the
of type I as type II. Write a system
of inequalities to describe the situation.
line y D #2x C 10 and on or above the line y D x # 2. It is the unshaded region in
et x be the number of type I that the Figure 7.10.
store sells in a wee and y be the number
of type II that it sells in a wee . Find the y
region described by the system of linear
inequalities.

y Ú -2x + 10
yÚx-2
y = x -2

y = -2x + 10

FIGURE Solution of a
system of linear inequalities.

Now ork Problem 9 G

E AM LE S S L I
y
2x + 3y … 60 Find the region described by
xÚ0 8
20 yÚ0 <2x C 3y " 60
x!0
: y!0
2x + 3y = 60
S This system relates to Inequality (1) at the beginning of the section. The first
x
30 inequality is equivalent to y " # 23 x C 20. The last two inequalities restrict the solution
FIGURE Solution of a to points that are both on or to the right of the y-axis and also on or above the x-axis.
system of linear inequalities. The desired region is unshaded in Figure 7.11 (and includes the bounding lines).
Now ork Problem 17 G
R BLEMS
! !
In Pr b ems s e the inequa ities 2x # 2 ! y 2y < 4x C 2
3x C 4y > 2 3x # 2y ! 12 2x " 3 # 2y y < 2x C 1
8 8̂
6x # 3y " 12 y > 6 # 2x x C 2y "
<x # y > 4 <
#x " 2y # 4 3x C 5y ! 12 2x C y "
x<2
: y > #5 :̂x ! 2
3x C y < 0 x C 4y < 4 y!2
! ! 8 8
3x # 2y < 6 2x C 3y > #6 <y < 2x C 4 <2x C y ! 6
x # 3y > 3x # y < 6 x ! #2 x"y
! ! :y < 1 : y " 5x C 2
2x C 3y " 6 2y # 3x < 6
8 8
x!0 x<0 < xCy>1 <2x # 3y > #12
! ! 3x # 5 " y 3x C y > #6
3x C y " 3 x#y<1 : :
xCy"2 y#x<1 y < 2x y>x
 Section 7.2 near Pro ra n 299
8̂ 8
5x C y " 10 Poc et-Size at 24 cm. et x be the number of Petit models and y
< <5y # 2x " 10 the number of Poc et-Size models produced at the unenburg
xCy" 5
4x # 6y " 12
:̂ x! 0 : y!0
factory per wee . The factory can produce at most 50 Petit and
y! 0 Poc et-Size models combined in a wee . oreover, each Petit has
If a c nsumer ants t spend n m re than P d ars t purchase 2 cameras and each Poc et-Size has 4 cameras but iant obile
quantities x and y f t pr ducts ha ing prices f p1 and p2 can only source 1 00 cameras per wee . Write inequalities to
d ars per unit respecti e y then p1 x C p2 y " P here describe this situation.
x y ! 0 In Pr b ems and nd ge metrica y the p ssib e Manufacturing A chair company produces two models of
c mbinati ns f purchases by determining the s uti n chairs. The Sequoia model ta es 3 wor er-hours to assemble and
1
f this system f r the gi en a ues f p1 ; p2 and P 2
wor er-hour to paint. The Saratoga model ta es 2 wor er-hours
p1 D 6, p2 D 4, P D 20 p1 D 7, p2 D 3, P D 25 to assemble and 1 wor er-hour to paint. The maximum number of
wor er-hours available to assemble chairs is 240 per day, and the
If a manufacturer wishes to purchase a t ta of no more than maximum number of wor er-hours available to paint chairs is
100 lb of product from suppliers A and , set up a system of 0 per day. Write a system of linear inequalities to describe the
inequalities that describes the possible combinations of quantities situation. et x represent the number of Sequoia models produced
that can be purchased from each supplier. S etch the solution in in a day and y represent the number of Saratoga models
the plane. produced in a day. Find the region described by this system of
Manufacturing The iant obile Phone Company linear inequalities.
produces two models of cell phones: the Petit at 1 cm and the

Objective L
o state t e nat re of near Sometimes we want to maximize or minimize a function, sub ect to certain restrictions
ro ra n ro e s to ntro ce
ter no o assoc ate t t e an on the natura domain of the function. We recall from Chapter 2 that the domain of a
to so e t e eo etr ca function f W X $ , without specific instructions to the contrary, is the set of all x in X
for which the rule f is defined. ut we also saw in Chapter 2 that we frequently want to
restrict the values of x further than is mathematically required, so as to capture aspects
of a practical problem. For example, prices are required to be n nnegati e numbers and
quantities should often be n nnegati e integers. The problems in this chapter all deal
with further restrictions on the domain that are called constraints, and in this chapter
they will be prescribed by what are called inear inequa ities as studied in the previous
section. For example, a manufacturer may want to maximize a profit function, sub ect
to production restrictions imposed by limitations on the use of machinery and labor,
with the latter given by linear inequalities.
We will now consider how to solve such problems when the function to be maxi-
This is a more restrictive use of the word
inear than we have employed thus far
mized or minimized is inear. A linear function in x and y has the form
Notice that we do not have a (nonzero) P D P.x; y/ D ax C by
constant term in P.
where a and b are constants. In the first instance, it should be noted that a linear function
in x and y is a particular ind of function of two variables as introduced in Section 2. ,
and the natural domain for such a function is the set .#1; 1/%.#1; 1/ of all ordered
pairs .x; y/ with both x and y in .#1; 1/. However, because of the ind of applications
we have in mind, the domain is nearly always restricted to Œ0; 1/ % Œ0; 1/, which is
to say that we restrict to x ! 0 and y ! 0. We will soon give examples of the further
A great deal of terminology is used in
discussing linear programming. It is a
linear constraints that appear in what will be called linear programming problems.
good idea to master this terminology as In a linear programming problem, the function to be maximized or minimized is
soon as it is introduced. called the ob ective function. Its domain is de ned t be the set of all solutions to
the system of linear constraints that are given in the problem. The set of all solutions
to the system of linear constraints is called the set of feasible points. Typically, there
are infinitely many feasible points (points in the domain of the ob ective function),
but the aim of the problem is to find a point that optimi es the value of the ob ective
function. To optimize is either to maximi e or to minimi e depending on the nature of
the problem.
We now give an example of a linear programming problem and a geometric
approach to solve such problems, when the ob ective function is a linear function of t
variables, as defined above. However, as soon as we have seen a few examples of such
problems, it becomes clear that in practice we need to be able to solve similar problems
300 C near Pro ra n

in many variables. ur geometric approach is not really practical for even three vari-
ables. In Section 7.3, a matrix approach will be discussed that essentially encodes the
geometric approach numerically, in a way that generalizes to many variables. It is to
be noted that the main matrix tools we will use are the e ementary r perati ns and
parametrized s uti ns.
We consider the following problem. A company produces two types of can openers:
manual and electric. Each requires in its manufacture the use of three machines: A, ,
and C. Table 7.1 gives data relating to the manufacture of these can openers. Each
manual can opener requires the use of machine A for 2 hours, machine for 1 hour,
and machine C for 1 hour. An electric can opener requires 1 hour on A, 2 hours on
, and 1 hour on C. Furthermore, suppose the maximum numbers of hours available
per month for the use of machines A, , and C are 1 0, 160, and 100, respectively.
The profit on a manual can opener is 4, and on an electric can opener it is 6. If the
company can sell all the can openers it can produce, how many of each type should it
ma e in order to maximize the monthly profit

Table .1
anual Electric Hours Available

A 2 hr 1 hr 1 0
1 hr 2 hr 160
C 1 hr 1 hr 100
Profit Unit 4 6

To solve the problem, let x and y denote the number of manual and electric can
openers, respectively, that are made in a month. Since the number of can openers made
is not negative,
x!0 and y!0
For machine A, the time needed for wor ing on x manual can openers is 2x hours, and
the time needed for wor ing on y electric can openers is 1y hours. The sum of these
times cannot be greater than 1 0, so
2x C y " 1 0
Similarly, the restrictions for machines and C give
x C 2y " 160 and x C y " 100
The profit is a function of x and y and is given by the pr t functi n
P D 4x C 6y
Summarizing, we want to maximize the bjecti e functi n
P D 4x C 6y
sub ect to the condition that x and y must be a solution of the system of constraints:
8̂
ˆ 2x C y " 1 0
ˆ
ˆ
ˆ
< x C 2y " 160
x C y " 100
ˆ
ˆ
ˆ
ˆ x!0
:̂
y!0
Thus, we have a linear programming problem. Constraints (5) and (6) are called
nonnegativity conditions. The region simultaneously satisfying constraints (2) (6) is
unshaded in Figure 7.12. Each point in this region represents a feasible point, and the
set of all these points is called the feasible region. We remar that, with obvious termi-
nology, the feasible region contains corner points where the constraint lines intersect.
We have labeled these as A, B, C, , and E. We repeat that the feasib e regi n is ust
 Section 7.2 near Pro ra n 301

y (Electric)

160
2x + y = 180

120

x + y = 100
80
E
A

40
Feasible x + 2y = 160
region
B
D C x (Manual)
0 40 80 120 160

FIGURE Feasible region.

2
P = 600 ( y = - 3 x + 100)
100
Maximum
80 profit line
E
Isoprofit
A lines
40

P = 300 B
D C
x
0 40 80 120 160

FIGURE Isoprofit lines and the feasible region.

another word for the d main of the ob ective function, in the context of linear program-
ming problems. Although there are infinitely many feasible points, we must find one
at which the ob ective function assumes a maximum value.
Since the ob ective function equation, P D 4x C 6y, is equivalent to
2 P
yD# xC
3 6
it defines a family of parallel lines, one for each possible value of P, each having a slope
of #2=3 and y-intercept (0, P=6). For example, if P D 600, then we obtain the line
2
y D # x C 100
3
shown in Figure 7.13. This line, called an isoprofit line, is an example of a e e cur e
as introduced in Section 2. . It gives all possible pairs .x; y/ that yield the profit, 600.
Note that this isoprofit line has no point in common with the feasible region, whereas
the isoprofit line for P D 300 has infinitely many such points. et us loo for the
member of the family of parallel lines that contains a feasible point and whose P-value
is maximum. his i be the ine h se y intercept is farthest fr m the rigin gi ing
a maximum a ue f P and that has at east ne p int in c mm n ith the feasib e
302 C near Pro ra n

regi n. It is not di cult to observe that such a line will contain the c rner p int A. Any
isoprofit line with a greater profit will contain no points of the feasible region.
From Figure 7.12, we see that A lies on both the line x C y D 100 and the line
x C 2y D 160. Thus, its coordinates may be found by solving the system
!
x C y D 100
x C 2y D 160
This gives x D 40 and y D 60. Substituting these values into the equation P D 4x C 6y,
we find that the maximum profit sub ect to the constraints is 520, which is obtained
by producing 40 manual can openers and 60 electric can openers per month.
We have finished our problem It is well to note that there are two important parts
to the conclusion: (1) the optimum value, in this case the maximum profit of 520,
and (2) the feasible point at which the optimum value is attained, in this case .40; 60/,
where the first coordinate is manual and the second is electric .
While Figures 7.12 and 7.13 are easily at hand, let us suppose that the profit func-
tion P D P.x; y/ had been such that the isoprofit lines were para e to the line segment
oining corner points A and B in Figures 7.12 and 7.13. (The interested reader may wish
to determine values a and b in P D ax C by for which the isoprofit lines do have this
property.) In that case, the isoprofit line that contains the line segment AB would be the
member of the family of parallel lines that contains a feasible point and whose P-value
is maximum. In that case, each point on AB would provide an optimal solution. We will
have a little more to say about this possibility of multiple optimal solutions below.
It is of course possible that the constraints of a linear programming problem are
such that in the process of imposing them we end up shading the entire x; y-plane.
In this case the set of feasible points, the feasible region, is the empty set, ;, and
any linear programming problem with an empty feasible region has no solution. If
the feasible region contains at least one point, then we say that the feasible region
is nonempty. There is another extreme situation to consider. In the problem we ust
completed about manufacturing can openers, the points in the feasible region could be
enclosed by a circle for example, the one centered at the origin with radius 100 will
do. When the feasible region can be enclosed by some circle we say that the feasible
region is bounded. Sometimes the constraints of a linear programming problem give
an unbounded feasible region. This means that no circle can be drawn to enclose all
of the feasible region. In other words, an unbounded feasible region is one that contains
points arbitrarily far from the origin. A linear programming problem with an unbounded
feasible region may fail to have an optimum solution.
However, with these possibilities in mind, it can be shown that

A linear function defined on a nonempty, bounded feasible region has both a max-
imum value and a minimum value. oreover, each of these values can be found at
a corner point.

We hasten to add that in practical linear programming problems that arise in busi-
ness applications, one is interested in either finding a maximum value of, say, a profit
function r a minimum value of, say, a cost function. In these situations the unwanted
extreme value is often trivial. The reader should see, by inspection of Figure 7.13, that
the profit function P of the can opener problem attains a minimum value, of 0, at the
corner point , with coordinates .0; 0/.
This statement gives us a way of finding an optimum solution without drawing iso-
profit lines, as we did previously: We simply evaluate the ob ective function at each of
the corner points of the feasible region and then choose the corner points at which the
function attains the desired optimum value. (We remar that if there are two ad acent
corner points at which the ob ective function attains an optimal value then that optimal
value is attained at a p ints n the ine segment j ining the c rner p ints in questi n.
ut note that this situation is still detected by our solution strategy.) f course, evaluat-
ing the ob ective function at the corner points requires that we first find the coordinates
of the corner points.
 Section 7.2 near Pro ra n 303

For example, in Figure 7.13 with corner points A, B, C, , and E, we found A before
to be .40; 60/. To find B, we see from Figure 7.12 that we must solve 2x C y D 1 0
and x C y D 100 simultaneously. This gives the point B D . 0; 20/. In a similar way,
we obtain all the corner points:

A D .40; 60/ B D . 0; 20/ C D . 0; 0/

D .0; 0/ E D .0; 0/

We now evaluate the ob ective function P D 4x C 6y at each point:

P.A/ D P.40; 60/ D 4.40/ C 6.60/ D 520

P.B/ D P. 0; 20/ D 4. 0/ C 6.20/ D 440
P.C/ D P. 0; 0/ D 4. 0/ C 6.0/ D 360
P. / D P.0; 0/ D 4.0/ C 6.0/ D 0
P.E/ D P.0; 0/ D 4.0/ C 6. 0/ D 4 0

Thus, P has a maximum value of 520 at A, where x D 40 and y D 60 (and a minimum

value of 0 at , where x D 0 and y D 0).
The optimum solution to a linear programming problem is given by the optimum
value of the ob ective function and the point where the optimum value of the ob ective
function occurs.

E AM LE S L
y
aximize the ob ective function P D 3x C y sub ect to the constraints

8 2x C y "
2x + y = 8 2x C 3y " 12
A = (0, 0)
B = (4, 0) x!0
C = (3, 2)
4 D = (0, 4) y!0
D
2x + 3y = 12
C S In Figure 7.14, the feasible region is nonempty and bounded. Thus, P attains
A B
x a maximum at one of the four corner points. The coordinates of A, B, and are obvious
4 6 on inspection. To find the coordinates of C, we solve the equations 2x C y D and
2x C 3y D 12 simultaneously, which gives x D 3; y D 2. Thus,
FIGURE A, B, C, and are
corner points of the feasible region.
A D .0; 0/ B D .4; 0/ C D .3; 2/ D .0; 4/

Evaluating P at these points, we obtain

P.A/ D P.0; 0/ D 3.0/ C 0 D 0

P.B/ D P.4; 0/ D 3.4/ C 0 D 12
P.C/ D P.3; 2/ D 3.3/ C 2 D 11
P. / D P.0; 4/ D 3.0/ C 4 D 4

Hence, the maximum value of P, sub ect to the constraints, is 12, and it occurs when
x D 4 and y D 0.
Now ork Problem 1 G
We remar that, for a linear programming problem with an unbounded feasible
region, if the pr b em has an ptima s uti n then that solution does occur at a corner
point. However, we caution that, for an unbounded feasible region, there may n t be
an ptima s uti n.
304 C near Pro ra n

R BLEMS
aximize inimize
P D 5x C 7y C D 5x C y

sub ect to sub ect to

2x # y ! #2
2x C 3y " 45
4x C 3y " 12
x # 3y ! 2
x # y D #1
x; y ! 0
x; y ! 0
aximize
P D 3x C 2y aximize
sub ect to D 0:4x # 0:2y
x C y " 70
sub ect to
x C 3y " 240
2x # 5y ! #3
x C 3y " 0
2x # y " 5
x; y ! 0
3x C y D 6
aximize
x; y ! 0
D 4x # 6y
inimize
sub ect to CDxCy
y"7 sub ect to
3x # y " 3
x C 2y ! 4
xCy ! 5
2x C y ! 4
x; y ! 0
x; y ! 0
inimize
inimize
C D 2x C y C D 2x C 2y
sub ect to sub ect to
x C 2y ! 0
x#y ! 0
3x C 2y ! 160
x!4
5x C 2y ! 200
x " 10
x; y ! 0
aximize
aximize
D 4x # 10y
D 10x C 2y
sub ect to sub ect to
x # 4y ! 4
2x # y " 2 x C 2y ! 4

x; y ! 0 x # 2y ! 0
x; y ! 0
inimize
D 20x C 30y inimize
D #2x C y
sub ect to
sub ect to
2x C y " 10 x!2
3x C 4y " 24 3x C 5y ! 15
x C 7y ! 56 x # y ! #3
x; y ! 0 x; y ! 0
 Section 7.2 near Pro ra n 305

Production for Maximum Profit A toy manufacturer Mineral Extraction A company extracts minerals from
preparing a production schedule for two new toys, truc s and ore. The numbers of pounds of minerals A and that can be
spinning tops, must use the information concerning its extracted from each ton of ores I and II are given in the following
construction times given in the following table: table, together with the costs per ton of the ores:
re I re II
achine A achine Finishing
ineral A 0 lb 160 lb
Truc 2 hr 3 hr 5 hr ineral 140 lb 40 lb
Spinning Top 1 hr 1 hr 1 hr Cost per ton 60 0

If the company must produce at least 4000 lb of A and 2000 lb

For example, each truc requires 2 hours on machine A. of , how many tons of each ore should be processed in order to
The available employee hours per wee are as follows: for minimize cost What is the minimum cost
operating machine A, 0 hours for , 50 hours for finishing,
Production Scheduling An oil company that has two
70 hours. If the profits on each truc and spinning top are 7 and
refineries needs at least 000, 14,000, and 5000 barrels of low-,
2, respectively, how many of each toy should be made per wee
medium-, and high-grade oil, respectively. Each day, Refinery I
in order to maximize profit What is the maximum profit
produces 2000 barrels of low-, 3000 barrels of medium-, and
Production for Maximum Profit A manufacturer 1000 barrels of high-grade oil, whereas Refinery II produces 1000
produces two types of external hard drives: ymemory and barrels each of low- and high- and 2000 barrels of medium-grade
ystorage. uring production, the devices require the services of oil. If it costs 25,000 per day to operate Refinery I and 20,000
both the assembly and pac aging departments. The numbers of per day to operate Refinery II, how many days should each
hours needed in each department are provided in the following refinery be operated to satisfy the production requirements
table: at minimum cost What is the minimum cost (Assume that a
minimum cost exists.)
ymemory ystorage Construction Cost A chemical company is designing a
plant for producing two types of polymers, P1 and P2 . The plant
Assembly 3 hr 7 hr must be capable of producing at least 520 units of P1 and 330
Pac aging 2 hr 3 hr units of P2 each day. There are two possible designs for the basic
reaction chambers that are to be included in the plant. Each
chamber of type A costs 300,000 and is capable of producing 40
The assembly department runs for 21 hours per day, and the units of P1 and 20 units of P2 per day type is a more expensive
pac aging department runs for 12 hours per day. If the company design, costing 400,000, and is capable of producing 40 units of
ma es a profit of on each ymemory unit and 14 on each P1 and 30 units of P2 per day. ecause of operating costs, it is
ystorage unit, how many of each should it ma e each day to necessary to have at least five chambers of each type in the plant.
maximize its profit and what is the resulting profit How many chambers of each type should be included to minimize
the cost of construction and still meet the required production
Diet Formulation A diet is to contain at least 16 units of
schedule
carbohydrates and 20 units of protein. Food A contains 2 units
of carbohydrates and 4 of protein food contains 2 units of Pollution Control ecause of new federal regulations on
carbohydrates and 1 of protein. If food A costs 1.20 per unit and pollution, a chemical company has introduced into its plant a new,
food costs 0. 0 per unit, how many units of each food should more expensive process to supplement or replace an older process
be purchased in order to minimize cost What is the minimum in the production of a particular chemical. The older process
cost discharges 25 grams of carbon dioxide and 50 grams of
particulate matter into the atmosphere for each liter of chemical
Fertili er utrients A produce grower is purchasing
produced. The new process discharges 15 grams of carbon dioxide
fertilizer containing three nutrients: A, , and C. The minimum
and 40 grams of particulate matter into the atmosphere for each
wee ly requirements are 0 units of A, 120 of , and 240 of C.
liter produced. The company ma es a profit of 40 cents per liter
There are two popular blends of fertilizer on the mar et. lend I,
and 15 cents per liter on the old and new processes, respectively.
costing a bag, contains 2 units of A, 6 of , and 4 of C.
If the government allows the plant to discharge no more than
lend II, costing 10 a bag, contains 2 units of A, 2 of , and
12,525 grams of carbon dioxide and no more than 20,000 grams
12 of C. How many bags of each blend should the grower
of particulate matter into the atmosphere each day, how many
buy each wee to minimize the cost of meeting the nutrient
liters of chemical should be produced daily, by each process, to
requirements
maximize daily profit What is the maximum daily profit
306 C near Pro ra n

Construction Discount The highway department has aximize

decided to add exactly 300 m of highway and exactly 200 m
of expressway to its road system this year. The standard price for D 14x # 3y
road construction is 2 million per ilometer of highway and
sub ect to
million per ilometer of expressway. nly two contractors,
company A and company , can do this ind of construction, so y ! 12:5 # 4x
the entire 500 m of road must be built by these two companies.
However, company A can construct at most 400 m of roadway y " :3 # x
(highway and expressway), and company can construct y ! 4:7 C 0: x
at most 300 m. For political reasons, each company must
be awarded a contract with a standard price of at least 300 million x; y ! 0
(before discounts). Company A offers a discount of 2000 per
ilometer of highway and 6000 per ilometer of expressway inimize
company offers a discount of 3000 for each ilometer of
P D 3:4y # 2:7x
highway and 5000 for each ilometer of expressway.
a et x and y represent the number of ilometers of highway sub ect to
and expressway, respectively, awarded to company A. Show that
the total discount received from both companies is given by :6x C 2:5y ! 33
D 1 00 # x C y 2:3x C y " 6:
where is in thousands of dollars. 2:5x # y ! 0:3
b The highway department wishes to maximize the total
discount . Show that this problem is equivalent to the following x; y ! 0
linear programming problem, by showing exactly how the first six
constraints arise:
aximize D 1 00 # x C y inimize
sub ect to D 17:3x # 14:4y
x C y " 400
sub ect to
x C y ! 500
2x C y ! 300 0:73x # y " #2:4
2x C y " 1 00 1:22x # y ! #5:1
x " 300 0:45x # y ! #12:4
y " 200
x; y ! 0
x; y ! 0
inimize
c Find the values of x and y that maximize .
In Pr b ems r und ans ers t t decima p aces C D 6x C 14y
aximize sub ect to
D 4x C y
14x C 7y ! 43
sub ect to
6x C 2y " 12 3x C 7y ! 21

2x C 3y ! 6 #x C y ! #5
x; y ! 0 x; y ! 0

Objective T S M
os o o t es e et o s Until now, we have solved linear programming problems by a geometric method. This
se to so e a stan ar a
near ro ra n ro e s method is not practical when the number of variables increases to three, and is not
et o ena es t e so t on of possible for four or more variables. Now we will loo at a different technique the
ro e s t at cannot e so e simplex method whose name is lin ed in more advanced discussions to a geometrical
eo etr ca
ob ect called a simplex.
The simplex method begins with a feasible corner point and tests whether the value
of the ob ective function at this point is optimal. If it is not, the method leads to another
corner point, which is at least as good and usually better. If this new point does not
produce an optimal value, we repeat the procedure until the simplex method does lead
to an optimal value, if one exists.
 Section 7.3 e e et o 307

esides being e cient, the simplex method has other advantages. For one, it is
completely mechanical. It uses matrices, elementary row operations, and basic arith-
metic. oreover, no graphs need to be drawn this allows us to solve linear program-
ming problems having any number of constraints and any number of variables.
In this section, we consider only so-called standard maximum linear program
ming problems. In other sections we will consider less restrictive problems. A standard
maximum problem is one that can be put in the following form:

S M L
aximize the linear function D c1 x1 C c2 x2 C & & & C cn xn sub ect to the constraints
9
a11 x1 C a12 x2 C & & & C a1n xn " b1 >
>
a21 x1 C a22 x2 C & & & C a2n xn " b2 > >
>
=
& & & &
& & & & >>
& & & & >>
>
;
am1 x1 C am2 x2 C & & & C amn xn " bm
where x1 ; x2 ; : : : ; xn and b1 ; b2 ; : : : ; bm are nonnegative.

It is helpful to formulate the problem in matrix notation so as to ma e its structure

more memorable. et
2 3
x1
6 x2 7
" # 6 7
6 & 7
C D c1 c2 & & & cn and X D 6 7
6 & 7
4 & 5
xn
Then the ob ective function can be written as
D .X/ ust recalls that is a functi n D .X/ D CX
of X.

Now if we write
2 3 2
3
a11 a12 & & & a1n b1
6 a21 a22 & & & a2n 7 6 b2 7
6 7 6 7
6 & & & 7 6 & 7
AD6
& 7
and BD6 7
6 & & 7 6 & 7
4 & & & 5 4 & 5
am1 am2 & & & amn bm

then we can say that a standard maximum linear programming problem is one that can
be put in the form

aximize D CX
!
AX " B
Note that B ! 0 is a condition on the data sub ect to
of the problem and is not a constraint
X !0
imposed on the variable X. where B!0

( atrix inequalities are to be understood li e matrix equality. The comparisons

refer to matrices of the same size and the inequality is required to hold for all corre-
sponding entries.)
ther types of linear programming problems will be discussed in Sections 7.4
and 7.5.
308 C near Pro ra n

Note that one feasible point for a standard maximum inear pr gramming pr b em
is always x1 D 0; x2 D 0; : : : ; xn D 0 and that at this feasible point the value of the
ob ective function is 0. Said otherwise, (the n % 1 matrix) 0 is feasible for a standard
problem and .0/ D 0 (where the last 0 is the number (1 % 1 matrix) 0).
The procedure that we follow here will We now apply the simplex method to the problem in Example 1 of Section 7.2,
be outlined later in this section. which can be written
maximize D 3x1 C x2
sub ect to the constraints
2x1 C x2 "
and
2x1 C 3x2 " 12
and
x1 ! 0; x2 ! 0
This problem is of standard form. We begin by expressing Constraints (2) and (3) as
equations. In (2), 2x1 Cx2 will equa if we add some nonnegative number s1 to 2x1 Cx2 ,
so that
2x1 C x2 C s1 D for some s1 ! 0
We call s1 a slac variable, since it ma es up for the slac on the left side of (2)
We mentioned slac variables in
Section 1.2 in the process of introducing to give us equality. Similarly, Inequality (3) can be written as an equation by using the
inequalities. slac variable s2 we have
2x1 C 3x2 C s2 D 12 for some s2 ! 0
The variables x1 and x2 are called decision variables.
Now we can restate the problem in terms of equations:

aximize D 3x1 C x2
sub ect to
2x1 C x2 C s1 D
and
2x1 C 3x2 C s2 D 12
where x1 ; x2 ; s1 , and s2 are nonnegative.

From Section 7.2, we now that the optimal value occurs at a corner point of the
feasible region in Figure 7.15. At each of these points, at least t of the variables
x1 ; x2 ; s1 , and s2 are 0, as the following listing indicates:

x2

2x1 + x2 = 8

(0, 4)
D

(3, 2) 2x1 + 3x2 = 12

C

A B
x1
(0, 0) (4, 0)

FIGURE ptimal value occurs at a corner point of the feasible region.

 Section 7.3 e e et o 309

At A, we have x1 D 0 and x2 D 0 (and s1 D and s2 D 12).

At B, x1 D 4 and x2 D 0. ut from Equation (5), 2.4/ C 0 C s1 D . Thus, s1 D 0.
(And from Equation (6) and x1 D 4 and x2 D 0 we get s2 D 4.)
At C, x1 D 3 and x2 D 2. ut from Equation (5), 2.3/ C 2 C s1 D . Hence, s1 D 0.
From Equation (6), 2.3/ C 3.2/ C s2 D 12. Therefore, s2 D 0.
At , x1 D 0 and x2 D 4. From Equation (6), 2.0/ C 3.4/ C s2 D 12. Thus, s2 D 0.
(And from Equation (5) and x1 D 0 and x2 D 4, we get s1 D 4.)
It can also be shown that any solution to Equations (5) and (6), such that at least
We remar ed earlier that there is a great t of the four variables x1 ; x2 ; s1 , and s2 are zero, corresponds to a corner point. Any
deal of terminology used in the
discussion of linear programing. In
such solution where at least two of these variables are zero is called a basic feasible
particular, there are many types of solution abbreviated FS. This number, 2, is determined by the number n of decision
ariab es. It is important to understand variables 2 in the present example. For any particular FS, the variables held at 0 are
that the variables called decision called nonbasic variables, and all the others are called basic variables for that FS.
variables x1 ; x2 ; : : : ; xn remain decision Since there is a total of n C m variables, the number of basic variables in the general
variables throughout the solution of a
problem, and the same remar applies to
system that arises from (1) is m, the number of constraints (other than those expressing
the slac variables s1 ; s2 ; : : : ; sm . In the nonnegativity). Thus, for the FS corresponding to item 3 in the preceding list, s1 and
process of examining the corner points s2 are the nonbasic variables and x1 and x2 are the basic variables, but for the FS
of the feasible region, we find solutions corresponding to item 4, the nonbasic variables are x1 and s2 and the basic variables
to the system in which at least n of the are x2 and s1 .
n C m variables are 0. Precisely n of these
are called nonbasic variables, and the
We will first find an initial FS, and hence, an initial corner point, and then deter-
remaining m are called basic variables. mine whether the corresponding value of can be increased by a different FS. Since
Which m of the n C m variables are basic x1 D 0 and x2 D 0 is a feasible point for this standard linear programming problem,
depends on the corner point under let us initially find the FS such that the decision variables x1 and x2 are nonbasic and,
consideration. Among other things, the hence, the slac variables s1 and s2 are basic. That is, we choose x1 D 0 and x2 D 0
procedure that we are describing provides
a mechanical way of eeping trac of
and find the corresponding values of s1 ; s2 , and . This can be done most conveniently
which variables, at any time, are basic. by matrix techniques, based on the methods developed in Chapter 6.
If we write Equation (4) as #3x1 # x2 C D 0, then Equations (5), (6), and (4)
form the linear system
8
< 2x1 C x2 C s1 D
2x1 C 3x2 C s2 D 12
:#3x # x C D 0
1 2

in the variables x1 , x2 , s1 , s2 , and . Thus, in general, when we add the ob ective func-
tion to the system that provides the constraints, we have m C 1 equations in n C m C 1
un nowns. In terms of an augmented coe cient matrix, called the initial simplex
table, we have
x1 x2 s1 s2 R
2 3
s1 2 1 1 0 0
s2 4 2 3 0 1 0 12 5
#3 #1 0 0 1 0
It is convenient to be generous with labels for matrices that are being used as simplex
tables. Thus, the columns in the matrix to the left of the vertical bar are labeled, naturally
enough, by the variables to which they correspond. We have chosen R as a label for
the column that provides the Right sides of the system of equations. We have chosen
to label the list of row labels. The first two rows correspond to the constraints, and
the last row, called the ob ective row, corresponds to the ob ective equation thus, the
horizontal separating line. Notice that if x1 D 0 and x2 D 0, then, from rows 1, 2, and
3, we can directly read off the values of s1 ; s2 , and : s1 D ; s2 D 12, and D 0.
Thus, the rows of this initia simplex table are labeled to the left by s1 , s2 , and . We
recall that, for the feasible point x1 D 0, x2 D 0, s1 and s2 are the basic variables. So
the column heading can be understood to stand for asic variables. ur initial basic
feasible solution is
x 1 D 0 x 2 D 0 s1 D s2 D 12
at which D 0.
310 C near Pro ra n

et us see if we can find a FS that gives a larger value of . The variables x1 and
x2 are nonbasic in the preceding FS. We will now loo for a FS in which one of
these variables is basic while the other remains nonbasic. Which one should we choose
as the basic variable et us examine the possibilities. From the -row of the preceding
matrix, D 3x1 Cx2 . If x1 is allowed to become basic, then x2 remains at 0 and D 3x1
thus, for each one-unit increase in x1 , increases by three units. n the other hand, if x2
is allowed to become basic, then x1 remains at 0 and D x2 hence, for each one-unit
increase in x2 , increases by one unit. Consequently, we get a greater increase in the
value of if x1 , rather than x2 , enters the basic-variable list. In this case, we call x1 the
entering variable. Thus, in terms of the simplex table below (which is the same as
the matrix presented earlier, except for some additional labeling), the entering variable
can be found by loo ing at the most negative of the numbers enclosed by the brace
in the -row. ( y m st negati e, we mean the negative indicator having the greatest
magnitude.) Since that number is #3 and appears in the x1 -column, x1 is the entering
variable. The numbers in the brace are called indicators.
entering
variable
#
x x 2 s1 s2 R
2 1 3
s1 2 1 1 0 0
s2 4 2 3 0 1 0 12 5
#3 #1 0 0 1 0
„ ƒ‚ …
indicators
et us summarize the information that can be obtained from this table. It gives a
FS where s1 and s2 are the basic variables and x1 and x2 are nonbasic. The FS is
s1 D (the right-hand side of the s1 -row), s2 D 12 (the right-hand side of the s2 -row),
x1 D 0, and x2 D 0. The #3 in the x1 -column of the -row indicates that if x2 remains
0, then increases three units for each one-unit increase in x1 . The #1 in the x2 -column
of the -row indicates that if x1 remains 0, then increases one unit for each one-unit
increase in x2 . The column in which the most negative indicator, #3, lies gives the
entering variable x1 that is, the variable that should become basic in the next FS.
In our new FS, the larger the increase in x1 (from x1 D 0), the larger is the increase
in . Now, by how much can we increase x1 Since x2 is still held at 0, from rows 1 and
2 of the simplex table, it follows that
s1 D # 2x1
and
s2 D 12 # 2x1
Since s1 and s2 are nonnegative, we have
# 2x1 ! 0
and
12 # 2x1 ! 0
12
From the first inequality, x1 " 2
D 4 from the second, x1 " 2
D 6. Thus, x1 must
12
be less than or equal to the smaller of the quotients 2 and ,
which is 2 . Hence, x1
2
can increase at most by 4 and since we want to maximize , it is desirable to increase
x1 from 0 to 4. However, in a FS, at least two variables must be 0. We already have
x2 D 0. Since s1 D # 2x1 , s1 must be 0 if we ma e x1 D 4. Therefore, we have a
new FS with x1 replacing s1 as a basic variable. That is, s1 will depart from the list
of basic variables in the previous FS and will be nonbasic in the new FS. We say
that s1 is the departing variable for the previous FS. In summary, for our new FS,
we want x1 and s2 as basic variables with x1 D 4 and x2 and s1 as nonbasic variables
(x2 D 0, s1 D 0). These requirements lead to s2 D 12 # 2x1 D 12 # 2.4/ D 4.
 Section 7.3 e e et o 311

efore proceeding, let us update our table. To the right of the following table, the
quotients 2 and 12
2
are indicated:

entering variable
(most negative indicator)
#
x1 x2 s1 s2 R u tients
2 3
departing s1 2 1 1 0 0 '2D4
variable s2 4 2 3 0 1 0 12 5 12 ' 2 D 6
(smallest #3 #1 0 0 1 0
quotient) „ ƒ‚ …

These quotients are obtained by dividing each entry in the first two rows of the
R-column by the entry in the corresponding row of the entering-variable column, that
is the x1 -column. Notice that the departing variable is in the same row as the sma er
quotient, ' 2.
Since x1 and s2 will be basic variables in our new FS, it would be convenient
to change our previous table by elementary row operations into a form in which the
values of x1 , s2 , and can be read off with ease ( ust as we were able to do with the
solution corresponding to x1 D 0 and x2 D 0). To do this, we want to find a matrix that
is equivalent to the preceding table but that has the form

x1 x2 s1 s2 R
2 3
x1 1 ‹ ‹ 0 0 ‹
s2 4 0 ‹ ‹ 1 0 ‹5
0 ‹ ‹ 0 1 ‹

where the question mar s represent numbers to be determined. Notice here that if x2 D
0 and s1 D 0, then x1 equals the number in row x1 of column R, s2 equals the number
in row s2 of column R, and is the number in row of column R. Thus, we must
transform the table
entering
variable
#
x1 x2 s1 s2 R
2 3
departing s1 2 1 1 0 0
variable s24 2 3 0 1 0 12 5
#3 #1 0 0 1 0

into an eqivalent matrix that has a 1 where the shaded entry appears and 0 s elsewhere in
the x1 -column. The shaded entry is called the pivot entry it is in the column of the
entering variable (called the pi t c umn) and the row of the departing variable (called
the pi t r ). y elementary row operations, we have

x1 x2 s1 s2
2 3
2 1 1 0 0
4 2 3 0 1 0 12 5
#3 #1 0 0 1 0
2 1 1 3
1 2 2
0 0 4
1
R
2 1 4 2 3 0 1 0 12 5
########!
#3 #1 0 0 1 0
2 1 1 3
1 2 2
0 0 4
#2R1 C R2 4 0 2 #1 1 0 45
########!
3R1 C R3 0 1
2
3
2
0 1 12
312 C near Pro ra n

Thus, we have a new simplex table:

x1 x2 s1 s2 R
2 1 1 3
x1 1 2 2
0 0 4
s26
40 2 #1 1 0 47
5
1 3
0 2 2
0 1 12
„ ƒ‚ …
indicators

For x2 D 0 and s1 D 0, from the first row, we have x1 D 4 from the second, we obtain
s2 D 4. These values give us the new FS. Note that we replaced the s1 located to the
left of the initial table (7) by x1 in our new table ( ), so that s1 departed and x1 entered.
From row 3, for x2 D 0 and s1 D 0, we get D 12, which is a larger value than we
had before. ( efore, we had D 0.)
In our present FS, x2 and s1 are nonbasic variables .x2 D 0; s1 D 0/. Suppose we
loo for another FS that gives a larger value of and is such that one of x2 or s1 is
basic. The equation corresponding to the -row is given by 12 x2 C 32 s1 C D 12, which
can be rewritten as

1 3
D 12 # x2 # s1
2 2

If x2 becomes basic and therefore s1 remains nonbasic, then

1
D 12 # x2 .since s1 D 0/
2

Here, each one-unit increase in x2 decreases by 12 unit. Thus, any increase in x2 would
ma e smaller than before. n the other hand, if s1 becomes basic and x2 remains
nonbasic, then, from Equation ( ),

3
D 12 # s1 .since x2 D 0/
2

Here each one-unit increase in s1 decreases by 32 units. Hence, any increase in s1

would ma e smaller than before. Consequently, we cannot move to a better FS. In
short, no FS gives a larger value of than the FS x1 D 4, s2 D 4, x2 D 0, and
s1 D 0 (which gives D 12).
In fact, since x2 ! 0 and s1 ! 0, and since the coe cients of x2 and s1 in
Equation ( ) are negative, is maximum when x2 D 0 and s1 D 0. That is, in ( ),
ha ing a n nnegati e indicat rs means that e ha e an ptimum s uti n.
In terms of our original problem, if

D 3x1 C x2

sub ect to

2x1 C x2 " 2x1 C 3x2 " 12 x1 ; x2 ! 0

then is maximized when x1 D 4 and x2 D 0, and the maximum value of is 12. (This
confirms our result in Example 1 of Section 7.2.) Note that the values of s1 and s2 do
not have to appear here.
et us outline the simplex method for a standard linear programming problem with
three decision variables and four constraints, not counting nonnegativity conditions.
(So here n D 3 and m D 4.) The outline suggests how the simplex method wor s
 Section 7.3 e e et o 313

for any number of decision variables (in general n) and any number of constraints (in
general m).

S M

aximize D c1 x1 C c2 x2 C c3 x3
sub ect to
a11 x1 C a12 x2 C a13 x3 " b1
a21 x1 C a22 x2 C a23 x3 " b2
a31 x1 C a32 x2 C a33 x3 " b3
a41 x1 C a42 x2 C a43 x3 " b4
where x1 ; x2 ; x3 and b1 ; b2 ; b3 ; b4 are nonnegative.
M
Set up the initial simplex table:
x1 x2 x3 s1 s2 s3 s4 R
2 3
s1 a11 a12 a13 1 0 0 0 0 b1
s2 6 a21 a22 a23 0 1 0 0 0 b2 7
s3 6 a31 a32 a33 0 0 1 0 0 b3 7
6 7
s4 4 a41 a42 a43 0 0 0 1 0 b4 5
#c1 #c2#c3 0 0 0 0 1 0
„ ƒ‚ …
indicators
There are four slac variables: s1 ; s2 ; s3 , and s4 one for each constraint.
If all the indicators in the last row are nonnegative, then has a maximum with
the current list of basic variables and the current value of . (In the case of the
initial simplex table, this gives x1 D 0; x2 D 0, and x3 D 0, with maximum value
of D 0.) If there are any negative indicators, locate and mar the column in which
the most negative indicator appears. This pi t c umn gives the entering variable.
(If more than one column contains the most negative indicator, the choice of pivot
column is arbitrary.)
ivide each p siti e entry above the ob ective row in the entering-variable col-
umn int the corresponding value of column R. (The p siti e restriction will be
explained after Example 1.)
ar the entry in the pivot column that corresponds to the smallest quotient in
step 3. This is the pivot entry, and the row in which it is located is the pi t r .
The departing variable is the one that labels the pivot row.
Use elementary row operations to transform the table into a new equivalent table
that has a 1 where the pivot entry was and 0 s elsewhere in that column.
In the basic variables column, , of this table, the entering variable replaces the
departing variable.
If the indicators of the new table are all nonnegative, we have an optimum solu-
tion. The maximum value of is the entry in the last row and last column. It occurs
when the basic variables as found in the basic variables column, , are equal to the
corresponding entries in column R. All other variables are 0. If at least one of the
indicators is negative, repeat the process, beginning with step 2 applied to the new
table.
314 C near Pro ra n

As an aid in understanding the simplex method, we should be able to interpret

certain entries in a table. Suppose that we obtain a table in which the last row is as
shown in the following array:
x1 x2 x3 s1 s2 s3 s4 R
2 3
& & & & & & & & &
6 & & & & & & & & &7
4 & & & & & & & & &5
a b c d e f g 1 h
We can interpret the entry b, for example, as follows: If x2 is nonbasic and were to
become basic, then, for each one-unit increase in x2 ,

if b < 0; increases by jbj units

if b > 0; decreases by jbj units
if b D 0; there is no change in

E AM LE T S M

aximize D 5x1 C 4x2 sub ect to

x1 C x2 " 20
2x1 C x2 " 35
#3x1 C x2 " 12

and x1 ; x2 ! 0.
S This linear programming problem fits the standard form. The initial simplex
table is
entering
variable
#
x x 2 s1 s2 s3 R u tient
2 1 3
s1 1 1 1 0 0 0 20 20 ' 1 D 20
6 7
6 35 7 35
departing s26 2 1 0 1 0 0 7 35 ' 2 D 2
variable 6 7
s34 #3 1 0 0 1 0 12 5 no quotient, #3 6> 0
#5 #4 0 0 0 1 0
„ ƒ‚ …
indicators
The most negative indicator, #5, occurs in the x1 -column. Thus, x1 is the entering vari-
able. The smaller quotient is 35
2
, so s2 is the departing variable. The pivot entry is 2.
Using elementary row operations to get a 1 in the pivot position and 0 s elsewhere in
its column, we have
x1 x2 s1 s2 s3
2 3
1 1 1 0 0 0 20
6 2 1 0 1 0 0 35 7
6 7
4 #3 1 0 0 1 0 12 5
#5 #4 0 0 0 1 0
2 3
1 1 1 1 0 0 0 j 20
R
2 2 j 35
########! 6 1 1
0 1
0 0 j 2 7
6 2 2
j 7
4 #3 1 0 0 1 0 j 12 5
j
#5 #4 0 0 0 1 j 0
 Section 7.3 e e et o 315

2 1 3
0 1 # 12 0 0 j 5
#1R2 C R1 2
j 2
########! 61 1
0 1
0 0 j 35 7
3R2 C R3 6 2 2 2 7
6 j 7
5R2 C R4 40 5
2
0 3
2
1 0 j 12
2
5
j
0 # 32 0 5
2
0 1 j 175
2
ur new table is
entering
variable
#
x1 x2 s1 s2 s3 R u tients
2 1 3 5
' 12 D 5
departing s1 0 2
1 # 12 0 0 j 5
2 2
variable 6 1 1
j 35 7 35 1
x1 6 1 2
0 2
0 0 j 2 7 2
' 2
D 35
6 j 7 12 5
s3 4 0 5
2
0 3
2
1 0 j 12
2
5 2
' 2
D 25 45
j
0 # 32 0 0 1 5
2
j 175
2
„ ƒ‚ …
indicators
Note that in column , which eeps trac of which variables are basic, x1 has replaced
s2 . Since we still have a negative indicator, # 32 , we must continue our process. Evi-
dently, # 32 is the most negative indicator and the entering variable is now x2 . The
smallest quotient is 5. Hence, s1 is the departing variable and 12 is the pivot entry. Using
elementary row operations, we have
x1 x2 s1 s2 s3 R
2 1 3
0 2
1 # 12 0 0 5
2
61 1
0 1
0 0 35 7
6 2 2 2 7
6 7
40 5
2
0 3
2
1 0 12
2 5
0 # 32 0 5
2
0 1 175
2
2 1 3
0 2
1 # 12 0 0 5
2
#1R1 C R2 6 1 0 #1 1 0 0 15 7
########! 64
7
#5R1 C R3 0 0 #5 4 1 0 52 5
3R1 C R4 0 0 3 1 0 1 5
2 3
0 1 2 #1 0 0 5
61 0 #1 1 0 0 15 7
2R1 6 7
########!4 0 0 #5 4 1 0 52 5
0 0 3 1 0 1 5
ur new table is
x1 x2 s1 s2 s3 R
2 3
x2 0 1 2 #1 0 0 5
x1 6 1 0 #1 1 0 0 15 7
s3 6
40 0 #5 4 1 0 52 7
5
0 3 0 1 0 1 5
„ ƒ‚ …
indicators
where x2 replaced s1 in column . Since all indicators are nonnegative, the maximum
value of is 5 and occurs when x1 D 15, x2 D 5 and x3 D 0 (and s1 D 0, s2 D 0, and
s3 D 52).
Now ork Problem 1 G
It is interesting to see how the values of got progressively better in successive
tables in Example 1. These are the entries in the last row and last column of each
316 C near Pro ra n

simplex table. In the initial table, we had D 0. From then on, we obtained D 175 2
D
7 12 and then D 5, the maximum.
In Example 1, no quotient was considered in the third row of the initial table. This
is the positivity requirement in Part 3 of the method outlined for the general 3-variable,
4-constraint problem and we now explain it. The FS for this table is

s1 D 20; s2 D 35; s3 D 12; x1 D 0; x2 D 0

35
where x1 is the entering variable. The quotients 20 and 2
re ect that, for the next
35
FS, we have x1 " 20 and x1 " 2
.
Since the third row represents the equation s3 D
12 C 3x1 # x2 , and x2 D 0, it follows that s3 D 12 C 3x1 . ut s3 ! 0, so 12 C 3x1 ! 0,
which implies that x1 ! # 123
D #4. Thus, we have

35
x1 " 20; x1 " ; and x1 ! #4
2
Hence, x1 can increase at most by 35 2
. The condition x1 ! #4 has no in uence in
determining the maximum increase in x1 . That is why the quotient 12=.#3/ D #4 is
not considered in row 3. In general, n qu tient is c nsidered f r a r if the entry in
the entering ariab e c umn is negati e r .
It is of course possible that when considering quotients for comparison there are
n qu tients. For the record, we note:

If no quotients exist in a simplex table, then the standard maximum linear program-
ming problem has an unbounded solution. This means that the ob ective function
does not attain a maximum value because it attains arbitrarily large values. ore
precisely, it means that, for every positive integer n, there is a point An in the feasible
region with .An / > n.

Although the simplex procedure that has been developed in this section applies
only to linear programming problems of standard maximum form, other forms may be
A L IT I adapted to fit this form. Suppose that a constraint has the form
The Toones Company has 30,000 a1 x1 C a2 x2 C & & & C an xn ! #b
for the purchase of materials to ma e
three types of P3 players. The com- where b > 0. Here the inequality symbol is ! , and the constant on the right side is
pany has allocated a total of 1200 hours
negati e. Thus, the constraint is not in standard form. However, multiplying both sides
of assembly time and 1 0 hours of pac -
by #1 gives
aging time for the players. The follow-
ing table gives the cost per player, the
number of hours per player, and the
#a1 x1 # a2 x2 # & & & # an xn " b
profit per player for each type:
which d es have the proper form. Accordingly, it may be necessary to rewrite a con-
straint before proceeding with the simplex method.
In a simplex table, several indicators may tie for being most negative. In this
case, we choose any one of these indicators to give the column for the entering vari-
able. i ewise, there may be several quotients that tie for being the smallest. We can
then choose any one of these quotients to determine the departing variable and pivot
entry. Example 2 will illustrate this situation. However, when a tie for the smallest quo-
tient exists, then the next simplex table will address a FS with a basic variable that
is 0 (along with all the nonbasic variables that are 0 by declaration). In this case we
say that the FS is degenerate or, better, that the linear programming problem has a
degeneracy. egenerate linear programming problems sometimes lead to cyc ing di -
culties with the simplex method. We might, for example, arrive at a FS, call it FS1 ,
Find the number of players of each type
the company should produce to maxi- proceed to FS2 , and FS3 , and return to FS1 . There are techniques to deal with such
mize profit. di culties which do not often arise in practice but they are beyond the scope of this
boo .
 Section 7.3 e e et o 317

E AM LE T S M

aximize D 3x1 C 4x2 C 32 x3 sub ect to

#x1 # 2x2 ! #10

2x1 C 2x2 C x3 " 10
x1 ; x2 ; x3 ! 0

S Constraint (10) does not fit the standard form. However, multiplying both
sides of inequality (10) by #1 gives
x1 C 2x2 " 10
which d es have the proper form. Thus, our initial simplex table is table I:

SIMPLE ABLE I
entering
variable
#
x x2 x3 s1 s2 R u tients
2 1 3
departing s1 1 2 0 1 0 0 10 10 ' 2 D 5
variable s24 2 2 1 0 1 0 10 5 10 ' 2 D 5
#3 #4 # 32 0 0 1 0
„ ƒ‚ …
indicators
The entering variable is x2 . Since there is a tie for the smallest quotient, we can choose
either s1 or s2 as the departing variable. et us choose s1 . The pivot entry is shaded.
Using elementary row operations, we get table II:

SIMPLE ABLE II
entering
variable
#
x 1 x 2 x 3 s1 s2 R u tients
2 1 3
x2 1 0 1
0 0 5 no quotient 0 6> 0
2 2
departing 6
s2 4 1 0 1 #1 1 0 7
05 0 ' 1 D 0
variable
#1 0 # 32 2 0 1 20
„ ƒ‚ …
indicators
Table II corresponds to a FS in which a basic variable, s2 , is zero. Thus, the FS
is degenerate. Since there are negative indicators, we continue. The entering variable
is now x3 , the departing variable is s2 , and the pivot is shaded. Using elementary row
operations, we get table III:

SIMPLE ABLE III

x1 x2 x3 s1 s2 R
2 1 1 3
x2 2
1 0 2
0 0 5
6 7
x3 4 1 0 1 #1 1 0 05
1 1 3
2
0 0 2 2
1 20
„ ƒ‚ …
indicators
Since all indicators are nonnegative, is maximized when x2 D 5, x3 D 0, and x1 D
s1 D s2 D 0. The maximum value is D 20. Note that this value is the same as the
value of in table II. In degenerate problems, it is possible to arrive at the same value of
318 C near Pro ra n

at various stages of the simplex process. Problem 7 as s for a solution to this example
by using s2 as the departing variable in the initial table.
Now ork Problem 7 G
In Section 7.2, where linear programming problems in 2 variables were solved
geometrically, we drew attention to the possibility of multiple solutions by pointing out
that the family of parallel isoprofit lines may in fact be parallel to one of the bounding
edges of the feasible region. When using the simplex method one can also detect the
possibility of multiple optimum solutions but we will not pursue the idea in this boo .

In a table that gives an optimum solution, a zero indicator for a nonbasic variable
suggests the possibility of multiple optimum solutions.

ecause of its mechanical nature, the simplex procedure is readily adaptable to

computers to solve linear programming problems involving many variables and con-
straints.

R BLEMS
se the simp ex meth d t s e the f ing pr b ems aximize
aximize P D x1 C 2x2

D x1 C 2x2 sub ect to

x1 # x2 " 1
sub ect to
x1 C 2x2 "
2x1 C x2 "
x1 C x2 " 5
2x1 C 3x2 " 12
x1 ; x2 ! 0
x1 ; x2 ! 0
aximize
aximize
D 2x1 # 6x2
D 2x1 C x2 sub ect to
sub ect to x1 # x2 " 4

#x1 C x2 " 4 #x1 C x2 " 4

x1 C x2 " 6 x1 C x2 " 6

x1 ; x2 ! 0 x1 ; x2 ! 0
Solve the problem in Example 2 by choosing s2 as the
aximize
departing variable in table I.
D #x1 C 2x2 aximize
sub ect to D 2x1 # x2 C x3
sub ect to
3x1 C 2x2 " 5
2x1 C x2 # x3 " 4
#x1 C 3x2 " 3
x1 C x2 C x3 " 2
x1 ; x2 ! 0
x1 ; x2 ; x3 ! 0
aximize
aximize
D 4x1 C 7x2 D 2x1 C x2 # x3
sub ect to sub ect to
2x1 C 3x2 " x1 C x2 " 1
x1 C 5x2 " 10 x1 # 2x2 # x3 ! #2
x1 ; x2 ! 0 x1 ; x2 ; x3 ! 0
 Section 7.3 e e et o 319

aximize aximize
P D 60x1 C 0x2 C 0x3 C 0x4
P D #2x1 C 3x2
sub ect to
sub ect to x1 # 2x2 " 2
x1 C x2 " 1 x1 C x2 " 5
x1 # x2 " 2 x3 C x4 " 4
x1 # x2 ! #3 x3 # 2x4 " 7
x1 " 5 x1 ; x2 ; x3 ; x4 ! 0
x1 ; x2 ! 0 aximize
D 3x1 C 2x2 # 2x3 # x4
aximize
sub ect to
D x1 C x2
x1 C x3 # x4 " 3
sub ect to x1 # x2 C x4 " 6

2x1 # x2 " 4 x1 C x2 # x3 C x4 " 5

#x1 C 2x2 " 6 x1 ; x2 ; x3 ; x4 ! 0

5x1 C 3x2 " 20 Freight Shipments A freight company handles shipments

by two corporations, A and , that are located in the same city.
2x1 C x2 " 10 Corporation A ships boxes that weigh 3 lb each and have a volume
x1 ; x2 ! 0 of 2 ft3 ships 1-ft3 boxes that weigh 5 lbs each. oth A and
ship to the same destination. The transportation cost for each box
aximize from A is 0.75, and from it is 0.50. The freight company has
a truc with 2400 ft3 of cargo space and a maximum capacity of
D 2x1 C x2 # 2x3 36, 00 lb. In one haul, how many boxes from each corporation
should be transported by this truc so that the freight company
sub ect to receives maximum revenue What is the maximum revenue
Production A company manufactures three products: ,
#2x1 C x2 C x3 ! #2
, and . Each product requires machine time and finishing time
x1 # x2 C x3 " 4 as shown in the following table:
x1 C x2 C 2x3 " 6 achine Time Finishing Time
x1 ; x2 ; x3 ! 0
1 hr 4 hr
aximize 2 hr 4 hr
3 hr hr
D x1 # 12x2 C 4x3

sub ect to The numbers of hours of machine time and finishing time
available per month are 00 and 5000, respectively. The unit
4x1 C 3x2 # x3 " 1 profit on , , and is 6, , and 12, respectively. What is the
maximum profit per month that can be obtained
x1 C x2 # x3 ! #2
Production A company manufactures three types of patio
#x1 C x2 C x3 ! #1 furniture: chairs, roc ers, and chaise lounges. Each requires wood,
plastic, and aluminum as shown in the following table:
x1 ; x2 ; x3 ! 0
Wood Plastic Aluminum
aximize
Chair 1 unit 1 unit 2 units
D 4x1 C 0x2 # x3
Roc er 1 unit 1 unit 3 units
sub ect to Chaise lounge 1 unit 2 units 5 units

x1 C x2 C x3 " 6
The company has available 400 units of wood, 500 units of
x1 # x2 C x3 " 10 plastic, and 1450 units of aluminum. Each chair, roc er, and
chaise lounge sells for 21, 24, and 36, respectively. Assuming
x1 # x2 # x3 " 4
that all furniture can be sold, determine a production order so that
x1 ; x2 ; x3 ! 0 total revenue will be maximum. What is the maximum revenue
320 C near Pro ra n

Objective A
o se art c a ar a es to an e To start using the simplex method, a basic feasib e s uti n, FS, is required. (We
a at on ro e s t at are not of
stan ar a for algebraically start at a c rner p int using the initial simplex table, and each subse-
quent table ta es us, algebraically, to another corner point until we reach the one at
which an optimum solution is obtained.) For a standard maximum linear programming
problem, we begin with the FS in which all decision variables are zero, the origin in
x1 ; x2 ; & & & ; xn -space, where n is the number of decision variables. However, for a max-
imization problem that is not of standard maximum form, the origin .0; 0; & & & ; 0/ may
not be a FS. In this section, we will learn how the simplex method is modified for use
in such situations.
et us consider the following problem:
aximize D x1 C 2x2
sub ect to
x1 C x2 "
x1 # x2 ! 1
x1 ; x2 ! 0
Since constraint (2) cannot be written as a1 x1 C a2 x2 " b, where b is nonnegative,
this problem cannot be put into standard form. Note that .0; 0/ is not a feasible point
because it does not satisfy constraint (2). ( ecause 0 # 0 D 0 ! 1 is fa se ) To solve
the problem, we begin by writing Constraints (1) and (2) as equations. Constraint (1)
becomes
x1 C x2 C s1 D
where s1 ! 0 is a slac variable. For Constraint (2), x1 # x2 will equal 1 if we subtract
a nonnegative slac variable s2 from x1 # x2 . That is, by subtracting s2 , we are ma ing
up for the surplus on the left side of (2), so that we have equality. Thus,
x1 # x2 # s2 D 1
where s2 ! 0. We can now restate the problem:
aximize D x1 C 2x2
sub ect to
x1 C x2 C s1 D
x1 # x2 # s2 D 1
x1 ; x2 ; s1 ; s2 ! 0
Since .0; 0/ is not in the feasible region, we do not have a FS in which
x1 D x2 D 0. In fact, if x1 D 0 and x2 D 0 are substituted into Equation (7), then
0 # 0 # s2 D 1, which gives s2 D #1, and now the problem is that this contradicts the
condition that s2 ! 0.
To get the simplex method started, we need an initial FS. Although none is obvi-
ous, there is an ingenious method to arrive at one arti cia y. It requires that we consider
a related linear programming problem called the artificial problem. First, a new equa-
tion is formed by adding a nonnegative variable t to the left side of the equation in which
the coe cient of the slac variable is #1. The variable t is called an artificial variable.
In our case, we replace Equation (7) by x1 # x2 # s2 C t D 1. Thus, Equations (6) and
(7) become
x1 C x2 C s1 D
x1 # x2 # s2 C t D 1
x1 ; x2 ; s1 ; s2 ; t ! 0
 Section 7.4 rt c a ar a es 321

An obvious solution to Equations ( ) and ( ) is found by setting x1 ; x2 , and s2 equal

to 0. This gives
x1 D x2 D s2 D 0 s1 D tD1
Note that these values do not satisfy Equation (7). However, it is clear that any solution
of Equations ( ) and ( ) f r hich t D 0 will give a solution to Equations (6) and (7),
and conversely.
We can e entua y force t to be 0 if we alter the original ob ective function. We
define the artificial ob ective function to be
D # t D x1 C 2x2 # t
where the constant is a very large positive number. We will not worry about the par-
ticular value of and will proceed to maximize by the simplex method. Since there
are m D 2 constraints (excluding the nonnegativity conditions) and n D 5 variables in
Equations ( ) and ( ), any FS must have at least n # m D 3 variables equal to zero.
We start with the following FS:
x1 D x2 D s2 D 0 s1 D tD1
In this initial FS, the nonbasic variables are the decision variables and the surplus
variable s2 . The corresponding value of is D x1 C 2x2 # t D # , which
is extremely negative since we assume that is a very large positive number. A
significant improvement in will occur if we can find another FS for which t D 0.
Since the simplex method see s better values of at each stage, we will apply it until
we reach such a FS, if possible. That solution will be an initial FS for the original
problem.
To apply the simplex method to the artificial problem, we first write Equation (10) as
#x1 # 2x2 C tC D0
The augmented coe cient matrix of Equations ( ), ( ), and (12) is
x1 x2 s1 s2 t
2 3
1 1 1 0 0 0
4 1 #1 0 #1 1 0 15
#1 #2 0 0 1 0
An initial FS is given by (11). Notice that, from row s1 , when x1 D x2 D s2 D 0,
we can directly read the value of s1 namely, s1 D . From row 2, we get t D 1. From
row 3, t C D 0. Since t D 1, D # . ut in a simplex table we want the value
of to appear in the last row and last column. This is not so in (13) thus, we modify
that matrix.
To do this, we transform (13) into an equivalent matrix whose last row has the form
x1 x2 s1 s2 t
‹ ‹ 0 ‹ 0 1 j ‹
That is, the in the t-column is replaced by 0. As a result, if x1 D x2 D s2 D 0, then
equals the last entry. Proceeding to obtain such a matrix, we have, by pivoting at
the shaded element in column t:
x1 x2 s1 s2 t R
2 3
1 1 1 0 0 0
4 1 #1 0 #1 1 0 15
#1 #2 0 0 1 0
x1 x2 s1 s2 t R
2 3
1 1 1 0 0 0
# R2 C R3 4
#########! 1 #1 0 #1 1 0 1 5
#1 # #2 C 0 0 1 #
322 C near Pro ra n

et us now chec things out. If x1 D 0; x2 D 0, and s2 D 0, then from row 1 we get

s1 D , from row 2, t D 1, and from row 3, D # . Thus, we now have initial
simplex table I:
SIMPLE ABLE I
entering
variable
#
x1 x2 s1 s2 t R u tients
2 3
s1 1 1 1 0 0 0 '1D
departing t 4 1 #1 0 #1 1 0 1 51'1D1
variable #1 # #2 C 0 0 1 #
„ ƒ‚ …
indicators
From this point, we can use the procedures of Section 7.3. Since is a large positive
number, the most negative indicator is #1 # . Thus, the entering variable is x1 . From
the quotients, we get t as the departing variable. The pivot entry is shaded. Using ele-
mentary row operations to get 1 in the pivot position and 0 s elsewhere in that column,
we get simplex table II:
SIMPLE ABLE II
entering
variable
#
x1 x2 s1 s2 t R u tients
2 3
departing s1 0 2 1 1 #1 0 '2D4
variable x1 4 1 #1 0 #1 1 0 1 5 no quotient
0 #3 0 #1 1 C 1 1
„ ƒ‚ …
indicators
From table II, we have the following FS:
s1 D ; x1 D 1; x2 D 0; s2 D 0; t D 0
Since t D 0, the values s1 D , x1 D 1, x2 D 0, and s2 D 0 form an initial FS for the
rigina problem The artificial variable has served its purpose. For succeeding tables,
we will delete the t-column (since we want to solve the original problem) and change
the s to s (since D for t D 0). From table II, the entering variable is x2 , the
departing variable is s1 , and the pivot entry is shaded. Using elementary row operations
(omitting the t-column), we get table III:
SIMPLE ABLE III
x x s s2 R
2 1 2 1 3
x2 0 1 12 1
2
0 4
6 7
x1 6 1 0 12 # 12 0 57
4 5
0 0 32 1
2
1 13
„ ƒ‚ …
indicators
Since all the indicators are nonnegative, the maximum value of is 13. It occurs
when x1 D 5 and x2 D 4.
Here is a summary of the procedure It is worthwhile to review the steps we performed to solve our problem:
involving artificial variables.
aximize D x1 C 2x2
sub ect to
x1 C x2 "
x1 # x2 ! 1
x1 ; x2 ! 0
 Section 7.4 rt c a ar a es 323

We write Inequality (14) as

x1 C x2 C s1 D

Since Inequality (15) involves the symbol !, and the constant on the right side is non-
negative, we write Inequality (15) in a form having both a surplus variable and an
artificial variable:

x1 # x2 # s2 C t D 1

The artificial ob ective equation to consider is D x1 C 2x2 # t, equivalently,

#x1 # 2x2 C tC D0

The augmented coe cient matrix of the system formed by Equations (16) (1 ) is

x x 2 s1 s2 t R
2 1 3
s1 1 1 1 0 0 0
t 4 1 #1 0 #1 1 0 15
#1 #2 0 0 1 0

Next, we replace the entry in the ob ective row and the artificial variable column, an
, by 0 using elementary row operations. The resulting simplex table I corresponds to
the initial FS of the artificial problem in which the decision variables, x1 and x2 , and
the surplus variable s2 are each 0:

SIMPLE ABLE I
A L IT I
x1 x2 s1 s2 t R
The HI Company manufactures 2 3
s1 1 1 1 0 0 0
two models of snowboards, standard
t 4 1 #1 0 #1 1 0 1 5
and deluxe, at two different manufac-
turing plants. The maximum output at #1 # #2 C 0 0 1 #
plant I is 1200 per month, while the
maximum output at plant II is 1000 per The basic variables s1 and t in column of the table correspond to the nondecision
month. ue to contractual obligations, variables in Equations (16) and (17) that have positive coe cients. We now apply the
the number of deluxe models produced simplex method until we obtain a FS in which the artificial variable t equals 0. Then
at plant I cannot exceed the number of we can delete the artificial variable column, change the s to s, and continue the
standard models produced at plant I by procedure until the maximum value of is obtained.
more than 200. The profit per standard
and deluxe snowboard manufactured at
plant I is 40 and 60, respectively, E AM LE A
while the profit per standard and deluxe
snowboard manufactured at plant II is Use the simplex method to maximize D 2x1 C x2 sub ect to
45 and 50, respectively. This month,
HI received an order for 1000 standard
x1 C x2 " 12
and 00 deluxe models. Find how many
of each model should be produced at x1 C 2x2 " 20
each plant to satisfy the order and max-
imize the profit. ( int et x1 represent #x1 C x2 ! 2
the number of standard models and x2 x1 ; x2 ! 0
represent the number of deluxe models
manufactured at plant I.)
S The equations for (1 ) (21) will involve two slac variables, s1 and s2 , for
the two " constraints, and a surplus variable s3 and an artificial variable t, for the !
constraint. We thus have

x1 C x2 C s1 D 12

x1 C 2x2 C s2 D 20

#x1 C x2 # s3 C t D 2
324 C near Pro ra n

We consider D # t D 2x1 C x2 # t as the artificial ob ective equation equiva-

lently,
#2x1 # x2 C tC D0
where is a large positive number. Now we construct the augmented coe cient matrix
of Equations (22) (25):
x1 x2 s1 s2 s3 t
2 3
1 1 1 0 0 0 0 12
6 1 2 0 1 0 0 0 20 7
6 7
4 #1 1 0 0 #1 1 0 25
#2 #1 0 0 0 1 0
To get simplex table I, we replace the in the ob ective row and the artificial variable
column by 0 by adding .# / times row 3 to row 4:
SIMPLE ABLE I
entering
variable
#
x1 x2 s1 s2 s3 t R u tients
2 3
s1 1 1 1 0 0 0 0 12 12 ' 1 D 12
s2 6
6 1 2 0 1 0 0 0 20 7 20 ' 2 D 10
7
departing t 4 #1 1 0 0 #1 1 0 2 5 2'1D 2
variable #2 C #1 # 0 0 0 1 #2
„ ƒ‚ …
indicators

The variables s1 , s2 , and t in column that is, the basic variables are the nondecision
variables with positive coe cients in Equations (22) (24). Since is a large positive
number, .#1 # / is the most negative indicator and, thus, the entering variable is x2 .
The smallest quotient is evidently 2 and, thus, the departing variable is t. The pivot
entry is shaded. Proceeding, we get simplex table II:
SIMPLE ABLE II
entering
variable
#
x1 x2 s1
s2 s3 t R u tients
2 3
departing s1 2 0 10 1 #1 0 10 10 ' 2 D 5
variable s2 6
6 3 0 01 2 #2 0 16 77 16 ' 3 D 5 3
1

x24 #1 1 00 #1 1 0 25
#3 0 00 #1 1 C 1 2
„ ƒ‚ …
indicators
The FS corresponding to table II has t D 0. Thus, we delete the t-column and change
s to s in succeeding tables. Continuing, we obtain table III:
SIMPLE ABLE III
x1 x2 s1 s2 s3 R
2 1 1 3
x1 1 0 2
0 2
0 5
6 3
s2 6 0 0 # 2 1 1
17
0 7
6 2
7
6
x24 0 1 1 1
0 #2 0 77
2 5
3 1
0 0 2
0 2
1 17
„ ƒ‚ …
indicators
 Section 7.4 rt c a ar a es 325

All indicators are nonnegative. Hence, the maximum value of is 17. It occurs when
x1 D 5 and x2 D 7.
Now ork Problem 1 G
E C
When an equa ity constraint of the form
a1 x1 C a2 x2 C & & & C an xn D b; where b ! 0
occurs in a linear programming problem, artificial variables are used in the simplex
method. To illustrate, consider the following problem:
aximize D x1 C 3x2 # 2x3
sub ect to
x1 C x2 # x3 D 6
x1 ; x2 ; x3 ! 0
Constraint (26) is already expressed as an equation, so no slac variable is necessary.
Since x1 D x2 D x3 D 0 is not a feasible solution, we do not have an obvious starting
point for the simplex procedure. Thus, we create an artificial problem by first adding
an artificial variable t to the left side of Equation (26):
x1 C x2 # x3 C t D 6
Here an obvious FS is x1 D x2 D x3 D 0, t D 6. The artificial ob ective function is
D # t D x1 C 3x2 # 2x3 # t
where is a large positive number. The simplex procedure is applied to this artificial
problem until we obtain a FS in which t D 0. This solution will give an initial FS for
the original problem, and we then proceed as before.
In general, the simplex method can be used to
maximize D c1 x1 C c2 x2 C & & & C cn xn
sub ect to
9
a11 x1 C a12 x2 C & & & C a1n xn f"; !; Dg b1 >
>
a21 x1 C a22 x2 C & & & C a2n xn f"; !; Dg b2 =
:: :: :: ::
: : : :>>
;
am1 x1 C am2 x2 C & & & C amn xn f"; !; Dg bm
and x1 ! 0, x2 ! 0, : : : ; xn ! 0. The symbolism f"; !; Dg means that one of the
relations " , ! , or D exists for a constraint.
For each bi < 0, multiply the corresponding inequality by #1 (thus changing the
sense of the inequality). If, with all bi ! 0, all constraints involve " , the problem is
of standard form and the simplex techniques of the previous section apply directly. If,
ith a bi ! 0, any constraint involves ! or D , we begin with an artificial problem,
which is obtained as follows.
Each constraint that contains " is written as an equation involving a slac vari-
able si (with coe cient C1):
ai1 x1 C ai2 x2 C & & & C ain xn C si D bi
Each constraint that contains ! is written as an equation involving a surplus variable
sj (with coe cient #1) and an artificial variable tj (with coe cient C1):
aj1 x1 C aj2 x2 C & & & C ajn xn # sj C tj D bj
Each constraint that contains D is rewritten as an equation with an artificial variable
tk inserted (with coe cient C1):
ak1 x1 C ak2 x2 C & & & C akn xn C tk D bk
326 C near Pro ra n

Should the artificial variables involved in this problem be, for example, t1 , t2 , and t3 ,
then the artificial ob ective function is
D # t1 # t2 # t3
where is a large positive number. An initial FS occurs when x1 D x2 D & & & D xn D 0
and each surp us variable equals 0.
After obtaining an initial simplex table, we apply the simplex procedure until we
arrive at a table that corresponds to a FS in which a artificial variables are 0. We
then delete the artificial variable columns, change s to s, and continue by using the
procedures of the previous section.

E AM LE A E C

Use the simplex method to maximize D x1 C 3x2 # 2x3 sub ect to

#x1 # 2x2 # 2x3 D #6
#x1 # x2 C x3 " #2
x1 ; x2 ; x3 ! 0
S Constraints (2 ) and (2 ) will have the forms indicated in (27) (that is, b s
positive) if we multiply both sides of each constraint by #1:
x1 C 2x2 C 2x3 D 6
x1 C x2 # x3 ! 2
Since Constraints (30) and (31) involve D and ! , two artificial variables, t1 and t2 ,
will occur. The equations for the artificial problem are
x1 C 2x2 C 2x3 C t1 D6
x1 C x2 # x3 # s2 C t2 D 2
Here the subscript 2 on s2 re ects the order of the equations. The artificial ob ective
function is D # t1 # t2 , equivalently,
#x1 # 3x2 C 2x3 C t1 C t2 C D0
where is a large positive number. The augmented coe cient matrix of Equations
(32) (34) is
x1 x2 x3 s2 t1 t2
2 3
1 2 2 0 1 0 0 6
4 1 1 #1 #1 0 1 0 25
#1 #3 2 0 1 0
We now use elementary row operations to replace the s in the ob ective row from a
the artificial variable columns, by 0. y adding # times row 1 to row 3 and adding
# times row 2 to row 3, we get initial simplex table I:
SIMPLE ABLE I
entering
variable
#
x1 x2 x3 s2 t 1 t2 R u tients
2 3
t1 1 2 2 0 1 0 0 6 6'2D3
departing t2 4 1 1 #1 #1 0 1 0 2 52'1D2
variable #1 # 2 #3 # 3 2# 0 0 1 #
„ ƒ‚ …
indicators
 Section 7.4 rt c a ar a es 327

Note that the effect of the previous step was to display the value of when t1 D 6,
t2 D 2, and all other variables are 0. Proceeding, we obtain simplex tables II and III:
SIMPLE ABLE II
entering
variable
#
x1 x2 x3 s2 t1 t2 R u tients
2 3
departing t1 #1 0 4 2 1 #2 0 2 2 ' 4 D 12
variable x24 1 1 #1 #1 0 1 0 2 5
2C 0 #1 # 4 #3 # 2 0 3C3 1 6#2
„ ƒ‚ …
indicators

SIMPLE ABLE III

entering
variable
#
x1 x2 x3 s2 t1 t2 R u tients
2 3
departing x3 # 14 0 1 1
2
1
4
# 12 0 1
2
1
' 1
D1
2 2
variable 6 7
x26 3
1 0 # 12 1 1
0 5 7
4 4 4 2 2 5
7
4
0 # 52 14 C
0 5
2
C 1 13
2
„ ƒ‚ …
indicators
For the FS corresponding to table III, the artificial variables t1 and t2 are both 0. We
now can delete the t1 - and t2 -columns and change s to s. Continuing, we obtain
simplex table IV:
SIMPLE ABLE IV
x x2 x3 s2 R
2 1 3
s2 # 12 0 2 1 0 1
6 7
x26 12 1 1 0 0 37
4 5
1
2
0 5 0 1
„ ƒ‚ …
indicators
Since all indicators are nonnegative, we have reached the final table. The maximum
value of is , and it occurs when x1 D 0, x2 D 3, and x3 D 0.
Now ork Problem 5 G

E F R
It is possible that the simplex procedure terminates and not all artificial variables are 0.
It can be shown that in this situation the feasib e regi n f the rigina pr b em is empty
and, hence, there is n ptimum s uti n. The following example will illustrate.

E AM LE A E F R

Use the simplex method to maximize D 2x1 C x2 sub ect to

#x1 C x2 ! 2
x1 C x2 " 1
x1 ; x2 ! 0
328 C near Pro ra n

S Since Constraint (35) is of the form a11 x1 C a12 x2 ! b1 , where b1 ! 0, an

artificial variable will occur. The equations to consider are

#x1 C x2 # s1 C t1 D 2
x1 C x2 C s2 D1

where s1 is a surplus variable, s2 is a slac variable, and t1 is artificial. The artificial

ob ective function is D # t1 , equivalently,

#2x1 # x2 C t1 C D0

The augmented coe cient matrix of Equations (37) (3 ) is

x1 x2 s1 s2 t1
2 3
#1 1 #1 0 1 0 2
4 1 1 0 1 0 0 15
#2 #1 0 0 1 0

The simplex tables are as follows:

SIMPLE ABLE I
entering
variable
#
x1 x2 s1 s2 t1 R u tients
2 3 2'1D2
t1 #1 1 #1 0 1 0 2
departing s2 4 1 1 0 1 0 0 1 5 1'1D1
variable #2 C #1 # 0 0 1 #2
„ ƒ‚ …
indicators
x2 SIMPLE ABLE II
x1 x 2 s1 s2 t1 R
2 3
t1 #2 0 #1 #1 1 0 1
-x1 + x2 = 2 x24 1 1 0 1 0 0 1 5
2 #1 C 2 0 1C 0 1 1#
„ ƒ‚ …
1 indicators
x1 + x2 = 1

x1
1 Since is a large positive number, the indicators in simplex table II are nonneg-
ative, so the simplex procedure terminates. The value of the artificial variable t1 is 1.
FIGURE Empty feasible Therefore, as previously stated, the feasible region of the original problem is empty
region (no solution exists). and, hence, no solution exists. This result can be obtained geometrically. Figure 7.16
shows the graphs of #x1 C x2 D 2 and x1 C x2 D 1 for x1 ; x2 ! 0. Since there is no
point .x1 ; x2 / that simultaneously lies above #x1 C x2 D 2 and below x1 C x2 D 1 such
that x1 ; x2 ! 0, the feasible region is empty and, thus, no solution exists.
Now ork Problem 9 G
In the next section we will use the simplex method on minimization problems.
 Section 7.4 rt c a ar a es 329

R BLEMS
se the simp ex meth d t s e the f ing pr b ems aximize
aximize
D x1 C 4x2 # x3
P D 3x1 C x2
sub ect to sub ect to
x1 C x2 " 6
x1 C x2 # x3 ! 5
#x1 C 2x2 ! 2
x1 C x2 C x3 " 3
x1 ; x2 ! 0
x1 # x2 C x3 D 7
aximize
x1 ; x2 ; x3 ! 0
D 3x1 C 4x2
sub ect to aximize
x1 C 2x2 "
x1 C 6x2 ! 12 D 3x1 # 2x2 C x3

x1 ; x2 ! 0 sub ect to
aximize x1 C x2 C x3 " 1
D x1 C 2x2 C 3x3
x1 # x2 C x3 ! 2
sub ect to
x1 C 2x2 C 2x3 " 6 x1 # x2 # x3 " #6
x1 # x2 # x3 ! 1 x1 ; x2 ; x3 ! 0
x1 ; x2 ; x3 ! 0
aximize
aximize
D x1 C 4x2
D x1 # x2 C 4x3
sub ect to sub ect to
x1 C x2 C x3 "
x1 C 2x2 "
x1 # 2x2 C x3 ! 6
x1 C 6x2 ! 12
x1 ; x2 ; x3 ! 0
x2 ! 2
aximize
x1 ; x2 ! 0
D 3x1 C 2x2 C x3
sub ect to aximize
x1 C x2 C x3 " 10
x1 # x2 # x3 D 6 P D #4x1 C 2x2

x1 ; x2 ; x3 ! 0 sub ect to

aximize 3x1 # 2x2 " 6

P D x1 C 2x2 C 3x3 #3x1 C 2x2 D 4
sub ect to x1 ! 2
x2 # 2x3 ! 5
x1 ; x2 ! 0
x1 C x2 C x3 D
x1 ; x2 ; x3 ! 0 aximize

aximize D 2x1 # x2
D x1 # 10x2
sub ect to
sub ect to
x1 # x2 " 1 x1 # 2x2 ! #12

x1 C 2x2 " #x1 C x2 ! 2

x1 C x2 ! 5 x1 C x2 ! 10

x1 ; x2 ! 0 x1 ; x2 ! 0
330 C near Pro ra n

Production A company manufactures two types of des s: achine A achine

standard and executive. Each type requires assembly and finishing
times as given in the following table: Product 1 hr 1 hr
Product 2 hr 1 hr
Assembly Finishing Profit
Time Time per Unit Product 2 hr 2 hr

Standard 1 hr 2 hr 40
The numbers of hours per wee that A and are available for
Executive 2 hr 3 hr 50 production are 40 and 30, respectively. The profit per unit on ,
, and is 50, 60, and 75, respectively. At least five units of
The profit on each unit is also indicated. The number of hours must be produced next wee . What should be the production order
available per wee in the assembly department is 200, and in the for that period if maximum profit is to be achieved What is the
finishing department it is 500. ecause of a union contract, the maximum profit
finishing department is guaranteed at least 300 hours of wor per
wee . How many units of each type should the company produce Investments The prospectus of an investment fund states
each wee to maximize profit that all money is invested in bonds that are rated A, AA, and
AAA no more than 30 of the total investment is in A and AA
Production A company manufactures three products: , bonds, and at least 50 is in AA and AAA bonds. The A, AA,
, and . Each product requires the use of time on machines A and AAA bonds yield , 7 , and 6 , respectively, annually.
and as given in the following table: etermine the percentages of the total investment that should be
committed to each type of bond so that the fund maximizes its
annual yield. What is this yield

Objective M
os o o to so e a So far we have used the simplex method to maximize ob ective functions. In general, to
n at on ro e a ter n t e
o ect e f nct on so t at a minimize a function it su ces to maximize the negative of the function. To understand
a at on ro e res ts why, consider the function f.x/ D x2 # 4. In Figure 7.17(a), observe that the mini-
mum value of f is #4, and it occurs when x D 0. Figure 7.17(b) shows the graph of
g.x/ D #f.x/ D #.x2 # 4/. This graph is the re ection through the x-axis of the graph
of f. Notice that the maximum value of g is 4 and occurs when x D 0. Thus, the mini-
mum value of x2 # 4 is the negative of the maximum value of #.x2 # 4/. That is,
min f D #max.#f/
Alternatively, thin of a point C on the positive half of the number line moving to the
left. As it does so, the point #C moves to the right. It is clear that if, for some reason, C
stops, then it stops at the minimum value that C encounters. If C stops, then so does #C,
at the maximum value encountered by #C. Since this value of #C is still the negative
of the value of C, we see that
min C D #max.#C/

y y

4 g(x) = -f (x)
= -(x 2 - 4)

x x

f(x) = x 2 - 4
-4

(a) (b)

FIGURE inimum value of f.x/ is equal to the negative of the maximum value of #f.x/.
 Section 7.5 n at on 331

The problem in Example 1 will be solved E AM LE M

more e ciently in Example 4 of
Section 7.6. Use the simplex method to minimize D x1 C 2x2 sub ect to

#2x1 C x2 ! 1
#x1 C x2 ! 2

x1 ; x2 ! 0

S To minimize , we can maximize # D #x1 #2x2 . Note that constraints (1)

and (2) each have the form a1 x1 Ca2 x2 ! b, where b ! 0. Thus, their equations involve
two surplus variables s1 and s2 , each with coe cient #1, and two artificial variables t1
and t2 , each with coe cient C1:

#2x1 C x2 # s1 C t1 D 1

#x1 C x2 # s2 C t2 D 2

Since there are t artificial variables, we maximize the ob ective function

D .# / # t1 # t2

where is a large positive number. Equivalently,

x1 C 2x2 C t1 C t2 C D0

The augmented coe cient matrix of Equations (3) (5) is

x1 x2 s1 s2 t1 t2
2 3
#2 1 #1 0 1 0 0 1
4 #1 1 0 #1 0 1 0 25
1 2 0 0 1 0
Proceeding, we obtain simplex tables I, II, and III:

SIMPLE ABLE I
entering
variable
#
x1 x2 s1 s2 t1 t2 R u tients
2 3
departing t1 #2 1 #1 0 1 0 0 1 1'1D1
variable t2 4 #1 1 0 #1 0 1 0 2 52'1D2
1C3 2#2 0 0 1 #3
„ ƒ‚ …
indicators

SIMPLE ABLE II
entering
variable
#
x1 x2 s1 s2 t1 t2 R u tients
2 3
x2 #2 1 #1 0 1 0 0 1
departing t2 4 1 0 1 #1 #1 1 0 1 51'1D1
variable 5# 0 2# #2 C 2 0 1 #2 #
„ ƒ‚ …
indicators
332 C near Pro ra n

SIMPLE ABLE III

x1 x2 s1 s2 # R
2 3
x2 #1 1 0 #1 0 2
s1 6
4 1 0 1 #1 0 1 7
5
# 3 0 0 2 1 #4
„ ƒ‚ …
indicators
From table II, it is evident that both t1 and t2 will be absent from the (basic vari-
ables) column in table III. Thus, they are both 0 and are no longer needed. Accordingly,
we have removed their columns from table III. oreover, with t1 D 0 D t2 , it follows
that D # and the -row and the -column can be relabelled as shown. Even
more, all the indicators are now nonnegative so that # is maximized when x1 D 0
and x2 D 2. It follows that has a minimum value of #.#4/ D 4, when x1 D 0 and
x2 D 2.
Now ork Problem 1 G

Here is an interesting example dealing E AM LE R E

with environmental controls.
A cement plant produces 2,500,000 barrels of cement per year. The ilns emit 2 g
of dust for each barrel produced. A governmental agency dealing with environmental
protection requires that the plant reduce its dust emissions to no more than 00,000 g
per year. There are two emission control devices available, A and . evice A reduces
emissions to 12 g per barrel, and its cost is 0.20 per barrel of cement produced. For
device , emissions are reduced to 15 g per barrel, and the cost is 0.25 per barrel
of cement produced. etermine the most economical course of action that the plant
should ta e so that it complies with the agency s requirement and also maintains its
annual production of exact y 2,500,000 barrels of cement.1

S We must minimize the annual cost of emission control. et x1 , x2 , and x3 be

the annual numbers of barrels of cement produced in ilns that use device A, device ,
and no device, respectively. Then x1 ; x2 ; x3 ! 0, and the annual emission control cost
(in dollars) is
1 1
CD x1 C x2 C 0x3
5 4
Since 2,500,000 barrels of cement are produced each year,
x1 C x2 C x3 D 2; 500; 000

1 This example is adapted from Robert E. ohn, A athematical odel for Air Pollution Control, Sch Science
and athematics 6 (1 6 ), 4 7 4.
 Section 7.5 n at on 333

The numbers of ilograms of dust emitted annually by the ilns that use device A,
device , and no device are 12 x1 , 15 x2 , and 2x3 , respectively. Since the total number of
ilograms of dust emission is to be no more than 00,000,
1 1
x1 C x2 C 2x3 " 00; 000
2 5
To minimize C sub ect to constraints (7) and ( ), where x1 ; x2 ; x3 ! 0, we first maximize
#C by using the simplex method. The equations to consider are
x1 C x2 C x3 C t1 D 2; 500; 000
and
1 1
x1 C x2 C 2x3 C s2 D 00; 000
2 5
where t1 and s2 are artificial and slac variables, respectively. The artificial ob ective
equation is D .#C/ # t1 , equivalently,
1 1
x1 C x2 C 0x3 C t1 C D 0
5 4
where is a large positive number. The augmented coe cient matrix of Equations
( ) (11) is
x1 x2 x3 s2 t1
2 3
1 1 1 0 1 0 2; 500; 000
6 1 1
2 1 0 0 00; 000 7
4 2 5 5
1 1
5 4
0 0 1 0
After determining the initial simplex table, we proceed and obtain (after three additional
simplex tables) the final table:
x1 x2 x3 s2 #C R
2 3
x2 0 1 #5 # 10
3
0 1; 500; 000
6 7
x1 6 1 0 6 10
0 1; 000; 000 7
4 3 5
1 1
#C 0 0 20
1 6
#575;000
„ ƒ‚ …
indicators
Note that was replaced by #C when t1 was no longer a basic variable. The final table
shows that the maximum value of #C is #575; 000 and occurs when x1 D 1; 000; 000,
x2 D 1; 500; 000, and x3 D 0. Thus, the minimum annual cost of the emission con-
trol is #.#575; 000/ D 575; 000. evice A should be installed on ilns producing
1,000,000 barrels of cement annually, and device should be installed on ilns pro-
ducing 1,500,000 barrels annually.
Now ork Problem 11 G
R BLEMS
se the simp ex meth d t s e the f ing pr b ems inimize
inimize C D x1 C 12x2
sub ect to
D 2x1 C 5x2 2x1 C 2x2 ! 1
x1 C 3x2 ! 2
sub ect to x1 ; x2 ! 0
inimize
x1 # x2 ! 7
D 12x1 C 6x2 C 3x3
2x1 C x2 ! sub ect to
x1 ; x2 ! 0 x1 # x2 # x3 ! 1
x1 ; x2 ; x3 ! 0
334 C near Pro ra n

inimize inimize
D x1 C x2 C 2x3 D 4x1 C 4x2 C 6x3
sub ect to sub ect to
x1 C 2x2 # x3 ! 4 x1 # x2 # x3 " 3
x1 ; x2 ; x3 ! 0 x1 # x2 C x3 ! 3
x1 ; x2 ; x3 ! 0

inimize Emission Control A cement plant produces 3,300,000

barrels of cement per year. The ilns emit 2 lb of dust for each
D 2x1 C 3x2 C x3 barrel produced. The plant must reduce its dust emissions to no
more than 1,000,000 lb per year. There are two devices available,
sub ect to
A and , that will control emissions. evice A will reduce
emissions to 12 lb per barrel, and the cost is 0.25 per barrel of
x1 C x2 C x3 " 6
cement produced. For device , emissions are reduced to 14 lb per
x1 # x3 " #4 barrel, and the cost is 0.40 per barrel of cement produced.
etermine the most economical course of action the plant should
x2 C x3 " 5
ta e so that it maintains an annual production of exactly
x1 ; x2 ; x3 ! 0 3,300,000 barrels of cement.

inimize Building Lots A developer can buy lots for 00,000 on Par
Place and 600,000 on Virginia Avenue. n each Par Place lot
D 5x1 C x2 C 3x3 she can build a two-star condominium building and on each
Virginia Avenue lot she can build a one-star condominium
sub ect to building. City Hall requires that her development have a total star
rating of at least 1 . City Hall also requires that her development
3x1 C x2 # x3 " 4 earn at least 27 civic improvement points. The developer will earn
three points for each lot on Virginia Avenue and one point for
2x1 C 2x3 " 5 each lot on Par Place. How many lots should the developer buy
x1 C x2 C x3 ! 2 on each of Par Place and Virginia Avenue to minimize her costs,
and what is her minimum cost
x1 ; x2 ; x3 ! 0
ransportation Costs A retailer has stores in Columbus
inimize and ayton and has warehouses in A ron and Springfield.
Each store requires delivery of exactly 150 V players.
C D #x1 C x2 C 3x3 In the A ron warehouse there are 200 V players, and in the
Springfield warehouse there are 150.
sub ect to

x1 C 2x2 C x3 D 4
x2 C x3 D 1
The transportation costs to ship V players from the
x1 C x2 " 6 warehouses to the stores are given in the following table:
x1 ; x2 ; x3 ! 0
Columbus ayton
inimize
A ron 5 7
D x1 # x2 Springfield 3 2
sub ect to
For example, the cost to ship a V player from A ron to the
Columbus store is 5. How should the retailer order the V
#x1 C x2 ! 4
players so that the requirements of the stores are met and the
x1 C x2 D 1 total transportation costs are minimized What is the minimum
transportation cost
x1 ; x2 ! 0
Parts Purchasing An auto manufacturer purchases
inimize alternators from two suppliers, and . The manufacturer has
D x1 C x2 C 5x3 two plants, A and , and requires delivery of exactly 7000
alternators to plant A and exactly 5000 to plant . Supplier
sub ect to charges 300 and 320 per alternator (including transportation
x1 C x2 C x3 ! cost) to A and , respectively. For these prices, requires that the
auto manufacturer order at least a total of 3000 alternators.
#x1 C 2x2 C x3 ! 2
However, can supply no more than 5000 alternators. Supplier
x1 ; x2 ; x3 ! 0 charges 340 and 2 0 per alternator to A and , respectively,
 Section 7.6 e a 335

and requires a minimum order of 7000 alternators. etermine how

the manufacturer should order the necessary alternators so that 48"
his total cost is a minimum. What is this minimum cost

15" 15" 15" 3"

FIGURE
Similarly, from a stoc roll, two 15-in.-wide rolls, one 10-in.-wide
roll, and one -in.-wide roll could be cut. Here the trim loss would
be in. The following table indicates the number of 15-in. and
10-in. rolls, together with trim loss, that can be cut from a
stoc roll:
Producing rapping Paper A paper company stoc s its !
holiday wrapping paper in 4 -in.-wide rolls, called stoc rolls, 15 in. 3 2 1
Roll width
and cuts such rolls into smaller widths, depending on customers 10 in. 0 1
orders. Suppose that an order for 50 rolls of 15-in.-wide paper and Trim loss 3
60 rolls of 10-in.-wide paper is received. From a stoc roll, the
company can cut three 15-in.-wide rolls and one 3-in.-wide roll.
(See Figure 7.1 .) Since the 3-in.-wide roll cannot be used in the a Complete the last two columns of the table. b Assume that
order, 3 in. is called the trim loss for this roll. the company has a su cient number of stoc rolls to fill the order
and that at east 50 rolls of 15-in.-wide and at east 60 rolls of
10-in.-wide wrapping paper will be cut. If x1 ; x2 ; x3 , and x4 are the
numbers of stoc rolls that are cut in a manner described by
columns 1 4 of the table, respectively, determine the values of
the x s so that the total trim loss is minimized. c What is the
minimum amount of total trim loss

Objective T
o rst ot ate an t en for a There is a fundamental principle, called dua ity that allows us to solve a maximization
e ne t e a of a near
ro ra n ro e problem by solving a related minimization problem. et us illustrate.

Table .
achine A achine Profit Unit

anual 1 hr 1 hr 10
Electric 2 hr 4 hr 24
Hours available 120 1 0

Suppose that a company produces two types of garden shears, manual and electric,
and each requires the use of machines A and in its production. Table 7.2 indicates
that manual shears require the use of A for 1 hour and for 1 hour. Electric shears
require A for 2 hours and for 4 hours. The maximum numbers of hours available
per month for machines A and are 120 and 1 0, respectively. The profit on manual
shears is 10, and on electric shears it is 24. Assuming that the company can sell all
the shears it can produce, we will determine the maximum monthly profit. If x1 and x2
are the numbers of manual and electric shears produced per month, respectively, then
we want to maximize the monthly profit function
P D 10x1 C 24x2
sub ect to
x1 C 2x2 " 120
x1 C 4x2 " 1 0
x1 ; x2 ! 0
336 C near Pro ra n

Writing Constraints (1) and (2) as equations, we have

x1 C 2x2 C s1 D 120
and
x1 C 4x2 C s2 D 1 0
where s1 and s2 are slac variables. In Equation (3), x1 C 2x2 is the number of hours
that machine A is used. Since 120 hours on A are available, s1 is the number of avail-
able hours that are n t used for production. That is, s1 represents unused capacity (in
hours) for A. Similarly, s2 represents unused capacity for . Solving this problem by
the simplex method, we find that the final table is
x1 x2 s1 s2 P R
2 3
x1 1 0 2 #1 0 60
x2 6
40 1 # 12 1
2
0 30 7
5
P 0 0 2 1 1320
„ ƒ‚ …
indicators
Thus, the maximum profit per month is 1320, which occurs when x1 D 60 and x2 D
30.
Now let us loo at the situation from a different point of view. Suppose that the
company wishes to rent out machines A and . What is the minimum monthly rental
fee they should charge Certainly, if the charge is too high, no one would rent the
machines. n the other hand, if the charge is too low, it may not pay the company to
rent them at all. bviously, the minimum rent should be 1320. That is, the minimum
the company should charge is the profit it could ma e by using the machines itself. We
can arrive at this minimum rental fee directly by solving a linear programming problem.
et F be the total monthly rental fee. To determine F, suppose the company assigns
values or worths to each hour of capacity on machines A and . et these worths be
y1 and y2 dollars, respectively, where y1 ; y2 ! 0. Then the monthly worth of machine
A is 120y1 , and for it is 1 0y2 . Thus,
F D 120y1 C 1 0y2
The total worth of machine time to produce a set of manual shears is 1y1 C 1y2 . This
should be at least equal to the 10 profit the company can earn by producing those
shears. If not, the company would ma e more money by using the machine time to
produce a set of manual shears. Accordingly,
1y1 C 1y2 ! 10
Similarly, the total worth of machine time to produce a set of electric shears should be
at least 24:
2y1 C 4y2 ! 24
Therefore, the company wants to
minimize F D 120y1 C 1 0y2
sub ect to
y1 C y2 ! 10
2y1 C 4y2 ! 24
y1 ; y2 ! 0
To minimize F, we maximize #F. Since constraints (6) and (7) have the form
a1 y1 C a2 y2 ! b, where b ! 0, we consider an artificial problem. If r1 and r2 are
surplus variables and t1 and t2 are artificial variables, then we want to maximize
D .#F/ # t1 # t2
 Section 7.6 e a 337

where is a large positive number, such that

y1 C y2 # r1 C t1 D 10

2y1 C 4y2 # r2 C t2 D 24
and the y s, r s, and t s are nonnegative. The final simplex table for this problem (with
the artificial variable columns deleted and changed to #F) is
y1 y2 r1 r2 #F R
2 1
3
y1 1 0 #2 2
0
6 7
y2 6 0 1 1 # 12 0 27
4 5
#F 0 0 60 30 1 #1320
„ ƒ‚ …
indicators
Since the maximum value of #F is #1320, the minimum value of F is
#.#1320/ D 1320 (as anticipated). It occurs when y1 D and y2 D 2. We have
therefore determined the optimum value of one linear programming problem (maxi-
mizing profit) by finding the optimum value of another problem (minimizing rental
fee).
The values y1 D and y2 D 2 could have been anticipated from the final table of
the maximization problem. In (5), the indicator in the s1 -column means that at the
optimum level of production, if s1 increases by one unit, then the profit P decreases
by . That is, 1 unused hour of capacity on A decreases the maximum profit by .
Thus, 1 hour of capacity on A is worth . We say that the shadow price of 1 hour of
capacity on A is . Now, recall that y1 in the rental problem is the worth of 1 hour
of capacity on A. Therefore, y1 must equal in the optimum solution for that prob-
lem. Similarly, since the indicator in the s2 -column is 2, the shadow price of 1 hour
of capacity on is 2, which is the value of y2 in the optimum solution of the rental
problem.
et us now analyze the structure of our two linear programming problems:
aximize inimize
P D 10x1 C 24x2 F D 120y1 C 1 0y2
sub ect to sub ect to
$ $
x1 C 2x2 " 120 y1 C y2 ! 10
x1 C 4x2 " 1 0 2y1 C 4y2 ! 24
and x1 ; x2 ! 0. and y1 ; y2 ! 0.

Note that in ( ) the inequalities are all ", but in ( ) they are all !. The coe cients of
the ob ective function in the minimization problem are the constant terms in ( ). The
constant terms in ( ) are the coe cients of the ob ective function of the maximization
problem. The coe cients of the y1 s in ( ) are the coe cients of x1 and x2 in the first
constraint of ( ) the coe cients of the y2 s in ( ) are the coe cients of x1 and x2
in the second constraint of ( ). The minimization problem is called the dua of the
maximization problem, and vice versa.
In general, with any given linear programming problem, we can associate another
linear programming problem called its dual. The given problem is called primal. If the
primal is a maximization problem, then its dual is a minimization problem. Similarly,
if the primal involves minimization, then the dual involves maximization.
Any primal maximization problem can be written in the form indicated in Table 7.3.
Note that there are no nonnegativity restrictions on the b s (nor any other restrictions).
Thus, if we have a maximization problem with an inequality constraint that involves
!, then multiplying both sides by #1 yields an inequality involving ". (With the for-
mulation we are n considering it is immaterial that multiplication by #1 may yield a
constant term that is negative.) oreover, if a constraint is an equality, it can be written
338 C near Pro ra n

in terms of two inequalities, one involving " and one involving !. This is a direct con-
sequence of the logical principle:
. D R/ if and only if . " R and ! R/
The corresponding dual minimization problem can be written in the form indicated
in Table 7.4. Similarly, any primal minimization problem can be put in the form of
Table 7.4, and its dual is the maximization problem in Table 7.3.

Table .
aximize D c1 x1 C c2 x2 C & & & C cn xn
sub ect to
9
a11 x1 C a12 x2 C & & & C a1n xn " b1 >
>
>
a21 x1 C a22 x2 C & & & C a2n xn " b2 >
>
>
=
& & & &
& & & & >
>
>
>
& & & & >
>
;
am1 x1 C am2 x2 C & & & C amn xn " bm
and x1 ; x2 ; : : : ; xn ! 0

Table .
inimize D b1 y1 C b2 y2 C & & & C bm ym
sub ect to
9
a11 y1 C a21 y2 C & & & C am1 ym ! c1 >
>
>
a12 y1 C a22 y2 C & & & C am2 ym ! c2 >
>
>
=
& & & &
& & & & >
>
>
>
& & & & >
>
;
a1n y1 C a2n y2 C & & & C amn ym ! cn

and y1 ; y2 ; : : : ; ym ! 0

et us compare the primal and its dual in Tables 7.3 and 7.4. For convenience,
when we refer to constraints, we will mean those in (10) or (11) we will not include
the nonnegativity conditions on the variables. bserve that if all the constraints in the
primal involve " .!/, then all the constraints in its dual involve ! ."/. The coe -
cients in the dual s ob ective function are the constant terms in the primal s constraints.
Similarly, the constant terms in the dual s constraints are the coe cients of the primal s
ob ective function. The coe cient matrix of the left sides of the dual s constraints is the
transp se of the coe cient matrix of the left sides of the primal s constraints. That is,
2 3 2 3T
a11 a21 & & & am1 a11 a21 & & & am1
6a12 a22 & & & am2 7 6 a21 a22 & & & a2n 7
6 7 6 7
6 & & & 7 6 & & & 7
6 & 7 D6
& 7
6 & & 7 6 & & 7
4 & & & 5 4 & & & 5
a1n a2n & & & amn am1 am2 & & & amn
If the primal involves n decision variables and m slac variables, then the dual
involves m decision variables and n slac variables. It should be noted that the dual of
the dua is the primal.
There is an important relationship between the primal and its dual:

If the primal has an optimum solution, then so does the dual, and the optimum value
of the primal s ob ective function is the same as that of its dual.
 Section 7.6 e a 339

oreover, suppose that the primal s ob ective function is

D c1 x1 C c2 x2 C & & & C cn xn
Then

If sj is the slac variable associated with the jth constraint in the dual, then the
indicator in the sj -column of the final simplex table of the dual is the value of xj in
the optimum solution of the primal.

Thus, we can solve the primal by merely solving its dual. At times this is more
convenient than solving the primal directly. The lin between the primal and the dual
can be expressed very succinctly using matrix notation. et
2 3
x1
6 x2 7
" # 6 7
6 & 7
C D c1 c2 & & & cn and X D 6 7
6 & 7
4 & 5
xn
Then the ob ective function of the primal problem can be written as
D CX
Furthermore, if we write
2 3 3 2
a11 a12 &&& a1n b1
6 a21 a22 &&& a2n 7 6 b2 7
6 7 6 7
6 & & & 7 6 & 7
AD6
& 7
and BD6 7
6 & & 7 6 & 7
4 & & & 5 4 & 5
am1 am2 &&& amn bm
then the system of constraints for the primal problem becomes
AX " B and X ! 0
where, as usual, we understand " .!/ between matrices of the same size to mean that
the inequality holds for each pair of corresponding entries. Now let
2 3
y1
6 y2 7
A L IT I 6 7
6 & 7
Find the dual of the following prob- D6 7
6 & 7
lem: Suppose that the What If Company 4 & 5
has 60,000 for the purchase of materi-
als to ma e three types of gadgets. The
ym
company has allocated a total of 2000 The dual problem has ob ective function given by
hours of assembly time and 120 hours
of pac aging time for the gadgets. The D BT
following table gives the cost per gad- and its system of constraints is
get, the number of hours per gadget, and
the profit per gadget for each type: A T ! CT and !0

E AM LE F M

Find the dual of the following:

aximize D 3x1 C 4x2 C 2x3
sub ect to
x1 C 2x2 C 0x3 " 10
2x1 C 2x2 C x3 " 10
and x1 ; x2 ; x3 ! 0
340 C near Pro ra n

S The primal is of the form of Table 7.3. Thus, the dual is

inimize D 10y1 C 10y2
sub ect to
y1 C 2y2 ! 3
2y1 C 2y2 ! 4
0y1 C y2 ! 2
and y1 ; y2 ! 0
Now ork Problem 1 G
E AM LE F M
A L IT I
Find the dual of the following prob- Find the dual of the following:
lem: A person decides to ta e two
different dietary supplements. Each sup- inimize D 4x1 C 3x2
plement contains two essential ingre-
dients, A and , for which there are sub ect to
minimum daily requirements, and each 3x1 # x2 ! 2
contains a third ingredient, C, that needs
to be minimized. x1 C x2 " 1
Supplement Supplement aily #4x1 C x2 " 3
1 2 Requirement

A 20 mg oz 6 mg oz mg and x1 ; x2 ! 0.
mg oz 16 mg oz 0 mg
C 6 mg oz 2 mg oz S Since the primal is a minimization problem, we want Constraints (13) and
(14) to involve !. (See Table 7.3.) ultiplying both sides of (13) and (14) by #1, we
get #x1 # x2 ! #1 and 4x1 # x2 ! #3. Thus, Constraints (12) (14) become
3x1 # x2 ! 2
#x1 # x2 ! #1
4x1 # x2 ! #3
The dual is

A L IT I aximize D 2y1 # y2 # 3y3

A company produces three inds sub ect to
of devices requiring three different pro-
duction procedures. The company has 3y1 # y2 C 4y3 " 4
allocated a total of 300 hours for proce- #y1 # y2 # y3 " 3
dure 1, 400 hours for procedure 2, and
600 hours for procedure 3. The follow- and y1 ; y2 ; y3 ! 0
ing table gives the number of hours per
device for each procedure:
Now ork Problem 3 G

E AM LE A S M

Use the dual and the simplex method to

aximize D 4x1 # x2 # x3
If the profit is 30 per device 1, 20 per
device 2, and 20 per device 3, then, sub ect to
using the dual and the simplex method,
3x1 C x2 # x3 " 4
find the number of devices of each ind
the company should produce to maxi- x1 C x2 C x3 " 2
mize profit.
and x1 ; x2 ; x3 ! 0.
 Section 7.6 e a 341

S The dual is
inimize D 4y1 C 2y2

sub ect to
3y1 C y2 ! 4
y1 C y2 ! #1
#y1 C y2 ! #1

and y1 ; y2 ! 0. To use the simplex method, we must get nonnegative constants in (16)
and (17). ultiplying both sides of these equations by #1 gives

#y1 # y2 " 1
y1 # y2 " 1

Since (15) involves !, an artificial variable is required. The equations corresponding

to (15), (1 ), and (1 ) are, respectively,

3y1 C y2 # s1 C t1 D 4
#y1 # y2 C s2 D1

and

y1 # y2 C s3 D1

where t1 is an artificial variable, s1 is a surplus variable, and s2 and s3 are slac variables.
To minimize , we maximize # . The artificial ob ective function is D .# /# t1 ,
where is a large positive number. After computations, we find that the final simplex
table is
y1 y2 s1 s2 s3 # R
2 3
y2 0 1 # 14 0 # 34 0 1
4
s2 6
60 0 # 12 1 # 12 0 5 7
7
6 2
7
y1641 0 # 14 0 1
4
0 5
4
7
5
3 1
# 0 0 2
0 2
1 # 11
2
„ ƒ‚ …
indicators
The maximum value of # is # 11 2
, so the minimum value of is 11
2
. Hence, the max-
11
imum value of is also 2 . Note that the indicators in the s1 -, s2 -, and s3 -columns are
3
2
, 0, and 12 , respectively. Thus, the maximum value of occurs when x1 D 32 , x2 D 0,
and x3 D 12 .
Now ork Problem 11 G
In Example 1 of Section 7.5 we used the simplex method to

inimize D x1 C 2x2
sub ect to
#2x1 C x2 ! 1
#x1 C x2 ! 2

This discussion, compared with that in and x1 ; x2 ! 0 The initial simplex table had 24 entries and involved two artificial
Example 1 of Section 7.5, shows that the variables. The table of the dual has only 1 entries, n arti cia ariab es, and is easier
dual problem may be easier to solve than to handle, as Example 4 will show. Thus, there may be a distinct advantage in solving
the primal problem.
the dual to determine the solution of the primal.
342 C near Pro ra n

E AM LE U S M

Use the dual and the simplex method to

inimize D x1 C 2x2
sub ect to
#2x1 C x2 ! 1
#x1 C x2 ! 2
and x1 ; x2 ! 0.
S The dual is
aximize D y1 C 2y2
sub ect to
#2y1 # y2 " 1
y1 C y2 " 2
and y1 ; y2 ! 0. The initial simplex table is table I:
SIMPLE ABLE I
entering
variable
#
y1 y2 s1 s2 R u tients
2 3
s1 #2 #1 1 0 0 1
departing s2 4 1 1 0 1 0 252'1D2
variable #1 #2 0 0 1 0
„ ƒ‚ …
indicators
Continuing, we get table II.
SIMPLE ABLE II
y1 y2 s1 s2 R
2 3
s1 #1 0 1 1 0 3
y24 1 1 0 1 0 25
1 0 0 2 1 4
„ ƒ‚ …
indicators
Since all indicators are nonnegative in table II, the maximum value of is 4. Hence,
the minimum value of is also 4. The indicators 0 and 2 in the s1 - and s2 -columns of
table II mean that the minimum value of occurs when x1 D 0 and x2 D 2.
Now ork Problem 9 G

R BLEMS
In Pr b ems nd the dua s n ts e aximize
aximize D 2x1 C x2 # x3
D 2x1 C 3x2 sub ect to
sub ect to 2x1 C 2x2 " 3
3x1 # x2 " 4 #x1 C 4x2 C 2x3 " 5
2x1 C 3x2 " 5 x1 ; x2 ; x3 ! 0
x1 ; x2 ! 0
 Section 7.6 e a 343

inimize In Pr b ems s e by using dua s and the simp ex meth d

D 2x1 # 3x2 C 5x3 inimize

sub ect to D 2x1 C 2x2 C 5x3

sub ect to
x1 # x2 C 2x3 ! 4
x1 # x2 C 2x3 ! 2
x1 C 4x2 # 3x3 ! #3
x1 ; x2 ; x3 ! 0 #x1 C 2x2 C x3 ! 3
x1 ; x2 ; x3 ! 0
inimize
inimize
D x1 C 12x2
sub ect to D 2x1 C 2x2

2x1 C 2x2 ! 1 sub ect to

x1 C 3x2 ! 2 x1 C 4x2 ! 2
x1 ; x2 ! 0 2x1 # x2 ! 2

aximize #3x1 C x2 ! 16

D x1 # x2 x1 ; x2 ! 0

sub ect to aximize

#x1 C 2x2 " 13 D 5x1 C 4x2

#x1 C x2 ! 3 sub ect to
x1 C x2 ! 11
2x1 C 3x2 " 6
x1 ; x2 ! 0
x1 C 4x2 " 10
aximize x1 ; x2 ! 0
D 2x1 C 5x2 # 2x3
aximize
sub ect to D 2x1 C 6x2
2x1 # 3x2 C x3 " 7 sub ect to
3x1 # 4x2 # x3 ! #1 3x1 C x2 " 12
x1 ; x2 ; x3 ! 0 x1 C x2 "
inimize x1 ; x2 ! 0

D 4x1 C 4x2 C 6x3 ; inimize

sub ect to D x1 C 6x2

sub ect to
x1 # x2 # x3 " 3;
2x1 C x2 ! 1
x1 # x2 C x3 ! 3;
#x1 # 3x2 " #27
x1 ; x2 ; x3 ! 0:
x1 ; x2 ! 0
inimize
inimize
D 2x1 C 3x2
D 2x1 C x2 C x3
sub ect to
sub ect to
#5x1 C 2x2 " #20 2x1 # x2 # x3 " 2
4x1 C 6x2 ! 15 #x1 # x2 C 2x3 ! 4
x1 ; x2 ! 0 x1 ; x2 ; x3 ! 0
344 C near Pro ra n

Advertising A firm is comparing the costs of advertising delivery truc s. The minimum needs are 12 units each of
in two media newspaper and radio. For every dollar s worth of refrigerated and nonrefrigerated space. Two standard types of
advertising, the following table gives the number of people, by truc s are available in the rental mar et. Type A has 2 units of
income group, reached by these media: refrigerated space and 1 unit of nonrefrigerated space. Type has
2 units of refrigerated space and 3 units of nonrefrigerated space.
The costs per mile are 0.40 for A and 0.60 for . Use the dual
Under ver
and the simplex method to find the minimum total cost per mile
40,000 40,000
and the number of each type of truc needed to attain it.
Newspaper 40 100 Labor Costs A company pays s illed and semis illed
Radio 50 25 wor ers in its assembly department 14 and per hour,
respectively. In the shipping department, shipping cler s are paid
per hour and shipping cler apprentices are paid 7.25
The firm wants to reach at least 0,000 persons earning under per hour. The company requires at least 0 wor ers in the
40,000 and at least 60,000 earning over 40,000. Use the dual assembly department and at least 60 in the shipping department.
and the simplex method to find the amounts that the firm should ecause of union agreements, at least twice as many semis illed
spend on newspaper and radio advertising so as to reach these wor ers must be employed as s illed wor ers. Also, at least twice
numbers of people at a minimum total advertising cost. What is as many shipping cler s must be employed as shipping cler
the minimum total advertising cost apprentices. Use the dual and the simplex method to find the
Delivery ruc Scheduling ecause of increased number of each type of wor er that the company must employ so
business, a catering service finds that it must rent additional that the total hourly wage paid to these employees is a minimum.
What is the minimum total hourly wage

Chapter 7 Review
I T S E
S Linear Inequalities in wo Variables
linear inequality open half-plane closed half-plane Ex. 2, p. 2 7
system of inequalities Ex. 3, p. 2
S Linear Programming
constraint linear function in x and y linear programming Ex. 1, p. 303
ob ective function feasible point nonnegativity conditions Ex. 1, p. 303
feasible region isoprofit line corner point multiple optimum solutions Ex. 1, p. 303
empty feasible region nonempty bounded unbounded feasible region Ex. 1, p. 303
S he Simplex Method
standard maximum linear programming problem Ex. 1, p. 314
slac variable decision variable basic feasible solution unbounded solution Ex. 1, p. 314
nonbasic variable basic variable simplex table ob ective row Ex. 2, p. 317
entering variable indicator departing variable pivot entry degeneracy Ex. 2, p. 317
S Artificial Variables
artificial problem artificial variable artificial ob ective function surplus variable Ex. 1, p. 323
S Minimi ation
min C D #max.#C/ Ex. 1, p. 331
S he Dual
shadow price dual primal Ex. 1, p. 33

S
The solution of a system of linear inequalities consists of all a nonempty feasible region is the corner-point method. The
points whose coordinates simultaneously satisfy all of the ob ective function is evaluated at each of the corner points of
inequalities. eometrically, it is the intersection of all of the the feasible region, and we choose a corner point at which
regions determined by the inequalities. the ob ective function is optimum.
inear programming involves maximizing or minimiz- For a problem involving more than two variables, the
ing a linear function (the ob ective function) sub ect to a corner-point method is either impractical or impossible.
system of constraints, which are linear inequalities or linear Instead, we use a matrix method called the simplex method,
equations. ne method for finding an optimum solution for which is e cient and completely mechanical.
 Chapter 7 e e 345

R
In Pr b ems s e the gi en inequa ity r system f inimize
inequa ities
D 2x C 3y
#3x C 2y > #6 5x # 2y C 10 ! 0
sub ect to
3x " #5 #x " #3
! ! xCy " 5
y # 3x < 6 x # 2y > 4
x # y > #3 xCy >1 2x C 5y " 10
! !
2x C y " 2 x>y 5x C y ! 20
#2x # y " 2 xCy<0 x; y ! 0
8 8
< 2x C y < #3 <2x # y > 5
inimize
3x # 2y > # x<3
: y!0 : y<7 D 2x C 2y
sub ect to
In Pr b ems d n t use the simp ex meth d
aximize xCy ! 4
#x C 3y " 1
D x # 2y
x"6
sub ect to x; y ! 0

y#x " 2 aximize

xCy " 4 D 4x C y
x"3 sub ect to
x; y ! 0 x C 2y "
3x C 2y " 12
aximize
x; y ! 0
D 3x C y
aximize
sub ect to
D 4x C y
2x C y " sub ect to
x"3 x C 2y ! 16
y!1 3x C 2y ! 24
x; y ! 0 x; y ! 0

inimize In Pr b ems use the simp ex meth d and p ssib y dua s

D 2x # y aximize

sub ect to D 2x1 C 3x2

x # y ! #2 sub ect to
xCy ! 1 x1 C 6x2 " 12
x # 2y " 2 x1 C 2x2 "
x; y ! 0 x1 ; x2 ! 0
aximize
aximize
D 1 x1 C 20x2
DxCy
sub ect to sub ect to
5x C 6y " 30 2x1 C 3x2 " 1
4x C 5y " 20 4x1 C 3x2 " 24
x # 2y ! 0 x2 " 5
x; y ! 0 x1 ; x2 ! 0
346 C near Pro ra n

inimize aximize

D 3x1 C 2x2 C x3 D 5x1 C 2x2 C 7x3

sub ect to sub ect to

4x1 # x2 " 2
x1 C 2x2 C 3x3 ! 5
# x1 C 2x2 C 5x3 " 2
x1 ; x2 ; x3 ! 0
x1 ; x2 ; x3 " 0
inimize inimize

D 3x1 C 5x2 D x1 C x2

sub ect to sub ect to

2x1 C 4x2 ! 2 x1 C x2 C 2x3 " 4
2x1 C 5x2 ! 4 x3 ! 1
x1 ; x2 ! 0 x1 ; x2 ; x3 ! 0

aximize In Pr b ems and s e by using dua s and the simp ex

meth d
D x1 C 2x2
inimize
sub ect to
D x1 C 2x2 C 3x3
x1 C x2 " 12
sub ect to
x1 C x2 ! 5
x1 " 10 x1 C x2 C x3 ! 4

x1 ; x2 ! 0 x1 C 2x2 C 3x3 ! 5
x1 ; x2 ; x3 ! 0
inimize
aximize
D 3x1 C x2
D x1 # 2x2
sub ect to
sub ect to
3x1 C 5x2 " 15
x1 C x2 ! 2 x1 # x2 " 3

x1 ; x2 ! 0 x1 C 2x2 " 4
4x1 C x2 ! 2
inimize
x1 ; x2 ! 0
D x1 C 2x2 C x3

sub ect to Production Order A company manufactures three

products: , , and . Each product requires the use of time on
x1 # x2 # x3 " #1 machines A and as given in the following table:
6x1 C 3x2 C 2x3 D 12
achine A achine
x1 ; x2 ; x3 ! 0
Product 1 hr 1 hr
aximize Product 2 hr 1 hr
Product 2 hr 2 hr
D 2x1 C 3x2 C 5x3

sub ect to
The numbers of hours per wee that A and are available for
x1 C x2 C 3x3 ! 5 production are 40 and 34, respectively. The profit per unit on ,
, and is 10, 15, and 22, respectively. What should be the
2x1 C x2 C 4x3 " 5 wee ly production order if maximum profit is to be obtained
x1 ; x2 ; x3 ! 0 What is the maximum profit
 Chapter 7 e e 347

Repeat Problem 31 assuming that the owner insists on a profit Diet Formulation A technician in a zoo must formulate a
of at least 300 per wee . diet from two commercial products, food A and food , for
Oil ransportation An oil company has storage facilities a certain group of animals. In 200 grams of food A there are
for heating fuel in cities A, , C, and . Cities C and are each 16 grams of fat, 32 grams of carbohydrate, and 4 grams of protein.
in need of exactly 500,000 gal of fuel. The company determines In 200 g of food there are grams of fat, 64 grams of
that A and can each sacrifice at most 600,000 gal to satisfy the carbohydrate, and 10 grams of protein. The minimum daily
needs of C and . The following table gives the costs per gallon to requirements are 176 grams of fat, 1024 grams of carbohydrate,
transport fuel between the cities: and 200 grams of protein. If food A costs cents per 100 grams
and food costs 22 cents per 100 grams, how many grams of each
food should be used to meet the minimum daily requirements at
To the least cost (Assume that a minimum cost exists.)
From C
In Pr b ems and d n t use the simp ex meth d R und
A 0.01 0.02 y ur ans ers t t decima p aces
0.02 0.04
inimize

D 4:2x # 2:1y
How should the company distribute the fuel in order to minimize
the total transportation cost What is the minimum transportation
cost sub ect to

Profit Imelda operates a home business selling two y " 3:4 C 1:2x
computer games: Space Raiders and reen iants. These
games are installed for Imelda by three friends, Nicolas, Harvey, y " #7:6 C 3:5x
and arl, each of whom must do some of the wor on the
installation of each game. The time that each must spend on each y " 1 :7 # 0:6x
game is given in the following table:
x; y ! 0

Nicolas Harvey arl aximize

Space Raiders 30 min 20 min 10 min D 12:4x C :3y

reen iants 10 min 10 min 50 min
sub ect to

Imelda s friends have other wor to do, but they each find that 1:4x C 1:7y " 15:
each wee they can spend at most 300 minutes wor ing on
#3:6x C 2:6y " #10:7
Imelda s games. Imelda ma es a profit of 10 on each sale of
Space Raiders and 15 on each sale of reen iants. How many #1:3x C 4:3y " #5:2
of each game should Imelda try to sell each wee to maximize
profit, and what is this maximum profit x; y ! 0
 ntro ct on
to Pro a t
an tat st cs
T
he term pr babi ity is familiar to most of us. It is not uncommon to hear
8.1 as c Co nt n Pr nc e
an Per tat ons such phrases as the probability of precipitation , the probability of ood-
ing , and the probability of receiving an A in a course . oosely spea ing,
8.2 Co nat ons an t er probability refers to a number that indicates the degree of li elihood that
Co nt n Pr nc es some future event will have a particular outcome. For example, before tossing a well-
8.3 a e aces an balanced coin, one does not now with certainty whether the outcome will be a head
ents or a tail. However, no doubt one considers these outcomes to be be equally li ely to
occur. This means that if the coin is tossed a large number of times, one expects that
8.4 Pro a t approximately half of the tosses will give heads. Thus, we say that the probability of a
8.5 Con t ona Pro a t head occurring on any toss is 12 D 50 .
an toc ast c Processes The study of probability forms the basis of the study of statistics. In statistics, we
8.6 n e en ent ents are concerned about ma ing an inference that is, a prediction or decision about a
population (a large set of ob ects under consideration) by using a sample of data drawn
8.7 a es or a from that population. For example, by drawing a sample of units from an assembly
line, we can statistically ma e an inference about a the units in a production run.
C er 8 e e
However, in the study of probability, we wor with a nown population and consider
the li elihood (or probability) of drawing a particular sample from it. For example,
if we deal five cards from a dec , we may be interested in the probability that it is a
pair , meaning that it contains two (but not three) cards of the same denomination.
The probability of a pair is
the number of possible pairs
the number of possible five card hands
Evidently, we need to be able to handle certain computations of the form the number
of : : : . Such computations can be more subtle than one might at first imagine. They
are called counting problems, and we begin our study of probability with them.
odern probability theory began with a very practica problem. If a game between
two gamblers is interrupted, the player who is ahead surely has a right to more than half
the pot of money being contested but not to all of it. How should the pot be divided
This problem was unsolved in 1654, when the Chevalier de r shared it with his
friend, the French mathematician and philosopher laise Pascal (1623 1662). Pascal,
in correspondence with Pierre de Fermat, solved this problem and we describe their
solution in Example of Section .4.

348
 Section 8. as c Co nt n Pr nc e an Per tat ons 349

Objective B C
o e e o an a a as c
Co nt n Pr nc e an to e ten t to
er tat ons B C
ater on, we will find that computing a probability may require us to calculate the num-
ber of elements in a set. ecause counting the elements individually may be extremely
tedious (or even prohibitive), we spend some time developing e cient counting tech-
niques. We begin by motivating the Basic Counting Principle, which is useful in solv-
ing a wide variety of problems.
Suppose a manufacturer wants to produce coffee brewers in 2-, -, and 10-cup
capacities, with each capacity available in colors of white, beige, red, and green. How
many types of brewers must the manufacturer produce To answer the question, it is
not necessary that we count the capacity color pairs one by one (such as 2-white and
-beige). Since there are three capacities, and for each capacity there are four colors,
the number of types is the product, 3 ! 4 D 12. We can systematically list the different
types by using the tree diagram of Figure .1. From the starting point, there are three
branches that indicate the possible capacities. From each of these branches are four
more branches that indicate the possible colors. This tree determines 12 paths, each
beginning at the starting point and ending at a tip. Each path determines a different
type of coffee brewer. We refer to the diagram as being a t e e tree: There is a level
for capacity and a level for color.
We can consider the listing of the types of coffee brewers as a two-stage procedure.
In the first stage we indicate a capacity and in the second a color. The number of types
of coffee brewers is the number of ways the first stage can occur (3), times the number
of ways the second stage can occur (4), which yields 3 ! 4 D 12. Suppose further that
the manufacturer decides to ma e all of the models available with a timer option that
allows the consumer to awa e with freshly brewed coffee. Assuming that this really is
an option, so that a coffee brewer either comes with a timer or without a timer, counting
the number of types of brewers now becomes a three-stage procedure. There are now
3 ! 4 ! 2 D 24 types of brewer.

Capacity Color Type

White 2, w
Beige 2, b
2
Red 2, r

Green 2, g
White 8, w

Beige 8, b
Start 8
Red 8, r

Green 8, g

White 10, w

Beige 10, b
10
Red 10, r

Green 10, g

Level 1 Level 2

FIGURE Two-level tree diagram for types of coffee brewers.

350 C ntro ct on to Pro a t an tat st cs

This multiplication procedure can be generalized into a asic Counting Principle:

B C
Suppose that a procedure involves a sequence of k stages. et n1 be the number
of ways the first can occur and n2 be the number of ways the second can occur.
Continuing in this way, let nk be the number of ways the kth stage can occur. Then
the total number of different ways the procedure can occur is
n1 ! n2 ! : : : ! nk

E AM LE T R

Two roads connect cities A and , four connect and C, and five connect C and .
(See Figure .2.) To drive from A, to , to C, and then to city , how many different
routes are possible
S Here we have a three-stage procedure. The first (A ! ) has two possi-
bilities, the second ( ! C) has four, and the third (C ! ) has five. y the asic
Counting Principle, the total number of routes is 2 ! 4 ! 5 D 40.
Now ork Problem 1 G

A
B

C D

FIGURE Roads connecting cities A, , C, .

E AM LE C T R

When a coin is tossed, a head (H) or a tail (T) may show. If a die is rolled, a 1, 2, 3, 4,
5, or 6 may show. Suppose a coin is tossed twice and then a die is rolled, and the result
is noted (such as H on first toss, T on second, and 4 on roll of die). How many different
results can occur
S Tossing a coin twice and then rolling a die can be considered a three-stage
procedure. Each of the first two stages (the coin toss) has two possible outcomes. The
third stage (rolling the die) has six possible outcomes. y the asic Counting Principle,
the number of different results for the procedure is
2 ! 2 ! 6 D 24

Now ork Problem 3 G

E AM LE A

In how many different ways can a quiz be answered under each of the following con-
ditions
a The quiz consists of three multiple-choice questions with four choices for each.
S Successively answering the three questions is a three-stage procedure. The
first question can be answered in any of four ways. i ewise, each of the other two
questions can be answered in four ways. y the asic Counting Principle, the number
of ways to answer the quiz is
4 ! 4 ! 4 D 43 D 64
 Section 8. as c Co nt n Pr nc e an Per tat ons 351

b The quiz consists of three multiple-choice questions (with four choices for each)
and five true false questions.
S Answering the quiz can be considered a two-stage procedure. First we can
answer the multiple-choice questions (the first stage), and then we can answer the
true false questions (the second stage). From part (a), the three multiple-choice ques-
tions can be answered in 4 ! 4 ! 4 ways. Each of the true false questions has two
choices ( true or false ), so the total number of ways of answering all five of them is
2 ! 2 ! 2 ! 2 ! 2. y the asic Counting Principle, the number of ways the entire quiz can
be answered is

.4 ! 4 ! 4/ .2 ! 2 ! 2 ! 2 ! 2/ D 43 ! 25 D 204
„ ƒ‚ … „ ƒ‚ …
multiple true false
choice

Now ork Problem 5 G

E AM LE L A

From the five letters A, , C, , and E, how many three-letter horizontal arrangements
(called words ) are possible if no letter can be repeated (A word need not ma e
sense.) For example, E and E are two acceptable words, but CAC is not.
S To form a word, we must successively fill the positions with
different letters. Thus, we have a three-stage procedure. For the first position, we can
choose any of the five letters. After filling that position with some letter, we can fill the
second position with any of the remaining four letters. After that position is filled, the
third position can be filled with any of the three letters that have not yet been used. y
If repetitions are allowed, the number of
words is 5 ! 5 ! 5 D 125. the asic Counting Principle, the total number of three-letter words is

5 ! 4 ! 3 D 60

Now ork Problem 7 G

In Example 4, we selected three different letters from five letters and arranged them in
an rder. Each result is called a permutati n f e etters taken three at a time. ore
generally, we have the following definition.

An ordered selection of r ob ects, without repetition, ta en from n distinct ob ects is

called a permutation f n bjects taken r at a time The number of such permutations
is denoted n Pr .

Thus, in Example 4, we found that

5 P3 D 5 ! 4 ! 3 D 60

y a similar analysis, we will now find a general formula for n Pr . In ma ing an ordered
arrangement of r ob ects from n ob ects, for the first position we may choose any one
of the n ob ects. (See Figure .3.) After the first position is filled, there remain n " 1
ob ects that may be chosen for the second position. After that position is filled, there
are n " 2 ob ects that may be chosen for the third position. Continuing in this way and
using the asic Counting Principle, we arrive at the following formula:
352 C ntro ct on to Pro a t an tat st cs

r objects

, , , ,

n n-1 n-2 n-r+1

Choices Choices Choices Choices

FIGURE An ordered arrangement of r ob ects selected from n ob ects.

The number of permutations of n ob ects ta en r at a time is given by

n Pr D n.n " 1/.n " 2/ ! ! ! .n " r C 1/
„ ƒ‚ …
r factors

The formula for n Pr can be expressed in terms of factorials. ultiplying the right side
See Section 2.2 for the definition of
fact ria . of Equation (1) by
.n " r/.n " r " 1/ ! ! ! .2/.1/
.n " r/.n " r " 1/ ! ! ! .2/.1/
gives
n.n " 1/.n " 2/ ! ! ! .n " r C 1/ ! .n " r/.n " r " 1/ ! ! ! .2/.1/
n Pr D
.n " r/.n " r " 1/ ! ! ! .2/.1/
The numerator is simply n , and the denominator is .n"r/Š. Thus, we have the following
result:

The number of permutations of n ob ects ta en r at a time is given by

nŠ
n Pr D
.n " r/Š

For example, from Equation (2), we have

7Š 7Š 7!6!5!4!3!2!1
7 P3 D D D D 210
.7 " 3/Š 4Š 4!3!2!1
any calculators can directly This calculation can be obtained easily with a calculator by using the factorial ey.
calculate n Pr . Alternatively, we can write
7Š 7 ! 6 ! 5 ! 4Š
D D 7 ! 6 ! 5 D 210
4Š 4Š
Notice how 7 was written so that the 4 s would cancel.

E AM LE C

A club has 20 members. The o ces of president, vice president, secretary, and treasurer
are to be filled, and no member may serve in more than one o ce. How many different
slates of candidates are possible
S We will consider a slate in the order of president, vice president, secretary,
and treasurer. Each ordering of four members constitutes a slate, so the number of
possible slates is 20 P4 . y Equation (1),

20 P4 D 20 ! 1 ! 1 ! 17 D 116,2 0
 Section 8. as c Co nt n Pr nc e an Per tat ons 353

Calculations with factorials tend to Alternatively, using Equation (2) gives

produce very large numbers. To avoid
over ow on a calculator, it is frequently 20Š 20Š 20 ! 1 ! 1 ! 17 ! 16Š
important to do some cance ati n before 20 P4 D D D
ma ing entries. .20 " 4/Š 16Š 16Š
D 20 ! 1 ! 1 ! 17 D 116,2 0

Note the large number of possible slates

Now ork Problem 11 G
E AM LE

A politician sends a questionnaire to her constituents to determine their concerns about

six important national issues: unemployment, the environment, taxes, interest rates,
national defense, and social security. A respondent is to select four issues of personal
concern and ran them by placing the number 1, 2, 3, or 4 after each issue to indicate
the degree of concern, with 1 indicating the greatest concern and 4 the least. In how
many ways can a respondent reply to the questionnaire
S A respondent is to ran four of the six issues. Thus, we can consider a reply
as an ordered arrangement of six items ta en four at a time, where the first item is the
issue with ran 1, the second is the issue with ran 2, and so on. Hence, we have a
permutation problem, and the number of possible replies is
6Š 6Š 6 ! 5 ! 4 ! 3 ! 2Š
6 P4 D D D D 6 ! 5 ! 4 ! 3 D 360
.6 " 4/Š 2Š 2Š

Now ork Problem 21 G

In case you want to find the number of permutations of n ob ects ta en all at a time,
setting r D n in Equation (2) gives
y definition, 0Š D 1. nŠ nŠ nŠ
n Pn D D D D nŠ
.n " n/Š 0Š 1
Each of these permutations is simply called a permutation of n ob ects.

The number of permutations of n ob ects is n .

For example, the number of permutations of the letters in the word SET is is 3Š D 6.
These permutations are

SET STE EST ETS TES TSE

E AM LE N L F

awyers Smith, ones, acobs, and ell want to form a legal firm and will name it by
using all four of their last names. How many possible names are there
S Since order is important, we must find the number of permutations of four
names, which is

4Š D 4 ! 3 ! 2 ! 1 D 24

Thus, there are 24 possible names for the firm.

Now ork Problem 19 G
354 C ntro ct on to Pro a t an tat st cs

R BLEMS
Production Process In a production process, a product goes code.) How many Canadian postal codes are possible What
through one of the assembly lines A, , or C and then goes percentage of these begin with 5W What percentage end with
through one of the finishing lines or E. raw a tree diagram that 1E6
indicates the possible production routes for a unit of the product. In Pr b ems determine the a ues
How many production routes are possible
6 P3 5 P1 6 P6
Air Conditioner Models A manufacturer produces air
P5
conditioners having 6000-, 000-, and 10,000- TU capacities. P4 7 P4 !4 P2
Each capacity is available with one- or two-speed fans. raw a P4
tree diagram that represents all types of models. How many types Compute 1000 without using a calculator. Now try it
are there with your calculator, using the factorial feature.
n Pr
etermine .
nŠ
In Pr b ems use any appr priate c unting meth d
ame of Firm Flynn, Peters, and Walters are forming an
advertising firm and agree to name it by their three last names.
How many names for the firm are possible
Softball If a softball league has six teams, how many
different end-of-the-season ran ings are possible Assume that
there are no ties.
Dice Rolls A red die is rolled, and then a green die is rolled.
raw a tree diagram to indicate the possible results. How many Contest In how many ways can a udge award first, second,
results are possible and third prizes in a contest having eight contestants
Coin oss A coin is tossed four times. raw a tree diagram Matching ype Exam n a history exam, each of six
to indicate the possible results. How many results are possible items in one column is to be matched with exactly one of eight
items in another column. No item in the second column can be
In Pr b ems use the Basic C unting Princip e
selected more than once. In how many ways can the matching
Course Selection A student must ta e a mathematics course, be done
a philosophy course, and a physics course. The available
Die Roll A die (with six faces) is rolled four times and the
mathematics classes are category theory, measure theory, real
outcome of each roll is noted. How many results are possible
analysis, and combinatorics. The philosophy possibilities are
logical positivism, epistemology, and modal logic. The available Coin oss A coin is tossed eight times. How many results
physics courses are classical mechanics, electricity and are possible if the order of the tosses is considered
magnetism, quantum mechanics, general relativity, and
Problem Assignment In a mathematics class with 12
cosmology. How many three-course selections can the student
students, the instructor wants homewor problems 1, 3, 5, and 7
ma e
put on the board by four different students. In how many ways can
Auto Routes A person lives in city A and commutes by the instructor assign the problems
automobile to city . There are five roads connecting A and .
a How many routes are possible for a round trip b How many Combination Loc A combination loc has 26 different
round-trip routes are possible if a different road is to be used for letters, and a sequence of three different letters must be selected
the return trip for the loc to open. How many combinations are possible
Dinner Choices At a restaurant, a complete dinner consists Student Questionnaire A university issues a questionnaire
of an appetizer, an entree, a dessert, and a beverage. The choices whereby each student must ran the four items with which he or
for the appetizer are soup and salad for the entree, the choices are she is most dissatisfied. The items are
chic en, fish, stea , and lamb for the dessert, the choices are tuition fees professors
cherries ubilee, fresh peach cobbler, chocolate tru e ca e, and par ing fees cafeteria food
blueberry roly-poly for the beverage, the choices are coffee, tea, dormitory rooms class sizes
and mil . How many complete dinners are possible
Multiple Choice Exam In how many ways is it possible The ran ing is to be indicated by the numbers 1, 2, 3 and 4, where
to answer a six-question multiple-choice examination if each 1 indicates the item involving the greatest dissatisfaction and 4 the
question has four choices (and one choice is selected for each least. In how many ways can a student answer the questionnaire
question) Die Roll A die is rolled three times. How many results are
rue False Exam In how many ways is it possible to possible if the order of the rolls is considered and the second roll
answer a 10-question true false examination produces a number less than 3
Canadian Postal Codes A Canadian postal code consists Letter Arrangements How many six-letter words from the
of a string of six characters, of which three are letters and three letters in the word EA W are possible if no letter is repeated
are digits, which begins with a letter and for which each letter is Letter Arrangements Using the letters in the word
followed by a (single) digit. (For readability, the string is bro en RE IT, how many four-letter words are possible if no letter is
into strings of three. For example, 5W 1E6 is a valid postal repeated
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 355

Boo Arrangements In how many ways can five of seven Club O cers A club has 12 members.
boo s be arranged on a boo shelf In how many ways can all a In how many ways can the o ces of president, vice president,
seven boo s be arranged on the shelf secretary, and treasurer be filled if no member can serve in more
Lecture all A lecture hall has five doors. In how many than one o ce
ways can a student enter the hall by one door and b In how many ways can the four o ces be filled if the
a Exit by a different door president and vice president must be different members
b Exit by any door Fraternity ames Suppose a fraternity is named by three
Po er and A po er hand consists of 5 cards drawn from a ree letters. (There are 24 letters in the ree alphabet.)
dec of 52 playing cards. The hand is said to be four of a ind if a How many names are possible
four of the cards have the same face value. For example, hands b How many names are possible if no letter can be used more
with four 10 s or four ac s or four 2 s are four-of-a- ind hands. than one time
How many such hands are possible
Bas etball In how many ways can a bas etball coach
assign positions to her five-member team if two of the members
are qualified for the center position and all five are qualified for all
the other positions
Car ames A European car manufacturer has three series
of cars A, S, and R in sizes 3, 4, 5, 6, 7, and . Each car it ma es
potentially comes in a omfort ( ), Progressiv (P), or a Techni
(T) trim pac age, with either an automatic (A) or a manual ( )
transmission, and a 2-, 3-, or 5-litre engine. The manufacturer
names its products with a string of letters and digits as suggested
by these attributes in the order given. For example, the
manufacturer spea s of the R4TA5. How many models can the
manufacturer name using these criteria
Merchandise Choice In a merchandise catalog, a C rac Baseball A baseball manager determines that, of his nine
is available in the colors of blac , red, yellow, gray, and blue. team members, three are strong hitters and six are wea . If the
When placing an order for one C rac , customers must indicate manager wants the strong hitters to be the first three batters in a
their first and second color choices. In how many ways can this batting order, how many batting orders are possible
be done Signal Flags When at least one of four ags colored red,
Fast Food Order Four students go to a pizzeria and order a green, yellow, and blue is arranged vertically on a agpole, the
margharita, a diavalo, a ree , and a meat-lover s one item for result indicates a signal (or message). ifferent arrangements give
each student. When the waiter brings the food to the table, she different signals.
forgets which student ordered which item and simply places one a How many different signals are possible if all four ags are
before each student. In how many ways can she do this used
Group Photograph In how many ways can three men and b How many different signals are possible if at least one ag is
two women line up for a group picture In how many ways can used
they line up if a woman is to be at each end

Objective C C
o sc ss co nat ons
er tat ons t re eate o ects C
an ass n ents to ce s
We continue our discussion of counting methods by considering the following. In a
20-member club the o ces of president, vice president, secretary, and treasurer are to
be filled, and no member may serve in more than one o ce. If these o ces, in the order
given, are filled by members A, , C, and , respectively, then we can represent this
slate by
A C
A different slate is
AC
These two slates represent different permutations of 20 members ta en four at a time.
Now, as a different situation, let us consider four-person c mmittees that can be formed
from the 20 members. In that case, the two arrangements
A C and AC
356 C ntro ct on to Pro a t an tat st cs

represent the same committee. Here the rder f isting the members is f n c ncern.
These two arrangements are considered to give the same c mbinati n of A, , C, and .
The classical definition of combination follows:

The important phrase here is ith ut

regard t rder. A selection of r elements, without regard to order and without repetition, selected
from n distinct elements, is called a c mbinati n f n e ements taken r at a time. The
number of such combinations is denoted n Cr , which can be read n choose r .

Another wording may be helpful. If we start with a set which has n elements and
select r elements from it, then we have an r-element subset of the original set. A c m
binati n f n e ements taken r at a time is precisely an r-element subset of the original
n-element set. It follows that n Cr is precisely the number of r-element subsets of an
n-element set.

E AM LE C C

ist all combinations and all permutations of the four letters

A; ; C; and

when they are ta en three at a time.

S The combinations are

A C A AC C

There are four combinations, so 4 C3 D 4. The permutations are

A C A AC C
AC A A C C
AC A CA C
CA A C A C
CA A AC C
C A A CA C

There are 24 permutations.

Now ork Problem 1 G
Again, we could say that the original set is fA, ,C, g and, as stressed in
Section 0.1, a set is determined by its e ements and neither rearrangements n r repe
titi ns in a isting a ect the set, so fA, ,C, g D f ,A, ,Cg, to give ust one possible
rearrangement. It is easy to see that the 3-element subsets of our 4-element set arise
by removal of one of the four elements and these are fA, ,Cg, fA, , g, fA,C, g, and
f ,C, g, classically nown as the combinations of the the four elements in fA, ,C, g
ta en three at a time. y contrast, note that each of the four combinations has six per-
mutations associated with it. The display above shows these permutations in the column
directly below the combination in question. In fact, it is better to display the combina-
tions as

fA, ,Cg fA, , g fA,C, g f ,C, g

because the parentheses tell us that rearrangements (and repetitions) are not considered.
The columnar display in Example 1 leads us to a formula for n Cr . For it shows
us that listing all the permutations accounted for by n Pr can be regarded as a two-step
procedure: First, list all the combinations accounted for by n Cr second, rearrange the
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 357

elements in each combination. For each r-element combination, there are rŠ ways to
rearrange its elements. Thus, by the asic Counting Principle of Section .1,

n Pr Dn Cr ! rŠ

Solving for n Cr gives

nŠ
n Pr .n " r/Š nŠ
n Cr D D D
rŠ rŠ rŠ.n " r/Š

The number of combinations of n ob ects ta en r at a time is given by

nŠ
n Cr D
any calculators can directly rŠ.n " r/Š
compute n Cr .

E AM LE C S

If a club has 20 members, how many different four-member committees are possible
S rder is not important because, no matter how the members of a commit-
tee are arranged, we have the same committee. Thus, we simply have to compute the
number of combinations of 20 ob ects ta en four at a time, 20 C4 :
20Š 20Š
20 C4 D D
4Š.20 " 4/Š 4Š16Š
bserve how 20Š was written so that the 20 ! 1 ! 1 ! 17 ! 16Š
D D 4 45
16Š s would cancel. 4 ! 3 ! 2 ! 1 ! 16Š
There are 4 45 possible committees.
Now ork Problem 9 G
It is important to remember that if a selection of ob ects is made and rder is
imp rtant, then permutati ns should be considered. If rder is n t imp rtant, consider
c mbinati ns. A ey aid to memory is that n Pr is the number of executive slates with r
ran s that can be chosen from n people, while n Cr is the number of committees with
r members that can be chosen from n people. An executive slate can be thought of as
a committee in which every individual has been ran ed. There are rŠ ways to ran the
members of a committee with r members. Thus, if we thin of of forming an executive
slate as a two-stage procedure, then, using the asic Counting Principle of Section .1,
we again get

n Pr D n Cr ! rŠ

E AM LE H

A p ker hand consists of 5 cards dealt from an ordinary dec of 52 cards. How many
different po er hands are there
S ne possible hand is

2 of hearts; 3 of diamonds; 6 of clubs;

4 of spades; ing of hearts

which we can abbreviate as

2H 3 6C 4S H
358 C ntro ct on to Pro a t an tat st cs

The order in which the cards are dealt does not matter, so this hand is the same as
H 4S 6C 3 2H
Thus, the number of possible hands is the number of ways that 5 ob ects can be selected
from 52, without regard to order. This is a combination problem. We have
52Š 52Š
52 C5 D D
5Š.52 " 5/Š 5Š47Š
52 ! 51 ! 50 ! 4 ! 4 ! 47Š
D
5 ! 4 ! 3 ! 2 ! 1 ! 47Š
52 ! 51 ! 50 ! 4 ! 4
D D 2,5 , 60
5!4!3!2
Now ork Problem 11 G
E AM LE M S C

A college promotion committee consists of five members. In how many ways can the
committee reach a ma ority decision in favor of a promotion

S A favorable ma ority decision is reached if, and only if,

exactly three members vote favorably,
or exactly four members vote favorably,
or all five members vote favorably
To determine the total number of ways to reach a favorable ma ority decision, we
add the number of ways that each of the preceding votes can occur.

S Suppose exactly three members vote favorably. The order of the members
is of no concern, and thus we can thin of these members as forming a combination.
Hence, the number of ways three of the five members can vote favorably is 5 C3 . Sim-
ilarly, the number of ways exactly four can vote favorably is 5 C4 , and the number of
ways all five can vote favorably is 5 C5 (which, of course, is 1). Thus, the number of
ways to reach a ma ority decision in favor of a promotion is
5Š 5Š 5Š
5 C3 C 5 C4 C 5 C5 D C C
3Š.5 " 3/Š 4Š.5 " 4/Š 5Š.5 " 5/Š
5Š 5Š 5Š
D C C
3Š2Š 4Š1Š 5Š0Š
5 ! 4 ! 3Š 5 ! 4Š
D C C1
3Š ! 2 ! 1 4Š ! 1
D 10 C 5 C 1 D 16
Now ork Problem 15 G

C S
The previous example leads naturally to some properties of combinations that are useful
in the study of probability. For example, we will show that

5 C0 C 5 C1 C 5 C2 C 5 C3 C 5 C4 C 5 C5 D 25

and, for any nonnegative integer n,

n C0 C n C1 C ! ! ! C n Cn!1 C n Cn D 2n
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 359

We can build on the last equation of Example 4 to verify the first of these equations:

5 C0 C 5 C1 C 5 C2 C 5 C3 C 5 C4 C 5 C5 D 5 C0 C 5 C1 C 5 C2 C 16
5Š 5Š 5Š
D C C C 16
0Š.5 " 0/Š 1Š.5 " 1/Š 2Š.5 " 2/Š
5Š 5Š 5Š
D C C C 16
0Š5Š 1Š4Š 2Š3Š
5 ! 4Š 5 ! 4 ! 3Š
D1C C C 16
4Š 2 ! 3Š
D 1 C 5 C 10 C 16
D 32
D 25

However, this calculation is not illuminating and would be impractical if we were to

adapt it for values of n much larger than 5.
Until this chapter we have we have primarily loo ed at sets in the context of sets
of numbers. However, in this chapter we have already seen that a combination is ust a
subset and the number n in n Cr , say, is ust the number of things under consideration.
In examples in the study of probability we will see a lot more of this. We often loo at
things li e sets of playing cards, sets of die rolls, sets of ordered pairs of dice rolls, and
sets of ways of doing things . Typically these sets are finite. If a set S has n elements,
we can, in principle, list its elements. For example, we might write

S D fs1 ; s2 ; : : : ; sn g

We recall, from Section 0.1, but now in more detail:

A subset E of S is a set with the property that e ery e ement f E is a s an e ement

f S. When this is the case we write E # S. Formally,
E # S if and only if, for all x; if x is an element of E then x is an element of S:

For any set S, we always have ; # S and S # S. If a set S has n elements, then any
subset of S has r elements, where 0 $ r $ n. The empty set, ;, is the only subset of
S that has 0 elements. The whole set, S, is the only subset of S that has n elements. A
general subset of S, containing r elements, where 0 $ r $ n is exactly a combination
of n ob ects ta en r at a time and the number of such combinations denoted n Cr is the
number f r e ement subsets f an n e ement set.
For any set S, we can form the set of a subsets of S. It is called the p er set of S
and sometimes denoted 2S . We claim that if S has n elements, then 2S has 2n elements.
This is easy to see. If

S D fs1 ; s2 ; ! ! ! ; sn g

then specification of a subset E of S can be thought of as a procedure involving n stages.

The first stage is to as , Is s1 an element of E the second stage is to as , Is s2 an
element of E . We continue to as such questions until we come to the nth stage
the last stage Is sn an element of E . bserve that each of these questions can be
answered in exactly two ways namely, yes or no. According to the asic Counting
Principle of Section .1, the total number of ways that specification of a subset of S can
occur is

! : : : ! …2 D 2n
2„ ! 2 ƒ‚
n factors
360 C ntro ct on to Pro a t an tat st cs

It follows that there are 2n subsets of an n-element set. It is convenient to write #.S/ for
the number of elements of set S. Thus, we have
#.2S / D 2#.S /
If #.S/ D n, then for each E in 2S , we have #.E/ D r, for some r satisfying 0 $ r $ n.
For each such r, let us write Sr for the subset of 2S consisting of all those elements E
with n.E/ D r. Thus, Sr is the set of all r-element subsets of the n-element set S. From
our observations in the last paragraph it follows that
#.Sr / D n Cr
Now we claim that
#.S0 / C #.S1 / C ! ! ! C #.Sn!1 / C #.Sn / D #.2S /
since every element E of 2S is in exact y one of the sets Sr . Substituting Equation (3),
for each 0 $ r $ n, and Equation (2) in Equation (4), we have Equation (1).

E AM LE AB C I

Establish the identity

n Cr C n CrC1 D nC1 CrC1

nŠ
S We can calculate using n Cr D :
rŠ.n " r/Š
nŠ nŠ
n Cr C n CrC1 D C
rŠ.n " r/Š .r C 1/Š.n " r " 1/Š
.r C 1/nŠ C .n " r/nŠ
D
.r C 1/Š.n " r/Š
..r C 1/ C .n " r//nŠ
D
.r C 1/Š.n " r/Š
.n C 1/nŠ
D
.r C 1/Š..n C 1/ " .r C 1//Š
.n C 1/Š
D
.r C 1/Š..n C 1/ " .r C 1//Š
D nC1 CrC1
S We can reason using the idea that n Cr is the number of r-element subsets
of an n-element set. et S be an n-element set that does not contain s" as an element.
Then S [ fs" g is an .n C 1/-element set. Now the .r C 1/-element subsets of S [ fs" g
are dis ointly of two inds:
those that contain s" as an element
those that do not contain s" as an element.
et us write S" for the .r C 1/-element subsets of S [ fs" g that contain s" and S for the
.r C 1/-element subsets of S [ fs" g which do not contain s" . Then,

nC1 CrC1 D #.S" / C #.S/

because every r C 1-element subset of S [ fs" g is in exactly one of S" or S. Now the
.r C 1/-element subsets of S [ fs" g that contain s" are in one-to-one correspondence
with the r-element subsets of S, so we have
#.S" / D n Cr
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 361

n the other hand, .r C 1/-element subsets of S [ fs" g that do not contain s" are in
one-to-one correspondence with the .r C 1/-element subsets of S, so
#.S/ D n CrC1
Assembling the last three displayed equations gives
nC1 CrC1 D n Cr C n CrC1
as required.
G
The first solution is good computational practice, but the second solution is illus-
trative of ideas and arguments that are often useful in the study of probability. The
identity we have ust established together with
n C0 D 1 D n Cn
for all n, allows us to generate Pasca s riang e
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
!
!
!
ou should convince yourself that the .r C 1/th entry in the .n C 1/th row of Pascal s
Triangle is n Cr .

R
In Section .1, we discussed permutations of ob ects that were all different. Now we
examine the case where some of the ob ects are ali e (or repeated). For example, con-
sider determining the number of different permutations of the seven letters in the word
SUCCESS
Here the letters C and S are repeated. If the two C s were interchanged, the resulting
permutation would be indistinguishable from SUCCESS. Thus, the number of distinct
permutations is not 7 , as it would be with 7 different ob ects. To determine the number
of distinct permutations, we use an approach that involves combinations.

S’s U C’s E S’s U C’s E

, , , , 2, 3, 6 1 5, 7 4

(a) (b)

FIGURE Permutations with repeated ob ects.

Figure .4(a) shows boxes representing the different letters in the word
SUCCESS. In these boxes we place the integers from 1 through 7. We place three
integers in the S s box (because there are three S s), one in the U box, two in the C s
box, and one in the E box. A typical placement is indicated in Figure .4(b). That place-
ment can be thought of as indicating a permutation of the seven letters in SUCCESS
namely, the permutation in which (going from left to right) the S s are in the second,
third, and sixth positions the U is in the first position and so on. Thus, Figure .4(b)
corresponds to the permutation
USSECSC
362 C ntro ct on to Pro a t an tat st cs

To count the number of distinct permutations, it su ces to determine the number

of ways the integers from 1 to 7 can be placed in the boxes. Since the order in which
they are placed into a box is not important, the S s box can be filled in 7 C3 ways. Then
the U box can be filled with one of the remaining four integers in 4 C1 ways. Then the
C s box can be filled with two of the remaining three integers in 3 C2 ways. Finally, the
E box can be filled with one of the remaining one integers in 1 C1 ways. Since we have
a four-stage procedure, by the asic Counting Principle the total number of ways to fill
the boxes or, equivalently, the number of distinguishable permutations of the letters in
SUCCESS is
7Š 4Š 3Š 1Š
7 C3 ! 4 C1 ! 3 C2 ! 1 C1 D ! ! !
3Š4Š 1Š3Š 2Š1Š 1Š0Š
7Š
D
3Š1Š2Š1Š
D 420

In summary, the word SUCCESS has four types of letters: S, U, C, and E. There are
three S s, one U, two C s, and one E, and the number of distinguishable permutations
of the seven letters is
7Š
3Š1Š2Š1Š
bserving the forms of the numerator and denominator, we can ma e the following
generalization:

R
The number of distinguishable permutations of n ob ects such that n1 are of
one type, n2 are of a second type, : : : , and nk are of a kth type, where
n1 C n2 C ! ! ! C nk D n, is
nŠ
n1 Šn2 Š ! ! ! nk Š

In problems of this ind there are often a number of different solutions to the same
problem. A solution that seems straightforward to one person may seem complicated
to another. Accordingly, we present another solution to the problem of counting the
number, , of different permutations of the letters of

SUCCESS

We will begin by tagging the letters so that they become distinguishable, thus
obtaining
S1 U1 C1 C2 E1 S2 S3

iving a permutation of these seven different letters can be described as a multistage

procedure. We can begin by permuting as if we can t see the subscripts, and by defini-
tion there are ways to accomplish this tas . For each of these ways, there are 3Š ways
to permute the three S s, for each of these, 1Š way to permute the one U, for each of
these, 2Š ways to permute the two C s, and for each of these, 1Š way to permute the one
E. According to the asic Counting Principle of Section .1, there are

! 3Š ! 1Š ! 2Š ! 1Š

ways to permute the seven different letters of S1 U1 C1 C2 E1 S2 S3 . n the other hand,

we already now that there are

7Š
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 363

permutations of seven different letters, so we must have

! 3Š ! 1Š ! 2Š ! 1Š D 7Š

From this we find

7Š
D
3Š1Š2Š1Š
in agreement with our earlier finding.

E AM LE L A R

For each of the following words, how many distinguishable permutations of the letters
are possible

a AP
S The word AP has six letters with repetition. We have one A, one P,
two s, and two s. Using Equation (5), we find that the number of permutations is
6Š
D1 0
1Š1Š2Š2Š
b ER
S None of the four letters in ER is repeated, so the number of permuta-
tions is

4 P4 D 4Š D 24

Now ork Problem 17 G

E AM LE N L F

A group of four lawyers, Smith, ones, Smith, and ell (the Smiths are cousins), want
to form a legal firm and will name it by using all of their last names. How many possible
names exist
S Each different permutation of the last four names is a name for the firm.
There are two Smiths, one ones, and one ell. From Equation (5), the number of
distinguishable names is
4Š
D 12
2Š1Š1Š

Now ork Problem 19 G

C
A B
At times, we want to find the number of ways in which ob ects can be placed into
compartments , or cells. For example, suppose that from a group of five people, three
2, 3, 5 1, 4 are to be assigned to room A and two to room . In how many ways can this be done
Figure .5 shows one such assignment, where the numbers 1; 2; : : : ; 5 represent the
FIGURE Assignment of people. bviously, the order in which people are placed into the rooms is of no concern.
people to rooms. The boxes (or cells) remind us of those in Figure .4(b), and, by an analysis similar to
the discussion of permutations with repeated ob ects, the number of ways to assign the
people is
5Š 5 ! 4 ! 3Š
D D 10
3Š2Š 3Š2Š
364 C ntro ct on to Pro a t an tat st cs

In general, we have the following principle:

A C
Suppose n distinct ob ects are assigned to k ordered cells with ni ob ects in cell
i.i D 1; 2; : : : ; k/, and the order in which the ob ects are assigned to cell i is of no
concern. The number of all such assignment is
nŠ
n1 Šn2 Š ! ! ! nk Š
where n1 C n2 C ! ! ! C nk D n.

We could reason: there are n1 Cn2 C###Cnk Cn1 ways to choose n1 ob ects to put in the
first cell, and for each of these ways there are n2 Cn3 C###Cnk Cn2 ways to choose n2 ob ects
to put in the second cell, and so on, giving, by the asic Counting Principle of Section
.1,

.n1 Cn2 C###Cnk Cn1 /.n2 Cn3 C###Cnk Cn2 / : : : .nk!1 Cnk Cnk!1 /.nk Cnk /
.n1 C n2 C ! ! ! C nk /Š .n2 C n3 C ! ! ! C nk /Š .nk!1 C nk /Š nk Š
D ! !!! !
n1 Š.n2 C n3 C ! ! ! C nk /Š n2 Š.n3 C n4 C ! ! ! C nk /Š nk!1 Šnk Š nk Š0Š
.n1 C n2 C ! ! ! C nk /Š
D
n1 Šn2 Š ! ! ! nk Š
which is the number in (6).

E AM LE A

A coach must assign 15 players to three vehicles to transport them to an out-of-town

game: 6 in a van, 5 in a station wagon, and 4 in an SUV. In how many ways can this
be done
S Here 15 people are placed into three cells (vehicles): 6 in cell 1, 5 in cell 2,
and 4 in cell 3. y Equation (2), the number of ways this can be done is
15Š
D 630,630
6Š5Š4Š
Now ork Problem 23 G
Example will show three different approaches to a counting problem. As we have
said, many counting problems have alternative methods of solution.

E AM LE A E

An artist has created 20 original paintings, and she will exhibit some of them in three
galleries. Four paintings will be sent to gallery A, four to gallery B, and three to gallery
C. In how many ways can this be done
S
et o The artist must send 4 C 4 C 3 D 11 paintings to the galleries, and the
that are not sent can be thought of as staying in her studio. Thus, we can thin of this
situation as placing 20 paintings into four cells:

4 in gallery A
4 in gallery
3 in gallery C
in the artist s studio
 Section 8.2 Co nat ons an t er Co nt n Pr nc es 365

From Equation (6), the number of ways this can be done is

20Š
D 1, 3 , 3 ,000
4Š4Š3Š Š
et o We can handle the problem in terms of a two-stage procedure and use
the asic Counting Principle. First, 11 paintings are selected for exhibit. Then, these
are split into three groups (cells) corresponding to the three galleries. We proceed as
follows.
Selecting 11 of the 20 paintings for exhibit (order is of no concern) can be done in
20 C11 ways. nce a selection is made, four of the paintings go into one cell (gallery A),
four go to a second cell (gallery ), and three go to a third cell (gallery C). y
11Š
Equation (6), this can be done in ways. Applying the asic Counting Princi-
4Š4Š3Š
ple gives the number of ways the artist can send the paintings to the galleries:
11Š 20Š 11Š
20 C11 ! D ! D 1, 3 , 3 ,000
4Š4Š3Š 11Š Š 4Š4Š3Š
et o Another approach to this problem is in terms of a three-stage procedure.
First, 4 of the 20 paintings are selected for shipment to gallery A. This can be done
in 20 C4 ways. Then, from the remaining 16 paintings, the number of ways 4 can be
selected for gallery is 16 C4 . Finally, the number of ways 3 can be sent to gallery C
from the 12 paintings that have not yet been selected is 12 C3 . y the asic Counting
Principle, the entire procedure can be done in
20Š 16Š 12Š 20Š
20 C4 !16 C4 !12 C3 D ! ! D
4Š16Š 4Š12Š 3Š Š 4Š4Š3Š Š
ways, which gives the previous answer, as expected
Now ork Problem 27 G

R BLEMS
In Pr b ems determine the a ues national chains through which the drin s are distributed, the
producer uses three colors of cardboard bands that hold the drin s
7 C4 6 C2 100 C100
together. How many different 3-pa s are possible
1;000;001 C1 5 P3 ! 4 C2 5 P2 ! 6 C4
Scoring on Exam In a 10-question examination, each
Verify that n Cr D n Cn!r . etermine n Cn . question is worth 10 points and is graded right or wrong.
Considering the individual questions, in how many ways can
Committee In how many ways can a five-member a student score 0 or better
committee be formed from a group of 1 people
eam Results A sports team plays 13 games. In how many
orse Race In a horse race, a horse is said to nish in ways can the outcomes of the games result in three wins, eight
the m ney if it finishes in first, second, or third place. For an losses, and two ties
eight-horse race, in how many ways can the horses finish in the Letter Arrangements How many distinguishable
money Assume no ties. arrangements of all the letters in the word ISSISSAU A are
possible
Math Exam n a 12-question mathematics examination, a
student must answer any questions. In how many ways can the Letter Arrangements How many distinguishable
questions be chosen (without regard to order) arrangements of all the letters in the word STREETS R are
possible
Cards From a dec of 52 playing cards, how many 4-card
hands are comprised solely of red cards Coin oss If a coin is tossed six times and the outcome of
each toss is noted, in how many ways can two heads and four tails
Quality Control A quality-control technician must select a occur
sample of 10 dresses from a production lot of 74 couture dresses. Die Roll A die is rolled six times and the order of the rolls
How many different samples are possible Express your answer in is considered. In how many ways can two 2 s, three 3 s, and one 4
terms of factorials. occur
Pac aging An energy drin producer ma es five types of
Scheduling Patients A physician s secretary must
energy drin s. The producer pac ages 3-pa s containing three
schedule eight o ce consultations. In how many ways can this be
drin s, no two of which are of the same type. To re ect the three
done
366 C ntro ct on to Pro a t an tat st cs

Baseball A ittle eague baseball team has 12 members orld Series A baseball team wins the World Series
and must play an away game. Three cars will be used for if it is the first team in the series to win four games. Thus, a series
transportation. In how many ways can the manager assign the could range from four to seven games. For example, a team
members to specific cars if each car can accommodate four winning the first four games would be the champion. i ewise, a
members team losing the first three games and winning the last four would
be champion. In how many ways can a team win the World
Pro ect Assignment The director of research and
Series
development for a company has nine scientists who are equally
qualified to wor on pro ects A, , and C. In how many ways can Subcommittee A committee has seven members, three
the director assign three scientists to each pro ect of whom are male and four female. In how many ways can a
subcommittee be selected if it is to consist exactly of
Identical Siblings A set of identical quadruplets, a set of
identical triplets, and three sets of identical twins pose for a group a three males
photograph. In how many ways can these 13 individuals line up in b four females
ways that are distinguishable in a photograph c two males and two females

rue False Exam A biology instructor includes several Subcommittee A committee has three male and five
true false questions on quizzes. From experience, a student female members. In how many ways can a subcommittee of four
believes that half of the questions are true and half are false. If be selected if at least two females are to serve on it
there are 10 true false questions on the next quiz, in how many Po er and A po er hand consists of 5 cards from a dec
ways can the student answer half of them true and the other of 52 playing cards. The hand is a full house if there are 3 cards
half false of one denomination and 2 cards of another. For example, three
Food Order A waiter ta es the following order from a 10 s and two ac s form a full house. How many full-house hands
lunch counter with nine people: three baconburgers, two are possible
veggieburgers, two tofuburgers, and two por belly delights. Upon Euchre and In euchre, only the denominations
returning with the food, he forgets who ordered what item and , 10, , , , and A from a standard 52-card dec are used.
simply places an item in front of each person. In how many ways A euchre hand consists of 5 cards from this reduced dec .
can the waiter do this a How many possible euchre hands are there b How many
Casewor er Assignment A social services o ce has euchre hands contain exactly four cards of the same suit
15 new clients. The supervisor wants to assign 5 clients to each c How many euchre hands contain exactly three cards of the
of three specific casewor ers. In how many ways can this be same suit
done
oc ey There are 11 members on a hoc ey team, and all
but one, the goalie, are qualified for the other five positions. In
how many ways can the coach form a starting lineup
Large Families arge families give rise to an enormous
number of re ati nships within the family that ma e growing up
with many siblings qualitatively different from life in smaller
families. Any two siblings within any family will have a
relationship of some sort that affects the life of the whole family.
In larger families, any three siblings or any four siblings will tend
to have a three-way or four-way relationship, respectively, that
affects the dynamics of the family, too. anet raunstein is third in
a family of 12 siblings: Claire, arbie, anet, Paul, lenn, ar ,
artha, aura, ulia, Carrie, Emily, and im. If we define a sibling
relationship to be any subset of the set of siblings of size greater ram Loading At a tourist attraction, two trams carry
than or equal to two, how many sibling relationships are there in sightseers up a picturesque mountain. ne tram can
anet s family How many sibling relationships are there in a accommodate seven people and the other eight. A busload of 1
family of three siblings tourists arrives, and both trams are at the bottom of the mountain.
bviously, only 15 tourists can initially go up the mountain. In
iring A company personnel director must hire six
how many ways can the attendant load 15 tourists onto the two
people: four for the assembly department and two for the shipping
trams
department. There are 10 applicants who are equally qualified to
wor in each department. In how many ways can the personnel Discussion Groups A history instructor wants to split a
director fill the positions class of 10 students into three discussion groups. ne group will
consist of four students and discuss topic A. The second and third
Financial Portfolio A financial advisor wants to create a
groups will discuss topics and C, respectively, and consist of
portfolio consisting of nine stoc s and five bonds. If ten stoc s
three students each.
and twelve bonds are acceptable for the portfolio, in how many
ways can the portfolio be created a In how many ways can the instructor form the groups
b If the instructor designates a group leader and a secretary
(different students) for each group, in how many ways can the
class be split
 Section 8.3 a e aces an ents 367

Objective S S E
o eter ne a sa e s ace an to
cons er e ents assoc ate t t o
re resent a sa e s ace an e ents S S
eans of a enn a ra o
ntro ce t e not ons of co e ent Inherent in any discussion of probability is the performance of an experiment (a proce-
n on an ntersect on dure) in which a particular result, or utc me involves chance. For example, consider
the experiment of tossing a coin. There are only two ways the coin can fall, a head (H)
or a tail (T), but the actual outcome is determined by chance. (We assume that the coin
does not land on its edge.) The set of possible outcomes,
fH, Tg
is called a samp e space for the experiment, and H and T are called samp e p ints.

A sample space S for an experiment is the set of all possible outcomes of the exper-
iment. The elements of S are called sample points. If there is a finite number of
sample points, that number is denoted #.S/, and S is said to be a finite sample
space.

When determining possible outcomes of an experiment, we must be sure that

The order in which sample points are they re ect the situation about which we are concerned. For example, consider the
listed in a sample space is of no concern. experiment of rolling a die and observing the top face. We could say that a sample
space is
S1 D f1; 2; 3; 4; 5; 6g
where the possible outcomes are the number of dots on the top face. However, other
possible outcomes are
odd number of dots appear (odd)
and even number of dots appear (even)
Thus, the set
S2 D fodd, eveng
is also a sample space for the experiment, so an experiment can have more than one
sample space.
If an outcome in S1 occurred, then we now which outcome in S2 occurred, but
the reverse is not true. To describe this asymmetry, we say that S1 is a m re primiti e
sample space than S2 . Usually, the more primitive a sample space is, the more questions
pertinent to the experiment it allows us to answer. For example, with S1 , we can answer
such questions as
id a 3 occur
id a number greater than 2 occur
id a number less than 4 occur
ut with S2 , we cannot answer these questions. As a rule of thumb, the more primitive
a sample space is, the more elements it has and the more detail it indicates. Unless
otherwise stated, when an experiment has more than one sample space, it will be our
practice to consider only a sample space that gives su cient detail to answer all perti-
nent questions relative to the experiment. For example, for the experiment of rolling a
die and observing the top face, it will be tacitly understood that we are observing the
number of dots. Thus, we will consider the sample space to be
S1 D f1; 2; 3; 4; 5; 6g
and will refer to it as the usua sample space for the experiment.
368 C ntro ct on to Pro a t an tat st cs

First Second Third Sample

First Second Sample
toss toss toss point
coin coin point

H HH H HHH
H
T HHT
H H
H HTH
T HT T
T HTT
Start Start
H THH
H TH H
T THT
T
T H TTH
T
T TT T TTT

FIGURE Tree diagram for toss of two coins. FIGURE Tree diagram for three tosses of a coin.

E AM LE S S T T C

Two different coins are tossed, and the result ( or ) for each coin is observed. eter-
mine a sample space.
S ne possible outcome is a head on the first coin and a head on the second,
which we can indicate by the ordered pair (H, H) or, more simply, HH. Similarly, we
indicate a head on the first coin and a tail on the second by HT, and so on. A sample
space is

S D fHH, HT, TH, TTg

A tree diagram is given in Figure .6 which illustrates further structure of this sample
space. We remar that S is also a sample space for the experiment of tossing a single
coin twice in succession. In fact, these two experiments can be considered one and the
same. Although other sample spaces can be contemplated, we ta e S to be the usua
sample space for these experiments.
Now ork Problem 3 G
E AM LE S S T T C
A L IT I
In 2016, on arch 26, Net ix had A coin is tossed three times, and the result of each toss is observed. escribe a sample
11 7 TV shows in its US catalogue. A space and determine the number of sample points.
viewer wanted to select two shows. How
many random choices did she have S ecause there are three tosses, we choose a sample point to be an ordered
trip e such as HHT, where each component is either H or T. y the asic Counting
Principle, the total number of sample points is 2 ! 2 ! 2 D . A sample space (the usua
one) is

S D fHHH, HHT, HTH, HTT, THH, THT, TTH, TTTg

and a tree diagram appears in Figure .7. Note that it is not necessary to list the entire
sample space to determine the number of sample points in it.
Now ork Problem 9 G
E AM LE S S B B

A bag contains four elly beans: one red, one pin , one blac , and one white. (See
Figure . .)
a A elly bean is withdrawn at random, its color is noted, and it is put bac in the
FIGURE Four colored bag. Then a elly bean is again randomly withdrawn and its color noted. escribe a
elly beans in a bag. sample space and determine the number of sample points.
 Section 8.3 a e aces an ents 369

S In this experiment we say that the two elly beans are withdrawn ith
rep acement. et R, P, , and W denote withdrawing a red, pin , blac , and white
elly bean, respectively. Then, our sample space consists of the sample points RW, P ,
R , WW, and so on, where (for example) RW represents the outcome that the first elly
bean withdrawn is red and the second is white. There are four possibilities for the first
withdrawal and, since that elly bean is placed bac in the bag, four possibilities for the
second withdrawal. y the asic Counting Principle, the number of sample points is
4 ! 4 D 16.
b etermine the number of sample points in the sample space if two elly beans are
selected in succession ith ut rep acement and the colors are noted.
S The first elly bean drawn can have any of four colors. Since it is n t returned
to the bag, the second elly bean drawn can have any of the three remaining colors.
Thus, the number of sample points is 4 ! 3 D 12. Alternatively, there are 4 P2 D 12
sample points.
Now ork Problem 7 G
E AM LE S S H

From an ordinary dec of 52 playing cards, a po er hand is dealt. escribe a sample

space and determine the number of sample points.
S A sample space consists of all combinations of 52 cards ta en 5 at a time.
From Example 3 of Section .2, the number of sample points is 52 C5 D 2,5 , 60.
Now ork Problem 13 G

E AM LE S S R T

A pair of dice is rolled once, and for each die, the number that turns up is observed.
escribe a sample space.
S Thin of the dice as being distinguishable, as if one were red and the other
green. Each die can turn up in six ways, so we can ta e a sample point to be an ordered
pair in which each component is an integer between 1 and 6, inclusive. For example,
(4, 6), (3, 2), and (2, 3) are three different sample points. y the asic Counting Prin-
ciple, the number of sample points is 6 ! 6, or 36.
Now ork Problem 11 G
E
At times, we are concerned with the outcomes of an experiment that satisfy a particular
condition. For example, we may be interested in whether the outcome of rolling a single
die is an even number. This condition can be considered as the set of outcomes f2; 4; 6g,
which is a subset of the sample space
S D f1; 2; 3; 4; 5; 6g
In general, any subset of a sample space is called an event for the experiment. Thus,
f2; 4; 6g
is the event that an even number turns up, which can also be described by
fx in Sjx is an even numberg
Note that although an event is a set, it may be possible to describe it verbally, as we ust
did. We often denote an event by E. When several events are involved in a discussion,
they may be denoted by E, F, G, , and so on, or by E1 ; E2 ; E3 , and so on.
370 C ntro ct on to Pro a t an tat st cs

An e ent E for an experiment is a subset of the sample space for the experiment. If
the outcome of the experiment is a sample point in E, then event E is said to ccur.

In the previous experiment of rolling a die, we saw that f2; 4; 6g is an event. Thus,
if the outcome is a 2, that event occurs. Some other events are
E D f1; 3; 5g D fx in Sjx is an odd numberg

S
F D f3; 4; 5; 6g D fx in Sjx % 3g
G D f1g
A sample space is a subset of itself, so it, too, is an event, called the certain event
E it must occur no matter what the outcome. An event, such as f1g, that consists of a
single sample point is called a simple event. We can also consider an event such as
fx in Sjx D 7g, which can be verbalized as 7 occurs . This event contains no sample
points, so it is the empty set ; (the set with no elements in it). In fact, ; is called the
impossible event, because it can never occur.
FIGURE Venn diagram for
sample space S and event E.
E AM LE E
S A coin is tossed three times, and the result of each toss is noted. The usual sample space
(from Example 2) is
fHHH, HHT, HTH, HTT, THH, THT, TTH, TTTg
E¿ E
etermine the following events.
a E D fone head and two tailsg.

E¿ is the shaded region S E D fHTT,THT,TTHg

FIGURE Venn diagram for b F D fat least two headsg.
the complement of E.
S F D fHHH, HHT, HTH, THHg
S
c G D fall headsg.
E F
S G D fHHHg
d I D fhead on first tossg.

S I D fHHH, HHT, HTH, HTTg

E F, union of E and F Now ork Problem 15 G

(a)
Sometimes it is convenient to represent a sample space S and an event E by a
S
Venn diagram, as in Figure . . The region inside the rectangle represents the sample
E F
points in S. (The sample points are not specifically shown.) The sample points in E
are represented by the points inside the circle. ecause E is a subset of S, the circular
E F
region cannot extend outside the rectangle.
With Venn diagrams, it is easy to see how events for an experiment can be used to
form other events. Figure .10 shows sample space S and event E. The shaded region
inside the rectangle, but outside the circle, represents the set of all sample points in S
that are not in E. This set is an event called the c mp ement f E and is denoted by E0 .
E F, intersection of E and F
Figure .11(a) shows two events, E and F. The shaded region represents the set of all
(b)
sample points either in E, or in F, or in both E and F. This set is an event called the uni n
FIGURE Representation of E and F and is denoted by E [ F. The shaded region in Figure .11(b) represents
of E [ F and E \ F. the event consisting of all sample points that are common to both E and F. This event
 Section 8.3 a e aces an ents 371

is called the intersecti n of E and F and is denoted by E \ F. In summary, we have the

following definitions.

et S be a sample space for an experiment with events E and F. The complement

of E, denoted by E0 , is the event consisting of all sample points in S that are not in
E. The union of E and F, denoted by E [ F, is the event consisting of all sample
points that are either in E, or in F, or in both E and F. The intersection of E and F,
denoted by E \ F, is the event consisting of all sample points that are common to
both E and F.

Note that if a sample point is in the event E [ F, then the point is in at least one of
the sets E and F. Thus, for the event E [ F to occur, at east ne of the events E and
F must occur, and conversely. n the other hand, if event E \ F occurs, then b th E
and F must occur, and conversely. If event E0 occurs, then E d es n t occur, and
conversely.

E AM LE C U I

iven the usual sample space

S D f1; 2; 3; 4; 5; 6g

for the rolling of a die, let E, F, and G be the events

E D f1; 3; 5g F D f3; 4; 5; 6g G D f1g

etermine each of the following events.

a E0
S Event E0 consists of those sample points in S that are not in E, so

E0 D f2; 4; 6g

We note that E0 is the event that an even number appears.

b E[F
S We want the sample points in E, or F, or both. Thus,

E [ F D f1; 3; 4; 5; 6g

c E\F
S The sample points common to both E and F are 3 and 5, so

E \ F D f3; 5g

d F\G
S Since F and G have no sample point in common,

F\GD;

e E [ E0
S Using the result of part (a), we have

E [ E0 D f1; 3; 5g [ f2; 4; 6g D f1; 2; 3; 4; 5; 6g D S

372 C ntro ct on to Pro a t an tat st cs

f E \ E0
S E \ E0 D f1; 3; 5g \ f2; 4; 6g D ;

Now ork Problem 17 G

The results of Examples 7(e) and 7(f) can be generalized as follows:

If E is any event for an experiment with sample space S, then

E [ E0 D S and E \ E0 D ;

Thus, the union of an event and its complement is the sample space the intersection
of an event and its complement is the empty set. These and other properties of events
are listed in Table .1.

Table .1 E
If E and F are any events for an experiment with sample space S, then
E[EDE
E\EDE
.E0 /0 D E (the complement of the complement
of an event is the event)
E [ E0 D S
E \ E0 D ;
E[SDS
E\SDE
E[;DE
E\;D;
E[FDF[E (commutative property of union)
E\FDF\E (commutative property of intersection)
.E [ F/0 D E0 \ F0 (the complement of a union is the
intersection of complements)
.E \ F/0 D E0 [ F0 (the complement of an intersection
is the union of complements)
E [ .F [ G/ D .E [ F/ [ G (associative property of union)
E \ .F \ G/ D .E \ F/ \ G (associative property of intersection)
E \ .F [ G/ D .E \ F/ [ .E \ G/ (distributive property of intersection
over union)
E [ .F \ G/ D .E [ F/ \ .E [ G/ (distributive property of union
over intersection)

S
When two events E and F have no sample point in common, that is,
E F
2 E\FD;
4 1
they are called mutua y exc usi e or disj int events. For example, in the rolling of a
6
die, the events
5 3
E D f2; 4; 6g and F D f1g
FIGURE utually exclusive
events. are mutually exclusive (see Figure .12).
 Section 8.3 a e aces an ents 373

Events E and F are said to be mutually exclusive events if and only if E \ F D ;.

When two events are mutually exclusive, the occurrence of one event means that
the other event cannot occur that is, the two events cannot occur simultaneously. An
event and its complement are mutually exclusive, since E \ E0 D ;.

E AM LE M E E

If E, F, and G are events for an experiment and F and G are mutually exclusive, show
that events E \ F and E \ G are also mutually exclusive.
S iven that F \ G D ;, we must show that the intersection of E \ F and
E \ G is the empty set. Using the properties in Table .1, we have

.E \ F/ \ .E \ G/ D .E \ F \ E/ \ G property 15
D .E \ E \ F/ \ G property 11
D .E \ F/ \ G property 2
D E \ .F \ G/ property 15
DE\; given
D; property

Now ork Problem 31 G

R BLEMS
In Pr b ems determine a samp e space f r the gi en In Pr b ems describe the nature f a samp e space f r the
experiment gi en experiment and determine the number f samp e p ints
Card Selection A card is drawn from a four-card dec Coin oss A coin is tossed six times in succession, and the
consisting of the of diamonds, of hearts, of clubs, and of faces showing are observed.
spades. Dice Roll Five dice are rolled, and the numbers that turn up
Euchre Dec A card is drawn from a euchre dec as are observed.
described in Problem 36 of Section .2.
Card and Die A card is drawn from an ordinary dec of
Die Roll and Coin osses A die is rolled and then a coin is 52 cards, and then a die is rolled.
tossed twice in succession.
Rabbit Selection From a hat containing eight
Dice Roll Two dice are rolled, and the sum of the numbers
distinguishable rabbits, five rabbits are pulled successively
that turns up is observed.
without replacement.
Digit Selection Two different digits are selected, in
succession, from those in the number 64 01 . Card Deal A four-card hand is dealt from a dec of
52 cards.
Genders of Children The genders of the first, second, third,
and fourth children of a four-child family are noted. ( et, for Letter Selection A four-letter word is formed by
example, denote that the first, second, third, and fourth successively choosing any four letters from the alphabet with
children are boy, girl, girl, boy, respectively.) replacement.
elly Bean Selection A bag contains three colored elly Supp se that S D f1; 2; 3; 4; 5; 6; 7; ; ; 10g is the samp e space
beans: one red, one white, and one blue. etermine a sample f r an experiment ith e ents
space if a three elly beans are selected with replacement and E D f1; 3; 5g F D f3; 5; 7; g G D f2; 4; 6; g
b three elly beans are selected without replacement.
Manufacturing Process A company ma es a product that
In Pr b ems determine the indicated e ents
goes through three processes during its manufacture. The first is
an assembly line, the second is a finishing line, and the third is an E[F G0
inspection line. There are four assembly lines (A, , C, and ), F0 \ G E0 [ G0
two finishing lines (E and F), and two inspection lines ( and H).
F0 .E [ F/0
For each process, the company chooses a line at random.
etermine a sample space. .F \ G/0 .F [ G/ \ E0
374 C ntro ct on to Pro a t an tat st cs

f the following events, which pairs are mutually exclusive Genders of Children A husband and wife have three
children. The outcome of the first child being a boy, the second a
E1 D f1; 2; 3g E2 D f3; 4; 5g
girl, and the third a girl can be represented by . etermine
E3 D f1; 2g E4 D f6; 7g each of the following:
Card Selection From a standard dec of 52 playing cards, a Sample space that describes all the orders of the possible
2 cards are drawn without replacement. Suppose E is the event genders of the children
that both cards are ac s, EC is the event that both cards are clubs, b The event that at least one child is a girl
and E3 is the event that both cards are 3 s. Which pairs of these c The event that at least one child is a boy
events are mutually exclusive d Is the event in part (c) the complement of the event in part (b)
Card Selection From a standard dec of 52 playing cards,
Arrivals Persons A, , and C enter a building at different
1 card is selected. Which pairs of the following events are
times. The outcome of A arriving first, second, and C third can
mutually exclusive
be indicated by A C. etermine each of the following:
E D fdiamondg a The sample space involved for the arrivals
F D fface cardg b The event that A arrives first
c The event that A does not arrive first
G D fblac g
Supplier Selection A grocery store can order fruits and
D fredg vegetables from suppliers U, V, and W meat from suppliers U, V,
, and and dry goods from suppliers V, W, , and . The
I D face of diamondsg
grocery store selects one supplier for each type of item. The
Dice A red and a green die are thrown, and the numbers on outcome of U being selected for fruits and vegetables, V for meat,
each are noted. Which pairs of the following events are mutually and W for dry goods can be represented by UVW.
exclusive a etermine a sample space.
E D fboth are eveng b etermine the event E that one supplier supplies all the
grocery store s requirements.
F D fboth are oddg c etermine E0 and give a verbal description of this event.
G D fsum is 2g If E and F are events for an experiment, prove that events
E \ F and E \ F0 are mutually exclusive.
D fsum is 4g
If E and F are events for an experiment, show that
I D fsum is greater than 10g
.E \ F/ [ .E \ F0 / D E
Coin oss A coin is tossed three times in succession, and
the results are observed. etermine each of the following: Note that from Problem 31, E \ F and E \ F0 are mutually
exclusive events. Thus, the foregoing equation expresses E as
a The usual sample space S
a union of mutually exclusive events.
b The event E that at least two heads occur
c The event F that at least one tail occurs
d E[F
e E\F
f .E [ F/0
g .E \ F/0

Objective
o e ne at s eant t e
ro a t of an e ent o e e o E S
for as t at are se n co tn
ro a t es as s s ace on We now introduce the basic concepts underlying the study of probability. Consider
e ro a e s aces tossing a well-balanced die and observing the number that turns up. The usual sample
space for the experiment is
S D f1; 2; 3; 4; 5; 6g
efore the experiment is performed, we cannot predict with certainty which of the six
possible outcomes (sample points) will occur. ut it does seem reasonable that each
outcome has the same chance of occurring that is, the outcomes are equally li ely.
This does not mean that in six tosses each number must turn up once. Rather, it means
that if the experiment were performed a large number of times, each outcome would
occur about 16 of the time.
To be more specific, let the experiment be performed n times. Each performance of
an experiment is called a trial. Suppose that we are interested in the event of obtaining
 Section 8.4 Pro a t 375

a 1 (that is, the simple event consisting of the sample point 1). If a 1 occurs in k of
these n trials, then the proportion of times that 1 occurs is k=n. This ratio is called the
relative frequency of the event. ecause getting a 1 is ust one of six possible equally
li ely outcomes, we expect that in the long run a 1 will occur 16 of the time. That is, as
n becomes very large, we expect the relative frequency k=n to approach 16 . The number
1
6
is ta en to be the probability of getting a 1 on the toss of a well-balanced die, which
is denoted P.1/. Thus, P.1/ D 16 . Similarly, P.2/ D 16 , P.3/ D 16 , and so on.
In this experiment, all of the simple events in the sample space (those consisting
of exactly one sample point) were understood to be equally li ely to occur. To describe
this equal li elihood, we say that S is an equipr bab e space.

A sample space S is called an equiprobable space if and only if all the simple events
are equally li ely to occur.

We remar that besides the phrase equa y ike y other words and phrases used in
the context of an equiprobable space are e ba anced fair unbiased, and at rand m.
For example, we may have a e ba anced die (as above), a fair coin, or unbiased
dice, or we may select a elly bean at rand m from a bag.
We now generalize our discussion of the die experiment to other (finite) equiprob-
able spaces.

If S is an equiprobable sample space with sample points (or outcomes), say S D

fs1 ; s2 ; : : : ; s g, then the pr babi ity f each simp e e ent fsi g is given by
1
P.si / D

for i D 1; 2; : : : ; . f course, P.si / is an abbreviation for P.fsi g/.

We remar that P.si / can be interpreted as the relative frequency of fsi g occurring in
the long run.
We can also assign probabilities to events that are not simple. For example, in the
die experiment, consider the event E of a 1 or a 2 turning up:

E D f1; 2g

ecause the die is well balanced, in n trials (where n is large) we expect that a 1 should
turn up approximately 16 of the time and a 2 should turn up approximately 16 of the time.
Thus, a 1 or 2 should turn up approximately 16 C 16 of the time, or 2
6
of the time. Hence,
it is reasonable to assume that the long-run relative frequency of E is 26 . For this reason,
2
we define 6
to be the probability of E and denote it P.E/.

1 1 2
P.E/ D C D
6 6 6

Note that P.E/ is simply the sum of the probabilities of the simple events that form E.
Equivalently, P.E/ is the ratio of the number of sample points in E (two) to the number
of sample points in the sample space (six).
376 C ntro ct on to Pro a t an tat st cs

If S is a finite equiprobable space for an experiment and E D fs1 ; s2 ; : : : ; sj g is an

event, then the probability of E is given by
P.E/ D P.s1 / C P.s2 / C ! ! ! C P.sj /
Equivalently,
#.E/
P.E/ D
#.S/
where #.E/ is the number of outcomes in E and #.S/ is the number of outcomes in S.

Note that we can thin of P as a function that assigns to each event E the probability
of E namely, P.E/. The probability of E can be interpreted as the relative frequency of
E occurring in the long run. Thus, in n trials, we would expect E to occur approximately
n ! P.E/ times, provided that n is large.

E AM LE C T

Two fair coins are tossed. etermine the probability that

a two heads occur

b at least one head occurs
S The usual sample space is

S D fHH, HT, TH, TTg

Since the four outcomes are equally li ely, S is equiprobable and #.S/ D 4.

a If E D fHHg, then E is a simple event, so

#.E/ 1
P.E/ D D
#.S/ 4

b et F D fat least one headg. Then

F D fHH, HT, THg

which has three outcomes. Thus,

#.F/ 3
P.F/ D D
#.S/ 4

Alternatively,

P.F/ D P.HH/ C P.HT/ C P.TH/

1 1 1 3
D C C D
4 4 4 4
Consequently, in 1000 trials of this experiment, we would expect F to occur approx-
imately 1000 ! 34 D 750 times.

Now ork Problem 1 G

 Section 8.4 Pro a t 377

E AM LE C

From an ordinary dec of 52 playing cards, 2 cards are randomly drawn without replace-
ment. If E is the event that one card is a 2 and the other a 3, find P.E/.
S We can disregard the order in which the 2 cards are drawn. As our sample
space S, we choose the set of all combinations of the 52 cards ta en 2 at a time. Thus, S
is equiprobable and #.S/ D 52 C2 . To find #.E/, we note that since there are four suits,
a 2 can be drawn in four ways and a 3 in four ways. Hence, a 2 and a 3 can be drawn
in 4 ! 4 ways, so
#.E/ 4!4 16
P.E/ D D D D
#.S/ C
52 2 1326 663

Now ork Problem 7 G

E AM LE F H H

Find the probability of being dealt a full house in a po er game. A full house is three of
one ind and two of another, such as three queens and two 10 s. Express your answer
in terms of n Cr .
S The set of all combinations of 52 cards ta en 5 at a time is an equiprob-
able sample space. (The order in which the cards are dealt is of no concern.) Thus,
#.S/ D 52 C5 . We now must find #.E/, where E is the event of being dealt a full house.
Each of the four suits has 13 inds, so three cards of one ind can be dealt in 13 ! 4 C3
ways. For each of these there are 12 ! 4 C2 ways to be dealt two cards of another ind.
Hence, a full house can be dealt in 13 ! 4 C3 ! 12 ! 4 C2 ways, and we have

#.E/ 13 ! 4 C3 ! 12 ! 4 C2 13 ! 12 ! 6 ! 4 6
P.full house/ D D D D
#.S/ 52 C5 4 ! 24 ! 17 ! 13 ! 10 4 ! 17 ! 5

which is about 0:144 .

Now ork Problem 13 G

E AM LE S S

From a committee of three males and four females, a subcommittee of four is to be

randomly selected. Find the probability that it consists of two males and two females.
S Since order of selection is not important, the number of subcommittees of
four that can be selected from the seven members is 7 C4 . The two males can be selected
in 3 C2 ways and the two females in 4 C2 ways. y the asic Counting Principle, the
number of subcommittees of two males and two females is 3 C2 ! 4 C2 . Thus,
3 C2! 4 C2
P.two males and two females/ D
7 C4
3Š 4Š
! 1
D 2Š1Š 2Š2Š D
7Š 35
4Š3Š

Now ork Problem 21 G

We now develop some properties of probability. et S be an equiprobable sample space

with outcomes that is, #.S/ D . (We assume a finite sample space throughout
378 C ntro ct on to Pro a t an tat st cs

this section.) If E is an event, then 0 $ #.E/ $ . ividing each member by #.S/ D

gives

#.E/
0$ $
#.S/

#.E/
ut D P.E/, so we have the following property:
#.S/

0 $ P.E/ $ 1

That is, the probability of an event is a number between 0 and 1, inclusive.

#.;/ 0
oreover, P.;/ D D D 0. Thus,
#.S/

P.;/ D 0

#.S/
Also, P.S/ D D D 1, so
#.S/

P.S/ D 1

Accordingly, the probability of the impossible event is 0, and the probability of the
certain event is 1.
Since P.S/ is the sum of the probabilities of the outcomes in the sample space, we
conclude that the sum of the probabilities of all the simple events for a sample space
is 1.
Now let us focus on the probability of the union of two events E and F. The event
E [ F occurs if and only if at least one of the events (E or F) occurs. Thus, P.E [ F/
is the probability that at east ne of the events E and F occurs. We now that

#.E [ F/
P.E [ F/ D
#.S/

Now
S
#.E [ F/ D #.E/ C #.F/ " #.E \ F/
E F
because #.E/ C #.F/ D #.E [ F/ C #.E \ F/. To see the truth of the last statement,
E F loo at Figure .13 and notice that E \ F is contained in b th E and F.
ividing both sides of Equation (1) for #.E[F/ by #(S) gives the following result:

U E
FIGURE E \ F is contained
in both E and F. If E and F are events, then
P.E [ F/ D P.E/ C P.F/ " P.E \ F/

Note that while we derived Equation (2) For example, let a fair die be rolled, and let E D f1; 3; 5g and F D f1; 2; 3g. Then
for an equiprobable sample space, the E \ F D f1; 3g, so
result is in fact a general one.
P.E [ F/ D P.E/ C P.F/ " P.E \ F/
3 3 2 2
D C " D
6 6 6 3
4
Alternatively, E [ F D f1; 2; 3; 5g, so P.E [ F/ D 6
D 23 .
 Section 8.4 Pro a t 379

If E and F are mutually exclusive events, then E \ F D ; so P.E \ F/ D P.;/ D 0.

Hence, from Equation (2), we obtain the following law:

A L M E E
If E and F are mutua y exc usi e events, then
P.E [ F/ D P.E/ C P.F/

For example, let a fair die be rolled, and let E D f2; 3g and F D f1; 5g. Then
E \ F D ;, so
2 2 2
P.E [ F/ D P.E/ C P.F/ D C D
6 6 3
The addition law can be extended to more than two mutually exclusive events. Two or
more events are mutually exclusive if and only if no two of them can occur at the same
time. That is, given any two of them, their intersection must be empty. For example, to
say the events E, F, and G are mutually exclusive means that
E\FDE\GDF\GD;
If events E, F, and G are mutually exclusive, then
P.E [ F [ G/ D P.E/ C P.F/ C P.G/
An event and its complement are mutually exclusive, so, by the addition law,
P.E [ E0 / D P.E/ C P.E0 /
ut P.E [ E0 / D P.S/ D 1. Thus,
1 D P.E/ C P.E0 /
so that
P.E0 / D 1 " P.E/
In order to find the probability of an equivalently,
event, sometimes it is more convenient
first to find the probability of its P.E/ D 1 " P.E0 /
complement and then subtract the result
from 1. See, especially, Example 6. Accordingly, if we now the probability of an event, then we can easily find the
probability of its complement, and vice versa. For example, if P.E/ D 14 , then
1
P.E0 / D 1 " 4
D 34 . P.E0 / is the probability that E does not occur.

E AM LE C

From a production run of 5000 light bulbs, 2 of which are defective, 1 bulb is selected
at random. What is the probability that the bulb is defective What is the probability
that it is not defective
S In a sense, this is tric question because the statement that 2 are defec-
2
tive means that 100 are defective , which in turn means that the chance of getting a
defective light bulb is 2 in a hundred , equivalently, that the probability of getting
a defective light bulb is 0:02. However, to reinforce the ideas we have considered so
far, let us suppose that the sample space S consists of the 5000 bulbs. Since a bulb
is selected at random, the possible outcomes are equally li ely. et E be the event of
selecting a defective bulb. The number of outcomes in E is 0:02 ! 5000 D 100. Thus,
#.E/ 100 1
P.E/ D D D D 0:02
#.S/ 5000 50
380 C ntro ct on to Pro a t an tat st cs

1
Alternatively, since the probability of selecting a particular bulb is 5000
and E contains
100 sample points, by summing probabilities we have
1
P.E/ D 100 ! D 0:02
5000
The event that the bulb selected is n t defective is E0 . Hence,

P.E0 / D 1 " P.E/ D 1 " 0:02 D 0:

Now ork Problem 17 G
The next example is a celebrated use of the rule P.E/ D 1 " P.E0 /. It is a case in
which P.E/ is di cult to calculate directly but P.E0 / is easy to calculate. ost people
find the result rather surprising.

E AM LE B S

For a random collection of n people, with n # 365, ma e the simplifying assumption

that all years consist of 365 days and calculate the probability that at least two of the n
people celebrate their birthday on the same day. Find the smallest value of n for which
this probability is greater than 50 . What happens if n > 365
S The sample space is the set S of all ways in which the birthdays of n people
can arise. It is convenient to assume that the people are labeled. There are 365 possibil-
ities for the birthday of person 1, and for each of these there are 365 possibilities for the
birthday of person 2. For each of these 3652 possibilities for the birthdays of persons
1 and 2, there are 365 possibilities for the birthday of person 3. y iterated use of the
asic Counting Principle, it is easy to see that #S D 365n . et En be the event that at
least 2 of the n people have their birthday on the same day. It is not easy to count En ,
but for .En /0 , the event that a n pe p e ha e their birthday n di erent days, we see
that there are 365 possibilities for the birthday of person 1, and for each of these there
are 364 possibilities for the birthday of person 2, and for each of these there are 363
possibilities for the birthday of person 3, and so on. Thus, #.En /0 D 365 Pn and it now
follows that
365 Pn
P.En / D 1 "
365n
We leave it as an exercise for the student to tabulate P.En / with the aid of a
programmable calculator. To do this it is helpful to note the recursion:
364 365 " n
P..E2 /0 / D and P..EnC1 /0 / D P..En /0 / !
365 365
from which we have
1 n C P.En /.365 " n/
P.E2 / D and P.EnC1 / D
365 365
If a programmable calculator is supplied with this recursive formula, it can be shown
that P.E22 / $ 0:4756 5 so that
22 C P.E22 /.365 " 22/ 22 C 0:4756 5.343/
P.E23 / D $ $ 0:5072 7 D 50:72 7
365 365
Thus, 23 is the smallest number n for which P.En / > 50 .
We note that if n > 365, there are more people than days in the year. At least two
people must share a birthday in this case. So, for n > 365, we have P.En / D 1.
Now ork Problem 45 G
 Section 8.4 Pro a t 381

E AM LE

A pair of well-balanced dice is rolled, and the number on each die is noted. etermine
the probability that the sum of the numbers that turns up is (a) 7, (b) 7 or 11, and
(c) greater than 3.
S Since each die can turn up in any of six ways, by the asic Counting
Principle the number of possible outcomes is 6 ! 6 D 36. ur sample space consists
of the following ordered pairs:
.1; 1/ .1; 2/ .1; 3/ .1; 4/ .1; 5/ .1; 6/
.2; 1/ .2; 2/ .2; 3/ .2; 4/ .2; 5/ .2; 6/
.3; 1/ .3; 2/ .3; 3/ .3; 4/ .3; 5/ .3; 6/
.4; 1/ .4; 2/ .4; 3/ .4; 4/ .4; 5/ .4; 6/
.5; 1/ .5; 2/ .5; 3/ .5; 4/ .5; 5/ .5; 6/
.6; 1/ .6; 2/ .6; 3/ .6; 4/ .6; 5/ .6; 6/
1
The outcomes are equally li ely, so the probability of each outcome is 36 . There are a
lot of characters present in the preceding list, since each of the 36 ordered pairs involves
five (a pair of parentheses, a comma, and 2 digits) for a total of 36 ! 5 D 1 0 characters.
The same information is conveyed by the following coordinatized boxes, requiring ust
12 characters and 14 lines.
1 2 3 4 5 6
1
2
3
4
5
6

a et E7 be the event that the sum of the numbers appearing is 7. Then

E7 D f.1; 6/; .2; 5/; .3; 4/; .4; 3/; .5; 2/; .6; 1/g
which has six outcomes (and can be seen as the rising diagonal in the coordinatized
boxes). Thus,
6 1
P.E7 / D D
36 6
b et E7 or 11 be the event that the sum is 7 or 11. If E11 is the event that the sum is
11, then
E11 D f.5; 6/; .6; 5/g
which has two outcomes. Since E7 or 11 D E7 [ E11 and E7 and E11 are mutually
exclusive, we have
6 2 2
P.E7 or 11 / D P.E7 / C P.E11 / D C D D
36 36 36
Alternatively, we can determine P.E7 or 11 / by counting the number of outcomes in
E7 or 11 . We obtain
E7 or 11 D f.1; 6/; .2; 5/; .3; 4/; .4; 3/; .5; 2/; .6; 1/; .5; 6/; .6; 5/g
which has eight outcomes. Thus,
2
P.E7 or 11 / D D
36
c et E be the event that the sum is greater than 3. The number of outcomes in E is
relatively large. Thus, to determine P.E/, it is easier to find E0 , rather than E, and
then use the formula P.E/ D 1 " P.E0 /. Here E0 is the event that the sum is 2 or 3.
We have
E0 D f.1; 1/; .1; 2/; .2; 1/g
382 C ntro ct on to Pro a t an tat st cs

which has three outcomes. Hence,

3 11
P.E/ D 1 " P.E0 / D 1 " D
36 12
Now ork Problem 27 G
E AM LE I G

btain Pascal and Fermat s solution to the problem of dividing the pot between two
gamblers in an interrupted game of chance, as mentioned in the introduction to this
chapter. We assume that the game consists of a sequence of rounds involving chance,
such as coin tosses, that each has an equal chance of winning, and that the overall
winner, who gets the pot, is the one who first wins a certain number of rounds. We
further assume that when the game was interrupted, Player 1 needed r more rounds
to win and Player 2 needed s more rounds to win. It is agreed that the pot should be
divided so that each player gets the value of the pot multiplied by the probability that
he or she would have won an uninterrupted game.
S We need only compute the probability that Player 1 would have won, for if
that is p, then the probability that Player 2 would have won is 1 " p. Now the game can
go at most r C s " 1 more rounds. To see this, observe that each round produces exactly
one winner, and let a be the number of the r C s " 1 rounds won by Player 1 and let b
be the number of the r C s " 1 rounds won by Player 2. So r C s " 1 D a C b. If neither
Player 1 nor Player 2 has won, then a # r " 1 and b # s " 1. ut in this case we have

r C s " 1 D a C b # .r " 1/ C .s " 1/ D r C s " 2

which is impossible. It is clear that after r C s " 2 there might not yet be an overall
winner, so we do need to consider a further r C s " 1 possible rounds from the time
of interruption. et n D r C s " 1. Now Player 1 will win if Player 2 wins k of the n
possible further n rounds, where 0 # k # s " 1. et Ek be the event that Player 2 wins
exact y k of the next n rounds. Since the events Ek , for k D 0; 1; ! ! ! s " 1, are mutually
exclusive, the probability that Player 1 will win is given by
X
s!1
P.E0 [ E1 [ ! ! ! [ Es!1 / D P.E0 / C P.E1 / C ! ! ! C P.Es!1 / D P.Ek /
kD0

It remains to determine P.Ek /. We may as well suppose that a round consists of ipping
a coin, with outcomes H and T. We further ta e a single-round win for Player 2 to be
an outcome of T. Thus, Player 2 will win exactly k of the next rounds if exactly k of
the next n outcomes are T s. The number of possible outcomes for the next n rounds
is of course 2n , by the multiplication principle. The number of these outcomes which
consists of exactly k T s is the number of ways of choosing k from among n. It follows
n Ck
that P.Ek / D n , and, substituting this value in Equation (3), we obtain
2

1 X
s!1

n Ck
2n kD0

Now ork Problem 29 G

F G
any of the properties of equiprobable spaces carry over to sample spaces that are
not equiprobable. To illustrate, consider the experiment of tossing two fair coins and
observing the number of heads. The coins can fall in one of four ways, namely,

HH HT TH TT
 Section 8.4 Pro a t 383

which correspond to two heads, one head, one head, and zero heads, respectively.
ecause we are interested in the number of heads, we can choose a sample space to be
S D f0; 1; 2g
However, the simple events in S are n t equally li ely to occur because of the four
possible ways in which the coins can fall: Two of these ways correspond to the one-
head outcome, whereas only one corresponds to the two-head outcome, and similarly
for the zero-head outcome. In the long run, it is reasonable to expect repeated trials to
result in one head about 24 of the time, zero heads about 14 of the time, and two heads
about 14 of the time. If we were to assign probabilities to these simple events, it is natural
to have
1 2 1 1
P.0/ D P.1/ D D P.2/ D
4 4 2 4
Although S is not equiprobable, these probabilities lie between 0 and 1, inclusive, and
their sum is 1. This is consistent with what was stated for an equiprobable space.
ased on our discussion, we define below a pr babi ity functi n for a general, finite
sample space S D fs1 ; s2 ; : : : ; s g. Recall that Œ0; 1! denotes the set of all real numbers
x with 0 # x # 1 and 2S denotes the set of all subsets of S equivalently, the set of all
events. We also write
[
k
Ej D E1 [ E2 [ : : : [ Ek
jD1

et S D fs1 ; s2 ; : : : ; s g be a sample space for an experiment. A function

P W 2S % Œ0; 1! is called a pr babi ity functi n if
P.;/ D 0
P.S/ D 1
For any collection E1 ; E2 ; : : : Ek of mutua y exc usi e events,
[
k X
k
P. Ej / D P.Ej /
jD1 jD1

Note that says, for mutually exclusive E1 ; E2 ; : : : Ek ,

P.E1 [ E2 [ : : : [ Ek / D P.E1 / C P.E2 / C : : : C P.Ek /:
Note too that fs1 g; fs2 g; : : : ; fs g are mutually exclusive events with
[
fsj g D S
jD1

It can be shown that

A pr babi ity functi n for a sample space S D fs1 ; s2 ; : : : ; s g can be equiva-

lently described by giving a function P W S % Œ0; 1! with
P.s1 / C P.s2 / C : : : P.s / D 1
and extending it to P W 2S % Œ0; 1! by requiring that
P.;/ D 0
and
P.fsj1 ; sj2 ; : : : sjk g/ D P.sj1 / C P.sj2 / C : : : P.sjk /
384 C ntro ct on to Pro a t an tat st cs

To illustrate the reformulated definition, consider the sample space for the previous
experiment of tossing two fair coins and observing the number of heads:
S D f0; 1; 2g
We c u d assign probabilities as follows:
P.0/ D 0:1 P.1/ D 0:2 P.2/ D 0:7
This would entail P.;/ D 0, P.f0g/ D 0:1, P.f1g/ D 0:2, P.f2g/ D 0:7, P.f0; 1g/ D 0:3,
P.f1; 2g/ D 0: , P.f0; 2g/ D 0: , and P.S/ D 1. Such a P satisfies the requirements
of a probability function. However, unless the coins were not in fact fair and had some
strange weighting, the assignment does not re ect the long-run interpretation of prob-
ability and, consequently, would not be acceptable from a practical point of view.
In general, for any probability function defined on a sample space the following
properties hold:
P.E0 / D 1 " P.E/
P.S/ D 1
P.E1 [ E2 / D P.E1 / C P.E2 / if E1 \ E2 D ;

E
We have seen how easy it is to assign probabilities to simple events when we have an
equiprobable sample space. For example, when a fair coin is tossed, we have S D fH,Tg
and P.H/ D P.T/ D 12 . These probabilities are determined by the intrinsic nature of
the experiment namely, that there are two possible outcomes that should have the
same probability because the outcomes are equally li ely. Such probabilities are called
the retica probabilities. However, suppose the coin is not fair. How can probabilities
then be assigned y tossing the coin a number of times, we can determine the rela-
tive frequencies of heads and tails occurring. For example, suppose that in 1000 tosses,
heads occurs 517 times and tails occurs 4 3 times. Then the relative frequencies of
517 4 3
heads and tails occurring are 1000 and 1000 , respectively. In this situation, the assign-
ment P.H/ D 0:517 and P.T/ D 0:4 3 would be reasonable. Probabilities assigned in
this way are called empirical probabilities. In general, probabilities based on sample
or historical data are empirical. Now suppose that the coin were tossed 2000 times,
and the relative frequencies of heads and tails occurring were 1023 2000
D 0:5115 and
77
2000
D 0:4 5, respectively. Then in this case, the assignment P.H/ D 0:5115 and
P.T/ D 0:4 5 would be acceptable. The latter probabilities may be more indicative
of the true nature of the coin than would be the probabilities associated with 1000
tosses.
In the next example, probabilities (empirical) are assigned on the basis of sam-
ple data.

E AM LE

An opinion poll of a sample of 150 adult residents of a town was conducted. Each
person was as ed his or her opinion about oating a bond issue to build a community
swimming pool. The results are summarized in Table .2.

Table .
Favor ppose Total

ale 60 20 0
Female 40 30 70
Total 100 50 150
 Section 8.4 Pro a t 385

Suppose an adult resident from the town is randomly selected. et be the event
male selected and F be the event selected person favors the bond issue . Find each
of the following:

a P. /
b P.F/
c P. \ F/
d P. [ F/

S We will assume that proportions that apply to the sample also apply to
the adult population of the town.

a f the 150 persons in the sample, 0 are males. Thus, for the adult population of
0
the town (the sample space), we assume that 150 are male. Hence, the (empirical)
probability of selecting a male is

0
P. / D D
150 15

b f the 150 persons in the sample, 100 favor the bond issue. Therefore,

100 2
P.F/ D D
150 3

c Table .2 indicates that 60 males favor the bond issue. Hence,

60 2
P. \ F/ D D
150 5

d To find P. [ F/, we use Equation (1):

P. [ F/ D P. / C P.F/ " P. \ F/
0 100 60 120 4
D C " D D
150 150 150 150 5

Now ork Problem 33 G

The probability of an event is sometimes expressed in terms of dds especially in

gaming situations.

The odds in favor of event E occurring are the ratio

P.E/
P.E0 /
p
provided that P.E0 / ¤ 0. dds are usually expressed as the ratio (or p q) of two
q
positive integers, which is read p to q .
386 C ntro ct on to Pro a t an tat st cs

E AM LE A E

A student believes that the probability of getting an A on the next mathematics exam is
0.2. What are the odds (in favor) of this occurring
S If E D gets an A , then P.E/ D 0:2 and P.E0 / D 1 " 0:2 D 0: . Hence,
the odds of getting an A are
P.E/ 0:2 2 1
D D D D1W4
P.E0 / 0: 4
That is, the odds are 1 to 4. (We remar that the odds against getting an A are 4 to 1.)
G
If the odds that event E occurs are a W b, then the probability of E can be easily
determined. We are given that
P.E/ a
D
1 " P.E/ b
Solving for P.E/ gives
bP.E/ D .1 " P.E//a clearing fractions
aP.E/ C bP.E/ D a
.a C b/P.E/ D a
a
P.E/ D
aCb

F
If the odds that event E occurs are a W b, then
a
P.E/ D
aCb

ver the long run, if the odds that E occurs are a W b, then, on the average, E should
occur a times in every a C b trials of the experiment.

E AM LE

A 1000 savings bond is one of the prizes listed on a contest brochure received in the
mail. The odds in favor of winning the bond are stated to be 1 : 10,000. What is the
probability of winning this prize
S Here a D 1 and b D 10,000. From the preceding rule,
a
P.winning prize/ D
aCb
1 1
D D
1 C 10,000 10,001
Now ork Problem 35 G
R BLEMS
In 4000 trials of an experiment, how many times should we If P.E/ D 14 ; P.F/ D 12 , and P.E \ F/ D 1 , find a P.E0 / and
expect event E to occur if P.E/ D 0:125 b P.E [ F/.
In 3000 trials of an experiment, how many times would you If P.E \ F/ D 0: 31, are E and F mutually exclusive
expect event E to occur if P.E0 / D 0:45
If P.E/ D 14 , P.E [ F/ D 12 , and P.E \ F/ D 1
12
, find P.F/.
If P.E/ D 0:5, P.F/ D 0:4, and P.E \ F/ D 0:1, find a P.E0 /
and b P.E [ F/.
 Section 8.4 Pro a t 387

Dice A pair of well-balanced dice is tossed. Find the elly Bean Selection Two bags contain colored elly beans.
probability that the sum of the numbers is a b 2 or 3 c 3, ag 1 contains three red and two green elly beans, and bag 2
4, or 5 d 12 or 13 e even f odd and g less than 10. contains four red and five green elly beans. A elly bean is
Dice A pair of fair dice is tossed. etermine the probability selected at random from each bag. Find the probability that
that at least one die shows a 1 or a 6. a both elly beans are red and b one elly bean is red and the
other is green.
Card Selection A card is randomly selected from a standard
dec of 52 playing cards. etermine the probability that the card Committee Selection From a group of three women and
is a the ing of hearts, b a diamond, c a ac , d red, e a four men, two persons are selected at random to form a committee.
heart or a club, f a club and a 4, g a club or a 4, h red and a Find the probability that the committee consists of women only.
ing, and i a spade and a heart. Committee Selection For the committee selection in
Coin and Die A fair coin and a fair die are tossed. Find the Problem 21, find the probability that the committee consists of
probability that a a head and a 5 show, b a head shows, c a 3 a man and a woman.
shows, and d a head and an even number show. Examination Score A student answers each question on a
Coin Die and Card A fair coin and a fair die are tossed, 10-question multiple-choice examination in a random fashion. For
and a card is randomly selected from a standard dec of 52 each of the questions there are 5 possible choices. If each question
playing cards. etermine the probability that the coin, die, and is worth 10 points, what is the probability that the student scores
card, respectively, show a a head, a 6, and the ace of spades 100 points
b a head, a 3, and a queen c a head, a 2 or 3, and a queen and Multiple Choice Examination n an eight-question,
d a head, an odd number, and a diamond. multiple-choice examination, there are four choices for
Coins Three fair coins are tossed. Find the probability that each question, only one of which is correct. If a student answers
a three heads show, b exactly one tail shows, c no more than each question in a random fashion, find the probability that the
two heads show, and d no more than one tail shows. student answers a each question correctly and b exactly four
questions correctly.
Card Selection Three cards from a standard dec of
52 playing cards are successively drawn at random without Po er and Find the probability of being dealt four of a
replacement. Find the probability that a all three cards are ind in a po er game. This simply means four of one ind and one
ac s and b all three cards are spades. other card, such as four queens and a 10. Express your answer
Card Selection Two cards from a standard dec of using the symbol n Cr .
52 playing cards are successively drawn at random with Suppose P.E/ D 15 , P.E [ F/ D 41
, and P.E \ F/ D 17 .
105
replacement. Find the probability that a both cards are ings
and b one card is a ing and the other is a heart. a Find P.F/ b Find P.E0 [ F/
Genders of Children Assuming that the gender of a int
person is determined at random, determine the probability that a
family with three children has a three girls, b exactly one boy, F D .E \ F/ [ .E0 \ F/
c no girls, and d at least one girl.
where E \ F and E0 \ F are mutually exclusive.
elly Bean Selection A elly bean is randomly ta en from
a bag that contains five red, nine white, and two blue elly beans. Faculty Committee The classification of faculty at a
Find the probability that the elly bean is a blue, b not red, college is indicated in Table .3. If a committee of three faculty
c red or white, d neither red nor blue, e yellow, and f red members is selected at random, what is the probability that it
or yellow. consists of a all females b a professor and two associate
professors
Stoc Selection A stoc is selected at random from a list
of 60 utility stoc s, 4 of which have an annual dividend yield of
6 or more. Find the probability that the stoc pays an annual Table . F C
dividend that yields a 6 or more and b less than 6 .
ale Female Total
Inventory A clothing store maintains its inventory of
sweaters so that 51 are 100 pure wool. If a tie is selected Professor 12 3 15
at random, what is the probability that it is a 100 pure wool Associate Professor 15 24
b not 100 pure wool Assistant Professor 1 26
Examination Grades n an examination given to 40 Instructor 20 15 35
students, 10 received an A, 25 a , 35 a C, 25 a , and
5 an F. If a student is selected at random, what is the probability Total 65 35 100
that the student a received an A, b received an A or a ,
c received neither a nor an F, and d did not receive an F
2
e Answer questions (a) (d) if the number of students that were Biased Die A die is biased so that P.1/ D 10
,
given the examination is un nown. P.2/ D P.3/ D P.4/ D P.5/ D 10 1
, and P.6/ D 4
the . If
10
die is tossed, find the probability of tossing an even number.
388 C ntro ct on to Pro a t an tat st cs

Interrupted Game A pair of gamblers is tossing a coin Digital Camcorder Sales A department store chain has
and calling so that exactly one of them wins each toss. There is a stores in the cities of Exton and Whyton. Each store sells three
pot of 25, which they agree will go to the first one to win 10 brands of camcorders, A, , and C. ver the past year, the
tosses. Their mothers arrive on the scene and call a halt to the average monthly unit sales of the camcorders was determined,
game when Shiloh has won 7 tosses and Caitlin has won 5. ater and the results are indicated in Table .5. Assume that future
Shiloh and Caitlin split the money according to the Pascal and sales follow the pattern indicated in the table.
Fermat formula. What is Shiloh s share of the pot a etermine the probability that a sale of a camcorder next
Interrupted Game Repeat Problem 2 for Shiloh and month is for brand .
Caitlin s next meeting, when the police brea up their game of 10 b Next month, if a sale occurs at the Exton store, find the
tosses for 50, with Shiloh having won 5 tosses and Caitlin only 2. probability that it is for brand C.
Biased Die When a biased die is tossed, the probabilities
Table . U S M
of 1 and 2 showing are the same. The probabilities of 3 and 4
showing are also the same, but are twice those of 1 and 2. The A C
probabilities of 5 and 6 showing are also the same, but are three
times those of 1 and 2. etermine P.1/. Exton 25 40 30
Whyton 20 25 30
For the sample space fa; b; c; d; e; f; gg, suppose that the
probabilities of a, b, c, d, and e are the same and that the
probabilities of f and g are the same. Is it possible to determine In Pr b ems f r the gi en pr babi ity nd the dds that E
P. f/ If it is also nown that P.fa; fg/ D 13 , what more can be i ccur
said 4 2
P.E/ D 5
P.E/ D 7
ax Increase A legislative body is considering a tax
increase to support education. A survey of 100 registered voters P.E/ D 0:7 P.E/ D 0:015
was conducted, and the results are indicated in Table .4. Assume
that the survey re ects the opinions of the voting population. If a In Pr b ems the dds that E i ccur are gi en Find
person from that population is selected at random, determine each PE
of the following (empirical) probabilities. 7:5 100 : 1 3:7 aWa
a P(favors tax increase)
b P(opposes tax increase) eather Forecast A television weather forecaster
c P(is a Republican with no opinion) reported a 7 chance of rain tomorrow. What are the odds that it
will rain tomorrow
Table . T I S If the odds of event E n t occurring are 3 W 5, what are the
Favor ppose No pinion Total odds that E does occur Repeat the question with the odds of
event E n t occurring being a W b.
emocrat 30 30 5 65
Birthday Surprise For En as in Example 6, calculate
Republican 10 15 5 30 P.E25 / as a percentage rounded to one decimal place.
ther 5 0 0 5
Birthday Surprise For En as in Example 6, calculate
Total 45 45 10 100 P.E30 / as a percentage rounded to one decimal place.

Objective C
o sc ss con t ona ro a t
ot a re ce sa e s ace
S
an t e or na s ace o ana e a
stoc ast c rocess t t e a of a
ro a t tree o e e o t e
C
enera t cat on a The probability of an event could be affected when additional related information about
for P.E \ F/
the experiment is nown. For example, if you guess at the answer to a multiple-choice
question having five choices, the probability of getting the correct answer is 15 .
However, if you now that answers A and are wrong and, thus, can be ignored, the
probability of guessing the correct answer increases to 13 . In this section, we consider
similar situations in which we want the probability of an event E when it is nown
that some other event F has occurred. This is called a conditional probability and
we write P.EjF/ for the conditional probability of E, given F . For instance, in the
situation involving the multiple-choice question, we have
1
P.guessing correct answerjA and eliminated/ D
3
 Section 8.5 Con t ona Pro a t an toc ast c Processes 389

To investigate the notion of conditional probability, we consider the following

situation. A fair die is rolled, and we are interested in the probability of the event

E D feven number showsg

The usual equiprobable sample space for this experiment is

S D f1; 2; 3; 4; 5; 6g

so

E D f2; 4; 6g

Thus,
#.E/ 3 1
P.E/ D D D
#.S/ 6 2
Now we change the situation a bit. Suppose the die is rolled out of our sight, and
then we are told that a number greater than 3 occurred. In light of this additional infor-
mation, what now is the probability of an even number To answer that question, we
reason as follows. The event F of a number greater than 3 is

F D f4; 5; 6g

Since F already occurred, the set of possible outcomes is no longer S it is F. That is,
F becomes our new sample space, called a reduced sample space or a subspace of S.
The outcomes in F are equally li ely, and, of these, only 4 and 6 are favorable to E
that is,

E \ F D f4; 6g

Since two of the three outcomes in the reduced sample space are favorable to an even
number occurring, we say that 23 is the c nditi na pr babi ity f an e en number gi en
that a number greater than 3 ccurred:
#.E \ F/ 2
P.EjF/ D D
#.F/ 3
The Venn diagram in Figure .14 illustrates the situation.

1 3
4 n(E F )
P (E | F ) =
2 5 n(F )
6 = 2
3
E F
Reduced
E F sample
space, F

FIGURE Venn diagram for conditional

probability.

If we compare the conditional probability P.EjF/ D 23 with the unconditional

probability P.E/ D 12 , we see that P.EjF/ > P.E/. This means that nowing that a
number greater than 3 occurred increases the li elihood that an even number occurred.
There are situations, however, in which conditional and unconditional probabilities are
the same. These are discussed in the next section.
390 C ntro ct on to Pro a t an tat st cs

In summary, we have the following generalization of Equation (1):

F C
If E and F are events associated with an equiprobable sample space and
F ¤ ;, then
#.E \ F/
P.EjF/ D
#.F/

Since E \ F and E0 \ F are dis oint events whose union is F, it is easy to see that
P.EjF/ C P.E0 jF/ D 1
from which we get
P.E0 jF/ D 1 " P.EjF/

E AM LE B B

A bag contains two blue elly beans (say, B1 and B2 ) and two white elly beans
( 1 and 2 ). If two elly beans are randomly ta en from the bag, without replace-
ment, find the probability that the second elly bean ta en is white, given that the first
one is blue. (See Figure .15.)
S For our equiprobable sample space, we ta e all ordered pairs, such as
.B1 ; 2 / and . 2 ; 1 /, whose components indicate the elly beans selected on the first
and on the second draw. et B and be the events
B D fblue on first drawg
D fwhite on second drawg
We are interested in
#. \ B/
P. jB/ D
#.B/
The reduced sample space B consists of all outcomes in which a blue elly bean is
drawn first:
B D f.B1 ; B2 /; .B1 ; 1 /; .B1 ; 2 /; .B2 ; B1 /; .B2 ; 1 /; .B2 ; 2 /g

Event \ B consists of the outcomes in B for which the second elly bean is white:
\ B D f.B1 ; 1 /; .B1 ; 2 /; .B2 ; 1 /; .B2 ; 2 /g

Draw two jelly beans

without replacement.

FIGURE Two white and two blue elly beans in a bag.

 Section 8.5 Con t ona Pro a t an toc ast c Processes 391

Since #.B/ D 6 and #. \ B/ D 4, we have

4 2
P. jB/ D D
6 3

Now ork Problem 1 G

Example 1 showed how e cient the use of a reduced sample space can be. Note
that it was not necessary to list all the outcomes either in the original sample space or
in event . Although we listed the outcomes in B, we could have found #.B/ by using
counting methods.
There are two ways in which the first elly bean can be blue, and three possibilities
for the second elly bean, which can be either the remaining blue elly bean or one of
the two white elly beans. Thus, #.B/ D 2 ! 3 D 6.
The number #. \ B/ could also be found by means of counting methods.

E AM LE S

In a survey of 150 people, each person was as ed his or her marital status and opin-
ion about oating a bond issue to build a community swimming pool. The results are
summarized in Table .6. If one of these persons is randomly selected, find each of the
following conditional probabilities.

Table . S
Favor (F) ppose (F 0 ) Total

arried . / 60 20 0
Single . 0/ 40 30 70
Total 100 50 150

a The probability that the person favors the bond issue, given that the person is
married.
S We are interested in P.Fj /. The reduced sample space . / contains
0 married persons, of which 60 favor the bond issue. Thus,

#.F \ / 60 3
P.Fj / D D D
#. / 0 4

b The probability that the person is married, given that the person favors the bond
issue.
S We want to find P. jF/. The reduced sample space .F/ contains
100 persons who favor the bond issue. f these, 60 are married. Hence,

#. \ F/ 60 3
P. jF/ D D D
#.F/ 100 5

Note that here P. jF/ ¤ P.Fj /. Equality is possible precisely if P. / D P.F/,

assuming that all of P. /, P.F/, and P. \ F/ are not equal to zero.

G
392 C ntro ct on to Pro a t an tat st cs

Another method of computing a conditional probability is by means of a formula

involving probabilities with respect to the rigina sample space. efore stating the for-
mula, we will provide some motivation. (The discussion that follows is oversimplified
in the sense that certain assumptions are tacitly made.)
We consider P.EjF/, in terms of the events F and E \ F and their respective prob-
abilities P.F/ and P.E \ F/. We assume that P.F/ ¤ 0. et the experiment associated
with our problem be repeated n times, where n is very large. Then the number f tria s
in hich F ccurs is appr ximate y n ! P.F/. f these, the number in hich e ent E
also ccurs is appr ximate y n ! P.E \ F/. For large n, we estimate P.EjF/ by the rel-
ative frequency of the number of occurrences of E \ F with respect to the number of
occurrences of F, which is approximately

n ! P.E \ F/ P.E \ F
D
n ! P.F P.F/

This result strongly suggests the formula that appears in the following formal de niti n
of conditional probability. (The definition applies to any sample space, whether or not
it is equiprobable.)

The c nditi na pr babi ity of an event E, given that event F has occurred, is denoted
P.EjF/ and is de ned by
P.E \ F/
P.EjF/ D if P.F/ ¤ 0
P.F/

Similarly,

P.F \ E/
P.FjE/ D if P.E/ ¤ 0
P.E/

We emphasize that the pr babi ities in Equati ns and are ith respect t the
rigina samp e space. Here we do n t deal directly with a reduced sample space.

E AM LE C

After the initial production run of a new style of steel des , a quality control technician
found that 40 of the des s had an alignment problem and 10 had both a defective
paint ob and an alignment problem. If a des is randomly selected from this run and it
has an alignment problem, what is the probability that it also has a defective paint ob
S et A and be the events

A D falignment problemg
D fdefective paint obg

We are interested in P. jA/, the probability of a defective paint ob, given an alignment
problem. From the given data, we have P.A/ D 0:4 and P. \ A/ D 0:1. Substituting
into Equation (3) gives

P. \ A/ 0:1 1
P. jA/ D D D
P.A/ 0:4 4

It is convenient to use Equation (3) to solve this problem, because we are given
probabilities rather than information about the sample space.
Now ork Problem 7 G
 Section 8.5 Con t ona Pro a t an toc ast c Processes 393

E AM LE G

If a family has two children, find the probability that both are boys, given that one of
the children is a boy. Assume that a child of either gender is equally li ely and that, for
example, having a girl first and a boy second is ust as li ely as having a boy first and
a girl second.
S et E and F be the events
E D fboth children are boysg
F D fat least one of the children is a boyg
We are interested in P.EjF/. etting the letter denote boy and denote girl , we
use the equiprobable sample space
SDf , , , g
where, in each outcome, the order of the letters indicates the order in which the children
are born. Thus,
EDf g FDf , , g and E\FDf g
From Equation (3),
1
P.E \ F/ 4 1
P.EjF/ D D D
P.F/ 3 3
4
Alternatively, this problem can be solved by using the reduced sample space F:
#.E \ F/ 1
P.EjF/ D D
#.F/ 3
Now ork Problem 9 G
Equations (3) and (4) can be rewritten in terms of products by clearing fractions.
This gives
P.E \ F/ D P.F/P.EjF/
and
P.F \ E/ D P.E/P.FjE/
y the commutative law, P.E \ F/ D P.F \ E/, so we can combine the preceding
equations to get an important law:

G M L
P.E \ F/ D P.E/P.FjE/
D P.F/P.EjF/

The general multiplication law states that the probability that two events b th
occur is equal to the probability that one of them occurs, times the conditional proba-
bility that the other one occurs, given that the first has occurred.

E AM LE A

A computer hardware company placed an ad for its new modem in a popular computer
magazine. The company believes that the ad will be read by 32 of the magazine s
readers and that 2 of those who read the ad will buy the modem. Assume that this is
true, and find the probability that a reader of the magazine will read the ad and buy the
modem.
394 C ntro ct on to Pro a t an tat st cs

S etting R denote the event read ad and B denote buy modem , we are
interested in P.R\B/. We are given that P.R/ D 0:32. The fact that 2 of the readers of
the ad will buy the modem can be written P.BjR/ D 0:02. y the general multiplication
law, Equation (5),
P.R \ B/ D P.R/P.BjR/ D .0:32/.0:02/ D 0:0064
Now ork Problem 11 G
S
The general multiplication law is also called the law of compound probability. The
reason is that it is extremely useful when applied to an experiment that can be expressed
as a sequence (or a compounding) of two or more other experiments, called trials or
stages. The original experiment is called a compound experiment, and the sequence
of trials is called a stochastic process. The probabilities of the events associated with
each trial (beyond the first) could depend on what events occurred in the preceding
trials, so they are conditional probabilities.
When we analyze a compound experiment, a tree diagram is extremely useful in
eeping trac of the possible outcomes at each stage. A complete path from the start to
a tip of the tree gives an outcome of the experiment.
The notion of a compound experiment is discussed in detail in the next example.
Read it carefully. Although the discussion is lengthy for the sa e of developing a new
idea, the actual computation ta es little time.

E AM LE C T

Two cards are drawn without replacement from a standard dec of cards. Find the
probability that the second card is red.
S The experiment of drawing two cards without replacement can be thought
of as a compound experiment consisting of a sequence of two trials: The first is drawing
a card, and the second is drawing a card after the first card has been drawn. The first
trial has two possible outcomes:
R1 D fred cardg or B1 D fblac cardg
(Here the subscript 1 refers to the first trial.) In Figure .16, these outcomes are
represented by the two branches in the first level of the tree. eep in mind that these
outcomes are mutually exclusive, and they are also exhausti e in the sense that there
are no other possibilities. Since there are 26 cards of each color, we have
26 26
P.R1 / D and P.B1 / D
52 52
These unc nditi na probabilities are written along the corresponding branches. We
appropriately call Figure .16 a probability tree.
Now, if a red card is obtained in the first trial, then, of the remaining 51 cards, 25
are red and 26 are blac . The card drawn in the second trial can be red .R2 / or blac
.B2 /. Thus, in the tree, the for at R1 has two branches: red and blac . The c nditi na
probabilities P.R2 jR1 / D 25 51
and P.B2 jR1 / D 26 51
are placed along these branches.
Similarly, if a blac card is obtained in the first trial, then, of the remaining 51 cards,
26 are red and 25 are blac . Hence, P.R2 jB1 / D 26 51
and P.B2 jB1 / D 25
51
, as indicated
alongside the two branches emanating from B1 . The complete tree has two levels (one
for each trial) and four paths (one for each of the four mutually exclusive and exhaustive
events of the compound experiment).
Note that the sum of the probabilities along the branches from the vertex Start
to R1 and B1 is 1:
26 26
C D1
52 52
 Section 8.5 Con t ona Pro a t an toc ast c Processes 395

Outcome Probability
P(R2 |R1)
25 R2 R1 R2 26 25 25
=
51 52 51 102

P(R1)
R1
26
52
26 26 26
51 B2 R1 B2 52 51
Start
26 R2 B1 R2 26 26 13
=
51 52 51 51
26
52
B1

25 26 25
51 B2 B1 B2
52 51
Trial 1 Trial 2
(First draw) (Second draw)

FIGURE Probability tree for compound experiment.

In general, the sum of the probabilities along all the branches emanating from a single
vertex to an outcome of that trial must be 1. Thus, for the vertex at R1 ,
25 26
C D1
51 51
and for the vertex at B1 ,
26 25
C D1
51 51
Now, consider the topmost path. It represents the event red on first draw and red
on second draw . y the general multiplication law,

26 25 25
P.R1 \ R2 / D P.R1 /P.R2 jR1 / D ! D
52 51 102
That is, the pr babi ity f an e ent is btained by mu tip ying the pr babi ities in the
branches f the path f r that e ent. The probabilities for the other three paths are also
indicated in the tree.
Returning to the original question, we see that two paths give a red card on the sec-
ond draw, namely, the paths for R1 \ R2 and B1 \ R2 . Therefore, the event second card
red is the union of two mutually exclusive events. y the addition law, the probability
of the event is the sum of the probabilities for the two paths:

26 25 26 26 25 13 1
P.R2 / D ! C ! D C D
52 51 52 51 102 51 2
Note how easy it was to find P.R2 / by using a probability tree.
Here is a summary of what we have done:

R2 D .R1 \ R2 / [ .B1 \ R2 /
P.R2 / D P.R1 \ R2 / C P.B1 \ R2 /
D P.R1 /P.R2 jR1 / C P.B1 /P.R2 jB1 /
26 25 26 26 25 13 1
D ! C ! D C D
52 51 52 51 102 51 2

Now ork Problem 29 G

396 C ntro ct on to Pro a t an tat st cs

E AM LE C

Two cards are drawn without replacement from a standard dec of cards. Find the
probability that both cards are red.
S Refer bac to the probability tree in Figure .16. nly one path gives a red
card on both draws, namely, that for R1 \ R2 . Thus, multiplying the probabilities along
this path gives the desired probability,
26 25 25
P.R1 \ R2 / D P.R1 /P.R2 jR1 / D ! D
52 51 102
Now ork Problem 33 G
E AM LE C C

A company uses one computer chip in assembling each unit of a product. The chips are
purchased from suppliers A, B, and C and are randomly pic ed for assembling a unit.
Twenty percent come from A, 30 come from B, and the remainder come from C. The
company believes that the probability that a chip from A will prove to be defective in
the first 24 hours of use is 0.03, and the corresponding probabilities for and C are
0.04 and 0.01, respectively. If an assembled unit is chosen at random and tested for
24 continuous hours, what is the probability that the chip in it is defective
S In this problem, there is a sequence of two trials: selecting a chip (A, B, or
C) and then testing the selected chip defective . / or nondefective . 0 / . We are given
the unconditional probabilities
P.A/ D 0:2 and P.B/ D 0:3
Since A, B, and C are mutually exclusive and exhaustive,
P.C/ D 1 " .0:2 C 0:3/ D 0:5
From the statement of the problem, we also have the conditional probabilities
P. jA/ D 0:03 P. jB/ D 0:04 P. jC/ D 0:01
We want to find P. /. To begin, we construct the two-level probability tree shown in
Figure .17. We see that the paths that give a defective chip are those for the events
A\ B\ C\

Probability

D (0.2) (0.03)
0.03

0.97
0.2 D'

D (0.3) (0.04)
0.04
0.3
Start B

0.96 D'

0.5 D (0.5) (0.01)

0.01

0.99
D'

FIGURE Probability tree for Example .

 Section 8.5 Con t ona Pro a t an toc ast c Processes 397

Since these events are mutually exclusive,

P. / D P.A \ / C P.B \ / C P.C \ /
D P.A/P. jA/ C P.B/P. jB/ C P.C/P. jC/
D .0:2/.0:03/ C .0:3/.0:04/ C .0:5/.0:01/ D 0:023
Now ork Problem 47 G
The general multiplication law can be extended so that it applies to more than two
events. For n events, we have
P.E1 \ E2 \ ! ! ! \ En /
D P.E1 /P.E2 jE1 /P.E3 jE1 \ E2 / ! ! ! P.En jE1 \ E2 \ ! ! ! \ En!1 /
(We assume that all conditional probabilities are defined.) In words, the probability that
two or more events all occur is equal to the probability that one of them occurs, times
the conditional probability that a second one occurs given that the first occurred,
times the conditional probability that a third occurs given that the first two occurred, and
so on. For example, in the manner of Example 7, the probability of drawing three red
cards from a dec without replacement is
26 25 24
P.R1 \ R2 \ R3 / D P.R1 /P.R2 jR1 /P.R3 jR1 \ R2 / D ! !
52 51 50
E AM LE B B

ag I contains one blac and two red elly beans, and ag II contains one pin elly
bean. (See Figure .1 .) A bag is selected at random. Then a elly bean is randomly
ta en from it and placed in the other bag. A elly bean is then randomly ta en from that
bag. Find the probability that this elly bean is pin .

Draw Draw
jelly bean jelly bean

Bag I Bag II
Select
bag

Draw Draw
jelly bean jelly bean

Bag II Bag I

FIGURE elly bean selections from bags.

S This is a compound experiment with three trials:

a Selecting a bag
b Ta ing a elly bean from the bag
c Putting the elly bean in the other bag and then ta ing a elly bean from that bag
398 C ntro ct on to Pro a t an tat st cs

1/2 R

R Probability
2/3
1/2 1 2 1
P 2 3 2
I
1/2 B
1/2
1/3
B
Start 1 1 1
1/2 P 2 3 2

1/2 2/4 R
1/4
II P B
1
1/4 P
1
1
1
2 4

Trial 1 Trial 2 Trial 3

(Select bag) (First draw) (Second draw)

FIGURE Three-level probability tree.

We want to find P(pin elly bean on second draw). We analyze the situation by
constructing a three-level probability tree. (See Figure .1 .) The first trial has two
equally li ely possible outcomes, ag I or ag II, so each has probability of 12 .
If ag I was selected, the second trial has two possible outcomes, red .R/ or
blac .B/, with conditional probabilities P.RjI/ D 23 and P.BjI/ D 13 . If ag II was
selected, there is one possible outcome, pin .P/, so P.PjII/ D 1. Thus, the second
level of the tree has three branches.
Now we turn to the third trial. If ag I was selected and a red elly bean ta en from
it and placed in ag II, then ag II contains one red and one pin elly bean. Hence,
at the end of the second trial, the for at vertex R has two branches, R and P, with
conditional probabilities

1 1
P.RjI \ R/ D and P.PjI \ R/ D
2 2

Similarly, the tree shows the two possibilities if ag I was initially selected and a blac
elly bean was placed into ag II. Now, if ag II was selected in the first trial, then
the pin elly bean in it was ta en and placed into ag I, so ag I contains two red,
one blac , and one pin elly bean. Thus, the for at P has three branches, one with
probability 24 and two with probability 14 .
We see that three paths give a pin elly bean on the third trial, so for each, we
multiply the probabilities along its branches. For example, the second path from the
top represents I ! R ! P the probability of this event is

P.I \ R \ P/ D P.I/P.RjI/P.PjI \ R/
1 2 1
D ! !
2 3 2

Adding the probabilities for the three paths gives

1 2 1 1 1 1 1 1
P.pin elly bean on second draw/ D ! ! C ! ! C !1!
2 3 2 2 3 2 2 4
1 1 1 3
D C C D
6 12

Now ork Problem 43 G

 Section 8.5 Con t ona Pro a t an toc ast c Processes 399

R BLEMS
iven the equiprobable sample space Table .
Area I Area II Area III Total
S D f1; 2; 3; 4; 5; 6; 7; ; g
Favor .F/ 46 35 44 125
and events pposed . / 22 15 10 47
No opinion . / 10 10 2
E D f1; 3g
Total 7 5 64 200
F D f1; 2; 4; 5; 6g
G D f5; 6; 7; ; g College Selection and Family Income A survey of 175
students resulted in the data shown in Table . , which shows the
type of college the student attends and the income level of the
find each of the following.
student s family. Suppose a student in the survey is randomly
a P.EjF/ b P.E0 jF/ c P.EjF0 / selected.
d P.FjE/ e P.EjF \ G/ a Find the probability that the student attends a public college,
iven the equiprobable sample space given that the student comes from a middle-income family.
b Find the probability that the student is from a high-income
S D f1; 2; 3; 4; 5g family, given that the student attends a private college.
c If the student comes from a high-income family, find the
and events probability that the student attends a private college.
d Find the probability that the student attends a public college
E D f1; 2g or comes from a low-income family.

F D f3; 4g Table .
College
G D f1; 2; 3g
Income Private Public Total
find each of the following.
High 14 11 25
a P.E/ b P.EjF/ c P.EjG/
d P.GjE/ e P.GjF0 / f P.E0 jF0 / iddle 25 55 0

If P.E/ > 0, find P.EjE/. ow 10 60 70

If P.E/ > 0, show P.;jE/ D 0. Total 4 126 175

If P.E0 jF/ D 0:62, find P.EjF/. Cola Preference A survey was ta en among cola drin ers
If F and G are mutually exclusive events with positive to see which of two popular brands people preferred. It was found
probabilities, find P.FjG/. that 45 li ed brand A, 40 li ed brand , and 20 li ed both
brands. Suppose that a person in the survey is randomly selected.
If P.E/ D 14 ; P.F/ D 13 , and P.E \ F/ D 16 , find each of the
following: a Find the probability that the person li ed brand A, given that
he or she li ed brand .
a P.EjF/ b P.FjE/ b Find the probability that the person li ed brand , given that
If P.E/ D 14 ; P.F/ D 13 , and P.EjF/ D 34 , find P.E [ F/. he or she li ed brand A.
int Use the addition law to find P.E [ F/. Quality Control f the smartphones produced by a
1
If P.E/ D P.E [ F/ D
, 7
,
and P.E \ F/ D find each of1
, well- nown firm, 17 have poor sound quality and 11 have
3 12 12
both poor sound quality and scratched screens. If a smartphone is
the following:
randomly selected from a shipment and it has poor sound quality
a P.FjE/ b P.F/ c P.EjF/
what is the probability that it has a scratched screen
d P.EjF0 / int Find P.E \ F0 / by using the identity
P.E/ D P.E \ F/ C P.E \ F0 /. In Pr b ems and assume that a chi d f either gender is
4 3 7 equa y ike y and that f r examp e ha ing a gir rst and a b y
If P.E/ D 5
, P.F/ D 10
, and P.E [ F/ D 10
, find P.FjE/.
sec nd is just as ike y as ha ing a b y rst and a gir sec nd
Gypsy Moth ecause of gypsy moth infestation of three
Genders of Offspring If a family has two children, what is
large areas that are densely populated with trees, consideration is
the probability that one child is a boy, given that at least one child
being given to aerial spraying to destroy larvae. A survey was
is a girl
made of the 200 residents of these areas to determine whether or
not they favor the spraying. The resulting data are shown in Genders of Offspring If a family has three children, find
Table .7. Suppose that a resident is randomly selected. et I be each of the following.
the event the resident is from Area I and so on. Find each of the a The probability that it has two girls, given that at least one
following: child is a boy
a P.F/ b P.FjII/ c P. jI/ b The probability that it has at least two girls, given that the
d P(III) e P.IIIj / f P.IIj 0 / oldest child is a girl
400 C ntro ct on to Pro a t an tat st cs

Coin oss If a fair coin is tossed three times in succession, Cards If two cards are randomly drawn without
find each of the following. replacement from a standard dec of cards, find the probability
a The probability of getting exactly two tails, given that the that the second card is a face card.
second toss is a tail Cards If two cards are randomly drawn without
b The probability of getting exactly two tails, given that the replacement from a standard dec , find the probability of getting
second toss is a head two ac s, given that the first card is a face card.
Coin oss If a fair coin is tossed four times in succession, a e p Call arbara Smith, a sales representative, is
find the odds of getting four tails, given that the first toss is a tail. staying overnight at a hotel and has a brea fast meeting with an
important client the following morning. She as ed the des to
Die Roll A fair die is rolled. Find the probability of getting
give her a 7 . . wa e-up call so she can be prompt for the
a number greater than 4, given that the number is even.
meeting. The probability that she will get the call is 0. . If she
Cards If a card is drawn randomly from a standard dec , gets the call, the probability that she will be on time is 0. .
find the probability of getting a spade, given that the card is blac . If the call is not given, the probability that she will be
Dice Roll If two fair dice are rolled, find the probability on time is 0.4. Find the probability that she will be on time
that two 1 s occur, given that at least one die shows a 1. for the meeting.
Dice Roll If a fair red die and a fair green die are rolled, axpayer Survey In a certain school district, a
find the probability that the sum is greater than , given that a 5 questionnaire was sent to all property taxpayers concerning
shows on the red die. whether or not a new high school should be built. f those
that responded, 60 favored its construction, 30 opposed it,
Dice Roll If a fair red die and a fair green die are rolled, and 10 had no opinion. Further analysis of the data concerning
find the probability of getting a total of 7, given that the green die the area in which the respondents lived gave the results in
shows an even number. Table . .
Dice Roll A fair die is rolled two times in succession.
Table .
a Find the probability that the sum is 7, given that the second
roll is neither a 3 nor a 5. Urban Suburban
b Find the probability that the sum is 7 and that the second roll
is neither a 3 nor a 5. Favor 45 55
Die Roll If a fair die is rolled two times in succession, find ppose 55 45
the probability of getting a total greater than , given that the first No opinion 35 65
roll is greater than 2.
Coin and Die If a fair coin and a fair die are thrown, find
the probability that the coin shows tails, given that the number on a If one of the respondents is selected at random,
the die is odd. what is the probability that he or she lives in an urban area
b If a respondent is selected at random, use the result of part
Cards If a card is randomly drawn from a dec of 52 cards,
(a) to find the probability that he or she favors the construction of
find the probability that the card is a ing, given that it is a heart.
the school, given that the person lives in an urban area.
Cards If a card is randomly drawn from a dec of 52
Mar eting A travel agency has a computerized telephone
cards, find the probability that the card is a heart, given that it is a
that randomly selects telephone numbers for advertising
face card (a ac , queen, or ing).
suborbital space trips. The telephone automatically dials the
Cards If two cards are randomly drawn without selected number and plays a prerecorded message to the recipient
replacement from a standard dec , find the probability that the of the call. Experience has shown that 2 of those called show
second card is not a heart, given that the first card is a heart. interest and contact the agency. However, of these, only 1.4
actually agree to purchase a trip.
In Pr b ems c nsider the experiment t be a c mp und
a Find the probability that a person called will contact the
experiment
agency and purchase a trip.
Cards If two cards are randomly drawn from a standard b If 100,000 people are called, how many can be expected to
dec , find the probability that both cards are aces if, contact the agency and purchase a trip
a the cards are drawn without replacement. Rabbits in a all at A tall hat contains four yellow and
b the cards are drawn with replacement. three red rabbits.
Cards If three cards are randomly drawn without a If two rabbits are randomly pulled from the hat without
replacement from a standard dec , find the probability of replacement, find the probability that the second rabbit pulled is
getting a ing, a queen, and a ac , in that order. yellow, given that the first rabbit pulled is red.
Cards If three cards are randomly drawn without b Repeat part (a), but assume that the first rabbit is replaced
replacement from a standard dec , find the probability of getting before the second rabbit is pulled.
the ace of spades, the ace of hearts, and the ace of diamonds, in elly Beans in a Bag ag 1 contains five green and two
that order. red elly beans, and ag 2 contains two green, two white, and
Cards If three cards are randomly drawn without three red elly beans. A elly bean is randomly ta en from ag 1
replacement from a standard dec , find the probability that all and placed into ag 2. If a elly bean is then randomly ta en from
three cards are ac s. ag 2, find the probability that the elly bean is green.
 Section 8.6 n e en ent ents 401

Balls in a Box ox 1 contains three red and two white with a chec for 1 in each. The remaining box contains one
balls. ox 2 contains two red and two white balls. A box is chosen envelope with a chec for 5000 inside and five envelopes with a
at random and then a ball is chosen at random from it. What is the chec for 1 inside each. If the contestant must select a box at
probability that the ball is white random and then randomly draw an envelope, find the probability
Balls in a Box ox 1 contains two red and three white that a chec for 5000 is inside.
balls. ox 2 contains three red and four white balls. ox 3 Quality Control A company uses one computer chip in
contains two red, two white, and two green balls. A box is chosen assembling each unit of a product. The chips are purchased from
at random, and then a ball is chosen at random from it. suppliers A, , and C and are randomly pic ed for assembling a
a Find the probability that the ball is white. unit. Ten percent come from A, 20 come from , and the
b Find the probability that the ball is red. remainder come from C. The probability that a chip from A will
c Find the probability that the ball is green. prove to be defective in the first 24 hours of use is 0.06, and the
corresponding probabilities for and C are 0.04 and 0.05,
elly Beans in a Bag ag 1 contains one green and one red respectively. If an assembled unit is chosen at random and tested
elly bean, and ag 2 contains one white and one red elly bean. for 24 continuous hours, what is the probability that the chip in it
A bag is selected at random. A elly bean is randomly ta en from will prove to be defective
it and placed in the other bag. A elly bean is then randomly drawn
from that bag. Find the probability that the elly bean is white. Quality Control A manufacturer of widgets has four
assembly lines: A, , C, and . The percentages of output
Dead Batteries s. Wood s lights went out in a recent produced by the lines are 30 , 20 , 35 , and 15 , respectively,
storm and she reached in the itchen drawer for 4 batteries for her and the percentages of defective units they produce are 6 , 3 ,
ashlight. There were 10 batteries in the drawer, but 5 of them 2 , and 5 . If a widget is randomly selected from stoc , what is
were dead. Fortunately, the ashlight wor ed with the ones she the probability that it is defective
randomly chose. ater, s Wood and r Wood discussed the
question of the probability of choosing 4 dead batteries. She Voting In a certain town, 45 of eligible voters are
argued, with obvious notation P. 1 \ 2 \ 3 \ 4 / D iberals, 30 are Conservatives, and the remainder are Social
P. 1 /P. 2 j 1 /P. 3 j 1 \ 2 /P. 4 j 1 \ 2 \ 3 / D emocrats. In the last provincial election, 20 of the iberals,
5 4 3 2 1 35 of the Conservatives, and 40 of the Social emocrats
5 C4
& & & D . He said the answer is . Who was voted.
10 7 42 10 C4
right a If an eligible voter is chosen at random, what is the
probability that he or she is a Social emocrat who voted
Quality Control An energy drin producer requires the b If an eligible voter is chosen at random, what is the
use of a can filler on each of its two product lines. The ellow probability that he or she voted
Cow line produces 36,000 cans per day, and the Half Throttle line
produces 60,000 cans per day. ver a period of time, it has been ob Applicants A restaurant has four openings for waiters.
found that the ellow Cow filler underfills 2 of its cans, whereas Suppose Allison, esley, Alan, Tom, Alaina, ronwen, Ellie, and
the Half Throttle filler underfills 1 of its cans. At the end of a Emmy are the only applicants for these obs, and all are equally
day, a can was selected at random from the total production. Find qualified. If four are hired at random, find the probability that
the probability that the can was underfilled. Allison, esley, Tom, and ronwen were chosen, given that Ellie
and Emmy were not hired.
Game Show A TV game show host presents the following
situation to a contestant. n a table are three identical boxes. ne Committee Selection Suppose six female and five male
of them contains two identical envelopes. In one is a chec for students wish to fill three openings on a campus committee on
5000, and in the other is a chec for 1. Another box contains cultural diversity. If three of the students are chosen at random for
two envelopes with a chec for 5000 in each and six envelopes the committee, find the probability that all three are female, given
that at least one is female.

Objective I E
o e e o t e not on of n e en ent In our discussion of conditional probability, we saw that the probability of an event
e ents an a t e s ec a can be affected by the nowledge that another event has occurred. In this section, we
t cat on a
consider the situation where the additional information has no effect. That is, the con-
ditional probability P.EjF/ and the unconditional probability P.E/ are the same. In this
discussion we assume that P.E/ ¤ 0 ¤ P.F/.
When P.EjF/ D P.E/, we say that E is independent of F. If E is independent of
F, it follows that F is independent of E (and vice versa). To prove this, assume that
P.EjF/ D P.E/. Then
P.E \ F/ P.F/P.E j F/ P.F/P.E/
P.FjE/ D D D D P.F/
P.E/ P.E/ P.E/
Independence of two events is defined by which means that F is independent of E. Thus, to prove independence, it su ces to
probabilities, not by a causal relationship. show that either P.EjF/ D P.E/ or P.FjE/ D P.F/, and when one of these is true, we
simply say that E and F are independent e ents.
402 C ntro ct on to Pro a t an tat st cs

et E and F be events with p siti e probabilities. Then E and F are said to be inde
pendent events if either
P.EjF/ D P.E/
or
P.FjE/ D P.F/
If E and F are not independent, they are said to be dependent events.
ependence does not imply causality.

Thus, with dependent events, the occurrence of one of the events d es affect the prob-
ability of the other. If E and F are independent events, it can be shown that the events
in each of the following pairs are also independent:
E and F0 E0 and F E0 and F0

E AM LE S T T E A I

A fair coin is tossed twice. et E and F be the events

E D .head on first toss/
F D .head on second toss/
etermine whether or not E and F are independent events.
S We suspect that they are independent, because one coin toss should not
in uence the outcome of another toss. To confirm our suspicion, we will compare
P.E/ with P.EjF/. For the equiprobable sample space S D fHH, HT, TH, TTg, we
have E D fHH, HTg and F D fHH, THg. Thus,
#.E/ 2 1
P.E/ D D D
#.S/ 4 2
#.E \ F #.fHHg/ 1
P.EjF/ D D D
#.F/ #.F/ 2
Since P.EjF/ D P.E/, events E and F are independent.
Now ork Problem 7 G
In Example 1 we suspected the result, and certainly there are other situations where
we have an intuitive feeling as to whether or not two events are independent. For exam-
ple, if a red die and green die are tossed, we expect (and it is indeed true) that the events
3 on red die and 6 on green die are independent, because the outcome on one die
should not be in uenced by the outcome on the other die. Similarly, if two cards are
drawn ith rep acement from a dec of cards, we would assume that the events first
card is a ac and second card is a ac are independent. However, suppose the cards
are drawn ith ut rep acement. ecause the first card drawn is not put bac in the dec ,
it should have an effect on the outcome of the second draw, so we expect the events
to be dependent. However, intuiti n can be unre iab e in determining hether e ents
E and F are independent r dependent . Ultimately, intuition can only be tested by
showing the truth (or falsity) of either Equation (1) or Equation (2).

E AM LE S S

In a study of smo ing and sinusitis, 4000 people were studied, with the results as given
in Table .10. Suppose a person from the study is selected at random. n the basis of
the data, determine whether or not the events having sinusitis ( ) and smo ing (S)
are independent events.
 Section 8.6 n e en ent ents 403
Table .10 S S
Smo er Nonsmo er Total

Sinusitis 432 101 1450

No sinusitis 52 2022 2550
Total 60 3040 4000

S We will compare P. / with P. jS/. The number P. / is the proportion of

the people studied that have sinusitis:
1450 2
P. / D D D 0:3625
4000 0
For P. jS/, the sample space is reduced to 60 smo ers, of which 432 have sinusitis:
432
P. jS/ D D D 0:45
We emphasize that dependency does not 60 20
imply causation, nor is it implied by Since P. jS/ ¤ P. /, having sinusitis and smo ing are dependent.
causation
Now ork Problem 9 G
The general multiplication law ta es on an extremely important form for indepen-
dent events. Recall that law:
P.E \ F/ D P.E/P.FjE/
D P.F/P.EjF/
If events E and F are independent, then P.FjE/ D P.F/, so substitution in the first
equation gives
P.E \ F/ D P.E/P.F/
The same result is obtained from the second equation. Thus, we have the following law:

S M L
If E and F are independent e ents then
P.E \ F/ D P.E/P.F/

Equation (3) states that if E and F are independent events, then the probability that
E and F both occur is the probability that E occurs times the probability that F occurs.
Note that Equation (3) is n t valid when E and F are dependent.

E AM LE S R

Suppose the probability of the event ob lives 20 more years (B) is 0. and the prob-
ability of the event oris lives 20 more years ( ) is 0. 5. Assume that B and are
independent events.
a Find the probability that both ob and oris live 20 more years.
S We are interested in P.B \ /. Since B and are independent events, the
special multiplication law applies:
P.B \ / D P.B/P. / D .0: /.0: 5/ D 0:6
b Find the probability that at least one of them lives 20 more years.
S Here we want P.B [ /. y the addition law,
P.B [ / D P.B/ C P. / " P.B \ /
404 C ntro ct on to Pro a t an tat st cs

From part (a), P.B \ / D 0:6 , so

P.B [ / D 0: C 0: 5 " 0:6 D 0: 7
c Find the probability that exactly one of them lives 20 more years.
S We first express the event
E D fexactly one of them lives 20 more yearsg
in terms of the given events, B and . Now, event E can occur in one of two mutua y
exc usi e ways: ob lives 20 more years but oris does not .B \ 0 /, or oris lives 20
more years but ob does not .B0 \ /. Thus,
0
E D .B \ / [ .B0 \ /
y the addition law (for mutually exclusive events),
0
P.E/ D P.B \ / C P.B0 \ /
To compute P.B \ 0 /, we note that, since B and are independent, so are B and 0
(from the statement preceding Example 1). Accordingly, we can use the multiplication
law and the rule for complements:
0
P.B \ / D P.B/P. 0 /
D P.B/.1 " P. // D .0: /.0:15/ D 0:12
Similarly,
P.B0 \ / D P.B0 /P. / D .0:2/.0: 5/ D 0:17
Substituting into Equation (4) gives
P.E/ D 0:12 C 0:17 D 0:2

Now ork Problem 25 G

In Example 3, it was assumed that events B and are independent. However, if
ob and oris are related in some way, it is possible that the survival of one of them
has a bearing on the survival of the other. In that case, the assumption of independence
is not ustified, and we could not use the special multiplication law, Equation (3).

E AM LE C

In a math exam, a student was given the following two-part problem. A card is drawn
rapidly from a dec of 52 cards. et , , and R be the events
D fheart dra ng
D fking dra ng
R D fred card dra ng
Find P. \ / and P. \ R/.
For the first part, the student wrote
13 4 1
P. \ / D P. /P. / D ! D
52 52 52
and for the second part, she wrote
13 26 1
P. \ R/ D P. /P.R/ D ! D
52 52
The answer was correct for P. \ / but not for P. \ R/. Why
 Section 8.6 n e en ent ents 405

S The reason is that the student assumed independence in b th parts by using

the special multiplication law to multiply unconditional probabilities when, in fact, that
assumption should n t have been made. et us examine the first part of the exam prob-
lem for independence. We will see whether P. / and P. j / are the same. We have
13 1
P. / D D
52 4
and
1
P. j / D one heart out of four ings
4
Since P. / D P. j /, events and are independent, so the student s procedure is
valid. For the second part, again we have P. / D 14 , but
13 1
P. jR/ D D 13 hearts out of 26 red cards
26 2
Since P. jR/ ¤ P. /, events and R are dependent, so the student should not have
multiplied the unconditional probabilities. However, the student would have been safe
by using the genera multiplication law that is,
13 1
P. \ R/ D P. /P.Rj / D !1D
52 4
equivalently,
26 13 1
P. \ R/ D P.R/P. jR/ D ! D
52 26 4
ore simply, observe that \RD , so
13 1
P. \ R/ D P. / D D
52 4

Now ork Problem 33 G

Equation (3) is often used as an alternative means of defining independent events,
and we will consider it as such:

Events E and F are independent if and only if

P.E \ F/ D P.E/P.F/

Putting everything together, we can say that to prove that events E and F, with
nonzero probabilities, are independent, only one of the following relationships has to
be shown:

P.EjF/ D P.E/

or

P.FjE/ D P.F/

or

P.E \ F/ D P.E/P.F/

In other words, if any one of these equations is true, then all of them are true if any is
false, then all of them are false, and E and F are dependent.
406 C ntro ct on to Pro a t an tat st cs

E AM LE

Two fair dice, one red and the other green, are rolled, and the numbers on the top faces
are noted. et E and F be the events
E D .number on red die is even/
F D .sum is 7/
Test whether P.E \ F/ D P.E/P.F/ to determine whether E and F are independent.
S ur usual sample space for the roll of two dice has 6 ! 6 D 36 equally li ely
outcomes. For event E, the red die can fall in any of three ways and the green die any
of six ways, so E consists of 3 ! 6 D 1 outcomes. Thus, P.E/ D 136 D 12 . Event F has
six outcomes:
F D f.1; 6/; .2; 5/; .3; 4/; .4; 3/; .5; 2/; .6; 1/g
where, for example, we ta e .1; 6/ to mean 1 on the red die and 6 on the green die.
6
Therefore, P.F/ D 36 D 16 , so
1 1 1
P.E/P.F/ D
! D
2 6 12
Now, event E \ F consists of all outcomes in which the red die is even and the sum is
7. Using Equation (5) as an aid, we see that
E \ F D f.2; 5/; .4; 3/; .6; 1/g
Thus,
3 1
P.E \ F/ D
D
36 12
Since P.E \ F/ D P.E/P.F/, events E and F are independent. This fact may not have
been obvious before the problem was solved.
Now ork Problem 17 G
E AM LE S

For a family with at least two children, let E and F be the events
E D .at most one boy/
F D .at least one child of each sex/
Assume that a child of either sex is equally li ely and that, for example, having
a girl first and a boy second is ust as li ely as having a boy first and a girl second.
etermine whether E and F are independent in each of the following situations:
a The family has exactly two children.
S We will use the equiprobable sample space
SDf , , , g
and test whether P.E \ F/ D P.E/P.F/. We have
EDf , , g FDf , g E\FDf , g
3 2 1 2 1
Thus, P.E/ D 4
, P.F/ D 4
D 2
, and P.E \ F/ D 4
D 2
. We as whether
‹
P.E \ F/ D P.E/P.F/
and see that
1 3 1 3
¤ ! D
2 4 2
so E and F are dependent events.
 Section 8.6 n e en ent ents 407

b The family has exactly three children.

S ased on the result of part (a), one may have an intuitive feeling that E and
F are dependent. Nevertheless, we must test this con ecture. For three children, we use
the equiprobable sample space
SDf , , , , , , , g
Again, we test whether P.E \ F/ D P.E/P.F/. We have
EDf , , , g
FDf , , , , , g
E\FDf , , g
4
Hence, P.E/ D D 12 , P.F/ D 6
D 43 , and P.E \ F/ D 3 , so
1 3 3
P.E/P.F/ D ! D D P.E \ F/
2 4
Therefore, we have the s me hat unexpected result that events E and F are indepen-
dent. Intuition cannot always be trusted.

Now ork Problem 27 G

We now generalize our discussion of independence to the case of more than two
events.

The events E1 ; E2 ; : : : ; En are said to be independent if and only if for each set of
two or more of the events, the probability of the intersection of the events in the set
is equal to the product of the probabilities of the events in that set.

For instance, let us apply the definition to the case of three events .n D 3/. We say
that E, F, and G are independent events if the special multiplication law is true for these
events, ta en two at a time and three at a time. That is, each of the following equations
must be true:
9
P.E \ F/ D P.E/P.F/ =
P.E \ G/ D P.E/P.G/ Two at a time
;
P.F \ G/ D P.F/P.G/
P.E \ F \ G/ D P.E/P.F/P.G/g Three at a time
As another example, if events E, F, G, and are independent, then we can assert such
things as
P.E \ F \ G \ / D P.E/P.F/P.G/P. /
P.E \ G \ / D P.E/P.G/P. /
and
P.F \ / D P.F/P. /
Similar conclusions can be made if any of the events are replaced by their complements.

E AM LE C

Four cards are randomly drawn, with replacement, from a dec of 52 cards. Find the
probability that the cards chosen, in order, are a ing ( ), a queen ( ), a ac ( ), and
a heart ( ).
408 C ntro ct on to Pro a t an tat st cs

S Since there is replacement, what happens on any draw does not affect the
outcome on any other draw, so we can assume independence and multiply the uncon-
ditional probabilities. We obtain
P. \ \ \ / D P. /P. /P. /P. /
4 4 4 13 1
D ! ! ! D
52 52 52 52 7
Now ork Problem 35 G

E AM LE A T

Personnel Temps, a temporary-employment agency, requires that each ob applicant

ta e the company s aptitude test, which has 0 accuracy.
a Find the probability that the test will be accurate for the next three applicants who
are tested.
S et A, B, and C be the events that the test will be accurate for applicants A,
, and C, respectively. We are interested in
P.A \ B \ C/
Since the accuracy of the test for one applicant should not affect the accuracy for any
of the others, it seems reasonable to assume that A, B, and C are independent. Thus,
we can multiply probabilities:
P.A \ B \ C/ D P.A/P.B/P.C/
D .0: /.0: /.0: / D .0: /3 D 0:512
b Find the probability that the test will be accurate for at least two of the next three
applicants who are tested.
S Here, at east t means exactly two or exactly three . In the first case, the
possible ways of choosing the two tests that are accurate are
A and B A and C B and C
In each of these three possibilities, the test for the remaining applicant is not accurate.
For example, choosing A and B gives the event A \ B \ C0 , whose probability is
P.A/P.B/P.C0 / D .0: /.0: /.0:2/ D .0: /2 .0:2/
Verify that the probability for each of the other two possibilities is also .0: /2 .0:2/.
Summing the three probabilities gives
P.exactly two accurate/ D 3Œ.0: /2 .0:2/! D 0:3 4
Using this result and that of part (a), we obtain
P.at least two accurate/ D P.exactly two accurate/ C P.three accurate/
D 0:3 4 C 0:512 D 0: 6
Alternatively, the problem could be solved by computing
1 " ŒP.none accurate/ C P.exactly one accurate/!
Why
Now ork Problem 21 G
We conclude with a note of caution: n t c nfuse independent e ents ith mutu
a y exc usi e e ents The concept of independence is defined in terms of probability,
whereas mutual exclusiveness is not. When two events are independent, the occurrence
of one of them does not affect the probability of the other. However, when two events
 Section 8.6 n e en ent ents 409

are mutually exclusive, they cannot occur simultaneously. Although these two concepts
are not the same, we can draw some conclusions about their relationship. If E and F
are mutually exclusive events ith p siti e pr babi ities then
P.E \ F/ D 0 ¤ P.E/P.F/ since P.E/ > 0 andP.F/ > 0
which shows that E and F are dependent. In short, mutua y exc usi e e ents ith p s
iti e pr babi ities must be dependent. Another way of saying this is that independent
e ents ith p siti e pr babi ities are n t mutua y exc usi e.

R BLEMS
1
If events E and F are independent with P.E/ D 3
and independent or dependent. (See page 5 of the uly 21, 1 1,
P.F/ D 34 , find each of the following. issue of SA A for the article Pests Now Appearing at a
Theater Near ou .)
a P.E \ F/ b P.E [ F/ c P.E j F/
d P.E0 j F/ e P.E \ F0 / f P.E [ F0 / Table .1 T
g P.E j F0 / ale Female Total
If events E, F, and G are independent with
Tal ers 60 10 70
P.E/ D 0:1; P.F/ D 0:3, and P.G/ D 0:6, find each
of the following. Crunchers 55 25 0
a P.E \ F/ b P.F \ G/ Seat ic ers 15 10 25
c P.E \ F \ G/ d P.E j .F \ G// Total 130 45 175
e P.E0 \ F \ G0 /
If events E and F are independent with P.E/ D 2
and Dice Two fair dice are rolled, one red and one green, and
7
the numbers on the top faces are noted. et event E be number
P.E \ F/ D 1 , find P.F/. on red die is neither 1 nor 2 nor 3 and event F be sum is 7 .
If events E and F are independent with P.E0 j F0 / D 14 , find etermine whether E and F are independent or dependent.
P.E/. Cards A card is randomly drawn from an ordinary dec
In Pr b ems and e ents E and F satisfy the gi en c nditi ns of 52 cards. et E and F be the events red card drawn and face
etermine hether E and F are independent r dependent card drawn respectively. etermine whether E and F are
independent or dependent.
P.E/ D 23 , P.F/ D 67 , P.E \ F/ D 4
7 Coins If two fair coins are tossed, let E be the event at
P.E/ D 0:2 ; P.F/ D 0:15; P.E \ F/ D 0:03 most one head and F be the event exactly one head . etermine
Stoc bro ers Six hundred investors were surveyed to whether E and F are independent or dependent.
determine whether a person who uses a full-service stoc bro er Coins If three fair coins are tossed, let E be the event at
has better performance in his or her investment portfolio than one most one head and F be the event at least one head and one
who uses a discount bro er. In general, discount bro ers usually tail . etermine whether E and F are independent or
offer no investment advice to their clients, whereas full-service dependent.
bro ers usually offer help in selecting stoc s but charge larger Chips in a Bowl A bowl contains seven chips numbered
fees. The data, based on the past 12 months, are given in from 1 to 7. Two chips are randomly withdrawn with replacement.
Table .11. etermine whether the event of having a full-service et E, F, and G be the events
bro er and the event of having an increase in portfolio value are
independent or dependent. E D 3 on first withdrawal
Table .11 F D 3 on second withdrawal
Increase ecrease Total G D sum is odd

Full service 320 0 400

a etermine whether E and F are independent or dependent.
iscount 160 40 200 b etermine whether E and G are independent or dependent.
Total 4 0 120 600 c etermine whether F and G are independent or dependent.
d Are E, F, and G independent
Cinema Offenses An observation of 175 patrons in a Chips in a Bowl A bowl contains six chips numbered
theater resulted in the data shown in Table .12. The table shows from 1 to 6. Two chips are randomly withdrawn. et E be the
three types of cinema offenses committed by male and female event of withdrawing two even-numbered chips and let F be
patrons. Crunchers include noisy eaters of popcorn and other the event of withdrawing two odd-numbered chips.
morsels, as well as cold-drin slurpers. etermine whether the a Are E and F mutually exclusive
event of being a male and the event of being a cruncher are b Are E and F independent
410 C ntro ct on to Pro a t an tat st cs

In Pr b ems and e ents E and F satisfy the gi en Survival Rates The probability that person A survives 15
c nditi ns etermine hether E and F are independent r more years is 34 , and the probability that person survives 15
dependent more years is 45 . Find the probability of each of the following.
P.EjF/ D 0:6, P.E \ F/ D 0:2, P.FjE/ D 0:4 Assume independence.
P.E j F/ D 23 ; P.E [ F/ D 17
1
; P.E \ F/ D 5
a A and both survive 15 years.
b survives 15 years, but A does not.
c Exactly one of A and survives 15 years.
In Pr b ems y u may make use f y ur intuiti n d At least one of A and survives 15 years.
c ncerning independent e ents if n thing t that e ect is e Neither A nor survives 15 years.
speci ed
Matching In his des , a secretary has a drawer containing
Dice Two fair dice are rolled, one red and one green. Find a mixture of two sizes of paper (A and ) and another drawer
the probability that the red die is a 4 and the green die is a number containing a mixture of envelopes of two corresponding sizes. The
greater than 4. percentages of each size of paper and envelopes in the drawers are
given in Table .13. If a piece of paper and an envelope are
randomly drawn, find the probability that they are the same size.

Table .1 E
rawers
Size Paper Envelopes

A 63 57
37 43
Die If a fair die is rolled three times, find the probability
that a 2 or 3 comes up each time.
Fitness Classes At a certain fitness center, the probability elly Beans in a Bag A bag contains five red, seven white,
that a member regularly attends an aerobics class is 14 . If two and six green elly beans. If two elly beans are randomly ta en
members are randomly selected, find the probability that both out with replacement, find each of the following.
attend the class regularly. Assume independence.
a The probability that the first elly bean is white and the
Monopoly In the game of onopoly, a player rolls two second is green.
fair dice. ne special situation that can arise is that the numbers b The probability that one elly bean is red and the other one is
on the top faces of the dice are the same (such as two 3 s). This white
result is called a double , and when it occurs, the player
continues his or her turn and rolls the dice again. The pattern Dice Suppose two fair dice are rolled twice. Find the
continues, unless the player is unfortunate enough to throw probability of getting a total of 7 on one of the rolls and a total of
doubles three consecutive times. In that case, the player goes to 12 on the other one.
ail. Find the probability that a player goes to ail in this way given elly Beans in a Bag A bag contains three red, two white,
that he has already rolled doubles twice in a row. four blue, and two green elly beans. If two elly beans are
Cards Three cards are randomly drawn, with replacement, randomly withdrawn with replacement, find the probability that
from an ordinary dec of 52 cards. Find the probability that the they have the same color.
cards drawn, in order, are an ace, a face card (a ac , queen, or Die Find the probability of rolling three consecutive
ing), and a spade. numbers in three throws of a fair die.
Die If a fair die is rolled seven times, find each of the ic ets in at Twenty tic ets numbered from 1 to 20
following. are placed in a hat. If two tic ets are randomly drawn with
a The probability of getting a number greater than 4 each time replacement, find the probability that the sum is 35.
b The probability of getting a number less than 4 each time Coins and Dice Suppose two fair coins are tossed and then
Exam Grades In a sociology course, the probability that two fair dice are rolled. Find each of the following.
ill gets an A on the final exam is 34 , and for im and inda, the a The probability that two tails and two 3 s occur
probabilities are 12 and 45 , respectively. Assume independence and b The probability that two heads, one 4, and one 6 occur
find each of the following. Carnival Game In a carnival game, a well-balanced
roulette-type wheel has 12 equally spaced slots that are numbered
a The probability that all three of them get an A on the exam
from 1 to 12. The wheel is spun, and a ball travels along the rim of
b The probability that none of them get an A on the exam
the wheel. When the wheel stops, the number of the slot in which
c The probability that, of the three, only inda gets an A
the ball finally rests is considered the result of the spin. If the
Die If a fair die is rolled four times, find the probability of wheel is spun three times, find each of the following.
getting at least one 1. a The probability that the first number will be 4 and the second
and third numbers will be 5
b The probability that there will be one even number and two
odd numbers
 Section 8.7 a es or a 411

Cards Three cards are randomly drawn, with replacement, Shooting Gallery At a shooting gallery, suppose ill, im,
from an ordinary dec of 52 cards. Find each of the following. and inda each ta e one shot at a moving target. The probability
that ill hits the target is 0.5, and for im and inda the
probabilities are 0.4 and 0.7, respectively. Assume independence
and find each of the following.
a The probability that none of them hit the target
b The probability that inda is the only one of them that hits
the target
c The probability that exactly one of them hits the target
d The probability that exactly two of them hit the target
e The probability that all of them hit the target
Decision Ma ing1 The president of eta Construction
a The probability of drawing, in order, a heart, a spade, and a Company must decide which of two actions to ta e, namely, to
red queen rent or to buy expensive excavating equipment. The probability
b The probability of drawing exactly three aces that the vice president ma es a faulty analysis and, thus,
c The probability that one red queen, one spade and one red ace recommends the wrong decision to the president is 0.04. To be
are drawn thorough, the president hires two consultants, who study the
d The probability of drawing exactly one ace problem independently and ma e their recommendations. After
Multiple Choice Exam A quiz contains 10 having observed them at wor , the president estimates that the
multiple-choice problems. Each problem has five choices for the first consultant is li ely to recommend the wrong decision with
answer, but only one of them is correct. Suppose a student probability 0.05, the other with probability 0.1. He decides to
randomly guesses the answer to each problem. Find each of the ta e the action recommended by a ma ority of the three
following by assuming that the guesses are independent. recommendations he receives. What is the probability that he will
ma e the wrong decision
a The probability that the student gets exactly three correct
answers
b The probability that the student gets at most three correct
answers
c The probability that the student gets four or more correct
answers

Objective B F
o so e a a es ro e o e eo In this section, we will be dealing with a two-stage experiment in which we now
a es for a the outcome of the second stage and are interested in the probability that a particular
outcome has occurred in the first stage.
To illustrate, suppose it is believed that of the total population (our sample space),
have a particular disease. Imagine also that there is a new blood test for detecting
the disease and that researchers have evaluated its effectiveness. ata from extensive
testing show that the blood test is not perfect: Not only is it positive for only 5 of
those who have the disease, but it is also positive for 3 of those who do not. Suppose
a person from the population is selected at random and given the blood test. If the result
is positive, what is the probability that the person has the disease
To analyze this problem, we consider the following events:
D .having the disease/
D .testing positive/
and their complements:
0
D .not having the disease/
0
D .testing negative/
We are given:
P. / D 0:0 P. j / D 0: 5 P. j 0 / D 0:03

1
Samuel oldberg, Pr babi ity an Intr ducti n (Prentice-Hall, Inc., 1 60, over Publications, Inc., 1 6),
p. 113. Adapted by permission of the author.
412 C ntro ct on to Pro a t an tat st cs

so by complementarity we also have,

P. 0 / D 1 " 0:0 D 0: 2 P. 0 j / D 0:05 P. 0 j 0 / D 0: 7
Figure .20 shows a two-stage probability tree that re ects this information. The first
stage ta es into account either having or not having the disease, and the second stage
shows possible test results.

T
| D)
P (T
0.95
D P (T
D) '|D
)
P( 8 0.05
0.0
T'

P( ) T
D' | D'
0.9 ) P (T
2 0.03
D' P (T
'|D
')
0.97
T'

FIGURE Two-stage probability tree.

We are interested in the probability that a person who tests positive has the disease.
That is, we want to find the conditional probability that occurred in the first stage,
given that occurred in the second stage:
P. j /
It is important to understand the difference between the conditional probabilities
P. j / and P. j /. The probability P. j /, which is gi en to us, is a typical con-
ditional probability, in that it deals with the probability of an outcome in the second
stage after an outcome in the first stage has occurred. However, with P. j /, we have
a reverse situation. Here we must find the probability of an outcome in the rst stage,
given that an outcome in the second stage occurred. This probability does not fit the
usual (and more natural) pattern of a typical conditional probability. Fortunately, we
have all the tools needed to find P. j /. We proceed as follows.
From the definition of conditional probability,
P. \ /
P. j / D
P. /
Consider the numerator. Applying the general multiplication law gives
P. \ / D P. /P. j /
D .0:0 /.0: 5/ D 0:076
which is indicated in the path through and in Figure .21. The denominator, P. /,
is the sum of the probabilities for all paths of the tree ending in . Thus,
0
P. / D P. \ / C P. \ /
D P. /P. j / C P. 0 /P. j 0 /
D .0:0 /.0: 5/ C .0: 2/.0:03/ D 0:1036
Hence,
P. \ /
P. j / D
P. /
probability of path through and
D
sum of probabilities of all paths to
0:076 760 1 0
D D D $ 0:734
0:1036 1036 25
 Section 8.7 a es or a 413

So the probability that the person has the disease, given that the test is positive, is
approximately 0.734. In other words, about 73.4 of people who test positive actually
have the disease. This probability was relatively easy to find by using basic principles
Equation (1) and a probability tree (Figure .21).

T P (D T)
| D)
P (T = P (D)P(T|D)
0.95 = (0.08) (0.95)
D P (T = 0.076
'|D
D) )
P( 8 0.05
0.0 T'

P( ) T P (D' T)
D | D'
0.9 ') P (T = P (D')P(T|D')
2 0.03 = (0.92) (0.03)
D' P (T
'|D
')
0.97
T'

FIGURE Probability tree to determine P. j /.

At this point, some terminology should be introduced. The unc nditi na proba-
bilities P. / and P. / are called prior probabilities, because they are given bef re
we have any nowledge about the outcome of a blood test. The conditional probability
P. j / is called a posterior probability, because it is found after the outcome, . /, of
the test is nown.
From our answer for P. j /, we can easily find the posterior probability of not
having the disease given a positive test result:
1 0 6
P. 0 j / D 1 " P. j / D 1 " D $ 0:266
25 25
f course, this can also be found by using the probability tree:
probability of path through 0 and
P. 0 j / D
sum of probabilities of all paths to
.0: 2/.0:03/ 0:0276 276 6
D D D D $ 0:266
0:1036 0:1036 1036 25
It is not really necessary to use a probability tree to find P. j /. Instead, a formula can
be developed. We now that
P. \ / P. /P. j /
P. j / D D
P. / P. /
Although we used a probability tree to express P. / conveniently as a sum of proba-
bilities, the sum can be found another way. Ta e note that events and 0 have two
properties: They are mutually exclusive and their union is the sample space S. Such
events are collectively called a partition of S. Using this partition, we can brea up
event into mutually exclusive pieces :
0
D \SD \. [ /
Then, by the distributive and commutative laws,
0
D. \ /[. \ /
Since and 0 are mutually exclusive, so are events \ and 0 \ . Thus, has
been expressed as a union of mutually exclusive events. In this form, we can find P. /
by adding probabilities. Applying the addition law for mutually exclusive events to
Equation (3) gives
0
P. / D P. \ / C P. \
D P. /P. j / C P. 0 /P. j 0 /
414 C ntro ct on to Pro a t an tat st cs

Substituting into Equation (2), we obtain

P. /P. j /
P. j / D
P. /P. j / C P. 0 /P. j 0 /
which is a formula for computing P. j /.
Equation (4) is a special case (namely, for a partition of S into two events) of the
following general formula, called Bayes’ formula, after Thomas ayes (1702 1761),
an 1 th-century English minister who discovered it:

B F
Suppose F1 ; F2 : : : ; Fn are n events that partition a sample space S. That is, the Fi s
are mutually exclusive and their union is S. Furthermore, suppose that E is any event
in S, where P.E/ > 0. Then the conditional probability of Fi given that event E has
occurred is expressed by
P.Fi /P.EjFi /
P.Fi jE/ D
P.F1 /P.EjF1 / C P.F2 /P.EjF2 / C ! ! ! C P.Fn /P.EjFn /
for each value of i, where i D 1; 2; : : : ; n.

ayes formula has had wide application in decision ma ing.

Rather than memorize the formula, a probability tree can be used to obtain P.Fi jE/.
Using the tree in Figure .22, we have
probability for path through Fi and E
P.Fi jE/ D
sum of all probabilities for paths to E

)
P (E | F 1 E P (F1)P (E | F1)
F1
1)

)
F

P (E | F 2 E P (F2)P (E | F2)
P(

P (F 2) F2

P (F )
i) P (E | F i E P (Fi )P (E | Fi )
Fi
P(
Fn
)

)
P(E | F n E P (Fn )P (E | Fn )
Fn

FIGURE Probability tree for P.Fi jE/.

E AM LE C

A digital camcorder manufacturer uses one microchip in assembling each camcorder it

produces. The microchips are purchased from suppliers A, , and C and are randomly
pic ed for assembling each camcorder. Twenty percent of the microchips come from
A, 35 come from , and the remainder come from C. ased on past experience,
the manufacturer believes that the probability that a microchip from A is defective is
0.03, and the corresponding probabilities for and C are 0.02 and 0.01, respectively.
A camcorder is selected at random from a day s production, and its microchip is found
to be defective. Find the probability that it was supplied a from A, b from , and
c from C. d From what supplier was the microchip most li ely purchased
 Section 8.7 a es or a 415

P(F | A)
A F (0.2) (0.03)
0.03

A)
P( 2
0.

P(B) P(F | B)
B F (0.35) (0.02)
0.35 0.02

P(
C
0.4 )
5

P(F | C)
C F (0.45) (0.01)
0.01

FIGURE ayes probability tree for Example 1.

S We define the following events:

A D .supplier A/
B D .supplier /
C D .supplier C/
F D .defective microchip/

We have

P.A/ D 0:2 P.B/ D 0:35 P.C/ D 0:45

and the conditional probabilities

P.FjA/ D 0:03 P.FjB/ D 0:02 P.FjC/ D 0:01

which are re ected in the probability tree in Figure .23. Note that the figure shows
only the portion of the complete probability tree that relates to event F. This is all
that actually needs to be drawn, and this abbreviated form is often called a Bayes’
probability tree.
For part (a), we want to find the probability of A given that F has occurred. That is,
probability of path through A and F
P.AjF/
sum of probabilities of all paths to F
.0:2/.0:03/
D
.0:2/.0:03/ C .0:35/.0:02/ C .0:45/.0:01/
0:006
D
0:006 C 0:007 C 0:0045
0:006 60 12
D D D $ 0:343
0:0175 175 35
This means that approximately 34.3 of the defective microchips come from
supplier A.
For part (b), we have

probability of path through B and F

P.BjF/ D
sum of probabilities of all paths to F
.0:35/.0:02/ 0:007 70 14
D D D D
0:0175 0:0175 175 35
416 C ntro ct on to Pro a t an tat st cs

For part (c),

probability of path through C and F
P.CjF/ D
sum of probabilities of all paths to F
.0:45/.0:01/ 0:0045 45
D D D D
0:0175 0:0175 175 35
For part (d), the greatest of P.AjF/, P.BjF/, and P.CjF/ is P.BjF/. Thus, the defec-
tive microchip was most li ely supplied by .
Now ork Problem 9 G
E AM LE B B

Two identical bags, ag I and ag II, are on a table. ag I contains one red and one
blac elly bean ag II contains two red elly beans. (See Figure .24.) A bag is selected
at random, and then a elly bean is randomly ta en from it. The elly bean is red. What
Bag I Bag II is the probability that the other elly bean in the selected bag is red
FIGURE iagram for S ecause the other elly bean could be red or blac , we might hastily con-
Example 2. 1
clude that the answer is . This is false. The question can be restated as follows: Find
2
the probability that the elly bean came from ag II, given that the elly bean is red. We
define the events
B1 D . ag I selected/
B2 D . ag II selected/
R D .red elly bean selected/
We want to find P.B2 jR/. Since a bag is selected at random,
1 1
P.B1 / D and P.B2 / D
2 2
From Figure .24, we conclude that
1
P.RjB1 / D
and P.RjB2 / D 1
2
We will show two methods of solving this problem, the first with a probability tree and
the second with ayes formula.
Method Probability ree Figure .25 shows a ayes probability tree for our
problem. Since all paths end at R,
probability for path through B2 and R
P.B2 jR/ D
sum of probabilities of all paths
!1" 1
.1/ 2 2
D ! 1 " ! 1 "2 ! 1 " D 3 D
2 2
C 2
.1/ 3
4

P(R | B1) 1 1
B1 1
R 2 2
2
)
P (B
1
1
2

P (B
2)
1
2
P(R | B2) 1
B2 R 2 1
1

FIGURE ayes probability tree for Example 2.

 Section 8.7 a es or a 417

1
Note that the unconditional probability of choosing ag II, namely, P.B2 / D ,
2
2
increases to , given that a red elly bean was ta en. An increase is reasonable: Since
3
there are only red elly beans in ag II, choosing a red elly bean should ma e it more
li ely that it came from ag II.
Method Bayes’ Formula ecause B1 and B2 partition the sample space, by ayes
formula we have
P.B2 /P.RjB2 /
P.B2 jR/ D
P.B1 /P.RjB1 / C P.B2 /P.RjB2 /
!1" 1
.1/ 2
D ! 1 " ! 1 "2 ! 1 " D 23 D
2 2
C 2 .1/ 4
3

Now ork Problem 7 G

R BLEMS
Suppose events E and F partition a sample space S, where E Disease esting A new test was developed for detecting
and F have probabilities amma s disease, which is believed to affect 3 of the
population. Results of extensive testing indicate that 6 of
4 3 persons who have this disease will have a positive reaction to the
P.E/ D P.F/ D
7 7 test, whereas 7 of those who do not have the disease will also
have a positive reaction.
If is an event such that a What is the probability that a randomly selected person who
has a positive reaction will actually have amma s disease
2 1 b What is the probability that a randomly selected person who
P. jE/ D P. jF/ D
3 has a negative reaction will actually have amma s disease
0 Earnings and Dividends f the companies in a particular
find the probabilities a P.Ej / and b P.Fj /.
sector of the economy, it is believed that 1 4 will have an increase
A sample space is partitioned by events E1 ; E2 , and E3 , whose in quarterly earnings. f those that do, 2 3 will declare a dividend.
probabilities are 15 ; 10
3
, and 12 , respectively. Suppose S is an event f those that do not have an increase, 1 10 will declare a
such that the following conditional probabilities hold: dividend. What percentage of companies that declare a dividend
will have an increase in quarterly earnings
2 7 1 elly Beans in a Bag A bag contains four red and two
P.S j E1 / D P.S j E2 / D P.S j E3 / D
5 10 2 green elly beans, and a second bag contains two red and
three green elly beans. A bag is selected at random and a elly
Find the probabilities P.E1 j S/ and P.E3 j S0 /. bean is randomly ta en from it. The elly bean is red. What is the
Voting In a certain precinct, 42 of the eligible voters are probability that it came from the first bag
registered emocrats, 33 are Republicans, and the remainder Balls in a Bowl owl I contains three red, two white, and
are Independents. uring the last primary election, 45 of the five green balls. owl II contains three red, six white, and nine
emocrats, 37 of the Republicans, and 35 of the Independents green balls. owl III contains six red, two white, and two green
voted. Find the probability that a person who voted is a emocrat. balls. A bowl is chosen at random, and then a ball is chosen at
Imported versus Domestic ires ut of 3000 tires in the random from it. The ball is red. Find the probability that it came
warehouse of a tire distributor, 2000 tires are domestic and 1000 from owl II.
are imported. Among the domestic tires, 40 are all-season Quality Control A manufacturing process requires the use
of the imported tires, 10 are all-season. If a tire is selected at of a robotic welder on each of two assembly lines, A and , which
random and it is an all-season, what is the probability that it is produce 300 and 500 units of product per day, respectively. ased
imported on experience, it is believed that the welder on A produces 2
defective units, whereas the welder on produces 5 defective
units. At the end of a day, a unit was selected at random from the
total production and was found to be defective. What is the
probability that it came from line A
Quality Control An automobile manufacturer has four
plants: A, , C, and . The percentages of total daily output that
are produced by the four plants are 35 , 20 , 30 , and 15 ,
respectively. The percentages of defective units produced by the
plants are estimated to be 2 , 5 , 3 , and 4 , respectively.
Suppose that a car on a dealer s lot is randomly selected and found
to be defective. What is the probability that it came from plant
a A b c C d
418 C ntro ct on to Pro a t an tat st cs

a e p Call arbara Smith, a sales representative, is Physics Exam After a physics exam was given, it turned
staying overnight at a hotel and has a brea fast meeting with an out that only 75 of the class answered every question. f those
important client the following morning. She as ed the front des who did, 0 passed, but of those who did not, only 50 passed.
to give her a 6 . . wa e-up call so she can be prompt for the If a student passed the exam, what is the probability that the
meeting. The probability that the des ma es the call is 0. . If the student answered every question (P.S.: The instructor eventually
call is made, the probability that she will be on time is 0. , but if reached the conclusion that the test was too long and curved the
the call is not made, the probability that she will be on time is 0.7. exam grades, to be fair and merciful.)
If she is on time for the meeting, what is the probability that the Giving p Smo ing In a 2004 survey of smo ers, 50
call was made predicted that they would still be smo ing five years later. Five
Candy Snatcher n a high shelf are two identical opaque years later, 0 of those who predicted that they would be
candy ars containing 50 raisin clusters each. The clusters in one smo ing did not smo e, and of those who predicted that they
of the ars are made with dar chocolate. In the other ar, 20 are would not be smo ing, 5 did not smo e. What percentage
made with dar chocolate and 30 are made with mil chocolate. of those who were not smo ing after five years had predicted that
(They are mixed well, however.) ob ones, who has a sudden they would be smo ing
craving for chocolate, reaches up and randomly ta es a raisin Alien Communication . . Cosmos, a scientist, believes
cluster from one of the ars. If it is made with dar chocolate, that the probability is 25 that aliens from an advanced civilization
what is the probability that it was ta en from the ar containing
on Planet are trying to communicate with us by sending
only dar chocolate
high-frequency signals to Earth. y using sophisticated
Physical Fitness Activity uring the wee of National equipment, Cosmos hopes to pic up these signals. The
Employee Health and Fitness ay, the employees of a large manufacturer of the equipment, Tre ee, Inc., claims that if aliens
company were as ed to exercise a minimum of three times that are indeed sending signals, the probability that the equipment will
wee for at least 20 minutes per session. The purpose was to detect them is 35 . However, if aliens are not sending signals, the
generate exercise miles . All participants who completed this
probability that the equipment will seem to detect such signals is
requirement received a certificate ac nowledging their 1
contribution. The activities reported were power wal ing, cycling, 10
. If the equipment detects signals, what is the probability that
and running. f all who participated, 13 reported power wal ing, 12 aliens are actually sending them
reported cycling, and 16 reported running. Suppose that the Calculus Grades In an honors Calculus I class, 60 of
students had an A average at midterm. f these, 70 ended up
probability that a participant who power wal s will complete
with a course grade of A, and of those who did not have an A
the requirement is 10 , and for cycling and running it is 23 and 13 , average at midterm, 60 ended up with a course grade of A.
respectively. What percentage of persons who completed the If one of the students is selected at random and is found to have
requirement do you expect reported power wal ing (Assume that received an A for the course, what is the probability that the
each participant got his or her exercise from only one activity.) student did not have an A average at midterm
Battery Reliability When the weather is extremely frigid, Movie Critique A well- nown pair of highly in uential
a motorist must charge his car battery during the night in order to movie critics have a popular TV show on which they review new
improve the li elihood that the car will start early the following movie releases and recently released videos. ver the past 10
morning. If he does not charge it, the probability that the car will years, they gave a Two Thumbs Up to 60 of movies that
not start is 45 . If he does charge it, the probability that the car turned out to be box-o ce successes they gave a Two Thumbs
will not start is 1 . Past experience shows that the probability that own to 0 of movies that proved to be unsuccessful. A new
movie, ath Guru, whose release is imminent, is considered
he remembers to charge the battery is 10 . ne morning, during a favorably by others in the industry who have previewed it in fact,
cold spell, he cannot start his car. What is the probability that he they give it a prior probability of success of 0 . Find the
forgot to charge the battery probability that it will be a success, given that the pair of TV
Automobile Satisfaction Survey In a customer critics give it a Two Thumbs own after seeing it. Assume that
all films are given either Two Thumbs Up or Two Thumbs
satisfaction survey, 35 of those surveyed had a apanese-made car, own .
1 3
10
a European-made car, and 10 an American-made car. f the
Balls in a Bowl owl 1 contains five green and four red
first group, 5 said they would buy the same ma e of car again,
balls, and owl 2 contains three green, one white, and three red
and for the other two groups the corresponding percentages are
balls. A ball is randomly ta en from owl 1 and placed in owl 2.
50 and 40 . What is the probability that a person who said he
A ball is then randomly ta en from owl 2. If the ball is green,
or she would buy the same ma e again had a apanese-made car
find the probability that a green ball was ta en from owl 1.
Mineral est Borings A geologist believes that the
probability that the rare earth mineral dalhousium occurs in the Ris y Loan In the loan department of The an of
reater Toronto region is 0.001. If dalhousium is present in that ontreal, past experience indicates that 25 of loan requests are
region, the geologist s test borings will have a positive result 0 considered by ban examiners to fall into the substandard class
of the time. However, if dalhousium is not present, a negative and should not be approved. However, the ban s loan reviewer,
result will occur 0 of the time. r. lac well, is lax at times and concludes that a request is not
a If a test is positive on a site in the region, find the probability in the substandard class when it is, and vice versa. Suppose that
that dalhousium is there. 15 of requests that are actually substandard are not considered
b If a test is negative on such a site, find the probability that substandard by lac well and that 10 of requests that are not
dalhousium is there. substandard are considered by lac well to be substandard and,
hence, not approved.
 Chapter 8 e e 419

a Find the probability that lac well considers that a request is Coins in Chests Each of three identical chests has two
substandard. drawers. The first chest contains a gold coin in each drawer. The
b Find the probability that a request is substandard, given that second chest contains a silver coin in each drawer, and the third
lac well considers it to be substandard. contains a silver coin in one drawer and a gold coin in the other.
c Find the probability that lac well ma es an error in A chest is chosen at random and a drawer is opened. There is a
considering a request. (An error occurs when the request is not gold coin in it. What is the probability that the coin in the other
substandard but is considered substandard, or when the request drawer of that chest is silver
is substandard but is considered to be not substandard.)

Chapter 8 Review
I T S E
S Basic Counting Principle and Permutations
tree diagram asic Counting Principle Ex. 1, p. 350
permutation, n Pr Ex. 5, p. 352
S Combinations and Other Counting Principles
combination, n Cr Ex. 2, p. 357
permutation with repeated ob ects cells Ex. 6, p. 363
S Sample Spaces and Events
sample space sample point finite sample space Ex. 1, p. 36
event certain event impossible event simple event Ex. 6, p. 370
Venn diagram complement, E0 union, [ intersection, \ Ex. 7, p. 371
mutually exclusive events Ex. , p. 373
S Probability
equally li ely outcomes trial relative frequency Ex. 1, p. 376
equiprobable space probability of event, P.E/ Ex. 2, p. 377
addition law for mutually exclusive events empirical probability Ex. 5, p. 37
odds Ex. , p. 3 4
S Conditional Probability and Stochastic Processes
conditional probability, P.E j F/ reduced sample space Ex. 1, p. 3 0
general multiplication law trial compound experiment Ex. 5, p. 3 3
probability tree Ex. 6, p. 3 4
S Independent Events
independent events dependent events Ex. 1, p. 402
special multiplication law Ex. 3, p. 403
S Bayes’ Formula
partition prior probability posterior probability Ex. 1, p. 414
ayes formula ayes probability tree Ex. 2, p. 416

S
It is important to now the number of ways a procedure can n ob ects ta en r at a time. The number of such permutations
occur. Suppose a procedure involves a sequence of k stages. is denoted n Pr and is given by
et n1 be the number of ways the first stage can occur, and
nŠ
n2 the number of ways the second stage can occur, and so on, n Pr D n.n " 1/.n " 2/ ! ! ! .n " r C 1/ D
with nk the number of ways the kth stage can occur. Then the „ ƒ‚ … .n " r/Š
r factors
number of ways the procedure can occur is
If the selection is made without regard to order, then it is
n1 ! n2 ! ! ! nk simply an r-element subset of an n-element set and is called
a combination of n ob ects ta en r at a time. The number of
such combinations is denoted n Cr and is given by
This result is called the asic Counting Principle.
nŠ
An ordered selection of r ob ects, without repetition, n Cr D
ta en from n distinct ob ects is called a permutation of the rŠ.n " r/Š
420 C ntro ct on to Pro a t an tat st cs

When some of the ob ects are repeated, the number of The probability that an event E occurs, given that event F
distinguishable permutations of n ob ects, such that n1 are of has occurred, is called a conditional probability. It is denoted
one type, n2 are of a second type, and so on, and nk are of a by P.E j F/ and can be computed either by considering a
kth type, is reduced equiprobable sample space and using the formula

nŠ #.E \ F/
P.E j F/ D
n1 Šn2 Š ! ! ! nk Š #.F/
or from the formula
where n1 C n2 C ! ! ! C nk D n.
The expression in Equation (5) can also be used to deter- P.E \ F/
P.E j F/ D
mine the number of assignments of ob ects to cells. If n dis- P.F/
tinct ob ects are placed into k ordered cells, with ni ob ects
in cell i, for i D 1; 2; : : : ; k, then the number of such assign- which involves probabilities with respect to the original sam-
ments is ple space.
To find the probability that two events both occur, we
nŠ can use the general multiplication law:
n1 Šn2 Š ! ! ! nk Š
P.E \ F/ D P.E/P.F j E/ D P.F/P.E j F/
where n1 C n2 C ! ! ! C nk D n. Here we multiply the probability that one of the events occurs
A sample space for an experiment is a set S of all possi- by the conditional probability that the other one occurs, given
ble outcomes of the experiment. These outcomes are called that the first has occurred. For more than two events, the cor-
sample points. A subset E of S is called an event. Two special responding law is
events are the sample space itself, which is a certain event,
and the empty set, which is an impossible event. An event P.E1 \ E2 \ ! ! ! \ En /
consisting of a single sample point is called a simple event. D P.E1 /P.E2 j E1 /P.E3 j E1 \ E2 / ! ! !
Two events are said to be mutually exclusive when they have
no sample point in common. P.En j E1 \ E2 \ ! ! ! \ En!1 /
A sample space whose outcomes are equally li ely is
The general multiplication law is also called the law of
called an equiprobable space. If E is an event for a finite
compound probability, because it is useful when applied to
equiprobable space S, then the probability that E occurs is
a compound experiment one that can be expressed as a
given by
sequence of two or more other experiments, called trials or
stages.
#.E/ When we analyze a compound experiment, a probability
P.E/ D tree is extremely useful in eeping trac of the possible out-
#.S/
comes for each trial of the experiment. A path is a complete
sequence of branches from the start to a tip of the tree. Each
If F is also an event in S, we have path represents an outcome of the compound experiment, and
the probability of that path is the product of the probabilities
P.E [ F/ D P.E/ C P.F/ " P.E \ F/ for the branches of the path.
P.E [ F/ D P.E/ C P.F/ for E and F mutually exclusive Events E and F are independent when the occurrence of
P.E0 / D 1 " P.E/ one of them does not affect the probability of the other
that is,
P.S/ D 1
P.E j F/ D P.E/ or P.F j E/ D P.F/
P.;/ D 0

For an event E, the ratio Events that are not independent are dependent.
If E and F are independent, the general multiplication
P.E/ P.E/ law simplifies to the special multiplication law:
D
P.E0 / 1 " P.E/ P.E \ F/ D P.E/P.F/

gives the odds that E occurs. Conversely, if the odds that E Here the probability that E and F both occur is the probabil-
occurs are a W b, then ity of E times the probability of F. The preceding equation
forms the basis of an alternative definition of independence:
a Events E and F are independent if and only if
P.E/ D
aCb
P.E \ F/ D P.E/P.F/
 Chapter 8 e e 421

Three or more events are independent if and only if for each when prior and conditional probabilities are nown, we can
set of two or more of the events, the probability of the inter- use ayes formula:
section of the events in that set is equal to the product of the
probabilities of those events. P.Fi j E/ D
A partition divides a sample space into mutually exclu-
P.Fi /P.E j Fi /
sive events. If E is an event and F1 ; F2 ; : : : ; Fn is a partition,
then, to find the conditional probability of event Fi , given E, P.F1 /P.E j F1 / C P.F2 /P.E j F2 / C ! ! ! C P.Fn /P.E j Fn /

A ayes-type problem can also be solved with the aid of a

ayes probability tree.

R
In Pr b ems determine the a ues Multiple Choice Exam Each question of a 10-question
P3 multiple-choice examination is worth 10 points and has four
r P1 C7 12 C5
choices, only one of which is correct. y guessing, in how many
License Plate A six-character license plate consists of three ways is it possible to receive a score of 0 or better
letters followed by three numbers. How many different license
plates are possible Letter Arrangement How many distinguishable
horizontal arrangements of the letters in ISSISSIPPI are
Dinner In a restaurant, a complete dinner consists of one possible
appetizer, one entree, and one dessert. The choices for the
appetizer are soup and salad for the entree, chic en, stea , Flag Signals Colored ags arranged vertically on a
lobster, and veal and for the dessert, ice cream, pie, and pudding. agpole indicate a signal (or message). How many different
How many complete dinners are possible signals are possible if two red, three green, and four white ags
are all used
Garage Door Opener The transmitter for an electric
garage-door opener transmits a coded signal to a receiver. The Personnel Agency A mathematics professor personnel
code is determined by 10 switches, each of which is either in an agency provides mathematics professors on a temporary basis to
on or off position. etermine the number of different codes universities that are short of staff. The manager has a pool of
that can be transmitted. 17 professors and must send four to alhousie University, seven
to St. ary s, and three to ount Saint Vincent. In how many
Baseball A baseball manager must determine a batting order ways can the manager ma e assignments
for his nine-member team. How many batting orders are possible
our Operator A tour operator has three vans, and
Softball A softball league has seven teams. In terms of first, each can accommodate seven tourists. Suppose 14 people arrive
second, and third place, in how many ways can the season end for a city sightseeing tour and the operator will use only two
Assume that there are no ties. vans. In how many ways can the operator assign the people to
rophies In a trophy case, nine different trophies are to be the vans
placed two on the top shelf, three on the middle, and four on the Suppose S D f1; 2; 3; 4; 5; 6; 7; g is the sample space and
bottom. Considering the order of arrangement on each shelf, in E1 D f1; 2; 3; 4; 5; 6g and E2 D f4; 5; 6; 7g are events for an
how many ways can the trophies be placed in the case experiment. Find a E1 [ E2 , b E1 \ E2 , c E1 0 [ E2 ,
Groups Eleven stranded wait-listed passengers surge to the d E1 \ E1 0 , and e .E1 \ E2 0 /0 . f Are E1 and E2 mutually
counter for boarding passes. ut there are only six boarding passes exclusive
available. How many different groups of passengers can board Die and Coin A die is rolled and then a coin is tossed.
Cards From a 52-card dec of playing cards, a five-card a etermine a sample space for this experiment. etermine the
hand is dealt. In how many ways can exactly two of the cards be events that b a 2 shows and c a head and an even number show.
of one denomination and exactly two be of another denomination Bags of elly Beans Three bags, labeled 1, 2, and 3, each
(Such a hand is called t pairs.) contain two elly beans, one red and the other green. A elly bean
Light Bulbs A carton contains 24 light bulbs, one of which is selected at random from each bag. a etermine a sample
is defective. a In how many ways can three bulbs be selected space for this experiment. etermine the events that b exactly
b In how many ways can three bulbs be selected if one is two elly beans are red and c the elly beans are the same
defective color.
Suppose that E and F are events for an experiment. If
P.E/ D 0:5, P.E [ F/ D 0:6, and P.E \ F/ D 0:1, find P.E0 \ F0 /.
Quality Control A manufacturer of computer chips
pac ages 10 chips to a box. For quality control, two chips
are selected at random from each box and tested. If any
one of the tested chips is defective, the entire box of chips
is re ected for sale. For a box that contains exactly one defective
chip, what is the probability that the box is re ected
422 C ntro ct on to Pro a t an tat st cs

Drugs Each of 100 white rats was in ected with one of four In Pr b ems and the dds that E i ccur are gi en Find
drugs, A, , C, or . rug A was given to 35 , to 25 , and C PE
to 15 . If a rat is chosen at random, determine the probability that 6:1 3:4
it was in ected with either C or . If the experiment is repeated on
a larger group of 300 rats but with the drugs given in the same Cards If a card is randomly drawn from a fair dec of 52
proportion, what is the effect on the previous probability cards, find the probability that it is not a face card (a ac , queen,
or ing), given that it is a heart.

Multiple Choice Exam Each question on a five-question Dice If two fair dice are rolled, find the probability that the
multiple-choice examination has four choices, only one of which sum is less than 7, given that a 6 shows on at least one of the dice.
is correct. If a student answers each question in a random fashion,
Movie and Sequel The probability that a particular movie
what is the probability that the student answers exactly two
will be successful is 0.55, and if it is successful, the probability
questions incorrectly
that a sequel will be made is 0.60. Find the probability that the
Cola Preference To determine the national preference of movie will be successful and followed by a sequel.
cola drin ers, an advertising agency conducted a survey of 200 of
Cards Three cards are drawn from a standard dec of
them. Two cola brands, A and , were involved. The results of the
cards. Find the probability that the cards are, in order, a queen, a
survey are indicated in Table .14. If a cola drin er is selected at
heart, and the ace of clubs if the cards are drawn with
random, determine the (empirical) probability that the person
replacement.
a i es both A and
b i es A, but not Dice If two dice are thrown, find each of the following.
a The probability of getting a total of 7, given that a 4 occurred
Table .1 C on at least one die
b The probability of getting a total of 7 and that a 4 occurred on
i e A only 70
at least one die
i e only 0
Die A fair die is tossed two times in succession. Find the
i e both A and 35 probability that the first toss is less than 4, given that the total is
i e neither A nor 15 greater than .
Total 200 Die If a fair die is tossed two times in succession, find
the probability that the first number is less than or equal to the
second number, given that the second number is less than 3.
elly Beans in a Bag A bag contains six red and six green
elly beans. Cards Three cards are drawn without replacement from a
standard dec of cards. Find the probability that the third card is a
a If two elly beans are randomly selected in succession with club.
replacement, determine the probability that both are red.
Seasoning Survey A survey of 600 adults was made to
b If the selection is made without replacement, determine the
determine whether or not they li ed the taste of a new seasoning.
probability that both are red.
The results are summarized in Table .15.
Dice A pair of fair dice is rolled. etermine the probability
that the sum of the numbers is a 2 or 7, b a multiple of 3, and
c no less than 7. Table .1 S S
Cards Three cards from a standard dec of 52 playing i e isli e Total
cards are randomly drawn in succession with replacement.
etermine the probability that a all three cards are blac and ale 0 40 120
b two cards are blac and the other is a diamond. Female 320 160 4 0
Cards Two cards from a standard dec of 52 playing cards Total 400 200 600
are randomly drawn in succession without replacement.
etermine the probability that a both are hearts and b one is an
ace and the other is a red ing. a If a person in the survey is selected at random, find the
probability that the person disli es the seasoning . 0 /, given that
In Pr b ems and f r the gi en a ue f P E nd the dds the person is a female .F/.
that E i ccur
b etermine whether the events D fli ing the seasoningg and
3 D fbeing a maleg are independent or dependent.
P.E/ D P.E/ D 0: 3
 Chapter 8 e e 423

Chips A bowl contains six chips numbered from 1 to 6. Bags of elly Beans ag I contains four red and two white
Two chips are randomly withdrawn with replacement. et E be elly beans. ag II contains two red and three white elly beans. A
the event of getting a 4 the first time and F be the event of getting bag is chosen at random, and then a elly bean is randomly ta en
a 4 the second time. from it.
a Are E and F mutually exclusive a What is the probability that the elly bean is white
b Are E and F independent b If the elly bean is white, what is the probability that it was
ta en from ag II
College and Family Income A survey of 175 students
resulted in the data shown in Table .16. The table shows the type Grade Distribution ast semester, the grade distribution
of college the student attends and the income level of the student s for a certain class ta ing an upper-level college course was
family. If a student is selected at random, determine whether the analyzed. It was found that the proportion of students receiving a
event of attending a public college and the event of coming from a grade of A was 0.4 and the proportion getting an A and being a
middle-class family are independent or dependent. graduate student was 0.1. If a student is randomly selected from
this class and is found to have received an A, find the probability
that the student is a graduate student.
Table .1 S S
Alumni Reunion At the most recent alumni day at Alpha
College University, 735 persons attended. f these, 603 lived within the
Income Private Public Total state, and 43 of them were attending for the first time. Among
the alumni who lived out of the state, 72 were attending for the
High 15 10 25 first time. That day a ra e was held, and the person who won had
also won it the year before. Find the probability that the winner
iddle 25 55 0
was from out of state.
ow 10 60 70
Quality Control A music company burns C s on two
Total 50 125 175 shifts. The first shift produces 3000 discs per day, and the second
produces 5000. From past experience, it is believed that of the
output produced by the first and second shifts, 1 and 2 are
If P.E/ D 14 ; P.F/ D 13 , and P.E j F/ D 16 , find P.E [ F/. scratched, respectively. At the end of a day, a disc was selected at
random from the total production.
Shrubs When a certain type of shrub is planted, the
a Find the probability that the C is scratched.
probability that it will ta e root is 0.7. If four shrubs are planted,
b If the C is scratched, find the probability that it came from
find each of the following. Assume independence.
the first shift.
a The probability that all of them ta e root
b The probability that exactly three of them ta e root Aptitude est In the past, a company has hired only
c The probability that at least three of them ta e root experienced personnel for its word-processing department.
ecause of a shortage in this field, the company has decided to
Antibiotic A certain antibiotic is effective for 75 of the
hire inexperienced persons and will provide on-the- ob training. It
people who ta e it. Suppose four persons ta e this drug. What is
has supplied an employment agency with a new aptitude test that
the probability that it will be effective for at least three of them
has been designed for applicants who desire such a training
Assume independence.
position. f those who recently too the test, 35 passed. In
Bags of elly Beans ag I contains three green and two order to gauge the effectiveness of the test, everyone who too the
red elly beans, and ag II contains four red, two green, and two test was put in the training program. f those who passed the test,
white elly beans. A elly bean is randomly ta en from ag I and 0 performed satisfactorily, whereas of those who failed, only
placed in ag II. If a elly bean is then randomly ta en from ag 30 did satisfactorily. If one of the new trainees is selected at
II, find the probability that the elly bean is red. random and is found to be satisfactory, what is the probability that
the person passed the exam
 t ona o cs
n Pro a t

A
s we saw in Chapter , probability can be used to solve the problem of
9.1 screte an o
ar a es an ecte dividing up the pot of money between two gamblers when their game is
a e interrupted. Now we might as a follow-up question: What are the chances
that a game will be interrupted in the first place
9.2 e no a str t on ur answer depends on the details, of course. If the gamblers now in advance
9.3 ar o C a ns that they will play a fixed number of rounds the interruption being scheduled in
advance, as it were then it might be fairly easy to calculate the probability that time
C er 9 e e will run out before they finish. r, if the amount of time available is un nown, we might
calculate the expected durati n of a complete game and an expected time before the next
interruption. Then, if the expected game length came out well under the expected time
to the next interruption, we could say that the probability of having to brea the game
off in midplay was low. ut if we wanted to give a more exact, numerical answer we
would have to do a more complicated calculation.
The ind of problem encountered here is not unique to gambling. In industry, man-
ufacturers need to now how li ely they are to have to interrupt a production cycle due
to equipment brea down. ne way they eep this probability low is by logging the
usage hours on each machine and replacing it as the hours approach mean time to
failure the expected value of the number of usage hours the machine provides in its
lifetime. edical researchers face a related problem when they consider the possibil-
ity of having to brea off an experiment because too many test sub ects drop out. To
eep this probability low, researchers often calculate an expected number of dropouts
in advance and include this number, plus a cushion, in the number of people recruited
for a study.
The idea of the expected value for a number the length of time until something
happens, or the number of people who drop out of a study is one of the ey concepts
of this chapter.

424
 Section . screte an o ar a es an ecte a e 425

Objective R E
o e e o t e ro a t str t on With some experiments, we are interested in events associated with numbers. For exam-
of a ran o ar a e an to re resent
t at str t on eo etr ca a ple, if two coins are tossed, our interest may be in the number of heads that occur. Thus,
ra or a sto ra o co te t e we consider the events
ean ar ance an stan ar
e at on of a ran o ar a e f0g f1g f2g
If we let X be a variable that represents the number of heads that occur, then the only
values that X can assume are 0, 1, and 2. The value of X is determined by the outcome
of the experiment, and hence, by chance. In general, a variable whose values depend
on the outcome of a random process is called a random variable. Usually, random
variables are denoted by capital letters, such as X, , or , and the values that these
variables assume are often denoted by the corresponding lowercase letters (x, y, or z).
Thus, for the number of heads .X/ that occur in the tossing of two coins, we can indicate
the possible values by writing
X D x; where x D 0; 1; 2
or, more simply,
X D 0; 1; 2

E AM LE R

a Suppose a die is rolled and X is the number that turns up. Then X is a random variable
and X D 1; 2; 3; 4; 5; 6.
b Suppose a coin is successively tossed until a head appears. If is the number of
such tosses, then is a random variable and
D y where y D 1; 2; 3; 4; : : :
Note that can assume infinitely many values.
c A student is ta ing an exam with a one-hour time limit. If X is the number of minutes
it ta es to complete the exam, then X is a random variable. The values that X can
assume form the interval (0, 60 . That is, 0 < X ! 60.

Now ork Problem 7 G

A random variable is called a discrete random variable if it assumes only a finite
number of values or if its values can be placed in one-to-one correspondence with the
positive integers. In Examples 1(a) and 1(b), X and are discrete. A random variable
is called a continuous random variable if it assumes all values in some interval or
intervals, such as X does in Example 1(c). In this chapter, we will be concerned with
discrete random variables Chapter 16 deals with continuous random variables.
If X is a random variable, the probability of the event that X assumes the value x
is denoted P.X D x/. Similarly, we can consider the probabilities of events, such as
X ! x and X > x. If X is discrete, then the function f that assigns the number P.X D x/
to each possible value of X is called the probability function or the distribution of
the random variable X. Thus,
f.x/ D P.X D x/
It may be helpful to verbalize this equation as f.x/ is the probability that X assumes
the value x .

E AM LE R

Suppose that X is the number of heads that appear on the toss of two well-balanced
coins. etermine the distribution of X.
426 C t ona o cs n Pro a t

S We must find the probabilities of the events X D 0, X D 1, and X D 2. The

usual equiprobable sample space is
S D fHH, HT, TH, TTg
T Hence,
x P.X D x/ the event X D 0 is fTTg
0 1=4 the event X D 1 is fHT, THg
1 2=4 the event X D 2 is fHHg
2 1=4
The probability for each of these events is given in the pr babi ity tab e in the margin.
If f is the distribution for X, that is, f.x/ D P.X D x/, then
1 1 1
f.0/ D f.1/ D f.2/ D
P(X = x) 4 2 4

1
G
2

In Example 2, the distribution f was indicated by the listing

1
4 1 1 1
f.0/ D f.1/ D f.2/ D
4 2 4
x
0 1 2 However, the probability table for X gives the same information and is an acceptable
FIGURE raph of the way of expressing the distribution of X. Another way is by the graph of the distribution,
distribution of X. as shown in Figure .1. The vertical lines from the x-axis to the points on the graph
merely emphasize the heights of the points. Another representation of the distribution
of X is the rectangle diagram in Figure .2, called the probability histogram for X.
P(X = x)
Here a rectangle is centered over each value of X. The rectangle above x has width 1
and height P.X D x/. Thus, its area is the probability 1 " P.X D x/ D P.X D x/. This
1 interpretation of probability as an area is important in Chapter 16.
2 Note in Example 2 that the sum of f.0/, f.1/, and f.2/ is 1:
1 1 1 1
4 f.0/ C f.1/ C f.2/ D C C D1
4 2 4
x This must be the case, because the events X D 0, X D 1, and X D 2 are mutually exclu-
0 1 2 sive and the union of all three is the sample space and P.S/ D 1. We can conveniently
FIGURE Probability histogram indicate the sum f.0/ C f.1/ C f.2/ by the summation notation
for X. X
f.x/
To review summation notation, see x
Section 1.5.
This usage differs slightly from that in Section
P 1.5, in that the upper and lower bounds
of summation are not given explicitly. Here x f.x/ means that we are to sum all terms
of the form f.x/, for a values of x under consideration (which in this case are 0, 1,
and 2). Thus,
X
f.x/ D f.0/ C f.1/ C f.2/
x

In general, for any distribution f, we have 0 ! f.x/ ! 1 for all x, and the sum of all
function values is 1. Therefore,
X
f.x/ D 1
x

This means that in any probability histogram, the sum of the areas of the rectangles
is 1.
The distribution for a random variable X gives the relative frequencies of the values
of X in the long run. However, it is often useful to determine the average value of X
in the long run. In Example 2, for instance, suppose that the two coins were tossed n
 Section . screte an o ar a es an ecte a e 427

times, which resulted in X D 0 occurring k0 times, X D 1 occurring k1 times, and

X D 2 occurring k2 times. Then the average value of X for these n tosses is
0 " k0 C 1 " k1 C 2 " k2 k0 k1 k2
D0" C1" C2"
n n n n
ut the fractions k0 =n; k1 =n and k2 =n are the relative frequencies of the events X D 0,
X D 1, and X D 2, respectively, that occur in the n tosses. If n is very large, then these
relative frequencies approach the probabilities of the events X D 0; X D 1, and X D 2.
Thus, it seems reasonable that the average value of X in the long run is
1 1 1
0 " f.0/ C 1 " f.1/ C 2 " f.2/ D 0 " C 1 " C 2 " D 1
4 2 4
This means that if we tossed the coins many times, the average number of heads appear-
ing per toss is very close to 1. We define the sum in Equation (1) to be the mean of X. It
is also called the expected value of X and the expectati n of X. The mean of X is often
denoted by ! D !.X/ (! is the P ree letter mu ) and also by E.X/. Note that from
Equation (1), ! has the form x xf.x/. In general, we have the following definition.

If X is a discrete random variable with distribution f, then the mean of X is given by

X
! D !.X/ D E.X/ D xf.x/
x

The mean of X can be interpreted as the average value of X in the long run. In fact, if
the values that X ta es on are x1 ; x2 ; : : : ; xn and these are equiprobable so that we have
1
f.xi / D , for i D 1; 2; : : : ; n, then,
n
Pn
X Xn
1 xi
!D xf.x/ D xi D iD1
x iD1
n n
which is the average in the usua sense f that rd of the numbers x1 ; x2 ; : : : ; xn . In
the general case, it is useful to thin of the mean, !, as a eighted a erage where the
weights are provided by the probabilities, f.x/. We emphasize that the mean does not
necessarily have to be an outcome of the experiment. In other words, ! may be different
from all the values x that the random variable X actually assumes. The next example
will illustrate.

E AM LE E G

An insurance company offers a 1 0,000 catastrophic fire insurance policy to home-

owners of a certain type of house. The policy provides protection in the event that such
a house is totally destroyed by fire in a one-year period. The company has determined
that the probability of such an event is 0.002. If the annual policy premium is 37 ,
find the expected gain per policy for the company.

S If an insured house does not suffer a catastrophic fire, the company gains
37 . However, if there is such a fire, the company loses 1 0;000 # 37 (insured
value of house minus premium), which is 17 ,621. If X is the gain (in dollars) to the
company, then X is a random variable that may assume the values 37 and #17 ,621.
(A loss is considered a negative gain.) The expected gain per policy for the company
is the expected value of X.

S If f is the probability function for X, then

f.#17 ;621/ D P.X D #17 ;621/ D 0:002
428 C t ona o cs n Pro a t

and
f.37 / D P.X D 37 / D 1 # 0:002 D 0:
The expected value of X is given by
X
E.X/ D xf.x/ D #17 ;621f.#17 ;621/ C 37 f.37 /
x

D #17 ;621.0:002/ C 37 .0: /D1

Thus, if the company sold many policies, it could expect to gain approximately 1 per
policy, which could be applied to such expenses as advertising, overhead, and profit.
Now ork Problem 19 G
Since E.X/ is the average value of X in the long run, it is a measure of what might
be called the centra tendency of X. However, E.X/ does not indicate the dispersi n or
spread of X from the mean in the long run. For example, Figure .3 shows the graphs
of two distributions, f and g, for the random variables X and . It can easily be demon-
strated that both X and have the same mean: E.X/ D 2 and E. / D 2. (Verify this
claim.) ut from Figure .3, X is more li ely to assume the value 1 or 3 than is ,
because f.1/ and f.3/ are 25 , whereas g.1/ and g.3/ are 15 . Thus, X has more li elihood
of assuming values away from the mean than does , so there is more dispersion for X
in the long run.

f (x) g(y)

3
5
2
5
1 1
5 5

x y
1 2 3 1 2 3
E(X ) = 2 E(Y ) = 2
(a) (b)
FIGURE Probability distributions.

There are various ways to measure dispersion for a random variable X. ne way
is to determine the long-run average of the absolute values of the deviations from the
mean ! that is, E.jX # !j/, which is the mean of the derived random variable jX # !j.
In fact, if g is a suitable function and X is a random variable, then D g.X/
P is another
random variable. oreover, it can be shown that if D g.X/, then E. / D x g.x/f.x/,
where f is the probability function for X. For example, if D jX # !j, then
X
E.jX # !j/ D jx # !jf.x/
x

However, while E.jX # !j/ might appear to be an obvious measure of dispersion, it is

not often used.
any other measures of dispersion can be considered, but two are most widely
accepted. ne is the variance, and the other is the standard deviation. The variance
of X, denoted by Var.X/, is the long-run average of the squares of the deviations of X
from !. In other words, for the variance we consider the random variable D .X # !/2
and we have

X
X
Var.X/ D E..X # !/2 / D .x # !/2 f.x/
x
 Section . screte an o ar a es an ecte a e 429

Since .X # !/2 is involved in Var.X/, and both X and ! have the same units of measure-
ment, the units for Var.X/ are those of X2 . For instance, in Example 3, X is in dollars
thus, Var.X/ has units of dollars squared. It is convenient to have a measure of disper-
p
sion in the same units as X. Such a measure is Var.X/, which is called the standard
de iati n f X and is denoted by " D " .X/ (" is the lowercase ree letter sigma ).

S X
p
" D " .X/ D Var.X/

Note that " has the property that

" 2 D Var.X/
oth Var.X/ D " 2 and " are measures of the dispersion of X. The greater the
value of Var.X/, or ", the greater is the dispersion. ne result of a famous theorem,
Chebyshe s inequa ity, is that the probability of X falling within two standard devia-
tions of the mean is at least 34 . This means that the probability that X lies in the interval
.! # 2"; ! C 2" / is greater than or equal to 34 . ore generally, for k > 1, Chebyshev s
inequality tells us that
k2 # 1
P.X 2 .! # k"; ! C k" // $
k2
To illustrate further, with k D 4, this means that, for any probabilistic experiment, at
2
least 4 4!1
2 D 15
16
D 3:75 of the data values lie in the interval .! # 4"; ! C 4" /. To lie
in the interval .! # 4"; ! C 4" / is to lie within four standard deviations of the mean .
We can write the formula for variance in Equation (2) in a different way. It is a
good exercise with summation notation.
X
Var.X/ D .x # !/2 f.x/
x
X
D .x2 # 2x! C !2 /f.x/
x
X
D .x2 f.x/ # 2x!f.x/ C !2 f.x//
x
X X X
D x2 f.x/ # 2! xf.x/ C !2 f.x/
x x x
X X X
D x2 f.x/ # 2!.!/ C !2 .1/ (since xf.x/ D ! and f.x/ D 1/
x x x

Thus, we have

X
Var.X/ D " 2 D . x2 f.x// # !2 D E.X2 / # E.X/2
x

This formula for variance is useful, since it often simplifies computations.

E AM LE M S

A bas et contains 10 balls, each of which shows a number. Five balls show 1 two
show 2 and three show 3. A ball is selected at random. If X is the number that shows,
determine !, Var(X), and ".
430 C t ona o cs n Pro a t

S The sample space consists of 10 equally li ely outcomes (the balls). The
values that X can assume are 1, 2, and 3. The events X D 1, X D 2, and X D 3
contain 5, 2, and 3 sample points, respectively. Thus, if f is the probability function
for X,
5 1
f.1/ D P.X D 1/ D D
10 2
2 1
f.2/ D P.X D 2/ D D
10 5
3
f.3/ D P.X D 3/ D
10
Calculating the mean gives
X
!D xf.x/ D 1 " f.1/ C 2 " f.2/ C 3 " f.3/
x

5 2 3 1
D1" C2" C3" D D
10 10 10 10 5
To find Var.X/, either Equation (2) or Equation (3) can be used. oth will be used here
so that we can compare the arithmetical computations involved. y Equation (2),
X
Var.X/ D .x # !/2 f.x/
x
! "2 ! "2 ! "2
D 1# f.1/ C 2 # f.2/ C 3 # f.3/
5 5 5
! "2 ! "2 ! "2
4 5 1 2 6 3
D # " C " C "
5 10 5 10 5 10
16 5 1 2 36 3
D " C " C "
25 10 25 10 25 10
0 C 2 C 10 1 0 1
D D D
250 250 25
y Equation (3),
X
Var.X/ D . x2 f.x// # !2
x
! "2
D .12 " f.1/ C 22 " f.2/ C 32 " f.3// #
5
5 2 3 1
D1" C4" C " #
10 10 10 25
5 C C 27 1 40 1
D # D #
10 25 10 25
1 1
D4# D
25 25
Notice that Equation (2) involves .x # !/2 , but Equation (3) involves x2 . ecause of
this, it is often easier to compute variances by Equation (3) than by Equation (2).
1
Since " 2 D Var.X/ D 25
,
the standard deviation is
r p
p 1 1
" D Var.X/ D D
25 5
Now ork Problem 1 G
 Section . screte an o ar a es an ecte a e 431

R BLEMS
In Pr b ems the distributi n f the rand m ariab e X is In Pr b ems determine E.X/ " 2 and " f r the rand m
gi en etermine ! Var.X/ and " In Pr b em c nstruct the ariab e X
pr babi ity hist gram In Pr b em graph the distributi n
Coin oss Three fair coins are tossed. et X be the number
f .0/ D 0:2, f .1/ D 0:3, f .2/ D 0:3, f .3/ D 0:2 of heads that occur.
f .4/ D 0:4; f .5/ D 0:6 Balls in a Bas et A bas et contains eight balls, each of
See Figure .4. which shows a number. Three balls show a 1 two show a 2 two
show a 3 and one shows a 4. A ball is randomly selected and the
f(x) number that shows, X, is observed.

5
Committee From a group of two women and three men, two
12 persons are selected at random to form a committee. et X be the
1
3
number of men on the committee.
1
4 elly Beans in a ar A ar contains two red and three
green elly beans. Two elly beans are randomly withdrawn in
succession with replacement, and the number of red elly beans,
X, is observed.
x
1 2 3 Marbles in a Bag A bag contains five red and three white
marbles. Two marbles are randomly withdrawn in succession
FIGURE without replacement. et X be the number of red marbles
See Figure .5. withdrawn. Find the distribution f for X.
Subcommittee From a state government committee
f (x) consisting of four Whigs and six Tories, a subcommittee of three
is to be randomly selected. et X be the number of Whigs in the
subcommittee. Find a general formula, in terms of combinations,
2 that gives P.X D x/, where x D 0; 1; 2; 3.
7
Ra e A charitable organization is having a ra e for a
1 single prize of 7000. Each ra e tic et costs 3, and 000 tic ets
7
have been sold.
x a Find the expected gain for the purchaser of a single tic et.
0 1 2 3 4 b Find the expected gain for the purchaser of two tic ets.
FIGURE Coin Game Consider the following game. ou are to toss
three fair coins. If three heads or three tails turn up, your friend
The random variable X has the following distribution: pays you 10. If either one or two heads turn up, you must pay
your friend 6. What are your expected winnings or losses per
game
x P.X D x/
Earnings A landscaper earns 200 per day when wor ing
3 and loses 30 per day when not wor ing. If the probability of
wor ing on any day is 47 , find the landscaper s expected daily
5 0.3
earnings.
6 0.2
Fast Food Restaurant A fast-food chain estimates that if
7 0.4 it opens a restaurant in a shopping center, the probability that the
restaurant is successful is 0.72. A successful restaurant earns an
annual profit of 120,000 a restaurant that is not successful loses
a Find P.X D 3/ b Find ! c Find " 2 36,000. What is the expected gain to the chain if it opens a
The random variable X has the following distribution: restaurant in a shopping center
Insurance An insurance company offers a hospitalization
x P.X D x/ policy to individuals in a certain group. For a one-year period, the
company will pay 100 per day, up to a maximum of five days,
2 0:1 for each day the policyholder is hospitalized. The company
estimates that the probability that any person in this group is
4 5a
hospitalized for exactly one day is 0.001 for exactly two days,
6 4a 0.002 for exactly three days, 0.003 for exactly four days, 0.004
and for five or more days, 0.00 . Find the expected gain per policy
to the company if the annual premium is 10.
a Find P.X D 4/ and P.X D 6/ b Find !.
432 C t ona o cs n Pro a t

Insurance Premium In Example 3, if the company wants

an expected gain of 50 per policy, determine the annual
premium.
Roulette In the game of roulette, there is a wheel with 37
slots numbered with the integers from 0 to 36, inclusive. A player
bets 1 (for example) and chooses a number. The wheel is spun
and a ball rolls on the wheel. If the ball lands in the slot showing
the chosen number, the player receives the 1 bet plus 35.
therwise, the player loses the 1 bet. Assume that all numbers
are equally li ely, and determine the expected gain or loss
per play.
Coin Game Suppose that you pay 2.50 to play a game in
which two fair coins are tossed. If n heads occur, you receive 2n
dollars. What is your expected gain (or loss) on each play The
game is said to be fair to you when your expected gain is 0.
What should you pay to play if this is to be a fair game
Demand The following table for a small car rental
company gives the probability that x cars are rented daily:

x 0 1 2 3 4 5 6 7

P.X D x/ 0.05 0.05 0.10 0.25 0.20 0.20 0.15 0.10 0.05

etermine the expected daily demand for their cars.

Objective T B
o e e o t e no a str t on
an re ate t to t e no a t eore
B T
ater in this section we will see that the terms in the expansion of a power of a bino-
mial are useful in describing the distributions of certain random variables. It is worth-
while, therefore, first to discuss the bin mia the rem which is a formula for expanding
.a C b/n , where n is a positive integer.
Regardless of n, there are patterns in the expansion of .a C b/n . To illustrate, we
consider the cube of the binomial a C b. y successively applying the distributive law,
we have
.a C b/3 D Œ.a C b/.a C b/#.a C b/
D Œa.a C b/ C b.a C b/#.a C b/
D Œaa C ab C ba C bb#.a C b/
D aa.a C b/ C ab.a C b/ C ba.a C b/ C bb.a C b/
D aaa C aab C aba C abb C baa C bab C bba C bbb
so that
.a C b/3 D a3 C 3a2 b C 3ab2 C b3
Three observations can be made about the right side of Equation (2). First, notice that
the number of terms is four, which is one more than the power to which a C b is
raised (3). Second, the first and last terms are the cubes of a and b the powers of a
decrease from left to right (from 3 to 0) and the powers of b increase (from 0 to 3).
Third, for each term, the sum of the exponents of a and b is 3, which is the power to
which a C b is raised.
et us now focus on the coe cients of the terms in Equation (2). Consider the
coe cient of the ab2 -term. It is the number of terms in Equation (1) that involve exactly
two b s, namely, 3. ut let us see hy there are three terms that involve two b s. Notice
 Section .2 e no a str t on 433

in Equation (1) that each term is the product of three numbers, each of which is either
a or b. ecause of the distributive law, each of the three a C b factors in .a C b/3
contributes either an a or b to the term. Thus, the number of terms involving one a and
two b s is equal to the number of ways of choosing two of the three factors to supply a
3Š
b, namely, 3 C2 D D 3. Similarly,
2Š1Š
the coe cient of the a3 -term is 3 C0
the coe cient of the a2 b-term is 3 C1
and
the coe cient of the b3 -term is 3 C3
eneralizing our observations, we obtain a formula for expanding .a C b/n called
the binomial theorem.

B T
If n is a positive integer, then
.a C b/n D n C0 an C n C1 an!1 b C n C2 an!2 b2 C " " " C n Cn!1 abn!1 C n Cn bn
X
n
n!i i
D n Ci a b
iD0
The numbers n Cr are also called binomial coe cients for this reason.

E AM LE B T

Use the binomial theorem to expand .q C p/4 .

S Here, n D 4; a D q, and b D p. Thus,
.q C p/4 D 4 C0 q4 C 4 C1 q3 p C 4 C2 q2 p2 C 4 C3 qp3 C 4 C4 p4
4Š 4 4Š 3 4Š 2 2 4Š 4Š 4
D q C q pC qp C qp3 C p
0Š4Š 1Š3Š 2Š2Š 3Š1Š 4Š0Š
Recalling that 0Š D 1, we have
.q C p/4 D q4 C 4q3 p C 6q2 p2 C 4qp3 C p4
G
oo bac now to the display of Pasca s riang e in Section .2, which provides a
memorable way to generate the binomial coe cients. For example, the numbers in the
.4C1/th row of Pascal s Triangle, 1 4 6 4 1, are the coe cients found in Example 1.

B
We now turn our attention to repeated trials of an experiment in which the outcome of
any trial does not affect the outcome of any other trial. These are referred to as inde
pendent trials. For example, when a fair die is rolled five times, the outcome on one
roll does not affect the outcome on any other roll. Here we have five independent trials
of rolling a die. Together, these five trials can be considered as a five-stage compound
experiment involving independent events, so we can use the special multiplication law
of Section .6 to determine the probability of obtaining specific outcomes of the trials.
To illustrate, let us find the probability of getting exactly two 4 s in the five rolls of
the die. We will consider getting a 4 as a success (S) and getting any of the other five
numbers as a fai ure (F). For example, the sequence
SSFFF
434 C t ona o cs n Pro a t

denotes getting

4, 4, followed by three other numbers

This sequence can be considered as the intersection of five independent events: success
on the first trial, success on the second, failure on the third, and so on. Since the prob-
ability of success on any trial is 16 and the probability of failure is 1 # 16 D 56 , by the
special multiplication law for the intersection of independent events, the probability of
the sequence SSFFF occurring is
! "2 ! "3
1 1 5 5 5 1 5
" " " " D
6 6 6 6 6 6 6

In fact, this is the probability for any particular order of the two S s and three F s. et us
determine how many ways a sequence of two S s and three F s can be formed. ut of
five trials, the number of ways of choosing the two trials for success is 5 C2 . Another way
to loo at this problem is that we are counting permutati ns ith repeated bjects, as in
5Š
Section .2, of the word SSFFF. There are D 5 C2 of these. So the probability
2Š " 3Š
of getting exactly two 4 s in the five rolls is
! "2 ! "3
1 5
5 C2
6 6

If we denote the probability of success by p and the probability of failure by q.D1 # p/,
then (3) ta es the form
2 3
5 C2 p q

which is the term involving p2 in the expansion of .q C p/5 .

ore generally, consider the probability of getting exactly x 4 s in n rolls of the
die. Then n # x of the rolls must be some other number. For a particular order, the
probability is

px qn!x

The number of possible orders is n Cx , which again we can see as the question of finding
the number of permutations of n symbols, where x of them are S (success) and the
remaining n # x are F (failure). According to the result in Section .2, on permutati ns
ith repeated bjects, there are

nŠ
D n Cx
xŠ " .n # x/Š

of these and, therefore,

P.X D x/ D n Cx px qn!x

which is a general expression for the terms in .q C p/n . In summary, the distribution
for X (the number of 4 s that occur in n rolls) is given by the terms in .q C p/n .
Whenever we have n independent trials of an experiment in which each trial has
only two possible outcomes (success and failure) and the probability of success in each
trial remains the same, the trials are called Bernoulli trials. ecause the distribution
of the number of successes corresponds to the expansion of a power of a binomial,
the experiment is called a binomial experiment, and the distribution of the number of
successes is called a binomial distribution.
 Section .2 e no a str t on 435

B
If X is the number of successes in n independent trials of a binomial experiment
with probability p of success and q of failure on any trial, then the distribution f for
X is given by
f.x/ D P.X D x/ D n Cx px qn!x
where x is an integer such that 0 ! x ! n and q D 1 # p. Any random variable with
this distribution is called a binomial random variable and is said to have a bino
mial distribution. The mean and standard deviation of X are given, respectively, by
p
! D np " D npq

E AM LE B
A L IT I
et X be the number of persons out Suppose X is a binomial random variable with n D 4 and p D 13 . Find the distribution
of four ob applicants who are hired. for X.
If the probability of any one applicant 1
being hired is 0.3, find the distribution S Here q D 1 # p D 1 # 3
D 23 . So we have
of X.
P.X D x/ D n Cx px qn!x x D 0; 1; 2; 3; 4

Thus,
! "0 ! "4
1 2 4Š 16 16 16
P.X D 0/ D 4 C0 D "1" D1"1" D
3 3 0Š4Š 1 1 1
! "1 ! "3
1 2 4Š 1 1 32
P.X D 1/ D 4 C1 D " " D4" " D
3 3 1Š3Š 3 27 3 27 1
P (X = x ) ! "2 ! "2
1 2 4Š 1 4 1 4
P.X D 2/ D 4 C2 D " " D6" " D
32 3 3 2Š2Š 27
81
! "3 ! "1
8 1 2 4Š 1 2 1 2
27 P.X D 3/ D 4 C3 D " " D4" " D
3 3 3Š1Š 27 3 27 3 1
16
81 ! "4 ! "0
1 2 4Š 1 1 1
8 P.X D 4/ D 4 C4 D " "1D1" "1D
81 3 3 4Š0Š 1 1 1
1
The probability histogram for X is given in Figure .6. Note that the mean ! for X is
81
0 1 2 3 4
x # $
np D 4 13 D 43 , and the standard deviation is
FIGURE inomial distribution,
r r p
n D 4, p D 13 . p 1 2 2 2
" D npq D 4" " D D
3 3 3

Now ork Problem 1 G

E AM LE A L T H E C T

A fair coin is tossed eight times. Find the probability of getting at least two heads.
S If X is the number of heads that occur, then X has a binomial distribution
with n D , p D 12 , and q D 12 . To simplify our wor , we use the fact that

P.X $ 2/ D 1 # P.X < 2/

436 C t ona o cs n Pro a t

Now,
P.X < 2/ D P.X D 0/ C P.X D 1/
! "0 ! " ! "1 ! "7
1 1 1 1
D C0 C C1
2 2 2 2
1 1 1
D1"1" C " " D
256 2 12 256
Thus,
247
P.X $ 2/ D 1 # D
256 256
A probability histogram for X is given in Figure .7.

P (X = x )
35
128

7
32

7
64

1
32

1 x
256
0 1 2 3 4 5 6 7 8

FIGURE inomial distribution, n D ; p D 12 .

Now ork Problem 17 G

E AM LE I T A

For a particular group of individuals, 20 of their income tax returns are audited each
year. f five randomly chosen individuals, what is the probability that exactly two will
have their returns audited
S We will consider this to be a binomial experiment with five trials (selecting
an individual). Actually, the experiment is not truly binomial, because selecting an
individual from this group affects the probability that another individual s return will be
audited. For example, if there are 5000 individuals, then 20 , or 1000, will be audited.
1000
The probability that the first individual selected will be audited is 5000
. If that event
occurs, the probability that the second individual selected will be audited is 4 . Thus,
the trials are not independent. However, we assume that the number of individuals is
large, so that for all practical purposes, the probability of auditing an individual remains
constant from trial to trial.
For each trial, the two outcomes are being audited and n t being audited. Here we
define a success as being audited. etting X be the number of returns audited, p D 0:2,
and q D 1 # 0:2 D 0: , we have
5Š
P.X D 2/ D 5 C2 .0:2/2 .0: /3 D .0:04/.0:512/
2Š3Š
D 10.0:04/.0:512/ D 0:204

Now ork Problem 15 G

 Section .3 ar o C a ns 437

R BLEMS
In Pr b ems determine the distributi n f f r the bin mia Coin A biased coin is tossed three times in succession.
rand m ariab e X if the number f tria s is n and the pr babi ity The probability of heads on any toss is 14 . Find the probability that
f success n any tria is p A s nd ! and " a exactly two heads occur and b two or three heads occur.
1 1
n D 2; p D 5
n D 3; p D 2
Cards From an ordinary dec of 52 playing cards, 7 cards
2 are randomly drawn in succession with replacement. Find the
n D 3; p D 3
n D 5, p D 0:3 probability that there are a exactly four hearts and b at least
four hearts.
In Pr b ems determine the gi en pr babi ity if X is a
bin mia rand m ariab e n is the number f tria s and p is the Quality Control In a large production lot of smartphones,
pr babi ity f success n any tria it is believed that 0:015 are defective. If a sample of 10 is
1 1 randomly selected, find the probability that less than 2 will be
P.X D 3/ n D 4, p D P.X D 2/I n D 5; p D
3 3 defective.
P.X D 2/I n D 4; p D 4
5
P.X D 4/I n D 7; p D 0:2 igh Speed Internet For a certain large population, the
probability that a randomly selected person has access to
4
P.X > 3/ n D 5, p D 0:3 P.X $ 3/ n D 4, p D 5 high-speed Internet is 0. . If four people are selected at random,
find the probability that at least three have access to high-speed
Coin A fair coin is tossed 11 times. What is the probability Internet.
that exactly eight heads occur
Baseball The probability that a certain baseball player gets
Multiple Choice Qui Each question in a six-question a hit is 0.300. Find the probability that if he goes to bat four times,
multiple-choice quiz has four choices, only one of which is he will get at least one hit.
correct. If a student guesses at all six questions, find the
Stoc s A financial advisor claims that 60 of the stoc s
probability that exactly three will be correct.
that he recommends for purchase increase in value. From a list of
Marbles A ar contains five red and seven green marbles. 200 recommended stoc s, a client selects 4 at random. etermine
Four marbles are randomly withdrawn in succession with the probability, rounded to two decimal places, that at least 2
replacement. etermine the probability that exactly two of the of the chosen stoc s increase in value. Assume that the selections
marbles withdrawn are green. of the stoc s are independent trials and that the number of stoc s
Cards From a dec of 52 playing cards, 4 cards are that increase in value has a binomial distribution.
randomly selected in succession ith rep acement. etermine Genders of Children If a family has five children, find the
the probability that at least two cards are ac s. probability that at least two are girls. (Assume that the probability
Quality Control A manufacturer produces electrical that a child is a girl is 12 .)
switches, of which 3 are defective. From a production run of If X is a binomially distributed random variable with n D 100
60,000 switches, five are randomly selected and each is tested.
and p D 13 , find Var.X/.
etermine the probability that the sample contains exactly three
defective switches. Round your answer to three decimal places. Suppose X is a binomially distributed random variable such
Assume that the four trials are independent and that the that ! D 2 and " 2 D 32 . Find P.X D 2/.
number of defective switches in the sample has a binomial Quality Control In a production process, the probability
distribution. of a defective unit is 0.06. Suppose a sample of 15 units is
Coin A coin is biased so that P. / D 0:2 and P. / D 0: . selected at random. et X be the number of defectives.
If X is the number of heads in three tosses, determine a formula a Find the expected number of defective units.
for P.X D x/. b Find Var.X/.
c Find P.X ! 1/. Round your answer to two decimal places.

Objective M C
o e e o t e not ons of a ar o We conclude this chapter with a discussion of a special type of stochastic process called
c a n an t e assoc ate trans t on
atr o n state ectors an t e a ark chain, after the Russian mathematician Andrei ar ov (1 56 1 22).
stea state ector
M C
A Mar ov chain is a sequence of trials of an experiment in which the possible
outcomes of each trial remain the same from trial to trial, are finite in number, and
have probabilities that depend only upon the outcome of the previous trial.

To illustrate a ar ov chain, we consider the following situation. Imagine that a small

town has only two service stations say, stations 1 and 2 that handle the servicing
needs of the town s automobile owners. (These customers form the population under
consideration.) Each time a customer needs car servicing, he or she must ma e a ch ice
of which station to use.
438 C t ona o cs n Pro a t

Thus, each customer can be placed into a category according to which of the two
stations he or she most recently chose. We can view a customer and the service stations
as a system. If a customer most recently chose station 1, we will refer to this as state 1
of the system. Similarly, if a customer most recently chose station 2, we say that the
system is currently in state 2. Hence, at any given time, the system is in one of its two
states. f course, over a period of time, the system may move from one state to the other.
For example, the sequence 1, 2, 2, 1 indicates that in four successive car servicings, the
system changed from state 1 to state 2, remained at state 2, and then changed to state 1.
This situation can be thought of as a sequence of trials of an experiment (choosing a
service station) in which the possible outcomes for each trial are the two states (station
1 and station 2). Each trial involves observing the state of the system at that time.
If we now the current state of the system, we realize that we cannot be sure of its
state at the next observation. However, we may now the ike ih d of its being in a
particular state. For example, suppose that if a customer most recently used station 1,
then the probability that the customer uses station 1 the next time is 0.7. (This means
that, of those customers who used station 1 most recently, 70 continued to use station
1 the next time and 30 changed to station 2.) Assume also that if a customer used
station 2 most recently, the probability is 0. that the customer also uses station 2 the
next time. We recognize these probabilities as being c nditi na probabilities. That is,
P.remaining in state 1 j currently in state 1/ D 0:7
P.changing to state 2 j currently in state 1/ D 0:3
P.remaining in state 2 j currently in state 2/ D 0:
P.changing to state 1 j currently in state 2/ D 0:2
These four probabilities can be organized in a square matrix D Πij # by ta ing entry
ij to be the probability of a customer being next in state i given that they are currently
in state j. Thus,
ij D P.being next in state i j currently in state j/
and in the specific case at hand we have
ext Current State
State State 1 State 2
% &
State 1 0:7 0:2
D
State 2 0:3 0:
atrix is called a transiti n matrix because it gives the probabilities of transition
from one state to another in ne step that is, as we go from one observation period to
For example, the sum of the entries in the next. The entries are called transiti n pr babi ities. We emphasize that the tran
column 1 of is 0:7 C 0:3 D 1. siti n matrix remains the same at e ery stage f the sequence f bser ati ns. Note
that all entries of the matrix are in the interval Œ0; 1#, because they are probabilities.
oreover, the sum of the entries in each column must be 1, because, for each current
state, the probabilities account for all possible transitions.
et us summarize our service station situation up to now. We have a sequence of
trials in which the possible outcomes (or states) are the same from trial to trial and are
finite in number (two). The probability that the system is in a particular state for a given
trial depends only on the state for the preceding trial. Thus, we have a so-called t
state ark chain. A ar ov chain determines a square matrix , called a transition
matrix.

T M
A transition matrix for a k-state ar ov chain is a k % k matrix D Πij # in which
the entry ij is the probability, from one trial to the next, of moving t state i fr m
state j. All entries are in Œ0; 1#, and the sum of the entries in each column is 1. We
can say
ij D P.next state is i j current state is j/
 Section .3 ar o C a ns 439

1 (0.6) (0.7) = 0.42

0.7
1 0.5
0.6 0.3
2 (0.6) (0.3) = 0.18

Start

1 (0.4) (0.2) = 0.08 0.5

0.4 0.2
2
0.8 (0.4) (0.8) = 0.32
2

Most Next
recent station
station

FIGURE Probability tree for two-state ar ov chain.

Suppose that when observations are initially made, 60 of all customers used sta-
tion 1 most recently and 40 used station 2. This means that, before any additional
trials (car servicings) are considered, the probabilities that a customer is in state 1
or 2 are 0.6 and 0.4, respectively. These probabilities are called initia state pr ba
bi ities and are collectively referred to as being the initia distributi n. They can be
represented by a column vector, called an initial state vector, which is denoted by X0 .
A subscript of 0 is used for the initial In this case,
state vector. % &
0:6
X0 D
0:4
We would li e to find the vector that gives the state probabilities for a customer s
next visit to a service station. This state vector is denoted by X1 . ore generally, a state
vector is defined as follows:

S
The state vector Xn for a k-state ar ov chain is a k-entry column vector in which
the entry xj is the probability of being in state j after the nth trial.

We can find the entries for X1 from the probability tree in Figure . . We see that the
probability of being in state 1 after the next visit is the sum
.0:7/.0:6/ C .0:2/.0:4/ D 0:5
and the probability of being in state 2 is
.0:3/.0:6/ C .0: /.0:4/ D 0:5
Thus,
% &
0:5
X1 D
0:5
The sums of products on the left sides of Equations (1) and (2) remind us of matrix
multiplication. In fact, they are the entries in the matrix X0 obtained by multiplying
the initial state vector on the left by the transition matrix:
% &% & % &
0:7 0:2 0:6 0:5
X1 D X0 D D
0:3 0: 0:4 0:5
This pattern of ta ing the product of a state vector and the transition matrix to
get the next state vector continues, allowing us to find state probabilities for future
observations. For example, to find X2 , the state vector that gives the probabilities for
each state after two trials (following the initial observation), we have
% &% & % &
0:7 0:2 0:5 0:45
X2 D X1 D D
0:3 0: 0:5 0:55
440 C t ona o cs n Pro a t

Thus, the probability of being in state 1 after two car servicings is 0.45. Note that, since
X1 D X0 , we can write
X2 D . X0 /
so that
2
X2 D X0
In general, the nth state vector Xn can be found by multiplying the previous state vector
Xn!1 on the left by .

If is the transition matrix for a ar ov chain, then the state vector Xn for the nth
trial is given by
Xn D Xn!1

Equivalently, we can find Xn by using only the initial state column vector X0 and the
transition matrix :
n
Here, we find Xn by using powers of . Xn D X0

et us now consider the situation in which we now the initial state of the system.
For example, ta e the case of observing initially that a customer has most recently
chosen station 1. This means the probability that the system is in state 1 is 1, so the
initial state vector must be
% &
1
X0 D
0
Suppose we determine X2 , the state vector that gives the state probabilities after the
next two visits. This is given by
% &2 % &
0:7 0:2 1
X2 D 2 X0 D
0:3 0: 0
% &% & % &
0:55 0:30 1 0:55
D D
0:45 0:70 0 0:45

Thus, for this customer, the probabilities of using station 1 or station 2 after two steps
are 0.55 and 0.45, respectively. bserve that these probabilities form the rst c umn
of 2 . n the other hand, if the system were initially in state 2, then the state vector
after two steps would be
% & % &% & % &
2 0 0:55 0:30 0 0:30
D D
1 0:45 0:70 1 0:70

Hence, for this customer, the probabilities of using station 1 or station 2 after two steps
are 0.30 and 0.70, respectively. bserve that these probabilities form the sec nd c umn
of 2 . ased on our observations, we now have a way of interpreting 2 : The entries in
1 2
% &
2 1 0:55 0:30
D
2 0:45 0:70
give the probabilities of moving to a state from another in t steps. In general, we
have the following:

If is a transition matrix, then for n the entry in row i and column j gives the
This gives the significance of the entries probability of being in state i after n steps, starting from state j.
in n .
 Section .3 ar o C a ns 441

E AM LE

A certain county is divided into three demographic regions. Research indicates that
each year 20 of the residents in region 1 move to region 2 and 10 move to region 3.
(The others remain in region 1.) f the residents in region 2, 10 move to region 1 and
10 move to region 3. f the residents in region 3, 20 move to region 1 and 10
move to region 2.

a Find the transition matrix for this situation.

S We have
Fr m Regi n
Regi n 1 2 3
2 3
1 0:7 0:1 0:2
D2 40:2 0: 0:15
3 0:1 0:1 0:7

Note that to find 11 we subtracted the sum of the other two entries in the first column
from 1. The entries 22 and 33 are found similarly.
b Find the probability that a resident of region 1 this year is a resident of region 1 next
year in two years.
S From entry 11 in transition matrix , the probability that a resident of
region 1 remains in region 1 after one year is 0.7. The probabilities of moving from
one region to another in two steps are given by 2 :
1 2 3
2 3
1 0:53 0:17 0:2
2
D 2 40:31 0:67 0:1 5
3 0:16 0:16 0:52

Thus, the probability that a resident of region 1 is in region 1 after two years is 0.53.
c This year, suppose 40 of county residents live in region 1, 30 live in region 2,
and 30 live in region 3. Find the probability that a resident of the county lives in
region 2 after three years.
S The initial state vector is
2 3
0:40
X0 D 40:305
0:30

The distribution of the population after three years is given by state vector X3 . From
Equation (3) with n D 3, we have
3 2
X3 D X0 D X0
2 32 32 3
0:7 0:1 0:2 0:53 0:17 0:2 0:40
D 40:2 0: 0:15 40:31 0:67 0:1 5 40:305
0:1 0:1 0:7 0:16 0:16 0:52 0:30
2 3
0:336
D 40:40245
0:260

This result means that in three years, 33.6 of the county residents live in region 1,
f course, X3 can be easily obtained with
a graphing calculator: Enter X0 and , 40.24 live in region 2, and 26.0 live in region 3. Thus, the probability that a
and then evaluate 3 X0 directly. resident lives in region 2 in three years is 0.4024.

G
442 C t ona o cs n Pro a t

S S
et us now return to our service station problem. Recall that if the initial state vector is
% &
0:6
X0 D
0:4

then
% &
0:5
X1 D
0:5
% &
0:45
X2 D
0:55

Some state vectors beyond the second are

% &% & % &
0:7 0:2 0:45 0:425
X3 D X2 D D
0:3 0: 0:55 0:575
% &% & % &
0:7 0:2 0:425 0:4125
X4 D X3 D D
0:3 0: 0:575 0:5 75
% &% & % &
0:7 0:2 0:4125 0:40625
X5 D X4 D D
0:3 0: 0:5 75 0:5 375
"
"
"
% &
0:40020
X10 D X &
0:5 0

These results strongly suggest, and it is indeed the case, that as the number of trials
increases the entries in the state vectors tend to get closer and closer to the correspond-
ing entries in the vector
% &
0:40
D
0:60

(Equivalently, it can be shown that the entries in each column of n approach the corre-
sponding entries in those of as n increases.) Vector has a special property. bserve
the result of multiplying on the left by the transition matrix :
% &% & % &
0:7 0:2 0:40 0:40
D D D
0:3 0: 0:60 0:60

We, thus, have

which shows that remains unchanged fr m tria t tria .

In summary, as the number of trials increases, the state vectors get closer and closer
to , which remains unchanged from trial to trial. The distribution of the population
between the service stations stabilizes. That is, in the long run, approximately 40 of
the population will have their cars serviced at station 1 and 60 at station 2. To describe
The steady-state vector is unique and this, we say that is the steady state vector of this process. It can be shown that the
does not depend on the initial steady-state vector is unique. (There is only one such vector.) oreover, does not
distribution. depend on the initial state vector X0 but depends only on the transition matrix . For
this reason, we say that is the steady state ect r f r .
What we need now is a procedure for finding the steady-state vector without
having to compute state vectors for large values of n. Fortunately, the previously stated
 Section .3 ar o C a ns 443
% &
q1
property that D can be used to find . If we let D , we have
q2
D DI
#I D0
. # I/ D 0
0" # " #1 % & % &
@ 0:7 0:2
#
1 0 A q1
D
0
0:3 0: 0 1 q2 0
% &% & % &
#0:3 0:2 q1 0
D
0:3 #0:2 q2 0
which suggests that can be found by solving the resulting system of linear equations
arising here in matrix form. Using the techniques of Chapter 6, we see immediately
that the coe cient matrix of the last equation reduces to
% &
3 #2
0 0
which suggests that there are infinitely many possibilities for the steady-state vector .
However, the entries of a state vector must add up to 1 so that the further equation
q1 C q2 D 1 must be added to the system. We arrive at
% &% & % &
3 #2 q1 0
D
1 1 q2 1
which is easily seen to have the unique solution
% & % &
q 0:4
D 1 D
q2 0:6
which confirms our previous suspicion.
We must point out that for ar ov chains in general, the state vectors do not always
approach a steady-state vector. However, it can be shown that a steady-state vector for
does exist, provided that is regu ar

A transition matrix is regular if there exists a positive integer power n for which
all entries of n are (strictly) positive.

nly regular transition matrices will be considered in this section. A ar ov chain

whose transition matrix is regular is called a regular Mar ov chain.
In summary, we have the following:

Suppose is the k%k transition matrix for a regular ar ov chain. Then the steady-
state column vector
2 3
q1
6q2 7
6 7
D 6 :: 7
4:5
qk
is the solution to the matrix equations
Œ1 1 """ 1# D1
. # Ik / D0
where in Equation (4) the (matrix) coe cient of is the row vector consisting of k
entries all of which are 1.
444 C t ona o cs n Pro a t

Equations (4) and (5) can always be combined into a single matrix equation:
"
D 0"
where " is the .kC1/%k matrix obtained by pasting the row Œ1 1 " " " 1# to the top
of the k%k matrix #Ik (where Ik is the k%k identity matrix) and 0" is the kC1-column
vector obtained by pasting a 1 to the top of the zero k-column vector. We can then find
by reducing the augmented matrix Π" j 0" #. The next example will illustrate.

E AM LE S S

For the demography problem of Example 1, in the long run, what percentage of county
residents will live in each region
S The population distribution in the long run is given by the steady-state vector
, which we now proceed to find. The matrix for this example was shown to be
2 3
0:7 0:1 0:2
40:2 0: 0:15
0:1 0:1 0:7
so that # I is 2 3
#0:3 0:1 0:2
4 0:2 #0:2 0:15
0:1 0:1 #0:3
"
and Πj 0" # is
2 3
1 1 1 1
6#0:3 0:1 0:2 07
6 7
4 0:2 #0:2 0:1 05
0:1 0:1 #0:3 0
which reduces to 2 3
1 0
0 5=16
60 0 7=167
1
6 7
40 1 0 1=45
0 0 0 0
2 3 2 3
5=16 0:3125
showing that the steady-state vector D 47=165 D 40:43755. Thus, in the long run,
1=4 0:2500
the percentages of county residents living in regions 1, 2, and 3 are 31.25 , 43.75 ,
and 25 , respectively.
Now ork Problem 37 G
R BLEMS
In Pr b ems can the gi en matrix be a transiti n matrix f r a In Pr b ems a transiti n matrix f r a ark chain is gi en
ark
" 1 chain# " # etermine the a ues f the etter entries
2 "2 # " #
2 3 0:1 1 3
b a b
# 32 1 0: 0 a 1 5
a
3 4 12
2 3 2 3
1 1 1 2 3 2 3
0:2 0:6 0:5 a b c
6 21 3
7 0:1 a a 6 7
6# 5 17 40:7 0:2 0:15 4 a 0:2 b5 6a 1
b7
4 4 35 4 4 5
3 1 1 0:1 0:2 0:2 0:2 b c
4 4 3 a a a
2 3 2 3
0:2 0:1 0:7 0:5 0:1 0:3 In Pr b ems determine hether the gi en ect r c u d be a
40:1 0:2 0 5 40:4 0:3 0:35 state ect r f r a ark chain
0:7 0:7 0:3 0:6 0:6 0:4 " # " #
0:4 1
0:6 0
 Section .3 ar o C a ns 445
2 3 2 3
0:2 0:1 Spread of Flu A u has attac ed a college dorm that has
40:75 41:15 200 students. Suppose the probability that a student having the u
0:5 0:2 will still have it 4 days later is 0.1. However, for a student who
does not have the u, the probability of having the u 4 days later
In Pr b ems a transiti n matrix and an initia state is 0.2.
ect r X0 are gi en C mpute the state ect rs X1 X2 and X3
" # " # a Find a transition matrix for this situation.
1 1 1
4
0 2 4
b If 120 students now have the u, how many students (to the
D 3 D 1 3 nearest integer) can be expected to have the u days from now
4
1 2 4
12 days from now
" # " #
1 Physical Fitness A physical-fitness center has found that,
0 2
X0 D X0 D 1 of those members who perform high-impact exercising on one
1 2
visit, 55 will do the same on the next visit and 45 will do
" # " # low-impact exercising. f those who perform low-impact
0:3 0:5 0:1 0: exercising on one visit, 75 will do the same on the next visit
D D
0:7 0:5 0: 0:1 and 25 will do high-impact exercising. n the last visit,
" # " # suppose that 65 of members did high-impact exercising and
0:4 0:2 35 did low-impact exercising. After two more visits, what
X0 D X0 D
0:6 0: percentage of members will be performing high-impact
2 3 exercising
0:2 0:1 0:4 ewspapers In a certain area, two daily newspapers are
D 40:1 0:5 0:25 available. It has been found that if a customer buys newspaper A
0:7 0:4 0:4 on one day, then the probability is 0.3 that he or she will change to
2 3
0:2 the other newspaper the next day. If a customer buys newspaper
X0 D 40:15 on one day, then the probability is 0.6 that he or she will buy the
0:7 same newspaper the next day.
2 3 a Find the transition matrix for this situation.
0 0:1 0:2 0:7 b Find the probability that a person who buys A on onday will
60:1 0:2 0:7 0 7 buy A on Wednesday.
D46 7
0:2 0:7 0 0:15 Video Rentals A video rental store has three locations in a
0:7 0 0:1 0:2 city. A video can be rented from any of the three locations and
2 3
1 returned to any of them. Studies show that videos are rented from
607 one location and returned to a location according to the
X0 D 4 7
6
05 probabilities given by the following matrix:
0

In Pr b ems a transiti n matrix is gi en Returned Rented from

a Compute 2
and . 3 to 1 2 3
2 3
b What is the probability of going to state 2 from state 1 after 1 0:7 0:1 0:1
two steps 2 40:2 0: 0:15
c What is the probability of going to state 1 from state 2 after 3 0:1 0 0:
three steps
" # " #
1 3 1 1
4 4 3 2 Suppose that 30 of the videos are initially rented from location
3 1 2 1
4 4 3 2 1, 30 from 2, and 40 from 3. Find the percentages of videos
2 3 2 3 that can be expected to be returned to each location:
0 0:5 0:3 0:2 0:1 0:4
41 0:4 a After this rental
0:35 40:1 0:5 0:25
b After the next rental
0 0:1 0:4 0:7 0:4 0:4
Voting In a certain city, voter preference was analyzed
In Pr b ems nd the steady state ect r f r the gi en according to party a liation: iberal, Conservative, and other. It
transiti was found that on a year-to-year basis, the probability that a voter
" n matrix
# " # switches to Conservative from iberal is 0 to other from iberal,
1 1 1 1
2 3 2 4 0.3 to iberal from Conservative, 0.1 to other from Conservative,
1 2 1 3
2 3 2 4 0.2 to iberal from other, 0.3 and to Conservative from other,
" # " # 0.1.
1 3 1 1
5 5 4 3 a Find a transition matrix for this situation.
4 2 3 2
5 5 4 3 b What is the probability that a current Conservative voter will
2 3 2 3 be iberal two years from now
0:2 0:1 0:4 0:1 0:2 0:1
40:1 c If 40 of the present voters are iberal and 30 are
0:5 0:25 40:4 0:3 0:55
Conservative, what percentage can be expected to be Conservative
0:7 0:4 0:4 0:5 0:5 0:4
one year from now
446 C t ona o cs n Pro a t

Demography The residents of a certain region are a Find a transition matrix for this situation.
classified as urban (U), suburban (S), or rural (R). A mar eting b Suppose that four people line up at a pop machine that is
firm has found that over successive 5-year periods, residents shift nown to have wor ed ust before they arrived. What is the
from one classification to another according to the probabilities probability that the fourth person will receive a pop (Assume
given by the following matrix: nobody ma es more than one attempt.)
c If there are 40 such pop machines on a university campus and
U S R they are not getting regular maintenance, how many, in the long
2 3 run, do you expect to wor properly
U 0:7 0:1 0:1
S 40:1 0: 0:15 Advertising A supermar et chain sells bread from
R 0:2 0:1 0: ba eries A and . Presently, A accounts for 50 of the chain s
daily bread sales. To increase sales, A launches a new advertising
a Find the probability that a suburban resident will be a rural campaign. The ba ery believes that the change in bread sales at
resident in 15 years. the chain will be based on the following transition matrix:
b Suppose the initial population of the region is 50 urban, 25
suburban, and 25 rural. etermine the expected population A
distribution in 15 years. "3 1
#
A 4 2
Long Distance elephone Service A ma or long-distance
1 1
telephone company (company A) has studied the tendency of 4 2
telephone users to switch from one carrier to another. The
company believes that over successive six-month periods, the a Find the steady-state vector.
probability that a customer who uses A s service will switch to a b In the long run, by what percentage can A expect to increase
competing service is 0.2 and the probability that a customer of present sales at the chain Assume that the total daily sales of
any competing service will switch to A is 0.3. bread at the chain remain the same.
a Find a transition matrix for this situation.
b If A presently controls 70 of the mar et, what percentage
can it expect to control six months from now
c What percentage of the mar et can A expect to control in the
long run
Automobile Purchases In a certain region, a study of car
ownership was made. It was determined that if a person presently
owns a Ford, then the probability that the next car the person buys
is also a Ford is 0.75. If a person does not presently own a Ford,
Ban Branches A ban with three branches, A, , and C,
then the probability that the person will buy a Ford on the next car
finds that customers usually return to the same branch for their
purchase is 0.35.
ban ing needs. However, at times a customer may go to a
a Find the transition matrix for this situation. different branch because of a changed circumstance. For example,
b In the long run, what proportion of car purchases in the region a person who usually goes to branch A may sometimes deviate
can be expected to be Fords and go to branch because the person has business to conduct in
Laboratory Mice Suppose 100 mice are in a the vicinity of branch . For customers of branch A, suppose that
two-compartment cage and are free to move between the 0 return to A on their next visit, 10 go to , and 10 go to
compartments. At regular time intervals, the number of mice in C. For customers of branch , suppose that 70 return to on
each compartment is observed. It has been found that if a mouse is their next visit, 20 go to A, and 10 go to C. For customers of
in compartment 1 at one observation, then the probability that the branch C, suppose that 70 return to C on their next visit, 20 go
mouse will be in compartment 1 at the next observation is 35 . If a to A, and 10 go to .
mouse is in compartment 2 at one observation, then the a Find a transition matrix for this situation.
probability that the mouse will be in compartment 2 at the next b If a customer most recently went to branch , what is the
observation is 25 . Initially, suppose that 50 mice are placed into probability that the customer returns to on the second ban visit
each compartment. from now
a Find the transition matrix for this situation. c Initially, suppose 200 customers go to A, 200 go to , and 100
b After two observations, what percentage of mice (rounded to go to C. n their next visit, how many can be expected to go to A
two decimal places) can be expected to be in each compartment To To C
c In the long run, what percentage of mice can be expected in d f the initial 500 customers, in the long run how many can be
each compartment expected to go to A To To C
Vending Machines If a pop machine fails to deliver, "1 #
1
people often warn bystanders, on t put your money in that Show that the transition matrix D 2
is regular.
1
thing I tried it and it didn t wor Suppose that if a vending 2
0
2
machine is wor ing properly one time, then the probability ( int Examine the entries in .)
that it will wor properly the next time is 0. 5. n the other 2 3
hand, suppose that if the machine is not wor ing properly one 001
time, then the probability that it will not wor properly the Show that the transition matrix D 4 0 1 0 5 is not regular.
next time is 0. 5. 100
 Chapter e e 447

Chapter 9 Review
I T S E
S Discrete Random Variables and Expected Value
discrete random variable probability function histogram Ex. 2, p. 425
mean, ! expected value, E.X/ Ex. 3, p. 427
variance, Var.X/ standard deviation, " Ex. 4, p. 42
S he Binomial Distribution
binomial theorem binomial coe cients Ex. 1, p. 433
ernoulli trials binomial experiment binomial distribution Ex. 2, p. 435
S Mar ov Chains
ar ov chain transition matrix, state vector, Xn Ex. 1, p. 441
regular transition matrix steady-state vector, Ex. 2, p. 444

S
If X is a discrete random variable and f is the function such where p is the probability of success on any trial and q D 1#p
that f.x/ D P.X D x/, then f is called the probability func- is the probability of failure. The mean ! and standard devia-
tion, or distribution, of X. In general, tion " of this X are given by
X
f.x/ D 1 p
! D np and "D npq
x

The mean, or expected value, of X is the long-run average of A binomial distribution is intimately connected with the
X and is denoted ! or E X : binomial theorem, which is a formula for expanding the nth
X power of a binomial, namely,
! D E.X/ D xf.x/
x X
n

The mean can be interpreted as a measure of the central ten- .a C b/n D n Ci a

n!i i
b
dency of X in the long run. A measure of the dispersion of X iD0
is the variance, denoted Var.X/ and given by
for n a positive integer.
X A ar ov chain is a sequence of trials of an experiment
2
Var.X/ D .x # !/ f.x/
in which the possible outcomes of each trial, which are called
x
states, remain the same from trial to trial, are finite in number,
equivalently, by and have probabilities that depend only upon the outcome of
the previous trial. For a k-state ar ov chain, if the proba-
X bility of moving to state i from state j from one trial to the
Var.X/ D . x2 f.x// # !2
next is written ij , then the k % k matrix D Πij # is called
x
the transition matrix for the ar ov chain. The entries in the
Another measure of dispersion of X is the standard deviation nth power of also represent probabilities the entry in the
": ith row and jth column of n gives the probability of moving
to state i from state j in n steps. A k-entry column vector in
p which the entry xj is the probability of being in state j after
" D Var.X/
the nth trial is called a state vector and is denoted Xn . The
initial state probabilities are represented by the initial state
If an experiment is repeated several times, then each per-
vector X0 . The state vector Xn can be found by multiplying
formance of the experiment is called a trial. The trials are
the previous state vector Xn!1 on the left by the transition
independent when the outcome of any single trial does not
matrix :
affect the outcome of any other. If there are only two possi-
ble outcomes (success and failure) for each independent trial, Xn D Xn!1
and the probabilities of success and failure do not change
from trial to trial, then the experiment is called a binomial Alternatively, Xn can be found by multiplying the initial state
experiment. For such an experiment, if X is the number of vector X0 by n :
successes in n trials, then the distribution f of X is called a n
binomial distribution, and Xn D X0
If the transition matrix is regular, that is, if there is a posi-
f.x/ D P.X D x/ D n Cx px qn!x
tive integer n such that all entries of n are strictly positive,
448 C t ona o cs n Pro a t

then, as the number of trials n increases, Xn gets closer and where " is the .kC1/%k matrix obtained by pasting the row
closer to a vector , called the steady-state vector of . If Œ1 1 " " " 1# to the top of the k % k matrix # Ik (where Ik
2 3 is the k % k identity matrix) and 0" is the k C 1-column vec-
q1 tor obtained by pasting a 1 to the top of the zero k-column
6q2 7
6 7 vector. Thus, we construct and reduce
D 6 :: 7
4:5 % &
qk 1"""1 1
#I 0
then the entries of indicate the long-run probability distri-
bution of the states. The vector can be found by solving which, if is regular, will result in
the matrix equation % &
I
"
D 0" 0 0

R
In Pr b ems and the distributi n f r the rand m ariab e X is In Pr b ems and determine the distributi n f f r the
gi en C nstruct the pr babi ity hist gram and determine ! bin mia rand m ariab e X if the number f tria s is n and the
ar X and " pr babi ity f success n any tria is p A s nd ! and "
1
f .1/ D 0:2, f .2/ D 0:5, f .3/ D 0:3 n D 4, p D 0:15 n D 5, p D
3
f .0/ D 16 ; f .1/ D 12 ; f .2/ D 1
3
In Pr b ems and determine the gi en pr babi ity if X is a
bin mia rand m ariab e n is the number f tria s and p is the
Coin and Die A fair coin and a fair die are tossed. et X be pr babi ity f success n any tria
the number of dots that show plus the number of heads. etermine 2 2
a the distribution f for X and b E.X/. P.X > 4/ n D 6, p D P.X > 2/ n D 6, p D
3 3
Cards Two cards from a standard dec of 52 playing cards
are randomly drawn in succession without replacement, and the Die A pair of fair dice is rolled five times. Find the
number of aces, X, is observed. etermine a the distribution f for probability that exactly three of the rolls result in a face
X and b E.X/. sum of 7.
Card Game In a game, a player pays 0.25 to randomly Planting Success The probability that a certain type of
draw 2 cards, with replacement, from a standard dec of bush survives planting is 0. . If four bushes are planted, what is
52 playing cards. For each ten that appears, the player receives 1. the probability that all of them die
What is the player s expected gain or loss ive your answer to
Coin A biased coin is tossed five times. The probability
the nearest cent.
that a head occurs on any toss is 25 . Find the probability that at
Gas Station Profits An oil company determines that the least two heads occur.
probability that a gas station located along the Trans-Canada
Highway is successful is 0.55. A successful station earns an elly Beans A bag contains three red, four green, and thee
annual profit of 160,000 a station that is not successful loses blac elly beans. Five elly beans are randomly withdrawn in
15,000 annually. What is the expected gain to the company if it succession with replacement. Find the probability that at least four
locates a station along the Trans-Canada Highway of the withdrawn elly beans are blac .

Mail Order Computers A mail-order computer company In Pr b ems and a transiti n matrix f r a ark chain is
offers a 30-day money-bac guarantee to any customer who is not gi en etermine the a ues f a b and c
completely satisfied with its product. The company realizes a 2 3 2 3
0:1 2a a a a a
profit of 200 for each computer sold, but assumes a loss of 100 4a b b5 4 b b a5
for shipping and handling for each unit returned. The probability
0:6 b c 0:4 c b
that a unit is returned is 0.0 .
a What is the expected gain for each unit shipped In Pr b ems and a transiti n matrix and an initia state
b If the distributor ships 4000 units per year, what is the ect r X0 f r a ark chain are gi en C mpute the state ect rs
expected annual profit X1 X2 and X3
2 3 2 3
Lottery In a certain lottery, you pay 4.00 to choose one of 0:1 0:3 0:1 0:4 0:1 0:1
41 million number combinations. If that combination is drawn, D 40:2 0:4 0:15 D 40:2 0:6 0:55
you win 50 million. What is your expected gain (or loss) per 0:7 0:3 0: 0:4 0:3 0:4
play
23 2 3
0:5 0:1
X0 D 4 05 X0 D 40:35
0:5 0:6
 Chapter e e 449

In Pr b ems and a transiti n matrix f ra ark chain Automobile Mar et For a particular segment of the
is gi en automobile mar et, the results of a survey indicate that 0 of
a Compute 2 and 3 . people who own a apanese car would buy a apanese car the next
b What is the probability of going to state 1 from state 2 after time and 20 would buy a non- apanese car. f owners of
two steps non- apanese cars, 40 would buy a non- apanese car the next
c What is the probability of going to state 2 from state 1 after time and 60 would buy a apanese car.
three steps a f those who currently own a apanese car, what percentage
will buy a apanese car two cars later
"1 2
# b If 60 of this segment currently own apanese cars and 40
5 5 own non- apanese cars, what will be the distribution for this
4 3
5 5
segment of the mar et two cars from now
c How will this segment be distributed in the long run
2 3
0 0:4 0:3 Voting Suppose that the probabilities of voting for
40 0:3 0:55 particular parties in a future election depend on the voting patterns
1 0:3 0:2 in the previous election. For a certain province where there is a
three-party political system, assume that these probabilities are
contained in the matrix
2 3
In Pr b ems and nd the steady state ect r f r the gi en 0:6 0:1 0:1
transiti n matrix f r a ark chain D Πij # D 40:1 0:7 0:15
"1 1# 0:3 0:2 0:
4 3
3 2 where ij is the probability that a voter will vote for party i in the
4 3 next election if he or she voted for party j in the last election.
2 3 a At the last election, 50 of the electorate voted for party 1,
0:4 0:4 0:3
40:3 30 for party 2, and 20 for party 3. What is the expected
0:2 0:35
percentage distribution of votes for the next election
0:3 0:4 0:4
b In the long run, what is the percentage distribution of votes
 0 ts
an Cont n t

T
he philosopher eno of Elea was fond of paradoxes about motion. His most
10.1 ts
famous one goes something li e this: The warrior Achilles agrees to run a race
10.2 ts Cont n e against a tortoise. Achilles can run 10 meters per second and the tortoise only
1 meter per second, so the tortoise gets a 10-meter head start. Since Achilles
10.3 Cont n t is so much faster, he should still win. ut by the time he has covered his first 10 meters
10.4 Cont n t e to and reached the place where the tortoise started, the tortoise has advanced 1 meter and
ne a t es is still ahead. And after Achilles has covered that 1 meter, the tortoise has advanced
another 0.1 meter and is still ahead. And after Achilles has covered that 0.1 meter,
C er 10 e e the tortoise has advanced another 0.01 meter and is still ahead. And so on. Therefore,
Achilles gets closer and closer to the tortoise but can never catch up.
eno s audience new that the argument was fishy. The position of Achilles at time
t after the race has begun is .10 m s/t. The position of the tortoise at the same time t is
.1 m s/t C 10 m. When these are equal, Achilles and the tortoise are side by side. To
solve the resulting equation
.10 m s/t D .1 m s/t C 10 m
for t is to find the time at which Achilles pulls even with the tortoise. The solution is
! "
t D 1 1 seconds, at which time Achilles will have run 1 1 s .10 m s/ D 11 1 meters.
What puzzled eno and his listeners is how it could be that
1 1 1
10 C 1 C C C ! ! ! D 11
10 100
where the left side represents an in nite sum and the right side is a finite result.
We have already loo ed brie y at exactly this situation in Section 1.5 and, actually,
put it to use in Section 5.6, in our examination of perpetuities. In 5.6 we spo e of imits
f sequences in a somewhat preliminary way and noted that the number e, named in
honour of Euler and which figured prominently in Chapter 4, is the limit of the sequence
# $
nC1 n
. eno s paradox is resolved by limits and in this chapter we go considerably
n
further with the topic.

450
 Section 0. ts 451

Objective L
o st ts an t e r as c In a par ing-lot situation, one may have to inch up to the car in front ith ut t uching
ro ert es
it. This notion of getting closer and closer to something, but yet not touching it, is
also important in mathematics and is involved in the concept of imit which lies at
the foundation of calculus. We will let a variable inch up to a particular value and
examine the effect this process has on the values of a function.
For example, consider the function
x3 " 1
f.x/ D
x"1
This function is not defined at x D 1, but that is the n y number at which it is not
defined. In particular, it is defined for all numbers as close as we li e to 1, and we are
free to examine the functions values f.x/ as x inches up to 1. Table 10.1 gives some
values of x that are slightly less than 1 and some that are slightly greater than 1, along
with the corresponding function values. Notice that as x ta es on values closer and
closer to 1, regardless of whether x approaches it fr m the eft .x < 1/ or fr m the right
.x > 1/, the corresponding values of f.x/ get closer and closer to one and only one
number, namely, 3. This is also clear from the graph of f in Figure 10.1. Notice there
y
that even though the function is not defined at x D 1 (as indicated by the hollow dot),
the function values get closer and closer to 3 as x gets closer and closer to 1. To express
this, we say that the limit of f.x/ as x approaches 1 is 3 and write
x3 - 1
f(x) =
x-1 x3 " 1
lim D3
x!1 x " 1
3
We can ma e f.x/ as close as we li e to 3, and eep it that close, by ta ing x su ciently
close to, but different from, 1. The limit of f at 1 exists, even though the value of f at 1
does not exist. Notice that saying the value of f at 1 does not exist is ust a clumsy
1 way of saying 1 is not in the domain of f .

x
1 Table 10.1
FIGURE x<1 x>1
x3 " 1 x f.x/ x f.x/
lim D 3.
x!1 x " 1
0. 2.44 1.2 3.64
0. 2.71 1.1 3.31
0. 5 2. 525 1.05 3.1525
0. 2. 701 1.01 3.0301
0. 5 2. 5025 1.005 3.015025
0. 2. 7001 1.001 3.003001

We can also consider the limit of a function as x approaches a number that is in the
domain. et us examine the limit of f.x/ D x C 3 as x approaches 2:
lim .x C 3/
x!2

bviously, if x is close to 2 (but not equal to 2), then x C 3 is close to 5. This is also
apparent from the table and graph in Figure 10.2. Thus,
lim .x C 3/ D 5
x!2

iven a function f and a number a, there may be two ways of associating a number to the
pair . f; a/. ne such number is the e a uati n f f at a namely, f.a/. It exists precisely
x3 " 1
when a is in the domain of f. For example, if f.x/ D , our first example, then
x"1
f.1/ does not exist. Another way of associating a number to the pair . f; a/ is the imit
452 C ts an Cont n t

y
x62 x72
x f(x) x f(x)

1.5 4.5 2.5 5.5 f(x) = x + 3

1.9 4.9 2.1 5.1 5

1.95 4.95 2.05 5.05

3
1.99 4.99 2.01 5.01

1.999 4.999 2.001 5.001

x
2

FIGURE lim .x C 3/ D 5.
x!2

f f.x/ as x appr aches a, which is denoted limx!a f.x/. We have given two examples.
Here is the general case.

he imit f f.x/ as x appr aches a is the number , written

lim f.x/ D
x!a

provided that we can ma e the values f.x/ as close as we li e to , and eep them
that close, by ta ing x su ciently close to, but different from, a. If there is no such
number, we say that the limit of f.x/ as x approaches a d es n t exist.

We emphasize that, when finding a limit, we are concerned not with what happens
to f.x/ when x equa s a, but only with what happens to f.x/ when x is c se t a.
In fact, even if f.a/ exists the preceding definition of limx!a f.x/ explicitly rules out
consideration of f.a/. In our second example, f.x/ D x C 3, we have f.2/ D 5 and also
limx!2 .x C 3/ D 5, but it is possible to have a function f and a number a for which both
f.a/ and limx!a f.x/ exist and are different. oreover, a limit must be independent of
the way in which x appr aches a; meaning the way in which x gets close to a. That
is, the limit must be the same whether x approaches a from the left or from the right
(for x < a or x > a, respectively).

E AM LE E L G

a Estimate limx!1 f.x/, where the graph of f is given in Figure 10.3(a).

y y

3
y = f(x) y = f(x)
2 2

x x
1 1
(a) (b)

FIGURE Investigation of limx!1 f .x/.

 Section 0. ts 453

S If we loo at the graph for values of x near 1, we see that f.x/ is near 2.
oreover, as x gets closer and closer to 1, f.x/ appears to get closer and closer to 2.
Thus, we estimate that
lim f.x/ D 2
x!1

b Estimate limx!1 f.x/, where the graph of f is given in Figure 10.3(b).

S Although f.1/ D 3, this fact has no bearing whatsoever on the limit of f.x/
as x approaches 1. We see that as x gets closer and closer to 1, f.x/ appears to get closer
and closer to 2. Thus, we estimate that
lim f.x/ D 2
x!1
Now ork Problem 1 G
Up to now, all of the limits that we have considered did indeed exist. Next we loo
at some situations in which a limit does not exist.

E AM LE L T N E

A L IT I a Estimate limx!!2 f.x/ if it exists, where the graph of f is given in Figure 10.4.
The greatest integer function, S As x approaches "2 from the left .x < "2/, the values of f.x/ appear to get
denoted f .x/ D bxc, is used every closer to 1. ut as x approaches "2 from the right .x > "2/; f.x/ appears to get closer
day by cashiers ma ing change for
to 3. Hence, as x approaches "2, the function values do not settle down to one and only
customers. This function tells the
amount of paper money for each
one number. We conclude that
amount of change owed. (For example,
lim f.x/ does not exist
if a customer is owed 5.25 in change, x!!2
he or she would get 5 in paper money
thus, b5:25c D 5.) Formally, bxc is
Note that the limit does not exist even though the function is defined at x D "2.
defined as the greatest integer less 1
than or equal to x. raph f, sometimes b Estimate lim 2 if it exists.
x!0 x
called a step function, on a graphing
calculator in the standard viewing S et f.x/ D 1=x2 . The table in Figure 10.5 gives values of f.x/ for some
rectangle. (It is in the numbers menu values of x near 0. As x gets closer and closer to 0, the values of f.x/ get larger and
it s called integer part .) Explore larger without bound. This is also clear from the graph. Since the values of f.x/ do not
this graph using TRACE. etermine approach a number as x approaches 0,
whether limx!a f .x/ exists.
1
lim does not exist
x!0 x2

y y

x f(x)
;1 1

; 0.5 4 f(x) = 12
3 x
; 0.1 100
2
; 0.01 10,000
1
; 0.001 1,000,000 1
x x
-1 1

1
FIGURE limx!!2 f .x/ does not FIGURE lim does not exist.
x!0 x2
exist.

Now ork Problem 3 G

454 C ts an Cont n t

With more complicated examples, computational equipment can be helpful for

determining if a limit exists and, if so, estimating its value. Consider the rational func-
Rational functions, quotients of
poynomial functions, were introduced tion
in Section 2.2. x3 C 2:1x2 " 10:2x C 4
f.x/ D
x2 C 2:5x "
and observe that f.2/ is not defined, because 22 C 2:5.2/ " D 0. However, we can try
to determine if limx!2 f.x/ exists by examining values of f.x/ for x c se t but di erent
fr m 2. It is tedious (but not impossible ) to calculate a table of function values f.x/,
for x close to 2. However, the screen shot from a programmable calculator given in
Figure 10.6 provides easily obtained evidence that the limit in question does exist and
suggests that the limit is approximately 1.57.

X Y1
1.9 1.4688
1.99 1.5592
1.999 1.5682
1.9999 1.5691
2.01 1.5793
2.001 1.5702
2.0001 1.5693
X=2.0001

FIGURE limx!2 f .x/ # 1:57.

Alternatively, we can estimate the limit from the graph of f. Figure 10.7 shows the
graph of f in the standard Œ"10; 10! $ Œ"10; 10! window of a graphing calculator.

10

-10 10

-10

FIGURE raph of f .x/ in standard window.

ooming and tracing around x D 2 produces the screen shot of Figure 10. , which
also suggests that the limit exists and is approximately 1.57.

X=1.9998753 . Y=1.5691055

FIGURE ooming and tracing around x D 2

gives limx!2 f .x/ # 1:57.
It is important to understand that neither calculator exercise pr es that the limit
exists. Actually, it is fairly easy to prove that our limit exists and that we have, exactly,
x3 C 2:1x2 " 10:2x C 4
lim D 1:56 23076
x!2 x2 C 2:5x "
See Problem 42.

L
To determine limits, we do not always want to compute function values or s etch
a graph. Alternatively, there are several properties of limits that we may be able to
 Section 0. ts 455

employ. The following properties should seem reasonable, and in fact they can be
proved, using a sharpened version of our definition of limx!a f.x/ D .

If f.x/ D c is a constant function, then

lim f.x/ D lim c D c
x!a x!a
For any positive integer n,
lim xn D an
x!a

E AM LE A L

a limx!2 7 D 7I limx!!5 7 D 7
b limx!6 x2 D 62 D 36
c limt!!2 t4 D ."2/4 D 16
Now ork Problem 9 G
Some other properties of limits are as follows:

If limx!a f.x/ and limx!a g.x/ exist and c is a constant then

˙ ˙
lim . f.x/ ! g.x// D lim f.x/ ! lim g.x/
x!a x!a x!a
hat is the imit f a sum di erence r pr duct is the sum di erence r pr d
uct respecti e y f the imits.

lim .cf.x// D c ! lim f.x/

x!a x!a
hat is the imit f a c nstant times a functi n is the c nstant times the imit f
the functi n.

A L IT I
The volume of helium in a spher- E AM LE A L
ical balloon (in cubic centimeters), as
a function of the radius r in centime- a lim .x2 C x/ D lim x2 C lim x Property 3
x!2 x!2 x!2
4
ters, is given by .r/ D "r3 . Find D2 C2D6 2
Property 2
3
limr!1 .r/.
b Property 3 can be extended to the limit of a finite number of sums, differences, and
products. For example,
lim .q3 " q C 1/ D lim q3 " lim q C lim 1
q!!1 q!!1 q!!1 q!!1
3
D ."1/ " ."1/ C 1 D 1
c lim Œ.x C 1/.x " 3/! D lim .x C 1/ ! lim .x " 3/ Property 3
x!2 x!2 x!2
! " ! "
D lim x C lim 1 ! lim x " lim 3
x!2 x!2 x!2 x!2

D .2 C 1/ ! .2 " 3/ D 3."1/ D "3

d lim 3x3 D 3 ! lim x3 Property 4
x!!2 x!!2

D 3."2/3 D "24
Now ork Problem 11 G
456 C ts an Cont n t

E AM LE L F

et f.x/ D cn xn C cn!1 xn!1 C ! ! ! C c1 x C c0 define a polynomial function. Then

lim f.x/ D lim .cn xn C cn!1 xn!1 C ! ! ! C c1 x C c0 /

x!a x!a

D cn ! lim xn C cn!1 ! lim xn!1 C ! ! ! C c1 ! lim x C lim c0

x!a x!a x!a x!a
n n!1
D cn a C cn!1 a C ! ! ! C c1 a C c0 D f.a/

Thus, we have the following property:

A L IT I
The revenue function for a certain If f is a polynomial function, then
product is given by R.x/ D 500x " 6x2 .
Find limx! R.x/. lim f.x/ D f.a/
x!a

In other words, if f is a polynomial and a is any number, then both ways of associating
a number to the pair . f; a/, namely, formation of the limit and evaluation, exist and
are equal.
Now ork Problem 13 G
The result of Example 5 allows us to find many limits simply by evaluation. For
example, we can find

lim .x3 C 4x2 " 7/

x!!3

by substituting "3 for x because x3 C 4x2 " 7 is a polynomial function:

lim .x3 C 4x2 " 7/ D ."3/3 C 4."3/2 " 7 D 2

x!!3

Similarly,

lim .2.h " 1// D 2.3 " 1/ D 4

h!3

We want to stress that we do not find limits simply by evaluating unless there is
a rule that covers the situation. We were able to find the previous two limits by evalu-
ation because we have a rule that applies to limits of polynomial functions. However,
indiscriminate use of evaluation can lead to errors. To illustrate, in Example 1(b) we
have f.1/ D 3, which is not limx!1 f.x/ in Example 2(a), f."2/ D 2, which is not
limx!!2 f.x/.
The next two limit properties concern quotients and roots.

If limx!a f.x/ and limx!a g.x/ exist and n is a positive integer then

f.x/ limx!a f.x/

lim D if lim g.x/ ¤ 0
x!a g.x/ limx!a g.x/ x!a

hat is the imit f a qu tient is the qu tient f imits pro i e t at the den m
inat r imit is n t 0
p q
n
lim f.x/ D n lim f.x/
x!a x!a
 Section 0. ts 457

E AM LE A L
2x2 C x " 3 limx!1 .2x2 C x " 3/ 2C1"3 0
a lim 3
D 3
D D D0
Note that in Example 6(a) the numerator x!1 x C4 limx!1 .x C 4/ 1C4 5
and denominator of the function are p q p
polynomials. In general, we can b lim t2 C 1 D lim.t2 C 1/ D 17
determine a limit of a rational function by t!4 t!4
evaluation, provided that evaluation of
the denominator does not give 0. p
3
q p p p
c lim x2 C 7 D 3 lim .x2 C 7/ D 3
16 D 3 ! 2 D 2 3 2
x!3 x!3

Now ork Problem 15 G

L A M
We now consider limits to which our limit properties do not apply and which cannot
be determined by evaluation. A fundamental result is the following:
The condition for equality of the limits
does not preclude the possibility that If f and g are two functions for which f.x/ D g.x/, for all x ¤ a, then
f .a/ D g.a/. The condition only concerns
x ¤ a. lim f.x/ D lim g.x/
x!a x!a
(meaning that if either limit exists, then the other exists and they are equal).

The result follows directly from the definition of imit since the value of
limx!a f.x/ depends only on those values f.x/ for x that are close to a. We repeat:
The evaluation of f at a, f.a/, or lac of its existence, is irrelevant in the
determination of limx!a f.x/ unless we have a specific rule that applies, such as in
the case when f is a polynomial.

A L IT I E AM LE F L
The rate of change of productivity p x2 " 1
(in number of units produced per hour) Find lim .
x!!1 x C 1
increases with time on the ob by the
function S As x ! "1, both numerator and denominator approach zero. ecause the
50.t2 C 4t/ limit of the denominator is 0, we cann t use Property 6. However, since what happens
p.t/ D to the quotient when x equals "1 is of no concern, we can assume that x ¤ "1 and
t2 C 3t C 20
simplify the fraction:
Find limt!2 p.t/.
x2 " 1 .x C 1/.x " 1/
D D x " 1 for x ¤ "1
xC1 xC1
x2 " 1
This algebraic manipulation (factoring and cancellation) of the original function
xC1
yields a new function x"1, which is the same as the original function for x ¤ "1. Thus,
the fundamental result displayed in the box at the beginning of this subsection applies
When both f .x/ and g.x/ approach 0 as
and we have
x ! a, then the limit x2 " 1
lim D lim .x " 1/ D "1 " 1 D "2
f .x/ x!!1 x C 1 x!!1
lim
x!a g.x/ Notice that, although the original function is not defined at "1, it d es have a limit as
is said to have the f rm 0=0. Similarly, x ! "1.
we spea of f rm k=0, for k ¤ 0 if f .x/
approaches k ¤ 0 as x ! a but g.x/ Now ork Problem 21 G
approaches 0 as x ! a.
In Example 7, the method of finding a limit by evaluation does not wor . Replacing
x by "1 gives 0=0, which has no meaning. When the meaningless form 0=0 arises,
algebraic manipulation (as in Example 7) may result in a function that agrees with the
original function, except possibly at the limiting value. In Example 7 the new function,
x " 1, is a polynomial and its limit can be found by evaluation.
458 C ts an Cont n t

In the beginning of this section, we found

x3 " 1
lim
x!1 x " 1

by examining a table of function values of f.x/ D .x3 " 1/=.x " 1/ and also by consid-
ering the graph of f. This limit has the form 0=0. Now we will determine the limit by
using the technique used in Example 7.

E AM LE F 0=0
There is frequently confusion about
which principle is being used in this x3 " 1
example and in Example 7. It is this: Find lim .
x!1 x " 1
If f .x/ D g.x/ for x ¤ a; S As x ! 1, both the numerator and denominator approach 0. Thus, we will
then lim f .x/ D lim g.x/: try to express the quotient in a different form for x ¤ 1. y factoring, we have
x!a x!a
x3 " 1 .x " 1/.x2 C x C 1/
D D x2 C x C 1 for x ¤ 1
x"1 .x " 1/
(Alternatively, long division would give the same result.) Therefore,
x3 " 1
lim D lim .x2 C x C 1/ D 12 C 1 C 1 D 3
x!1 x " 1 x!1

as we showed before.
Now ork Problem 23 G
E AM LE F 0=0
A L IT I
f.x C h/ " f.x/
The length of a material increases If f.x/ D x2 C 1, find lim .
as it is heated up according to the equa- h!0 h
tion D 125 C 2x. The rate at which the S
length is increasing is given by f.x C h/ " f.x/ Œ.x C h/2 C 1! " .x2 C 1/
lim D lim
125 C 2.x C h/ " .125 C 2x/ h!0 h h!0 h
lim
h!0 h Here we treat x as a constant because h, not x, is changing. As h ! 0, both the numer-
Calculate this limit. ator and denominator approach 0. Therefore, we will try to express the quotient in a
different form, for h ¤ 0. We have
Œ.x C h/2 C 1! " .x2 C 1/ Œx2 C 2xh C h2 C 1! " x2 " 1
lim D lim
h!0 h h!0 h
2xh C h2
D lim
h!0 h
The expression h.2x C h/
D lim
f .x C h/ " f .x/ h!0 h
h D lim .2x C h/
h!0
is called a di erence qu tient. The limit
of the difference quotient lies at the heart D 2x
of differential calculus. We will encounter
many such limits in Chapter 11.
h.2x C h/
Note: It is the fourth equality above, lim D lim .2x C h/, that uses the
h!0 h h!0
h.2x C h/
fundamental result. When and 2x C h are considered as functi ns f h, they
h
are seen to be equal, for all h ¤ 0. It follows that their limits as h approaches 0 are equal.
Now ork Problem 35 G
AS L
We conclude this section with a note concerning a most important limit, namely,
lim .1 C x/1=x
x!0
 Section 0. ts 459

Figure 10. shows the graph of f.x/ D .1 C x/1=x . Although f(0) does not exist, as
x ! 0 it is clear that the limit of .1 C x/1=x exists. It is approximately 2.71 2 and
is denoted by the letter e. This, you may recall, is the base of the system of natural
logarithms. The limit
This limit will be used in Chapter 12. lim .1 C x/1=x D e
x!0

can actually be considered the definition of e. It can be shown that this agrees with the
definition of e that we gave in Section 4.1.
f(x)

x (1 + x)1/x x (1 + x)1/x

0.5 2.2500 -0.5 4.0000

3
0.1 2.5937 -0.1 2.8680
2
0.01 2.7048 -0.01 2.7320
f(x) = (1 + x)1/x
0.001 2.7169 -0.001 2.7196 1

x
1

FIGURE limx!0 .1 C x/1=x D e.

R BLEMS
In Pr b ems use the graph f f t estimate each imit if it raph of f appears in Figure 10.12.
exists a limx!!1 f .x/ b limx!1 f .x/ c limx!2 f .x/
raph of f appears in Figure 10.10.
a limx!0 f .x/ b limx!1 f .x/ c limx!2 f .x/ y

y 3
2 y = f(x)
y = f(x)
x
-1 1 2
1

x FIGURE
1 2

raph of f appears in Figure 10.13.

FIGURE a limx!!1 f .x/ b limx!0 f .x/ c limx!1 f .x/

raph of f appears in Figure 10.11.

y
a limx!!1 f .x/ b limx!0 f .x/ c limx!1 f .x/

y y = f(x)

-1 1
x
1 1
-1

x
-1 1

FIGURE FIGURE
460 C ts an Cont n t

In Pr b ems use y ur ca cu at r t c mp ete the tab e and .x C h/2 " x2

use y ur resu ts t estimate the gi en imit Find lim by treating x as a constant.
h!0 h
3.x C h/2 C 7.x C h/ " 3x2 " 7x
3x2 C 2x " 1 Find lim by treating x as a
lim h!0 h
x!!1 xC1 constant.
x "0: "0: "0: "1:001 "1:01 "1:1 f .x C h/ " f .x/
f .x/ In Pr b ems nd lim
h!0 h
f .x/ D 7 C 7x f .x/ D 2x C 3
x2 "
lim f .x/ D x2 " 3 f .x/ D x2 C x C 1
x!!3 x C 3

x "3:1 "3:01 "3:001 "2: "2: "2: f .x/ D x3 " 4x2 f .x/ D 3 " 2x C x2
f.x/ p
x"2"2
Find lim ( int First rationalize the numerator
x!6 x"6
2x " 1
lim by multiplying both the numerator and denominator by
x!0 x p
x " 2 C 2.)
x "0:001 "0:0001 0:0001 0.001 0.01 0.1 x2 C x C c
f.x/ Find the constant c so that lim 2 exists. For that
x!3 x " 5x C 6
value of c, determine the limit. ( int Find the value of c for
p which x " 3 is a factor of the numerator.)
1Ch"1
lim
h!0 h Power Plant The maximum theoretical e ciency of a
power plant is given by
h "0:1 "0:01 "0:001 0.001 0.01 0.1
f.x/ h " c
ED
h
In Pr b ems nd the imits
where h and c are the absolute temperatures of the hotter and
lim 16 lim 2x colder reservoirs, respectively. Find a lim c !0 E and
x!2 x!3
b lim c ! h E.
2
lim .t " 5/ limt!1=3 .2t C 7/
t!!5 Satellite When a 3200-lb satellite revolves about the earth
4r " 3 in a circular orbit of radius r ft, the total mechanical energy E of
lim .3x3 " 4x2 C 2x " 3/ lim the earth satellite system is given by
x!!2 r! 11
t"2 x2 C 6 7:0 $ 1017
lim lim ED" ft-lb
t!!3 t C 5 x!!6 x " 6 r
t z2 " 5z " 4
lim lim Find the limit of E as r ! 7:5 $ 107 ft.
t!0 t3"tC7 z!0 z2 C 1
p p
lim p2 C p C 5 lim yC3 In Pr b ems use a graphing ca cu at r t graph the
p!4 y!15
functi ns and then estimate the imits R und y ur ans ers t t
x2 C 2x x"1 decima p aces
lim limx!1 x3 C 2:1x2 " 10:2x C 4
x!!2 x C 2 x"1 lim lim xx
x!2 x2 C 2:5x " x!0
x2 " x " 2 t3 C 3t2
lim lim p
x!2 x"2 t!0 t3 " 4t2 x " 10 x C 21 x3 C x2 " 5x C 3
lim p lim 3
x2 " x " 6 t2 " 4 x! 3" x x!1 x C 2x2 " 7x C 4
lim lim
x!3 x"3 t!2 t " 2
ater Purification The cost of purifying water is given by
xC2 x2 " 2x 50;000
lim lim CD " 6500, where p is the percent of impurities
x!!2 x2 " 4 x!0 x p
x2 " x C 20 x4 " 1 remaining after purification. raph this function on your graphing
lim lim calculator, and determine limp!0 C. iscuss what this means.
x!4 x2 " 3x " 4 x!!3 x2 C x C 15
Profit Function The profit function for a certain business
3x2 " x " 10 x2 " x " 2
lim lim is given by P.x/ D 225x " 3:2x2 " 701. raph this function on a
x!2 x2 C 5x " 14 x!2 x2 " 4 graphing calculator, and use the evaluation function to determine
.2 C h/2 " 22 .x C 2/2 " 4 limx!40:3 P.x/, using the rule about the limit of a polynomial
lim lim function.
h!0 h x!0 x
 Section 0.2 ts Cont n e 461

Objective L C
o st one s e ts n n te ts
an ts at n n t S L
y
Figure 10.14 shows the graph of a function f. Notice that f.x/ is not defined when x D 0.
As x approaches 0 fr m the right f.x/ approaches 1. We write this as
y = f(x) lim f.x/ D 1
1 x!0C
x
-1 n the other hand, as x approaches 0 fr m the eft f .x/ approaches "1, and we write
lim f.x/ D "1
x!0!
FIGURE limx!0 f .x/ does imits li e these are called one sided limits. From the preceding section, we now that
not exist. the limit of a function as x ! a is independent of the way x approaches a. Thus, the
limit will exist if and only if both one-sided limits exist and are equal. We, therefore,
conclude that

f (x)
lim f.x/ does not exist
x!0
p
As another example of a one-sided limit, consider f.x/ D x " 3 as x approaches 3.
f (x) = x-3
Since f is defined only when x % 3, we can spea of the limit of f.x/ as x approaches
2 3 from the right. If x is slightly greater than 3, then x " 3 is a positive number that is
1
p
close to 0, so x " 3 is close to 0. We conclude that
3 6
x p
lim x " 3 D 0
x!3C
p
FIGURE limx!3C x " 3 D 0. This limit is also evident from Figure 10.15.

I L
In the previous section, we considered limits of the form 0=0 that is, limits where
both the numerator and denominator approach 0. Now we will examine limits where
the denominator approaches 0, but the numerator approaches a number different from
y 0. For example, consider
1
lim
x!0 x2

y=
1 ,
x 0
Here, as x approaches 0, the denominator approaches 0 and the numerator approaches 1.
x2 et us investigate the behavior of f.x/ D 1=x2 when x is close to 0. The number x2 is
positive and also close to 0. Thus, dividing 1 by such a number results in a very large
1 number. In fact, the closer x is to 0, the larger the value of f.x/. For example, see the
x table of values in Figure 10.16, which also shows the graph of f. Clearly, as x ! 0
1
-1
both from the left and from the right, f.x/ increases without bound. Hence, no limit
x f (x) exists at 0. We say that as x ! 0; f.x/ becomes positively infinite, and symbolically we
;1 1
express this infinite limit by writing
;0.5 4 1
lim D C1 D 1
;0.1 100 x!0 x2
;0.01 10,000
;0.001 1,000,000 If limx!a f.x/ does not exist, it may be for a reason other than that the values f.x/
become arbitrarily large as x gets close to a. For example, loo again at the situation in
FIGURE lim
1
D 1.
Example 2(a) of Section 10.1. Here we have
x!0 x2
lim f.x/ does not exist but lim f.x/ ¤ 1
x!!2 x!!2

The use of the equality sign in this Consider now the graph of y D f.x/ D 1=x for x ¤ 0. (See Figure 10.17.) As x
situation does not mean that the limit approaches 0 from the right, 1=x becomes positively infinite as x approaches 0 from
exists. n the contrary, it is a way of the left, 1=x becomes negatively infinite. Symbolically, these infinite limits are written
saying specifically that there is no limit
and hy there is no limit. 1 1
lim D1 and lim! D "1
x!0C x x!0 x
462 C ts an Cont n t

x f (x)
0.01 100 y = 1x , x 0
0.001 1000
0.0001 10,000
x
-0.01 -100 0
-0.001 -1000
-0.0001 -10,000

1
FIGURE lim does not exist.
x!0 x

Either one of these facts implies that

1
lim does not exist
x!0 x

E AM LE I L

Find the limit (if it exists).

2
a lim
x!!1C xC1
-1 S As x approaches "1 from the right (thin of values of x such as "0: ; "0: ,
-0.99 -0.9 and so on, as shown in Figure 10.1 ), x C 1 approaches 0 but is always positive. Since
we are dividing 2 by positive numbers approaching 0, the results, 2=.xC1/, are positive
FIGURE x ! "1C . numbers that are becoming arbitrarily large. Thus,
2
lim D1
x!!1C xC1
and the limit does not exist. y a similar analysis, we can show that
2
lim D "1
x!!1! xC1
xC2
b lim
x!2 x2 " 4
S As x ! 2, the numerator approaches 4 and the denominator approaches 0.
Hence, we are dividing numbers near 4 by numbers near 0. The results are numbers
that become arbitrarily large in magnitude. At this stage, we can write
xC2
lim does not exist
x!2 x2 " 4

However, let us see if we can use the symbol 1 or "1 to be more specific about does
not exist . Notice that
xC2 xC2 1
lim 2 D lim D lim
x!2 x " 4 x!2 .x C 2/.x " 2/ x!2 x " 2

Since
1 1
lim D1 and lim D "1
x!2C x"2 x!2! x"2
xC2
lim is neither 1 nor "1.
x!2 x2 " 4

Now ork Problem 31 G

 Section 0.2 ts Cont n e 463

Example 1 considered limits of the form k=0, where k ¤ 0. It is important to

distinguish the form k=0 from the form 0=0, which was discussed in Section 10.1.
These two forms are handled differently.

E AM LE F L
t"2
Find lim .
t!2 t2 " 4

S As t ! 2, b th numerator and denominator approach 0 (form 0=0). Thus,

we first simplify the fraction, for t ¤ 2, as we did in Section 10.1, and then ta e the limit:
t"2 t"2 1 1
lim D lim D lim D
t!2 t2 " 4 t!2 .t C 2/.t " 2/ t!2 t C 2 4
Now ork Problem 37 G
L I
Now let us examine the function
1
We can obtain f.x/ D
x
1 1
lim and lim as x becomes infinite, first in a positive sense and then in a negative sense. From
x!1 x x!!1 x
Table 10.2, we can see that as x increases without bound through positive values, the
without the benefit of a graph or a table.
values of f.x/ approach 0. i ewise, as x decreases without bound through negative
ividing 1 by a large positive number
results in a small positive number, and values, the values of f.x/ also approach 0. These observations are also apparent from
as the divisors get arbitrarily large, the the graph in Figure 10.17. There, moving to the right along the curve through pos-
quotients get arbitrarily small. A similar itive x-values, the corresponding y-values approach 0 through positive values. Simi-
argument can be made for the limit as larly, moving to the left along the curve through negative x-values, the corresponding
x ! "1.
y-values approach 0 through negative values. Symbolically, we write
1 1
lim D 0 and lim D0
x!1 x x!!1 x

oth of these limits are called imits at in nity.

Table 10. B f .x/ x ! ˙1
x f.x/ x f.x/

1000 0.001 "1000 "0.001

10,000 0.0001 "10,000 "0.0001
100,000 0.00001 "100,000 "0.00001
1,000,000 0.000001 "1,000,000 "0.000001

A L IT I
E AM LE L I
The demand function for a certain Find the limit (if it exists).
10;000 4
product is given by p.x/ D , a lim
.x C 1/2 x!1 .x " 5/3
where p is the price in dollars and x is
the quantity sold. raph this function on S As x becomes very large, so does x " 5. Since the cube of a large number is
your graphing calculator in the window also large, .x " 5/3 ! 1. ividing 4 by very large numbers results in numbers near
Œ0; 10! $ Œ0; 10;000!. Use the TRACE 0. Thus,
function to find limx!1 p.x/. eter- 4
mine what is happening to the graph and lim D0
x!1 .x " 5/3
what this means about the demand func- p
tion. b lim 4"x
x!!1
S As x gets negatively infinite, 4 " x becomes positively infinite. ecause
square roots of large numbers are large numbers, we conclude that
p
lim 4"xD1
x!!1 G
464 C ts an Cont n t

In our next discussion we will need a certain limit, namely, limx!1 1=x p , where
p > 0. As x becomes very large, so does x p . ividing 1 by very large numbers results
in numbers near 0. Thus, limx!1 1=x p D 0. In general,
1 1
lim D 0 and, if p is such that 1=x p is defined for x < 0, lim D0
x!1 x p x!!1 x p

for p > 0. For example,

1 1
lim p D lim 1=3 D 0
x!1 3
x x!1 x
et us now find the limit of the rational function
4x2 C 5
f.x/ D 2
2x C 1
as x ! 1. (Recall from Section 2.2 that a rational function is a quotient of polynomi-
als.) As x gets larger and larger, b th the numerator and denominator of any rational
function become infinite in absolute value. However, the form of the quotient can be
changed, so that we can draw a conclusion as to whether or not it has a limit. To do this,
we divide both the numerator and denominator by the greatest power of x that occurs
in the denominator. Here it is x2 . This gives
4x2 C 5 4x2 5
2 C 2
4x C 5 x 2 x 2 x
lim D lim D lim
x!1 2x2 C 1 x!1 2x2 C 1 x!1 2x2 1
2 2
C 2
x x x
5 1
4C 2 lim 4 C 5 ! lim 2
x x!1 x!1 x
D lim D
x!1 1 1
2C 2 lim 2 C lim 2
x x!1 x!1 x
f(x) Since limx!1 1=x p D 0 for p > 0,
5 4x2 C 5 4 C 5.0/ 4
lim D D D2
2 x!1 2x2 C 1 2C0 2
f(x) = 4x 2 + 5
2x + 1 Similarly, the limit as x ! "1 is 2. These limits are clear from the graph of f in
Figure 10.1 .
For the preceding function, there is an easier way to find limx!1 f.x/. For arge
2 values of x, in the numerator the term involving the greatest power of x, namely, 4x2 ,
dominates the sum 4x2 C 5, and the dominant term in the denominator, 2x2 C 1, is 2x2 .
x
Thus, as x ! 1; f.x/ can be approximated by .4x2 /=.2x2 /. As a result, to determine
-1 1 the limit of f.x/, it su ces to determine the limit of .4x2 /=.2x2 /. That is,
4x2 C 5 4x2
FIGURE limx!1 f .x/ D 2 lim D lim D lim 2 D 2
x!1 2x2 C 1 x!1 2x2 x!1
and limx!!1 f .x/ D 2.
as we saw before. In general, we have the following rule:

L I R F
If f.x/ is a rati na functi n and an xn and bm xm are the terms in the numerator and
denominator, respectively, with the greatest powers of x, then
an xn
lim f.x/ D lim
x!˙1 x!˙1 bm xm

et us apply this rule to the situation where the degree of the numerator is greater than
the degree of the denominator. For example,
# $
x4 " 3x x4 1
lim D lim D lim " x3 D 1
x!!1 5 " 2x x!!1 "2x x!!1 2
 Section 0.2 ts Cont n e 465

(Note that in the next-to-last step, as x becomes very negative, so does x3 moreover,
" 12 times a very negative number is very positive.) Similarly,
# $
x4 " 3x 1
lim D lim " x3 D "1
x!1 5 " 2x x!1 2
From this illustration, we ma e the following conclusion:

If the degree of the numerator of a rati na functi n is greater than the degree of
the denominator, then the function has no limit as x ! 1 and no limit as x ! "1.

E AM LE L I R F
A L IT I
The yearly amount of sales y of Find the limit (if it exists).
a certain company (in thousands of
dollars) is related to the amount the x2 " 1
a lim
company spends on advertising, x (in x!1 7 " 2x C x2
thousands of dollars), according to the S
equation
500x
x2 " 1 x2 1 1
y.x/ D lim D lim D lim D
x!1 7 " 2x C x2 x!1 x2 x!1
x C 20
raph this function on your graph- x
b lim
ing calculator in the window x!!1 .3x " 1/2
Œ0; 1000! $ Œ0; 550!. Use TRACE
to explore limx!1 y.x/, and determine
S
what this means to the company. x x x
lim D lim D lim
x!!1 .3x " 1/2 x!!1 x2 " 6x C 1 x!!1 x2

1 1 1 1
D lim D ! lim D .0/ D 0
x!!1 x x!!1 x

x5 " x4
c lim
x!1 x4 " x3 C 2

S Since the degree of the numerator is greater than that of the denominator,
there is no limit. ore precisely,
x5 " x4 x5
lim D lim D lim x D 1
x!1 x4 " x3 C 2 x!1 x4 x!1
Now ork Problem 21 G
The preceding technique applies only to x2 " 1 x2
limits of rational functions at in nity. To find lim , we cannot simply determine the limit of . That simplifi-
x!0 7 " 2x C x2 x2
cation applies only in case x ! 1 or x ! "1. Instead, we have
x2 " 1 limx!0 x2 " 1 0"1 1
lim 2
D 2
D D"
x!0 7 " 2x C x limx!0 7 " 2x C x 7"0C0 7
et us now consider the limit of the polynomial function f.x/ D x2 " 2x as x ! 1:
lim . x2 " 2x/
x!1

ecause a polynomial is a rational function with denominator 1, we have

x2 " 2x x2
lim . x2 " 2x/ D lim D lim D lim x2
x!1 x!1 1 x!1 1 x!1

That is, the limit of x2 " 2x as x ! 1 is the same as the limit of the term involving
the greatest power of x, namely, x2 . As x becomes very large, so does x2 . Thus,
lim . x2 " 2x/ D lim x2 D 1
x!1 x!1
466 C ts an Cont n t

In general, we have the following:

As x ! 1 (or x ! "1), the limit of a p yn mia functi n is the same as the limit
of its term that involves the greatest power of x.

A L IT I
E AM LE L I F
The cost, C, of producing x units of a limx!!1 .x3 " x2 C x " 2/ D limx!!1 x3 . As x becomes very negative, so does
a certain product is given by C.x/ D x3 . Thus,
50;000C200xC0:3x2 . Use your graph-
ing calculator to explore limx!1 C.x/ lim .x3 " x2 C x " 2/ D lim x3 D "1
x!!1 x!!1
and determine what this means.
3 3
b limx!!1 ."2x C x/ D limx!!1 "2x D 1, because "2 times a very negative
number is very positive.
Now ork Problem 9 G
The technique of focusing on dominant terms to find limits as x ! 1 or x ! "1
is valid for rati na functi ns but it is not necessarily valid for other types of functions.
o not use dominant terms when a For example, consider
function is not rational. !p "
lim x2 C x " x
x!1
p
Notice that x2 C x " x is not a rational function. It is inc rrect to infer that because
x2 dominates in x2 C x, the limit in (1) is the same as
!p "
lim x2 " x D lim .x " x/ D lim 0 D 0
x!1 x!1 x!1

It can be shown (see Problem 62) that the limit in (1) is not 0, but is 12 .
The ideas discussed in this section will now be applied to a case-defined function.

E AM LE L C F
% 2
A L IT I x C 1 if x % 1
If f.x/ D , find the limit (if it exists).
3 if x < 1
A plumber charges 100 for the first
hour of wor at your house and 75 for a limx!1C f.x/
every hour (or fraction thereof) after-
S Here x gets close to 1 from the right. For x > 1, we have f.x/ D x2 C 1.
ward. The function for what an x-hour
visit will cost you is
Thus,
8̂ lim f.x/ D lim .x2 C 1/
ˆ 100 if 0 < x & 1 x!1C x!1C
<
175 if 1 < x & 2
f .x/ D
ˆ 250 if 2 < x & 3 If x is greater than 1, but close to 1, then x2 C 1 is close to 2. Therefore,
:̂
325 if 3 < x & 4
lim f.x/ D lim .x2 C 1/ D 2
Find limx!1 f .x/ and limx!2:5 f .x/. x!1C x!1C

b limx!1! f.x/
S Here x gets close to 1 from the left. For x < 1; f.x/ D 3. Hence,
lim f.x/ D lim! 3 D 3
x!1! x!1

c limx!1 f.x/
S We want the limit as x approaches 1. However, the rule of the function
depends on whether x % 1 or x < 1. Thus, we must consider one-sided limits. The
limit as x approaches 1 will exist if and only if both one-sided limits exist and are the
same. From parts (a) and (b),
lim f.x/ ¤ lim! f.x/ since 2 ¤ 3
x!1C x!1
 Section 0.2 ts Cont n e 467

Therefore,
lim f.x/ does not exist.
x!1

d limx!1 f.x/
S For very large values of x, we have x % 1, so f.x/ D x2 C 1. Thus,
lim f.x/ D lim .x2 C 1/ D lim x2 D 1
x!1 x!1 x!1

e limx!!1 f.x/
S For very negative values of x, we have x < 1, so f.x/ D 3. Hence,
lim f.x/ D lim 3 D 3
x!!1 x!!1
All the limits in parts (a) through (c) should be obvious from the graph of f in
Figure 10.20.

f(x)

3
x2 + 1, if x Ú 1
f(x) =
3, if x 6 1
2

x
1

FIGURE raph of a case-defined function.

Now ork Problem 57 G

R BLEMS
For the function f given in Figure 10.21, find the following For the function f given in Figure 10.22, find the following
limits. If the limit does not exist, so state, or use the symbol 1 or limits. If the limit does not exist, so state, or use the symbol 1 or
"1 where appropriate. "1 where appropriate.

f(x)
f(x)

2
2 1
1
x x
-1 1 1 2

FIGURE
FIGURE
a limx!0! f .x/ b limx!0C f .x/ c limx!0 f .x/
d limx!!1 f .x/ e limx!1 f .x/ f limx!1 f .x/
a limx!!1 f .x/ b limx!!1! f .x/ c limx!!1C f .x/
g limx!2C f .x/
d limx!!1 f .x/ e limx!0! f .x/ f limx!0C f .x/
g limx!0 f .x/ h limx!1! f .x/ i limx!1C f .x/ In each f Pr b ems nd the imit If the imit d es n t exist
limx!1 f .x/ limx!1 f .x/ s state r use the symb 1 r "1 here appr priate
lim .x C 7/ lim .1 " x2 / lim 5x
x!5C x!!1C x!!1

6x 5
lim "6 lim lim
x!!1 x!0! x4 x!3 x"2
468 C ts an Cont n t
p %
lim x2 lim .t " 1/3 lim h"1 x if x < 0
x!!1 t!1 h!1C g.x/ D
"x if x > 0
p "3
lim 5"h lim lim 21=2 a limx!0C g.x/ b limx!0! g.x/ c limx!0 g.x/
h!5! x!!3! xC3 x!0!
d limx!1 g.x/ e limx!!1 g.x/
p p
lim .4 x " 1/ lim .x 4 " x2 / % 2
x!1C x!2! x if x < 0
g.x/ D
p p "x2 if x > 0
3
lim x C 10 lim " 5 " 3x lim p a limx!0C g.x/ b limx!0! g.x/ c limx!0 g.x/
x!1 x!!1 x!1 x
d limx!1 g.x/ e limx!!1 g.x/
"6 x"5 2x " 4
lim p lim lim Average Cost If c is the total cost in dollars to produce
x!1 5x 3 x x!1 2x C 1 x!1 3 " 2x
q units of a product, then the average cost per unit for an output
3x2 C 2 r3 of q units is given by c D c=q. Thus, if the total cost equation is
lim lim c D 5000 C 6q, then
x!!1 2x3 C 5x " 7 r!1 r2 C 1

3t3 C 2t2 C t " 1 4x2 5000

lim lim cD C6
t!1 5t2 " 5 x!1 3x3 " x2 C 2 q
7 2
lim lim For example, the total cost of an output of 5 units is 5030, and
x!1 2x C 1 x!1 .3x C 2/2
the average cost per unit at this level of production is 1006. y
3 " 4x " 2x3 3 " 2x " 2x3 finding limq!1 c, show that the average cost approaches a level
lim lim
x!1 5x3 " x C 1 x!!1 7 " 5x3 C 2x2 of stability if the producer continually increases output. What is
the limiting value of the average cost S etch the graph of the
xC3 3x 3 2
"2
lim lim ! lim average-cost function.
x!3C x2 " x!!3 " x2 !1 7 2 C3
Average Cost Repeat Problem 5 , given that the fixed cost
4 " 3x3 6 " 4x2 C x3 is 12,000 and the variable cost is given by the function c D 7q.
lim 3 lim
x!1 x " 1 x!1 4 C 5x " 7x2 Population The population of a certain small city t years
from now is predicted to be
2x " x2 5x2 C 14x " 3
lim lim !
x!!1 x2 C 1 x " 64 x!!3 x2 C 3x 5000
D 40; 000 "
2
t C 4t " 5 2
x " 3x C 1 tC3
lim lim
t!1 4t2 " 2t " 2 x!1 x2 C 1 Find the population in the long run that is, find limt!1 .
# $
3
3x " x 2 1 Show that
lim lim 2 "
x!!1 2x C 1 x!2! x"2 !p " 1
lim x2 C x " x D
x5 C 2x3 " 1 x2 C 1 x!1 2
lim " 5 lim p
x!!1 x " 4x2 x!!5! x2 " 25 ( int Rationalize the numerator by multiplying the expression
p
x 5 x2 C x " x by
lim p lim 2 p
x!!2C 16 " x4 x!0 x C x
C

# $ x2 C x C x
1 "3 p
lim x2 C lim x.x " 1/!1 lim x2 C x C x
x!!1 x x!1 x!3 x " 3
# $ # $ Then express the denominator in a form such that x is a factor.)
"5 7
lim lim " lim jx " 1j ost Parasite Relationship For a particular host parasite
x!1C 1 " x x!3 x"3 x!1
relationship, it was determined that when the host density
ˇ ˇ # $
ˇ1ˇ x2 C 2 3 2x2 (number of hosts per unit of area) is x, the number of hosts
lim ˇˇ ˇˇ lim lim " 2 parasitized over a period of time is
x!0 x x!1 x2 x!1 x x C1
00x
In Pr b ems nd the indicated imits If the imit d es n t yD
10 C 45x
exist s state r use the symb 1 r "1 here appr priate
% Calculate lim y and interpret the result.
2 if x & 2 x!1
f .x/ D % p
1 if x > 2
2 " x if x < 2
a limx!2C f .x/ b limx!2! f .x/ c limx!2 f .x/ If f .x/ D 3 , determine the value of
x C k.x C 1/ if x % 2
d limx!1 f .x/ e limx!!1 f .x/ the constant k for which limx!2 f .x/ exists.
% In Pr b ems and use a ca cu at r t e a uate the gi en
2 " x if x & 3
f .x/ D functi n hen x D 1 0.5, 0.2, 0.1, 0.01, 0.001, and 0.0001. Fr m
"1 C 3x " x2 if x > 3
y ur resu ts specu ate ab ut limx!0C f .x/
a limx!3C f .x/ b limx!3! f .x/ c limx!3 f .x/
d limx!1 f .x/ e limx!!1 f .x/ f .x/ D x2x f .x/ D e1=x
 Section 0.3 Cont n t 469
p % 2
raph f .x/ D 4x2 " 1. Use the graph to estimate 2x C 3 if x < 2
raph f .x/ D . Use the graph to estimate
limx!1=2C f .x/. 2x C 5 if x % 2
p each of the following limits if it exists:
x2 " 4
raph f .x/ D . Use the graph to estimate a limx!2! f .x/ b limx!2C f .x/ c limx!2 f .x/
xC2
limx!!3! f .x/ if it exists.

Objective C
o st cont n t an to n o nts any functions have the property that there is no brea in their graphs. For example,
of scont n t for a f nct on
compare the functions
%
x if x ¤ 1
f.x/ D x and g.x/ D
2 if x D 1
whose graphs appear in Figures 10.23 and 10.24, respectively. The graph of f is unbro-
f(x)
en, but the graph of g has a brea at x D 1. Stated another way, if you were to trace
No break
both graphs with a pencil, you would have to lift the pencil off the graph of g when
in graph x D 1, but you would not have to lift it off the graph of f. These situations can be
f(x) = x
expressed by limits. As x approaches 1, compare the limit of each function with the
1
value of the function at x D 1:
x
1
lim f.x/ D 1 D f.1/
x!1

whereas
FIGURE Continuous at 1.
lim g.x/ D 1 ¤ 2 D g.1/
x!1
g(x)
In Section 10.1 we stressed that given a function f and a number a, there are two impor-
Break tant ways to associate a number to the pair . f; a/. ne is simple evaluation, f.a/, which
x, if x Z 1
in 2 g(x) = 2, if x = 1
exists precisely if a is in the domain of f. The other is limx!a f.x/, whose existence and
graph
1 determination can be more challenging. For the functions f and g above, the limit of f
x as x ! 1 is the same as f.1/, but the limit of g as x ! 1 is n t the same as g.1/. For
1 these reasons, we say that f is c ntinu us at 1 and g is disc ntinu us at 1.

FIGURE iscontinuous at 1.
A function f is continuous at a if and only if the following three conditions are met:
f.a/ exists
limx!a f.x/ exists
f.a/ D limx!a f.x/

If f is not continuous at a, then f is said to be discontinuous at a, and a is called a point

of discontinuity of f.

E AM LE A C

a Show that f.x/ D 5 is continuous at 7.

f(x) S We must verify that the preceding three conditions are met. First, f.7/ D 5,
f(x ) = 5
so f is defined at x D 7. Second,
5 lim f.x/ D lim 5 D 5
x!7 x!7
x
7 Thus, f has a limit as x ! 7. Third,
lim f.x/ D 5 D f.7/
FIGURE f is continuous at 7. x!7

Therefore, f is continuous at 7. (See Figure 10.25.)

470 C ts an Cont n t

g(x) b Show that g.x/ D x2 " 3 is continuous at "4.

S The function g is defined at x D "4 W g."4/ D 13. Also,

13 lim g.x/ D lim .x2 " 3/ D 13 D g."4/

g(x) = x2 - 3 x!!4 x!!4

x Therefore, g is continuous at "4. (See Figure 10.26.)

G
-4
Now ork Problem 1

FIGURE g is continuous at "4. We say that a function is continuous on an interval if it is continuous at each
point there. In this situation, the graph of the function is connected over the interval.
For example, f.x/ D x2 is continuous on the interval 2, 5 . In fact, in Example 5
of Section 10.1, we showed that, for any polynomial function f, for any number a,
limx!a f.x/ D f.a/. This means that

A polynomial function is continuous at every point.

It follows that such a function is continuous on every interval. We say that a function is
continuous on its domain if it is continuous at each point in its domain. If the domain
of such a function is the set of all real numbers, we may simply say that the function is
continuous.

E AM LE C F

The functions f.x/ D 7 and g.x/ D x2 " x C 3 are polynomial functions. Therefore,
they are continuous on their domains. For example, they are continuous at 3.
Now ork Problem 13 G
When is a function discontinuous Suppose that a function f is defined on an open
interval containing a, except possibly at a itself. Then f is discontinuous at a if

f.a/ does not exist (f is not defined at a)

or
limx!a f.x/ does not exist (f has no limit as x ! a/
or
f.a/ and limx!a f.x/ both exist but are different (f.a/ ¤ limx!a f.x/)

In Figure 10.27, we can find points of discontinuity by inspection.

y y y

f(a)

x x x
a a a

Defined at a Defined at a Not defined at a

but no limit and limit as but defined for all
as x a x a exists, but nearby values of a
limit is not f(a)

FIGURE iscontinuities at a.
 Section 0.3 Cont n t 471

f(x)
E AM LE

a et f.x/ D 1=x. (See Figure 10.2 .) Note that f is not defined at x D 0, but it is
f(x) = x1 defined for all other x nearby. Thus, f is discontinuous at 0. oreover,
1
limx!0C f.x/ D 1 and limx!0! f.x/ D "1. A function is said to have an
x
infinite discontinuity at a when at least one of the one-sided limits is either 1
1 or "1 as x ! a. Hence, f has an in nite disc ntinuity at x D 0.
8
< 1 if x > 0
b et f.x/ D 0 if x D 0 .
:"1 if x < 0

FIGURE Infinite discontinuity (See Figure 10.2 .) Although f is defined at x D 0; limx!0 f.x/ does not exist. Thus,
at 0. f is discontinuous at 0.
f(x) Now ork Problem 29 G
1, if x 7 0
f(x) = 0, if x = 0 The following property indicates where the discontinuities of a rational function occur:
-1, if x 6 0
1 R F
A rational function is discontinuous at points where the denominator is 0 and is
x
continuous otherwise. Thus, a rational function is continuous on its domain.
-1

E AM LE L R F
FIGURE iscontinuous For each of the following functions, find all points of discontinuity.
case-defined function.
x2 " 3
a f.x/ D
x2 C 2x "
S This rational function has denominator
x2 C 2x " D .x C 4/.x " 2/
xC1
The rational function f .x/ D is
xC1 which is 0 when x D "4 or x D 2. Thus, f is discontinuous only at "4 and 2.
continuous on its domain but it is not
defined at "1. It is discontinuous at "1.
xC4
b h.x/ D 2
The graph of f is a horizontal straight line x C4
with a hole in it at "1. S For this rational function, the denominator is never 0. (It is always positive.)
Therefore, h has no discontinuity.
Now ork Problem 19 G

E AM LE L C F

For each of the following functions, find all points of discontinuity.

%
x C 6 if x % 3
a f.x/ D
x2 if x < 3
S The cases defining the function are given by polynomials, which are con-
tinuous, so the only possible place for a discontinuity is at x D 3, where the separation
of cases occurs. We now that f.3/ D 3 C 6 D . So because
lim f.x/ D lim .x C 6/ D
x!3C x!3C

and
lim f.x/ D lim! x2 D
x!3! x!3

we can conclude that limx!3 f.x/ D D f.3/ and the function has no points of discon-
tinuity. We can reach the same conclusion by inspecting the graph of f in Figure 10.30.
472 C ts an Cont n t
%
xC2 if x > 2
b f.x/ D
x2 if x < 2
S Since f is not defined at x D 2, it is discontinuous at 2. Note, however, that

lim f.x/ D lim! x2 D 4 D lim x C 2 D lim f.x/

x!2! x!2 x!2C x!2C

shows that limx!2 f.x/ exists. (See Figure 10.31.)

f (x) f(x)

4
9 x + 2, if x 7 2
x + 6, if x Ú 3 f(x) =
f (x) = x2, if x 6 2
x2, if x 6 3

x x
3 2

FIGURE Continuous case-defined function. FIGURE iscontinuous at 2.

Now ork Problem 31 G

E AM LE US F
f(x)
The post-o ce function
8̂
111 3 if 0 < x & 1
ˆ
< 63 if 1 < x & 2
87 c D f.x/ D
ˆ 7 if 2 < x & 3
:̂
63 111 if 3 < x & 4
39
gives the cost c (in cents) of mailing, first class, an item of weight x (ounces), for
0 < x & 4, in uly 2006. It is clear from its graph in Figure 10.32 that f has disconti-
x
1 2 3 4 nuities at 1, 2, and 3 and is constant for values of x between successive discontinuities.
Such a function is called a step functi n because of the appearance of its graph.
FIGURE US Post- ce
function. Now ork Problem 35 G
There is another way to express continuity besides that given in the definition. If
we ta e the statement
This method of expressing continuity
at a is used frequently in mathematical
proofs. lim f.x/ D f.a/
x!a

and replace x by a C h, then as x ! a, we have h ! 0 and as h ! 0, we have x ! a.

y
y = f (x) It follows that limx!a f.x/ D limh!0 f.a C h/, provided the limits exist (Figure 10.33).
Thus, the statement
f(a + h)
lim f.a C h/ D f.a/
h!0
as x a,
then h 0
f(a)
h
assuming both sides exist, also defines continuity at a.
y observing the graph of a function, we may be able to determine where a dis-
a a+h
x continuity occurs. However, technological devices have their limitations. For example,
x the function

FIGURE iagram for x"1

f.x/ D
continuity at a. x2 " 1
 Section 0.3 Cont n t 473

is discontinuous at "1 and at 1 neither of these numbers are in the domain of f. The
discontinuity at "1 is clear, but that at 1 may not be obvious because the graph of f is
1
the same as the graph of except that it has a hole at x D 1. The screen shot
xC1
from a graphing calculator in Figure 10.34 illustrates the di culty.

5
Table 10.
S
Price Unit, p uantity Wee , q
-5 5
20 0
10 5
5 15
-5
4 20
2 45 FIGURE The discontinuity at 1 is not apparent
x"1
1 5 from the graph of f .x/ D 2 .
x "1

ften, it is helpful to describe a situation by a continuous function. For example,

the demand schedule in Table 10.3 indicates the number of units of a particular prod-
uct that consumers will demand per wee at various prices. This information can be
given graphically, as in Figure 10.35(a), by plotting each quantity price pair as a point.
Clearly, the graph does not represent a continuous function. Furthermore, it gives us
no information as to the price at which, say, 35 units would be demanded. However, if
we connect the points in Figure 10.35(a) by a smooth curve see Figure 10.35(b) , we
get a so-called demand curve. From it, we could guess that at about 2.50 per unit, 35
units would be demanded.
Frequently, it is possible and useful to describe a graph, as in Figure 10.35(b), by
means of an equation that defines a continuous function, f. Such a function not only
gives us a demand equation, p D f.q/, which allows us to anticipate corresponding
prices and quantities demanded, but also permits a convenient mathematical analysis
of the nature and basic properties of demand. f course, some care must be used in
wor ing with equations such as p D f.q/. athematically, f may be defined when
p p
q D 37, but from a practical standpoint, a demand of 37 units could be meaningless
p
to our particular situation. For example, if a unit is an egg, then a demand of 37 eggs
ma e no sense.
We remar that functions of the form f.x/ D xa , for fixed a, are continuous on
their domains. In particular, (square) root functions are continuous. Also, exponential
functions and logarithmic functions are continuous on their domains. Thus, exponential

p p

20 20

15 15

10 10

5 5
2.5
q q
25 50 75 100 25 35 50 75 100
(a) (b)

FIGURE Viewing data via a continuous function.

474 C ts an Cont n t

functions have no discontinuities, while a logarithmic function has a discontinuity only

at 0 (which is an infinite discontinuity). any more examples of continuous functions
are provided by the observation that if f and g are continuous on their domains, then
the composite function f ı g, given by f ı g.x/ D f.g.x// is continuous on its domain.
For example, the function
s # $
x2 C 1
f.x/ D ln
x"1
is continuous on its domain. etermining the domain of such a function may, of course,
be fairly involved.

R BLEMS
% %
In Pr b ems use the de niti n f c ntinuity t sh that the 3 if x % 0 3x C 5 if x % "2
gi en functi n is c ntinu us at the indicated p int f .x/ D f .x/ D
"2 if x < 0 2 if x < "2
x"3 % %
f .x/ D x3 " 5xI x D 2 f .x/ D I x D "3 0 if x & 1 x"3 if x > 2
5x f .x/ D f .x/ D
x"1 if x > 1 3 " 2x if x < 2
p x
g.x/ D 2 " 3xI x D 0 f .x/ D D3
4 % 2
xC3 p x C 1 if x > 2
h.x/ D x D "3 f .x/ D 3 xI x D "1 f .x/ D
x if x < 2
x"3
8
In Pr b ems determine hether the functi n is c ntinu us < 5
if x % "1
at the gi en p ints f .x/ D x.x " 2/
: 5x " 1 if x < "1
xC4 x2 " 4x C 4
f .x/ D I "2; 0 f .x/ D I 2; "2
x"2 6 elephone Rates Suppose the long-distance rate for a
x"5 3 telephone call from Hazleton, Pennsylvania to os Angeles,
g.x/ D 2 "5; 5 h.x/ D 2 3; "3 California, is 0.0 for the first minute or fraction thereof and
x " 25 x C
% 0.04 for each additional minute or fraction thereof. If y D f .t/ is
x C 2 if x % 2 a function that indicates the total charge y for a call of t minutes
f .x/ D I 2; 0
x2 if x < 2
8 duration, s etch the graph of f for 0 < t & 3 12 . Use your graph to
<1 determine the values of t, where 0 < t & 3 12 , at which
if x ¤ 0
f .x/ D x I 0; "1 discontinuities occur.
: 0 if x D 0
The greatest integer functi n f .x/ D bxc, is defined to be the
In Pr b ems gi e a reas n hy the functi n is c ntinu us greatest integer less than or equal to x, where x is any real number.
n its d main For example, b3c D 3, b1: c D 1, b 14 c D 0, and b"4:5c D "5.
5=7 " .1=4/x2
f .x/ D 2x2 " 3 f .x/ D S etch the graph of this function for "3:5 & x & 3:5. Use your
23=5 s etch to determine the values of x at which discontinuities occur.
p
f .x/ D ln. 3 x/ f .x/ D x.1 " x/
Inventory S etch the graph of
In Pr b ems nd a p ints f disc ntinuity 8
<"100x C 600 if 0 & x < 5
f .x/ D 3x2 " 3 h.x/ D x " 2 y D f .x/ D "100x C 1100 if 5 & x < 10
17=11 x2 C 5x " 2 :"100x C 1600 if 10 & x < 15
f .x/ D f .x/ D
x " 23 x2 "
.2x2 " 3/3 A function such as this might describe the inventory y of a
g.x/ D f .x/ D "1 company at time x. Is f continuous at 2 At 5 At 10
15 2
x2 C 6x C xC2 raph g.x/ D e!1=x . ecause g is not defined at x D 0, g is
f .x/ D 2 g.x/ D discontinuous at 0. ased on the graph of g, is
x C 2x " 15 x2 C x
x"3 2x " 3 % 2
h.x/ D 3 f .x/ D e!1=x if x ¤ 0
x " x 3 " 2x f .x/ D
0 if x D 0
x x4
p.x/ D 2 f .x/ D 4 continuous at 0
x C1 x "1

Objective C A I
o e e o tec n es for so n In Section 1.2, we solved linear inequalities. We now turn our attention to showing
non near ne a t es
how the notion of continuity can be applied to solving a nonlinear inequality such as
x2 C 3x " 4 < 0. The ability to do this will be important in our study of calculus.
 Section 0.4 Cont n t e to ne a t es 475

y Recall (from Section 2.5) that the x-intercepts of the graph of a function g are
y = g (x) precisely the roots of the equation g.x/ D 0. Hence, from the graph of y D g.x/ in
Figure 10.36, we conclude that r1 ; r2 , and r3 are roots of g.x/ D 0 and any other roots
(r2, 0) will give rise to x-intercepts (beyond what is actually shown of the graph). Assume that
(r1, 0) (r3, 0)
x in fact all the roots of g.x/ D 0, and hence, all the x-intercepts, are shown. Note further
from Figure 10.36 that the three roots determine four open intervals on the x-axis:
FIGURE r1 ; r2 , and r3 are ."1; r1 / .r1 ; r2 / .r2 ; r3 / .r3 ; 1/
roots of g.x/ D 0.
To solve x2 C 3x " 4 > 0, we let
f(x)
f.x/ D x2 C 3x " 4 D .x C 4/.x " 1/
ecause f is a polynomial function, it is continuous. The roots of f.x/ D 0 are "4 and 1
f(x) = x2 + 3x - 4
hence, the graph of f has x-intercepts ."4; 0/ and .1; 0/. (See Figure 10.37.) The roots
x determine three intervals on the x-axis:
-4 1
."1; "4/ ."4; 1/ .1; 1/

- 25 Consider the interval ."1; "4/. Since f is continuous on this interval, we claim
4
that either f.x/ > 0 or f.x/ < 0 thr ugh ut the interval. If this were not the case, then
FIGURE "4 and 1 are f.x/ would indeed change sign on the interval. y the continuity of f, there would be a
roots of f .x/ D 0. point where the graph intersects the x-axis for example, at .x0 ; 0/. (See Figure 10.3 .)
ut then x0 would be a root of f.x/ D 0. However, this cannot be, because there is no
root less than "4. Hence, f.x/ must be strictly positive or strictly negative on ."1; "4/.
f(x) 7 0 A similar argument can be made for each of the other intervals.
(x0, 0)
) To determine the sign of f.x/ on any one of the three intervals, it su ces to deter-
f(x) 6 0 -4 mine its sign at any point in the interval. For instance, "5 is in ."1; "4/ and
f."5/ D 6 > 0 Thus, f.x/ > 0 on ."1; "4/
FIGURE Change of sign
for a continuous function. Similarly, 0 is in ."4; 1/, and
f.0/ D "4 < 0 Thus, f.x/ < 0 on ."4; 1/
Finally, 3 is in .1; 1/, and
f.3/ D 14 > 0 Thus, f.x/ > 0 on .1; 1/
(See the sign chart in Figure 10.3 .) Therefore,
x2 C 3x " 4 > 0 on ."1; "4/ and .1; 1/
so we have solved the inequality. These results are obvious from the graph in
Figure 10.37. The graph lies above the x-axis, meaning that f.x/ > 0, on ."1; "4/
and on .1; 1/.
In more complicated examples it will be useful to exploit the multiplicative nature
of signs. We noted that f.x/ D x2 C 3x " 4 D .x C 4/.x " 1/. Each of x C 4 and
x " 1 has a sign chart that is simpler than that of x2 C 3x " 4. Consider the sign chart
in Figure 10.40. As before, we placed the roots of f.x/ D 0 in ascending order, from
left to right, so as to subdivide ."1; 1/ into three open intervals. This forms the top
line of the box. irectly below the top line we determined the signs of x C 4 on the
three subintervals. We now that for the linear function x C 4 there is exactly one root
of the equation x C 4 D 0, namely, "4. We placed a 0 at "4 in the row labeled x C 4.
y the argument illustrated in Figure 10.3 , it follows that the sign of the function
x C 4 is constant on ."1; "4/ and on ."4; 1/ and two evaluations of x C 4 settle the
distribution of signs for x C 4. From ."5/ C 4 D "1 < 0, we have x C 4 negati e
on ."1; "4/, so we entered a " sign in the ."1; "4/ space of the x C 4 row. From

f (x ) 7 0 f (x ) 6 0 f (x ) 7 0

-4 1

FIGURE Simple sign chart for x2 C 3x " 4.

476 C ts an Cont n t

-q -4 1 q
x+4 - 0 + +

x-1 - - 0 +

f(x) + 0 - 0 +

FIGURE Sign chart for x2 C 3x " 4.

.0/ C 4 D 4 > 0, we have x C 4 p siti e on ."4; 1/. Since ."4; 1/ has been further
subdivided at 1, we entered a C sign in each of the ."4; 1/ and .1; 1/ spaces of the
x C 4 row. In a similar way we constructed the row labeled x " 1.
Now the bottom row is obtained by ta ing, for each component, the product of the
entries above. Thus, we have .x C 4/.x " 1/ D f.x/, ."/."/ D C, 0.any number/ D 0,
.C/."/ D ", .any number/0 D 0, and .C/.C/ D C. Sign charts of this ind are
useful whenever a continuous function can be expressed as a product of several simpler,
continuous functions, each of which has a simple sign chart. In Chapter 13 we will rely
heavily on such sign charts.

E AM LE S I

Solve x2 " 3x " 10 > 0.

S If f.x/ D x2 " 3x " 10, then f is a polynomial (quadratic) function and, thus,
is continuous everywhere. To find the real roots of f.x/ D 0, we have

x2 " 3x " 10 D 0
.x C 2/.x " 5/ D 0
x D "2; 5

The roots "2 and 5 determine three intervals:

."1; "2/ ."2; 5/ .5; 1/

In the manner of the last example, we construct the sign chart in Figure 10.41. We see
that x2 " 3x " 10 > 0 on ."1; "2/ [ .5; 1/.

-q -2 5 q
x+2 - 0 + +

x-5 - - 0 +

f(x) + 0 - 0 +

FIGURE Sign chart for x2 " 3x " 10.

Now ork Problem 1 G

A L IT I
An open box is formed by cutting
a square piece out of each corner of an E AM LE S I
-inch by 10-inch piece of metal. If each
side of the cut-out squares is x inches Solve x.x " 1/.x C 4/ & 0.
long, the volume of the box is given by
S If f.x/ D x.x " 1/.x C 4/, then f is a polynomial function and, hence, con-
.x/ D x. "2x/.10"2x/. This problem
tinuous everywhere. The roots of f.x/ D 0 are (in ascending order) "4, 0, and 1 and
ma es sense only when this volume is
positive. Find the values of x for which lead to the sign chart in Figure 10.42.
the volume is positive. From the sign chart, noting the endpoints required, x.x " 1/.x C 4/ & 0 on
."1; "4! [ Œ0; 1!.
 Section 0.4 Cont n t e to ne a t es 477

-q -4 0 1 q
x - - 0 + +

x-1 - - - 0 +

x+1 - 0 + + +

f(x) - 0 + 0 - 0 +

FIGURE Sign chart for x.x " 1/.x C 4/.

Now ork Problem 11 G

The sign charts we have described are certainly not limited to solving polynomial
inequalities. The reader will have noticed that we used thic er vertical lines at the end-
points, "1 and 1, of the chart. These symbols do not denote real numbers, let alone
points in the domain of a function. We extend the thic vertical line convention to sin-
gle out isolated real numbers that are not in the domain of the function in question. The
next example will illustrate.

E AM LE S R F I

x2 " 6x C 5
Solve % 0.
x
S et
x2 " 6x C 5 .x " 1/.x " 5/
f.x/ D D
x x
For a rational function f D g=h, we solve the inequality by considering the intervals
determined by both the roots of g.x/ D 0 and the roots of h.x/ D 0. bserve that the
roots of g.x/ D 0 are the roots of f.x/ D 0 because the only way for a fraction to be
0 is for its numerator to be 0. n the other hand, the roots of h.x/ D 0 are precisely
the points at which f is not defined and these are also precisely the points at which f is
discontinuous. The sign of f may change at a root and it may change at a discontinuity.
Here the roots of the numerator are 1 and 5 and the root of the denominator is 0. In
ascending order these give us 0, 1, and 5, which determine the open intervals
."1; 0/ .0; 1/ .1; 5/ .5; 1/
These, together with the observation that 1=x is a fact r of f, lead to the sign chart in
Figure 10.43.
Here, the first two rows of the sign chart are constructed as before. In the third
row we have placed a $ sign at 0 to indicate that the factor 1=x is not defined at 0.
The bottom row, as before, is constructed by ta ing the products of the entries above.
bserve that a product is not defined at any point at which any of its factors is not
defined. Hence, we also have a $ entry at 0 in the bottom row.
From the bottom row of the sign chart we can read that the solution of .x!1/.x!5/
x
%
0 is .0; 1! [ Œ5; 1!. bserve that 1 and 5 are in the solution and 0 is not.

-q 0 1 5 q
x-1 - - 0 + +

x-5 - - - 0 +

1/x - * + + +

f(x) - * + 0 - 0 +

.x " 1/.x " 5/

FIGURE Sign chart for .
x
478 C ts an Cont n t

f(x)

1 x
5 10
2
f(x) = x - 6x
x
+5
f(x) Ú 0
-10

x2 " 6x C 5
FIGURE raph of f .x/ D .
x
2
In Figure 10.44 we have graphed f.x/ D x !6xC5
x
, and we can confirm visually that
the solution of the inequality f.x/ % 0 is precisely the set of all real numbers at which
the graph lies on or above the x-axis.
Now ork Problem 17 G
A sign chart is not always necessary, as the following example shows.

E AM LE S N I
2
a Solve x C 1 > 0.
S The equation x2 C 1 D 0 has no real roots. Thus, the continuous function
2
f.x/ D x C 1 has no x-intercepts. It follows that either f.x/ is always positive or f.x/ is
always negative. ut x2 is always positive or zero, so x2 C 1 is always positive. Hence,
the solution of x2 C 1 > 0 is ."1; 1/.
b Solve x2 C 1 < 0.
S From part (a), x2 C 1 is always positive, so x2 C 1 < 0 has no solution,
meaning that the set of solutions is ;, the empty set.
Now ork Problem 7 G
We conclude with a nonrational example. The importance of the function intro-
duced will become clear in later chapters.

E AM LE S N F I

Solve x ln x " x % 0.
S et f.x/ D x ln x " x D x.ln x " 1/, which, being a product of continuous
functions, is continuous. From the fact red form for f we see that the roots of f.x/ D 0
0 e q
are 0 and the roots of ln x " 1 D 0. The latter is equivalent to ln x D 1, which is
equivalent to eln x D e1 , since the exponential function is one-to-one. However, the last
x + +
equality says that x D e. The domain of f is .0; 1/ because ln x is defined only for
ln x - 1 - 0 + x > 0. The domain dictates the top line of our sign chart in Figure 10.45.
The first row of Figure 10.45 is straightforward. For the second row, we placed a 0
f(x) - 0 + at e, the only root of ln x " 1 D 0. y continuity of ln x " 1, the sign of ln x " 1 on .0; e/
and on .e; 1/ can be determined by suitable evaluations. For the first we evaluate at 1 in
FIGURE Sign chart for .0; e/ and get ln 1"1 D 0"1 D "1 < 0. For the second we evaluate at e2 in .e; 1/ and
x ln x " x. get ln e2 " 1 D 2 " 1 D 1 > 0. The bottom row is, as usual, determined by multiplying
the others. From the bottom row of Figure 10.45 the solution of x ln x " x % 0 is
evidently Œe; 1/.
Now ork Problem 35 G
 Chapter 0 e e 479

R BLEMS
In Pr b ems s e the inequa ities by the technique or shop Participation Computech is offering a
discussed in this secti n wor shop on web page design to ey personnel at Pear
x2 " 3x " 4 > 0 x2 " x C 15 > 0 Corporation. The price per person is 60, and Pear Corporation
guarantees that at least 20 people will attend. Suppose Computech
x2 " 3x " 10 & 0 15 " 2x " x2 % 0 offers to reduce the charge for e eryb dy by 1.00 for each person
over the 20 who attends. How should Computech limit the size of
3x2 " 17x C 10 < 0 x2 " 4 < 0
the group so that the total revenue it receives will not be less than
x2 C 4 < 0 2x2 " x " 2 & 0 that received for 20 persons
.x C 1/.x " 2/.x C 7/ & 0 .x C 3/.x C 1/.x " 1/ & 0 raph f .x/ D x3 C 7x2 " 5x C 4. Use the graph to determine
the solution of
"x.x " 5/.x C 4/ > 0 .x C 2/2 > 0
x3 C 7x2 " 5x C 4 & 0
x3 C 4x % 0 .x C 3/2 .x2 " 4/ < 0
x3 " 3x2 C 2x & 0 x3 C 6x2 C x < 0
3x2 " 0:5x C 2
2 raph f .x/ D . Use the graph to determine the
x x "1 6:2 " 4:1x
<0 <0 solution of
x2 " x
3 5x 3x2 " 0:5x C 2
%0 >0 >0
xC1 x2 " 6x " 7 6:2 " 4:1x
x2 " x " 6 x2 C 4x " 5 A n e ay f s ing a n n inear inequa ity ike f .x/ > 0 is by
%0 &0 examining the graph f g.x/ D f .x/=jf .x/j h se range c nsists
x2 C 4x " 5 x2 C 3x C 2
n y f 1 and "1
3 3x C 2
&0 &0 %
x2 C 6x C 5 .x " 1/2 f .x/ 1 if f .x/ > 0
g.x/ D D
2x2 C 5x % 3 x4 " 16 % 0 jf . f .x/j "1 if f .x/ < 0

Revenue Suppose that consumers will purchase q units of

a product when the price of each unit is 2 " 0:2q dollars. How he s uti n f f .x/ > 0 c nsists f a inter a s f r hich
many units must be sold for the sales revenue to be at least 750 g.x/ D 1 sing this technique s e the inequa ities in
Pr b ems and
Forest Management A lumber company owns a forest that x2 C x " 1
is of rectangular shape, 1 mi $ 2 mi. The company wants to cut a 6x2 " x " 2 > 0 <0
x2 C x " 6
uniform strip of trees along the outer edges of the forest. At most,
raph x ln x " x. oes the function appear to be continuous
how wide can the strip be if the company wants at least 1 165
mi2 of
oes the graph support the conclusions of Example 5 At what
forest to remain
value does the function appear to have a minimum value
Container Design A container manufacturer wishes to Can lim f .x/ be estimated
ma e an open box by cutting a 3-in. by 3-in. square from each x!0C
2
corner of a square sheet of aluminum and then turning up the raph e!x . oes the function appear to be continuous Can
sides. The box is to contain at least 1 2 cubic inches. Find the the conclusion be confirmed by invo ing facts about continuous
dimensions of the smallest square sheet of aluminum that can functions At what value does the function appear to have a
be used. maximum value

Chapter 10 Review
I T S E
S Limits
limx!a f.x/ D Ex. , p. 45
S Limits Continued
limx!a! f.x/ D limx!aC f.x/ D limx!a f.x/ D 1 limx!a f.x/ D 1 Ex. 1, p. 462
limx!1 f.x/ D limx!!1 f.x/ D Ex. 3, p. 463
S Continuity
continuous at a discontinuous at a Ex. 3, p. 471
continuous on an interval continuous on its domain Ex. 4, p. 471
S Continuity Applied to Inequalities
sign chart Ex. 1, p. 476
480 C ts an Cont n t

S
The notion of limit is fundamental for calculus. To say that If f.x/ increases without bound as x ! a, then we write
limx!a f.x/ D means that the values of f.x/ can be made limx!a f.x/ D 1. Similarly, if f.x/ decreases without bound,
as close to the number as we li e by ta ing x su ciently we have limx!a f.x/ D "1. To say that the limit of a func-
close to, but different from, a. If limx!a f.x/ and limx!a g.x/ tion is 1 (or "1) does not mean that the limit exists. Rather,
exist and c is a constant, then it is a way of saying that the limit does not exist and hy there
is no limit. f course, limx!a f.x/ does not exist does not
imply limx!a f.x/ D 1 .
lim c D c
x!a There is a rule for evaluating the limit of a rational func-
lim xn D an tion as x ! ˙1. If f.x/ is a rational function and an xn and
x!a
˙ ˙
bm xm are the terms in the numerator and denominator, respec-
lim . f.x/ ! g.x// D lim f.x/ ! lim g.x/ tively, with the greatest powers of x, then
x!a x!a x!a
lim .cf.x// D c ! lim f.x/
x!a x!a
a n xn
f.x/ limx!a f.x/ lim f.x/ D lim
lim D if lim g.x/ ¤ 0 x!˙1 x!˙1 bm xm
x!a g.x/ limx!a g.x/ x!a
p q
n
lim f.x/ D n lim f.x/
x!a x!a
In particular, as x ! ˙1, the limit of a polynomial is the
If f is a polynomial function, then limx!a f.x/ D f.a/.
same as the limit of the term that involves the greatest power
of x. This means that, for a nonconstant polynomial, the limit
as x ! ˙1 is either ˙1 or #1.
Property 7, saying that limits of polynomial functions can be
calculated by evaluation, must be used with care. With many
other functions f, attempting to find limx!a f.x/ by evalua- A function f is continuous at a if and only if
tion at a can lead to meaningless expressions of the form 0=0. f.a/ exists
In such cases, algebraic manipulations may be needed to find limx!a f.x/ exists
another function g that agrees with f, for x ¤ a, and for which
the limit can be determined, perhaps by e a uati n. f.a/ D limx!a f.x/
If f.x/ approaches as x approaches a from the right,
then we write limx!aC f.x/ D . If f.x/ approaches as
x approaches a from the left, we write limx!a! f.x/ D . eometrically this means that the graph of f has no brea at
These limits are called one-sided limits. x D a. If a function is not continuous at a, then the function is
The infinity symbol 1, which does not represent a num- said to be discontinuous at a. Polynomial functions and ratio-
ber, is used in describing limits. The statement nal functions are continuous on their domains. Thus, polyno-
mial functions have no discontinuities and a rational function
lim f.x/ D is discontinuous only at points where its denominator is zero.
x!1
To solve the inequality f.x/ > 0 (or f.x/ < 0), we first
means that as x increases without bound, the values of f.x/ find the real roots of f.x/ D 0 and the values of x for which
approach the number . A similar statement applies for the f is discontinuous. These values determine intervals, and on
situation when x ! "1, which means that x is decreasing each interval, f.x/ is either always positive or always nega-
without bound. In general, if p > 0, then tive. To find the sign on any one of these intervals, it su ces
to find the sign of f.x/ at any point there. After the signs are
1 1 determined for all intervals and assembled on a sign chart, it
lim D 0 and lim D0 is easy to give the solution of f.x/ > 0
x!1 xp x!!1 xp

R
In Pr b ems nd the imits if they exist If the imit d es n t x3 C 4x2 x2 " 7x C 10
exist s state r use the symb 1 r "1 here appr priate lim lim
x!!4 x2 C 2x " x!2 x2 C x " 6
2x2 " 3x C 1 2
lim .3x2 C 4x " 2/ lim x2 C 1
x!1 x!0 2x2 " 2 lim lim
x!1 xC1 x!1 2x2
x2 " 16 2x C 3 3x C 17 1
lim lim lim lim
x!4 x2 " 4x x!!4 x2 " 4 x!1 11x " 7 x!!1 x4
x2 " 1 3t " 4 x6
lim .x C h/ lim lim lim
h!0 x!1 2x2 C x " 3 t!4 t " 4 x!!1 x5
 Chapter 0 e e 481
p
xC3 lim
4
1 Using the definition of c ntinuity, show that the function
lim x!16 x"5
x!!1 1 " x
f .x/ D 2 is continuous at x D 5.
x C2
x2 " 1 x2 " 2x " 15
lim lim State whether f .x/ D x2 =5 is continuous at each real number.
x!1 .3x C 2/2 x!5 x"5
ive a reason for your answer.
xC3 2"x 2 3
lim lim State whether f .x/ D !x !epxCln 2 is continuous everywhere.
x!3! x2 " x!2 x"2 2
ive a reason for your answer.
p
3 p
lim x lim y"5 In Pr b ems nd the p ints f disc ntinuity if any f r
x!1 y!5C
each functi n
x100 C .1=x4 / ex2 " x4 x2 0
lim 6
lim f .x/ D f .x/ D
x!1 e"x x!!1 31x " 2x3 x2
xC3
! 2
x if 0 $ x < 1 x"1 f .x/ D .2 " 3x/3
lim f .x/ if f .x/ D f .x/ D
x!1 x if x > 1 2x2 C 3
!
x C 5 if x < 2 " x3 2x C 6
lim f .x/ if f .x/ D f .x/ D f .x/ D
x!2 if x % 2 x2 " x " 6 x3 C x
p ! !
x2 " 16 2x C 3 if x > 2 1=x if x < 1
lim ( int For x > 4, f .x/ D f .x/ D
x!4C 4"x 3x C 5 if x $ 2 1 if x % 1
p p p
x2 " 16 D x " 4 x C 4.)
In Pr b ems s e the gi en inequa ities
x2 C x " 12 x"3 p 2
lim p ( int For x > 3, p D x " 3.) x C 4x " 12 > 0 2x2 C 10x " 12 $ 0
x!3C x"3 x"3
x5 $ 7x4 x3 C x2 C 14x < 0
f .x C h/ " f .x/
If f .x/ D x " 2; find lim . xC5 x.x C 5/.x C /
h!0 h <0 <0
x2 " 1 3
f .x C h/ " f .x/
If f .x/ D 2x2 " 3; find lim . x2 " 6x x2 "
h!0 h %0 $0
ost Parasite Relationship For a particular host parasite x2 " 3x " 4 x2 " 16
relationship, it was determined that when the host density x3 C 3x2 " 1 x C 1
(number of hosts per unit of area) is x, then the number of hosts raph f .x/ D . Use the graph to
parasitized over a certain period of time is x3 " 2x2 C x " 2
estimate limx!2 f .x/.
" # p
2
y D 21 1 " xC3"2
2 C 5x raph f .x/ D . From the graph, estimate
x"1
If the host density were to increase without bound, what value limx!1 f .x/.
would y approach raph f .x/ D x ln x. From the graph, estimate the one-sided
Predator Prey Relationship For a particular predator limit limx!0C f .x/.
prey relationship, it was determined that the number y of prey ex " 1
consumed by an individual predator over a period of time was a raph f .x/ D 2x . Use the graph to estimate
e " ex
function of the prey density x (the number of prey per unit of limx!0 f .x/.
area). Suppose
raph f .x/ D x3 " x2 C x " 6. Use the graph to determine the
10x solution of
y D f .x/ D
1 C 0:1x x3 " x2 C x " 6 % 0
If the prey density were to increase without bound, what value x5 " 4
would y approach raph f .x/ D 3 . Use the graph to determine the
x C1
solution of
Using the definition of c ntinuity, show that the function
f .x/ D x C 3 is continuous at x D 2. x5 " 4
$0
x3 C 1
 fferent at on
overnment regulations generally limit the number of fish ta en from a given
11.1 e er at e
fishing ground by commercial fishing boats in a season. This prevents
11.2 es for erent at on overfishing, which depletes the fish population and leaves, in the long run,
fewer fish to catch.
11.3 e er at e as a ate From a strictly commercial perspective, the ideal regulations would maximize the
of C an e
number of fish available for the year-to-year fish harvest. The ey to finding those ideal
11.4 e Pro ct e an regulations is a mathematical function called the reproduction curve. For a given fish
t e ot ent e habitat, this function estimates the fish population a year from now, P.n C 1/, based on
the population now, P.n/, assuming no external interventions, such as fishing or in ux
11.5 e C an e
of predators.
C er 11 e e The figure to the bottom left shows a typical reproduction curve. Also graphed is
the line P.nC1/ D P.n/, the line along which the populations P.nC1/ and P.n/ would
be equal. Notice the intersection of the curve and the straight line at point A. This is
where, because of habitat crowding, the population has reached its maximum sustain-
able size. A population that is this size one year will be the same size the next year.
For any point on the horizontal axis, the distance between the reproduction curve
and the line P.n C 1/ D P.n/ represents the sustainable harvest: the number of fish
that could be caught, after the spawn have grown to maturity, so that in the end the
population is bac at the same size it was a year ago.
Commercially spea ing, the optimal population size is the one where the dis-
tance between the reproduction curve and the line P.n C 1/ D P.n/ is the greatest.
P (n + 1)
This condition is met where the slopes of the reproduction curve and the line
P.n C 1/ D P.n/ are equal. The slope of P.n C 1/ D P.n/ is, of course, 1. Thus,
for a maximum fish harvest year after year, regulations should aim to eep the fish
A
population fairly close to P0 .
A central idea here is that of the slope of a curve at a given point. That idea is the
cornerstone concept of this chapter.
Now we begin our study of calculus. The ideas involved in calculus are completely
different from those of algebra and geometry. The power and importance of these ideas
and their applications will become clear later in the boo . In this chapter we introduce
the deri ati e of a function and the important rules for finding derivatives. We also
P (n) show how the derivative is used to analyze the rate of change of a quantity, such as the
P0 rate at which the position of a body is changing.

482
 Section . e er at e 483

Objective T
o e e o t e ea of a tan ent ne to The main problem of differential calculus deals with finding the slope of the tangent
a c r e to e ne t e s o e of a c r e
an to e ne a er at e an e ta ine at a point on a curve. In high school geometry a tangent line, or tangent, to a circle is
eo etr c nter retat on o co te often defined as a line that meets the circle at exactly one point (Figure 11.1). However,
er at es sn t e t e n t on this idea of a tangent is not very useful for other inds of curves. For example, in
Figure 11.2(a), the lines 1 and 2 intersect the curve at exactly one point P. Although
we would not thin of 2 as the tangent at this point, it seems natural that 1 is. In
Figure 11.2(b) we intuitively would consider 3 to be the tangent at point P, even though
3 intersects the curve at other points.

y y
L2
L1

P P
Tangent lines L3

x x
FIGURE Tangent lines to
a circle.
L1 is a tangent line L3 is a tangent
at P, but L2 is not. line at P.
(a) (b)

FIGURE Tangent line at a point.

From these examples, we see that the idea of a tangent as simply a line that inter-
y
sects a curve at only one point is inadequate. To obtain a suitable definition of tangent
line, we use the limit concept and the geometric notion of a secant ine. A secant line
Q is a line that intersects a curve at two or more points.
oo at the graph of the function y D f.x/ in Figure 11.3. We wish to define the
tangent line at point P. If is a different point on the curve, the line P is a secant
y = f(x)
line. If moves along the curve and approaches P from the right (see Figure 11.4),
typical secant lines are P 0 , P 00 , and so on. As approaches P from the left, typical
P secant lines are P 1 , P 2 , and so on. In b th cases the secant ines appr ach the same
imiting p siti n. This common limiting position of the secant lines is defined to be the
Secant line
tangent line to the curve at P. This definition seems reasonable and applies to curves
x
in general, not ust circles.
A curve does not necessarily have a tangent line at each of its points. For example,
the curve y D jxj does not have a tangent at .0; 0/. As can be seen in Figure 11.5,
FIGURE Secant line P .
a secant line through .0; 0/ and a nearby point to its right on the curve must always

y
PQ
Q PQ¿
PQ–
PQ‡

y
PQ1
PQ2
P y= x
PQ3
Limiting position
(tangent at P)
y = - x, x 6 0 y = x, x 7 0
x x
(0,0)

FIGURE No tangent line

FIGURE The tangent line is a limiting position of secant lines. to graph of y D jxj at .0; 0/.
484 C erent at on

be the line y D x. Thus, the limiting position of such secant lines is also the line y D x.
However, a secant line through .0; 0/ and a nearby point to its left on the curve must
always be the line y D !x. Hence, the limiting position of such secant lines is also the
line y D !x. Since there is no common limiting position, there is no tangent line at
.0; 0/.
Now that we have a suitable definition of a tangent to a curve at a point, we can
define the s pe f a cur e at a point.

The slope of a curve at a point P is the slope, if it exists, of the tangent line at P.

Since the tangent at P is a limiting position of secant lines P , we consider the

slope of the tangent to be the limiting value of the slopes of the secant lines as
approaches P. For example, let us consider the curve f.x/ D x2 and the slopes of some
secant lines P , where P D .1; 1/. For the point D .2:5; 6:25/, the slope of P
(see Figure 11.6) is
rise 6:25 ! 1
mP D D D 3:5
run 2:5 ! 1

Q
(2.5, 6.25)

mPQ = 6.25 - 1 = 3.5

2.5 - 1

y = f(x) = x2
Tangent line
P
(1, 1)
x

FIGURE Secant line to f .x/ D x2 through .1; 1/ and .2:5; 6:25/.

Table 11.1 includes other points on the curve, as well as the corresponding slopes
of P . Notice that as approaches P, the i slopes of the secant lines seem to approach 2.
Thus, we expect the slope of the indicated tangent line at .1; 1/ to be 2. This will be
confirmed later, in Example 1. ut first, we wish to generalize our procedure.

Table 11.1 S S L C
f .x/ D x2 P D .1; 1/
Slope of P

(2.5, 6.25) .6:25 ! 1/=.2:5 ! 1/ D 3.5

(2, 4) .4 ! 1/=.2 ! 1/ D 3
(1.5, 2.25) .2:25 ! 1/=.1:5 ! 1/ D 2.5
(1.25, 1.5625) .1:5625 ! 1/=.1:25 ! 1/ D 2.25
(1.1, 1.21) .1:21 ! 1/=.1:1 ! 1/ D 2.1
(1.01, 1.0201) .1:0201 ! 1/=.1:01 ! 1/ D 2.01
 Section . e er at e 485
y

Q (z, f(z))

y = f(x) f(z) - f(a)

(a, f(a)) f(z) - f(a)

mPQ = z-a
P
z-a=h

x
a z

FIGURE Secant line through P and .

For the curve y D f.x/ in Figure 11.7, we will find an expression for the slope at
the point P D .a; f.a//. If D .z; f.z//, the slope of the secant line P is
f.z/ ! f.a/
mP D
z!a
If the difference z ! a is called h, then we can write z as a C h. Here, we must have
h ¤ 0, for if h D 0, then z D a, and no secant line exists. Accordingly,
f.z/ ! f.a/ f.a C h/ ! f.a/
mP D D
z!a h
Which of these two forms for mP is most convenient depends on the nature of the
function f. As moves along the curve toward P, z approaches a. This means that h
approaches zero. The limiting value of the slopes of the secant lines which is the slope
of the tangent line at .a; f.a// is

f.z/ ! f.a/ f.a C h/ ! f.a/

mtan D lim D lim
z!a z!a h!0 h

Again, which of these two forms is most convenient which limit is easiest to
determine depends on the nature of the function f. In Example 1, we will use this
limit to confirm our previous expectation that the slope of the tangent line to the curve
f.x/ D x2 at .1; 1/ is 2.

E AM LE F S T L

Find the slope of the tangent line to the curve y D f.x/ D x2 at the point .1; 1/.
S The slope is the limit in Equation (1) with f.x/ D x2 and a D 1:
f.1 C h/ ! f.1/ .1 C h/2 ! .1/2
lim D lim
h!0 h h!0 h
1 C 2h C h2 ! 1 2h C h2
D lim D lim
h!0 h h!0 h
h.2 C h/
D lim D lim .2 C h/ D 2
h!0 h h!0

Therefore, the tangent line to y D x2 at .1; 1/ has slope 2. (Refer to Figure 11.6.)
Now ork Problem 1 G
486 C erent at on

We can generalize Equation (1) so that it applies to any point .x; f.x// on a curve.
Replacing a by x gives a function, called the deri ati e x of f, whose input is x and
whose output is the slope of the tangent line to the curve at .x; f.x//, provided that the
tangent line exists and has a slope. (If the tangent line exists but is ertica , then it has
no slope.) We, thus, have the following definition, which forms the basis of differential
calculus:

The derivative of a function f is the function denoted f 0 (read f prime ) and

defined by
f.z/ ! f.x/ f.x C h/ ! f.x/
f 0 .x/ D lim D lim
z!x z!x h!0 h
provided that this limit exists. If f 0 .a/ can be found (while perhaps not all f 0 .x/ can
be found) f is said to be i erentia le at a, and f 0 .a/ is called the derivative of f at
a or the derivative of f with respect to x at a. The process of finding the derivative is
called i erentiation.

In the definition of the derivative, the expression

f.z/ ! f.x/ f.x C h/ ! f.x/
D
z!x h
where z D x C h, is called a difference quotient. Thus, f 0 .x/ is the limit of a difference
quotient.

E AM LE U F

If f.x/ D x2 , find the derivative of f.

Calculating a derivative via the definition S Applying the definition of a derivative gives
requires precision. Typically, the
f.x C h/ ! f.x/
difference quotient requires considerable f 0 .x/ D lim
manipulation before the limit step is h!0 h
ta en. This requires that each written step
be preceded by limh!0 to .x C h/2 ! x2 x2 C 2xh C h2 ! x2
D lim D lim
ac nowledge that the limit step is still h!0 h h!0 h
pending. bserve that after the limit step 2
is ta en, limh!0 is no longer be present. 2xh C h h.2x C h/
D lim D lim D lim .2x C h/ D 2x
h!0 h h!0 h h!0
bserve that, in ta ing the limit, we treated x as a constant, because it was h, not x,
that was changing. Also, note that f 0 .x/ D 2x defines a function of x, which we can
interpret as giving the slope of the tangent line to the graph of f at .x; f.x//. For example,
if x D 1, then the slope is f 0 .1/ D 2 " 1 D 2, which confirms the result in Example 1.
Now ork Problem 3 G
esides the notation f 0 .x/, other common ways to denote the derivative of y D f.x/
dy
The notation , which is called eibniz at x are
dx
n tati n, should not be thought of as a dy
pronounced dee y; dee x or dee y by dee x
fraction, although it loo s li e one. It is dx
a single symbol for a derivative. We d
have not yet attached any meaning to . f.x// dee f.x/; dee x or dee by dee x of f.x/
individual symbols, such as dy and dx. dx
0
y y prime
xy dee x of y
x . f.x// dee x of f.x/

ecause the derivative gives the slope of the tangent line, f 0 .a/ is the slope of the
line tangent to the graph of y D f.x/ at .a; f.a//.
 Section . e er at e 487

Two other notations for the derivative of f at a are

dy ˇˇ
ˇ and y0 .a/
dx xDa

E AM LE F E T L

If f.x/ D 2x2 C 2x C 3, find an equation of the tangent line to the graph of f at .1; 7/.
S

S We will first determine the slope of the tangent line by computing the
derivative and evaluating it at x D 1. Using this result and the point .1; 7/ in a
point-slope form gives an equation of the tangent line.

We have
f.x C h/ ! f.x/
f 0 .x/ D lim
h!0 h
.2.x C h/2 C 2.x C h/ C 3/ ! .2x2 C 2x C 3/
D lim
h!0 h
2x2 C 4xh C 2h2 C 2x C 2h C 3 ! 2x2 ! 2x ! 3
D lim
h!0 h
4xh C 2h2 C 2h
D lim D lim .4x C 2h C 2/
h!0 h h!0

So
f 0 .x/ D 4x C 2
and
f 0 .1/ D 4.1/ C 2 D 6
Thus, the tangent line to the graph at .1; 7/ has slope 6. A point-slope form of this
tangent is
y ! 7 D 6.x ! 1/
which in slope-intercept form is
y D 6x C 1

Now ork Problem 25 G

E AM LE F S C

Find the slope of the curve y D 2x C 3 at the point where x D 6.

S The slope of the curve is the slope of the tangent line. etting y D f.x/ D
2x C 3, we have
dy f.x C h/ ! f.x/ .2.x C h/ C 3/ ! .2x C 3/
D lim D lim
dx h!0 h h!0 h
2h
D lim D lim 2 D 2
h!0 h h!0

Since dy=dx D 2, the slope when x D 6, or in fact at any point, is 2. Note that the curve
is a straight line and thus has the same slope at each point.
Now ork Problem 19 G
488 C erent at on

E AM LE AF T L
d p
Find . x/.
dx
p
S etting f.x/ D x, we have
p p
d p f.x C h/ ! f.x/ xCh! x
. x/ D lim D lim
dx h!0 h h!0 h
Rationalizing numerators or As h ! 0, both the numerator and denominator approach zero. This can be avoided by
denominators of fractions is rationalizing the numerat r:
often helpful in calculating limits. p p
p p p p
xCh! x xCh! x xChC x
D :p p
h h xChC x
.x C h/ ! x h
D p p D p p
y h. x C h C x/ h. x C h C x/
Therefore,
Tangent y= x d p h 1 1 1
line at . x/ D lim p p D lim p p Dp p D p
(0, 0) dx h!0 h. x C h C x/ xChC x xC x 2 x
x p p
Note that the original function, x, is defined for x # 0, but its derivative, 1=.2 x/,
p
FIGURE Vertical tangent is defined only when x > 0. The reason for this is clear from the graph of y D x in
line at .0; 0/. Figure 11. . When x D 0, the tangent is a vertical line, so its slope is not defined.
Now ork Problem 17 G
p
In Example 5 we saw that the function y D x is not differentiable when x D 0,
because the tangent line is vertical at that point. It is worthwhile to mention that y D jxj
also is not differentiable when x D 0, but for a different reason: There is n tangent
line at all at that point. (Refer to Figure 11.5.) oth examples show that the domain of
f 0 may be strictly contained in the domain of f.
To indicate a derivative, eibniz notation is often useful because it ma es it con-
Variables other than x and y are often venient to emphasize the independent and dependent variables involved. For example,
more natural in applied problems. Time if the variable p is a function of the variable q, we spea of the derivative of p with
denoted by t, quantity by q, and price by respect to q, written dp=dq.
p are obvious examples. Example 6
illustrates.

E AM LE F p R q
A L IT I 1 dp
If p D f.q/ D , find .
If a ball is thrown upward at a speed 2q dq
of 40 ft s from a height of 6 feet, its
height in feet after t seconds is S We will do this problem first using the h ! 0 limit (the only one we have
used so far) and then using r ! q to illustrate the other variant of the limit.
D 6 C 40t ! 16t2 ! "
dp d 1 f.q C h/ ! f.q/
Find d =dt. D D lim
dq dq 2q h!0 h

1 1 q ! .q C h/
!
2.q C h/ 2q 2q.q C h/
D lim D lim
h!0 h h!0 h
q ! .q C h/ !h
D lim D lim
h!0 h.2q.q C h// h!0 h.2q.q C h//

!1 1
D lim D! 2
h!0 2q.q C h/ 2q
 Section . e er at e 489

We also have
dp f.r/ ! f.q/
D lim
dq r!q r!q
1 1 q!r
!
2r 2q 2rq
D lim D lim
r!q r ! q r!q r ! q

!1 !1
D lim D 2
r!q 2rq 2q

We leave it you to decide which form leads to the simpler limit calculation in this case.
Note that when q D 0 the function is not defined, so the derivative is also not even
defined when q D 0.
Now ork Problem 15 G
eep in mind that the derivative of y D f.x/ at x is nothing more than a limit,
namely,

f.x C h/ ! f.x/
lim
h!0 h
equivalently,

f.z/ ! f.x/
lim
z!x z!x
whose use we have ust illustrated. Although we can interpret the derivative as a func-
tion that gives the slope of the tangent line to the curve y D f.x/ at the point .x; f.x//,
this interpretation is simply a geometric convenience that assists our understanding.
The preceding limit may exist aside from any geometric considerations at all. As we
will see later, there are other useful interpretations of the derivative.
In Section 11.4, we will ma e technical use of the following relationship between
differentiability and continuity. However, it is of fundamental importance and needs to
be understood from the outset.

If f is differentiable at a, then f is continuous at a.

To establish this result, we will assume that f is differentiable at a. Then f 0 .a/ exists, and

f.a C h/ ! f.a/
lim D f 0 .a/
h!0 h

Consider the numerator f.a C h/ ! f.a/ as h ! 0. We have

! "
f.a C h/ ! f.a/
lim . f.a C h/ ! f.a// D lim "h
h!0 h!0 h
f.a C h/ ! f.a/
D lim " lim h
h!0 h h!0
0
D f .a/ " 0 D 0

Thus, limh!0 f.a C h/ ! f.a// D 0. This means that f.a C h/ ! f.a/ approaches 0 as
h ! 0. Consequently,

lim f.a C h/ D f.a/

h!0
490 C erent at on

y As stated in Section 10.3, this condition means that f is continuous at a. The foregoing,
then, proves that f is continuous at a when f is differentiable there. ore simply, we say
y = f(x) that differentiability at a point implies continuity at that point.
If a function is not continuous at a point, then it cannot have a derivative there. For
example, the function in Figure 11. is discontinuous at a. The curve has no tangent at
that point, so the function is not differentiable there.
x
a
E AM LE C
FIGURE f is not continuous
at a, so f is not differentiable at a. a et f.x/ D x2 . The derivative, 2x, is defined for all values of x, so f.x/ D x2 must
be continuous for all values of x.
1
b The function f.p/ D is not continuous at p D 0 because f is not defined there.
2p
Thus, the derivative does not exist at p D 0.

G
y The converse of the statement that differentiability implies continuity is fa se. That
is, continuity does not imply differentiability. In Example , we give a function that is
continuous at a point, but not differentiable there.

E AM LE C N I
f(x) = x
The function y D f.x/ D jxj is continuous at x D 0. (See Figure 11.10.) As we
x
mentioned earlier, there is no tangent line at x D 0. Thus, the derivative does not exist
Continuous at x = 0, but there. This shows that continuity does n t imply differentiability.
not differentiable at x = 0

FIGURE Continuity does not

G
imply differentiability.
Finally, we remar that while differentiability of f at a implies continuity of f at a,
the derivative function, f 0 , is not necessarily continuous at a. Unfortunately, the classic
example is constructed from a function not considered in this boo .

R BLEMS
In Pr b ems and a functi n f and a p int P n its graph are dy dy
if y D 3x C 5 if y D !5x
gi en dx dx
a Find the s pe f the secant ine P f r each p int d d # x$
D .x; f .x// h se x a ue is gi en in the tab e R und y ur .5 ! 7x/ 1!
dx dx 2
ans ers t f ur decima p aces.
b se y ur resu ts fr m part a t estimate the s pe f the f 0 .x/ if f .x/ D 3 f 0 .x/ if f .x/ D 7:01
tangent ine at P. d 2
.x C 4x ! / y0 if y D x2 C 5x C 7
f .x/ D x3 C 3; P D .!2; !5/ dx
dp d 2
x-value of !3 !2:5 !2:2 !2:1 !2:01 !2:001 if p D 3q2 C 2q C 1 .x ! x ! 3/
dq dx
mP
6 dC
y0 if y D if C D 7 C 2q ! 3q2
f .x/ D ln x, P D .1; 0/ x dq
p 3
x-value of 2 1.5 1.2 1.1 1.01 1.001 f 0 .x/ if f .x/ D 5x 0
.x/ if .x/ D
x!2
mP
Find the slope of the curve y D x2 C 4 at the point .!2; /.

In Pr b ems use the de niti n f the deri ati e t nd each Find the slope of the curve y D 1 ! x2 at the point .1; 0/.
f the f ing Find the slope of the curve y D 4x2 ! 5 when x D 0.
p
f 0 .x/ if f .x/ D x f 0 .x/ if f .x/ D 4x ! 1 Find the slope of the curve y D 5x when x D 20.
 Section .2 es for erent at on 491

In Pr b ems nd an equati n f the tangent ine t the In Pr b ems and use the imit f a di erence qu tient
cur e at the gi en p int de niti n t estimate f 0 .x/ at the indicated a ues f x R und
y ur ans ers t three decima p aces
y D x C 4 (3, 7) y D 3x2 ! 4I .1; !1/
f .x/ D x ln x ! x x D 1, x D e
y D x2 C 2x C 3I .1; 6/ y D .x ! 7/2 (6, 1)
x2 C 4x C 2
5 5 f .x/ D I x D 2; x D !4
yD .2; 1) yD I .2; !1/ x3 ! 3
xC3 1 ! 3x Find an equation of the tangent line to the curve f .x/ D x2 C x
Ban ing Equations may involve derivatives of functions. at the point .!2; 2/. raph both the curve and the tangent line.
In an article on interest rate deregulation, Christofi and Agapos1 Notice that the tangent line is a good approximation to the curve
solve the equation near the point of tangency.
! "! " The derivative of f .x/ D x3 ! x C 2 is f 0 .x/ D 3x2 ! 1. raph
! dC both the function f and its derivative f 0 . bserve that there are two
rD r !
1C! d points on the graph of f where the tangent line is horizontal. For
the x-values of these points, what are the corresponding values
for ! (the ree letter eta ). Here r is the deposit rate paid by of f 0 .x/ Why are these results expected bserve the intervals
commercial ban s, r is the rate earned by commercial ban s, where f 0 .x/ is positive. Notice that tangent lines to the graph of f
C is the administrative cost of transforming deposits into have positive slopes over these intervals. bserve the interval
return-earning assets, is the savings deposits level, and ! is where f 0 .x/ is negative. Notice that tangent lines to the graph of f
the deposit elasticity with respect to the deposit rate. Find !. have negative slopes over this interval.

In Pr b ems and use the numerica deri ati e feature f In Pr b ems and erify the identity .z ! x/
% Pn!1 i n!1!i &
y ur graphing ca cu at r t estimate the deri ati es f the x z D z n
! xn f r the indicated a ues f n and
iD0
functi ns at the indicated a ues R und y ur ans ers t three ca cu ate the deri ati e using the z ! x f rm f the de niti n f
decima p aces the deri ati e in Equati n
p
f .x/ D 2x2 C 3xI x D 1; x D 2 n D 4, n D 3, n D 2 f 0 .x/ if f .x/ D 2x4 C x3 ! 3x2
x
f .x/ D e .4x ! 7/I x D 0; x D 1:5 n D 5, n D 3 f 0 .x/ if f .x/ D 2x5 ! 5x3

Objective R
o e e o t e as c r es for ifferentiating a function by direct use of the definition of derivative can be tedious.
erent at n constant f nct ons an
o er f nct ons an t e co nn However, if a function is constructed from simpler functions, then the derivative of
r es for erent at n a constant the more complicated function can be constructed from the derivatives of the sim-
t e of a f nct on an a s pler functions. Usually, we need to now only the derivatives of a few basic functions
of t o f nct ons
and ways to assemble derivatives of constructed functions from the derivatives of their
components. For example, if functions f and g have derivatives f 0 and g0 , respectively,
then f C g has a derivative given by . f C g/0 D f 0 C g0 . However, some ru es are
less intuitive. For example, if f " g denotes the function whose value at x is given by
. f " g/.x/ D f.x/ " g.x/, then . f " g/0 D f 0 " g C f " g0 . In this chapter we study most
such combining rules and some basic rules for calculating derivatives of certain basic
functions.
We begin by showing that the derivative of a constant function is zero. Recall that
the graph of the constant function f.x/ D c is a horizontal line (see Figure 11.11),
which has a slope of zero at each point. This means that f 0 .x/ D 0 regardless of x. As
f(x) a formal proof of this result, we apply the definition of the derivative to f.x/ D c:

f.x C h/ ! f.x/ c!c

c f(x) = c f 0 .x/ D lim D lim
h!0 h h!0 h
Slope is zero 0
everywhere D lim D lim 0 D 0
x h!0 h h!0

FIGURE The slope of a

constant function is 0. 1
A. Christofi and A. Agapos, Interest Rate eregulation: An Empirical ustification, Re ie f Business and
Ec n mic Research , no. 1 (1 4), 3 4 .
492 C erent at on

Thus, we have our first rule:

BASIC RULE C
If c is a constant, then
d
.c/ D 0
dx
That is, the derivative of a constant function is zero.

E AM LE C F
d
a .3/ D 0 because 3 is a constant function.
dx p
b If g.x/ D 5, then g0 .x/ D 0 because g is a constant function. For example, the
derivative of g when x D 4 is g0 .4/ D 0.
c If s.t/ D .1, 3 ,623/ 07:4 , then ds=dt D 0.
Now ork Problem 1 G
The next rule gives a formula for the derivative of x raised to a constant power
that is, the derivative of f.x/ D xa , where a is an arbitrary real number. A function
of this form is called a power function. For example, f.x/ D x2 is a power function.
While the rule we record is valid for all real a, we will establish it only in the case where
a is a positive integer, n. The rule is so central to differential calculus that it warrants
a detailed calculation if only in the case where a is a positive integer, n. Whether we
use the h ! 0 form of the definition of derivative or the z ! x form, the calculation of
dxn
is instructive and provides good practice with summation notation, whose use is
dx
more essential in later chapters. We provide a calculation for each possibility. We must
either expand .x C h/n , to use the h ! 0 form of Equation (2) from Section 11.1, or
factor zn ! xn , to use the z ! x form.
For the first of these we recall the bin mia the rem of Section .2:
X n
.x C h/n D n Ci x
n!i i
h
iD0

where the n Ci are the binomial coe cients, whose precise descriptions, except for
n C0 D 1 and n C1 D n, are not necessary here (but are given in Section .2). For
the second we have
!
X
n!1
.z ! x/ xi zn!1!i D zn ! xn
iD0

which is easily verified by carrying out the multiplication using the rules for manipu-
lating summations given in Section 1.5. In fact, we have
!
X
n!1 X
n!1 X
n!1
.z ! x/ xi zn!1!i D z xi zn!1!i ! x xi zn!1!i
iD0 iD0 iD0

X
n!1 X
n!1
D xi zn!i ! xiC1 zn!1!i
iD0 iD0
! !
X
n!1 X
n!2
n i n!i iC1 n!1!i n
D z C xz ! x z Cx
iD1 iD0

D zn ! xn
where the reader should chec that the two summations in the second-to-last line really
do cancel as shown.
 Section .2 es for erent at on 493

There is a lot more to calculus than this BASIC RULE xa

important rule. If a is any real number, then
d a
.x / D axa!1
dx
That is, the derivative of a constant power of x is the exponent times x raised to a
power one less than the given power.

For n, a positive integer, if f.x/ D xn , the definition of the derivative gives

f.x C h/ ! f.x/ .x C h/n ! xn

f 0 .x/ D lim D lim
h!0 h h!0 h

y our previous discussion on expanding .x C h/n ,

X
n
n!i i
n Ci x h ! xn
iD0
f 0 .x/ D lim
h!0 h
X
n
n!i i
n Ci x h
.1/ iD1
D lim
h!0 h
X
n
n!i i!1
h n Ci x h
.2/ iD1
D lim
h!0 h
.3/ X
n
n!i i!1
D lim n Ci x h
h!0
iD1
!
.4/ X
n
D lim nxn!1 C n Ci x
n!i i!1
h
h!0
iD2
.5/
D nxn!1

where we ustify the further steps as follows:

The i D 0 term in the summation is n C0 xn h0 D xn , so it cancels with the separate,

last, term: !xn .
We are able to extract a common factor of h from each term in the sum.
This is the crucial step. The expressions separated by the equal sign are limits as
h ! 0 of functions of h that are equal for h ¤ 0.
The i D 1 term in the summation is n C1 xn!1 h0 D nxn!1 . It is the only one that
does not contain a factor of h, and we separated it from the other terms.
Finally, in determining the limit we made use of the fact that the isolated term is
independent of h, while all the others contain h as a factor and so have limit 0 as
h ! 0.

Now, using the z ! x limit for the definition of the derivative and f.x/ D xn ,
we have

f.z/ ! f.x/ zn ! x n
f 0 .x/ D lim D lim
z!x z!x z!x z ! x
494 C erent at on

y our previous discussion on factoring zn ! xn , we have

!
X
n!1
i n!1!i
.z ! x/ xz
iD0
f 0 .x/ D lim
z!x z!x

.1/ X
n!1
D lim xi zn!1!i
z!x
iD0

.2/ X
n!1
D xi xn!1!i
iD0

.3/ X
n!1
D xn!1
iD0

.4/
D nxn!1
where this time we ustify the further steps as follows:
Here the crucial step comes first. The expressions separated by the equal sign are
limits as z ! x of functions of z that are equal for z ¤ x.
The limit is given by evaluation because the expression is a polynomial in the
variable z.
An obvious rule for exponents is used.
Each term in the sum is xn!1 , independent of i, and there are n such terms.

E AM LE x

d 2
a y asic Rule 2, .x / D 2x2!1 D 2x.
dx
b If F.x/ D x D x1 , then F0 .x/ D 1 " x1!1 D 1 " x0 D 1. Thus, the derivative of x with
respect to x is 1.
c If f.x/ D x!10 , then f 0 .x/ D !10x!10!1 D !10x!11 .
Now ork Problem 3 G
When we apply a differentiation rule to a function, sometimes the function must
first be rewritten so that it has the proper form for that rule. For example, to differentiate
1
f.x/ D 10 we would first rewrite f as f.x/ D x!10 and then proceed as in Example 2(c).
x

E AM LE R F F xa

p p
a To differentiate y D x, we rewrite x as x1=2 so that it has the form xa . Thus,
dy 1 1 1 1
D x.1=2/!1 D x!1=2 D 1=2 D p
dx 2 2 2x 2 x
which agrees with our limit calculation in Example 5 of Section 11.1.
1
b et h.x/ D p . To apply asic Rule 2, we must rewrite h.x/ as h.x/ D x!3=2 so
x x
that it has the form xa . We have
d 3 3
h0 .x/ D .x!3=2 / D ! x.!3=2/!1 D ! x!5=2
dx 2 2
Now ork Problem 39 G
 Section .2 es for erent at on 495

Now that we can say immediately that the derivative of x3 is 3x2 , the question arises
as to what we could say about the derivative of a mu tip e of x3 , such as 5x3 . ur next
rule will handle this situation of differentiating a constant times a function.

C MBINING RULE C F R
If f is a differentiable function and c is a constant, then cf.x/ is differentiable and
d
.cf.x// D cf 0 .x/
dx
That is, the derivative of a constant times a function is the constant times the deriva-
tive of the function.

Pr f If g.x/ D cf.x/, applying the definition of the derivative of g gives

g.x C h/ ! g.x/ cf.x C h/ ! cf.x/
g0 .x/ D lim D lim
h!0 h h!0 h
! "
f.x C h/ ! f.x/ f.x C h/ ! f.x/
D lim c " D c " lim
h!0 h h!0 h
D cf 0 .x/

E AM LE C T F

ifferentiate the following functions.

a g.x/ D 5x3
S Here, g is a constant (5) times a function .x3 /. So
d d
.5x3 / D 5 .x3 / Combining Rule 1
dx dx
D 5.3x3!1 / D 15x2 asic Rule 2
13q
b f.q/ D
5
S

S We first rewrite f as a constant times a function and then apply

asic Rule 2.

13q 13 13
ecause D q, f is the constant times the function q. Thus,
5 5 5
13 d
f 0 .q/ D .q/ Combining Rule 1
5 dq
13 13
D "1D asic Rule 2
5 5

0:25
c yD p5 2
x
S We can express y as a constant times a function:
1
y D 0:25 " p
5 2
D 0:25x!2=5
x
496 C erent at on

Hence,
d !2=5
y0 D 0:25 .x / Combining Rule 1
dx
! "
2 !7=5
D 0:25 ! x D !0:1x!7=5 asic Rule 2
5
Now ork Problem 7 G
In differentiating f .x/ D .4x/3 , asic
The next rule involves derivatives of sums and differences of functions.
Rule 2 cannot be applied directly. It
applies to a power of the variable x, n t to C MBINING RULE S R
a power of an expression involving x, If f and g are differentiable functions, then f C g and f ! g are differentiable and
such as 4x. To apply our rules, write
d
f .x/ D .4x/3 D 43 x3 D 64x3 . Thus, . f.x/ C g.x// D f 0 .x/ C g0 .x/
dx
d 3
f 0 .x/ D 64 .x / D 64.3x2 / D 1 2x2 : and
dx
d
. f.x/ ! g.x// D f 0 .x/ ! g0 .x/
dx
That is, the derivative of the sum (difference) of two functions is the sum (difference)
of their derivatives.

Pr f For the case of a sum, if F.x/ D f.x/ C g.x/, applying the definition of the
derivative of F gives
F.x C h/ ! F.x/
F0 .x/ D lim
h!0 h
. f.x C h/ C g.x C h// ! . f.x/ C g.x//
D lim
h!0 h
. f.x C h/ ! f.x// C .g.x C h/ ! g.x//
D lim regrouping
h!0 h
! "
f.x C h/ ! f.x/ g.x C h/ ! g.x/
D lim C
h!0 h h
ecause the limit of a sum is the sum of the limits,
f.x C h/ ! f.x/ g.x C h/ ! g.x/
F0 .x/ D lim C lim D f 0 .x/ C g0 .x/
h!0 h h!0 h
The proof for the derivative of a difference of two functions now follows from the sum
rule and Combining Rule 1 by writing f.x/ ! g.x/ D f.x/ C .!1/g.x/. We encourage
the reader to write the details.
Combining Rule 2 can be extended to the derivative of any number of sums and
differences of functions. For example,
d
. f.x/ ! g.x/ ! h.x/ C k.x// D f 0 .x/ ! g0 .x/ ! h0 .x/ C k0 .x/
dx

E AM LE S F
A L IT I
If the revenue function for a certain ifferentiate the following
p functions.
product is r.q/ D 50q ! 0:3q2 , find the a F.x/ D 3x5 C x
derivative of this function, also nown p
S Here, F is the sum of two functions, 3x5 and x. Therefore,
as the marginal revenue.
d d
F0 .x/ D .3x5 / C .x1=2 / Combining Rule 2
dx dx
d d
D 3 .x5 / C .x1=2 / Combining Rule 1
dx dx
1 1
D 3.5x4 / C x!1=2 D 15x4 C p asic Rule 2
2 2 x
 Section .2 es for erent at on 497

z4 5
b f.z/ D ! 1=3
4 z
S To apply our rules, we will rewrite f in the form f.z/ D 14 z4 ! 5z!1=3 .
Since f is the difference of two functions,
! "
0 d 1 4 d
f .z/ D z ! .5z!1=3 / Combining Rule 2
dz 4 dz
1 d 4 d
D .z / ! 5 .z!1=3 / Combining Rule 1
4 dz dz
! "
1 1
D .4z3 / ! 5 ! z!4=3 asic Rule 2
4 3
5
D z3 C z!4=3
3
c y D 6x3 ! 2x2 C 7x !
S
dy d d d d
D .6x3 / ! .2x2 / C .7x/ ! . /
dx dx dx dx dx
d 3 d 2 d d
D 6 .x / ! 2 .x / C 7 .x/ ! . /
dx dx dx dx
D 6.3x2 / ! 2.2x/ C 7.1/ ! 0
D 1 x2 ! 4x C 7

Now ork Problem 47 G

In Examples 6 and 7, we need to rewrite E AM LE F
the given function in a form to which our
rules apply. Find the derivative of f.x/ D 2x.x2 ! 5x C 2/ when x D 2.
S We multiply and then differentiate each term:
f.x/ D 2x3 ! 10x2 C 4x
f 0 .x/ D 2.3x2 / ! 10.2x/ C 4.1/
D 6x2 ! 20x C 4
f 0 .2/ D 6.2/2 ! 20.2/ C 4 D !12
Now ork Problem 75 G
E AM LE F E T L

Find an equation of the tangent line to the curve

3x2 ! 2
yD
x
when x D 1.
S
dy
S First we find , which gives the slope of the tangent line at any point.
dx
dy
Evaluating when x D 1 gives the slope of the required tangent line. We then
dx
determine the y-coordinate of the point on the curve when x D 1. Finally, with
the slope and both coordinates of the point determined, we use point-slope form to
obtain an equation of the tangent line.
498 C erent at on

Rewriting y as a difference of two functions, we have

3x2 2
yD ! D 3x ! 2x!1
x x
Thus,
dy 2
D 3.1/ ! 2..!1/x!2 / D 3 C 2
dx x
The slope of the tangent line to the curve when x D 1 is
ˇ
dy ˇˇ 2
ˇ D3C 2 D5
dx xD1 1
3x2 ! 2
To find the y-coordinate of the point on the curve where x D 1, we evaluate y D
x
at x D 1. This gives
To obtain the y-value of the point on the 3.1/2 ! 2
curve when x D 1, evaluate the rigina
yD D1
1
function at x D 1.
Hence, the point .1; 1/ lies on both the curve and the tangent line. Therefore, an equation
of the tangent line is
y ! 1 D 5.x ! 1/
In slope-intercept form, we have
y D 5x ! 4

Now ork Problem 81 G

R BLEMS
In Pr b ems di erentiate the functi ns 5 5 3
% 6 &2=3 k.x/ D !2x2 C x C 11 f .x/ D x C x7
f .x/ D " f .x/ D 3 7 5
7 7
x 2x
y D x17 f .x/ D x21 p.x/ D C f .x/ D x2=7
7 3
0
yDx y D x2:1 f .x/ D 2x!14=5 y D x3=4 C 2x5=3
f .x/ D x 2
y D 7x 6 p
y D 4x2 ! x!3=5 y D 11 x
g. / D 7 .x/ D x e p p
5
y D x3 f .r/ D 6 3 r
3 6
p p
yD 5
x f .p/ D 3p4 y D 4 x2 f .x/ D x!6
s5 x7 f .s/ D 2s!3 f .x/ D x!6 C x!4 C x!2
f .s/ D yD
30 7 1
f .x/ D x C 3 f .x/ D 5x ! e f .x/ D 100x!3 C 10x1=2 yD
x
f .x/ D 4x2 ! 2x C 3 G.x/ D 7x3 ! 5x2 3
f .x/ D yD
x4 x5
g.p/ D p4 ! 3p3 ! 1 f .t/ D !13t2 C 14t C 1
p 1 4
y D x4 ! 3 x y D ! x4 C ln 2 yD g.x/ D
2x3 3x3
y D 11x5 C 12x3 ! 5x 1 3
yD 2 f .t/ D 3
.r/ D r ! 7r6 C 3r2 C 1 x 5t
7 x2 2
f .x/ D 2.13 ! x4 / .t/ D e.t7 ! 53 / g.x/ D f .x/ D C 2
x 2 x
13 ! x4 3.x3 ! 2x/ x3 3
g.x/ D f .x/ D ˆ.x/ D ! 3 f .x/ D ! x1=3 C 5x!2=5
3 4 3 x
x2 1
h.x/ D 4x4 C x3 ! C x f .z/ D 5z3=4 ! 62 ! z1=4 q.x/ D p
3
2 x2
 Section .3 e er at e as a ate of C an e 499

5 2 1 Find all points on the curve

f .x/ D p yD p yD p
6 5
x x 2 x 5 2
p yD x ! x3
yDx x 3 3 3
f .x/ D .2x /.4x / 2 2
2 3
where the tangent line is horizontal.
f .x/ D 3x .2x ! 3x/ f .x/ D x3 .3x6 ! 5x2 C 4/
p p Repeat Problem 5 for the curve
f .x/ D x3 .3x/2 s.x/ D x. 5 x C 7x C 2/
x6 x2
yD ! C1
.x/ D x!2=3 .x C 5/ f .x/ D x2=7 .x3 C 5x C 2/ 6 2
3q2 C 4q ! 2 !5 Find all points on the curve
f .q/ D f. / D
q 5 y D x2 ! 5x C 3
where the slope is 1.
f .x/ D .x ! 1/.x C 2/ f .x/ D x2 .x ! 2/.x C 4/
Repeat Problem 7 for the curve
x C x2 7x3 C x
.x/ D f .x/ D p y D x5 ! 4x C 13
x 6 x
p 1
If f .x/ D x C p , evaluate the expression
F r each cur e in Pr b ems nd the s pes at the indicated x
p ints x!1
p ! f 0 .x/
y D 3x2 C 4x ! I .0; ! /; .2; 12/; .!3; 7/ 2x x
Economics Eswaran and otwal2 consider agrarian
y D 3 C 5x ! 3x3 .0; 3/, . 12 ; 41 /, .2; !11/
economies in which there are two types of wor ers, permanent
y D 4 when x D !4; x D 7; x D 22 and casual. Permanent wor ers are employed on long-term
p contracts and may receive benefits, such as holiday gifts and
y D 2x ! 2 x when x D , x D 16, x D 25
emergency aid. Casual wor ers are hired on a daily basis and
In Pr b ems nd an equati n f the tangent ine t the perform routine and menial tas s, such as weeding, harvesting,
cur e at the indicated p int and threshing. The difference z in the present-value cost of hiring
a permanent wor er over that of hiring a casual wor er is given by
1 ! x2
y D 4x2 C 5x C 6 .1; 15/ yD .4; !3/ z D .1 C b/ p !b c
5
1 p where p and c are wage rates for permanent labor and casual
yD .2; 14 / y D ! 3 x . ; !2/ labor, respectively, b is a constant, and p is a function of c .
x2
Eswaran and otwal claim that
Find an equation of the tangent line to the curve ' (
dz d p b
D .1 C b/ !
d c d c 1Cb
y D 2 C 3x ! 5x2 C 7x3
Verify this.
when x D 1. Find an equation of the tangent line to the graph of
Repeat Problem 3 for the curve y D x3 ! 2x C 1 at the point .1; 0/. raph both the function and
the tangent line on the same screen.
p p
x.2 ! x2 / Find an equation of the tangent line to the graph of y D 3 x,
yD at the point .! ; !2/. raph both the function and the tangent line
x
on the same screen. Notice that the line passes through .! ; !2/
when x D 4. and the line appears to be tangent to the curve.

Objective T R C
o ot ate t e nstantaneo s rate of We have given a geometric interpretation of the derivative as being the slope of the
c an e of a f nct on eans of
e oc t an to nter ret t e er at e tangent line to a curve at a point. Historically, an important application of the derivative
as an nstantaneo s rate of c an e involves the motion of an ob ect traveling in a straight line. This gives us a convenient
o e e o t e ar na conce t way to interpret the derivative as a rate f change.
c s fre ent se n s ness
an econo cs To denote the change in a variable, such as x, the symbol #x (read delta x )
is commonly used. For example, if x changes from 1 to 3, then the change in x is
#x D 3 ! 1 D 2. The new value of x.D 3/ is the old value plus the change, which

2
. Eswaran and A. otwal, A Theory of Two-Tier abor ar ets in Agrarian Economies,
he American Ec n mic Re ie 75, no. 1 (1 5), 162 77.
500 C erent at on

is 1C#x. Similarly, if t increases by #t, the new value is tC#t. We will use #-notation
s = t2
s
in the discussion that follows.
0 1 9 Suppose an ob ect moves along the number line in Figure 11.12 according to the
t=1 t=3 equation
FIGURE otion along s D f.t/ D t2
a number line.
where s is the position of the ob ect at time t. This equation is called an equati n f
m ti n, and f is called a position function. Assume that t is in seconds and s is in
meters. At t D 1 the position is s D f.1/ D 12 D 1, and at t D 3 the position is
s D f.3/ D 32 D . ver this two-second time interval, the ob ect has a change in
position, or a disp acement of ! 1 D m, and the a erage e city of the ob ect
is defined as
displacement
ave D
length of time interval

D D4ms
2
To say that the average velocity is 4 m s from t D 1 to t D 3 means that, n the
a erage the position of the ob ect changed by 4 m to the right each second during
that time interval. et us denote the changes in s-values and t-values by #s and #t,
respectively. Then the average velocity is given by
#s
ave D D 4 m s .for the interval t D 1 to t D 3/
#t
The ratio #s=#t is also called the average rate of change of s with respect to t over
the interval from t D 1 to t D 3.
Now, let the time interval be only 1 second long (that is, #t D 1). Then, for the
sh rter interval from t D 1 to t D 1 C #t D 2, we have f.2/ D 22 D 4, so
#s f.2/ ! f.1/ 4!1
ave D D D D3ms
#t #t 1
ore generally, over the time interval from t D 1 to t D 1 C #t, the ob ect moves
from position f(1) to position f.1 C #t/. Thus, its displacement is
#s D f.1 C #t/ ! f.1/
Since the time interval has length #t, the ob ect s average velocity is given by
#s f.1 C #t/ ! f.1/
ave D D
#t #t
If #t were to become smaller and smaller, the average velocity over the interval from
t D 1 to t D 1 C #t would be close to what we might call the instantane us e city
at time t D 1 that is, the velocity at a p int in time .t D 1/ as opposed to the velocity
over an inter a of time. For some typical values of #t between 0.1 and 0.001, we get
the average velocities in Table 11.2, which the reader can verify.

Table 11.
ength of
Time Interval Time Interval Average Velocity
#s f.1 C #t/ ! f.1/
#t t D 1 to t D 1 C #t D
#t #t
0.1 t D 1 to t D 1:1 2.1 ms
0.07 t D 1 to t D 1:07 2.07 m s
0.05 t D 1 to t D 1:05 2.05 m s
0.03 t D 1 to t D 1:03 2.03 m s
0.01 t D 1 to t D 1:01 2.01 m s
0.001 t D 1 to t D 1:001 2.001 m s
 Section .3 e er at e as a ate of C an e 501

The table suggests that as the length of the time interval approaches zero, the aver-
age velocity approaches the value 2 m s. In other words, as #t approaches 0, #s=#t
approaches 2 m s. We define the limit of the average velocity as #t ! 0 to be the
instantane us e city (or simply the velocity), , at time t D 1. This limit is also
called the instantane us rate f change of s with respect to t at t D 1:
#s f.1 C #t/ ! f.1/
D lim ave D lim D lim
!t!0 #t
!t!0 !t!0 #t
If we thin of #t as h, then the limit on the right is simply the derivative of s with
respect to t at t D 1. Thus, the instantaneous velocity of the ob ect at t D 1 is ust ds dt
at t D 1. ecause s D t2 and
ds
D 2t
dt
the velocity at t D 1 is
ˇ
ds ˇˇ
D D 2.1/ D 2 m s
dt ˇ tD1
which confirms our previous conclusion.
In summary, if s D f.t/ is the position function of an ob ect moving in a straight
line, then the average velocity of the ob ect over the time interval Œt; t C #t$ is given by
#s f.t C #t/ ! f.t/
ave D D
#t #t
and the velocity at time t is given by
f.t C #t/ ! f.t/ ds
D lim D
!t!0 #t dt
Selectively combining equations for , we have
ds #s
D lim
dt !t!0 #t

ecause # is the uppercase ree letter corresponding to d, this equation provides

motivation for the otherwise bizarre eibniz notation for derivatives.

E AM LE F A

Suppose the position function of an ob ect moving along a number line is given by
s D f.t/ D 3t2 C 5, where t is in seconds and s is in meters.
a Find the average velocity over the interval 10, 10.1 .
b Find the velocity when t D 10.
S
a Here t D 10 and #t D 10:1 ! 10 D 0:1. So we have
#s f.t C #t/ ! f.t/
ave D D
#t #t
f.10 C 0:1/ ! f.10/
D
0:1
f.10:1/ ! f.10/
D
0:1
311:03 ! 305 6:03
D D D 60.3 m s
0:1 0:1
b The velocity at time t is given by
ds
D D 6t
dt
502 C erent at on

When t D 10, the velocity is

ˇ
ds ˇˇ
D 6.10/ D 60 m s
dt ˇtD10
Notice that the average velocity over the interval 10, 10.1 is close to the velocity
at t D 10. This is to be expected because the length of the interval is small.

Now ork Problem 1 G

ur discussion of the rate of change of s with respect to t applies equally well to
any function y D f.x/. This means that we have the following:

If y D f.x/, then
8̂
average rate of change
f.x C #x/ ! f.x/ <
#y of y with respect to x
D D
#x #x :̂over the interval from
x to x C #x
and
)
dy #y instantaneous rate of change
D lim D
dx !x!0 #x of y with respect to x

ecause the instantaneous rate of change of y D f.x/ at a point is a derivative, it is also

the s pe f the tangent ine to the graph of y D f.x/ at that point. For convenience,
we usually refer to the instantaneous rate of change simply as the rate of change. The
interpretation of a derivative as a rate of change is extremely important.
et us now consider the significance of the rate of change of y with respect to x.
From Equation (2), if #x (a change in x) is close to 0, then #y=#x is close to dy=dx.
That is,
#y dy
$
#x dx
Therefore,
dy
#y $ #x
dx
That is, if x changes by #x, then the change in y, #y, is approximately dy=dx times the
change in x. In particular,
dy
if x changes by 1, an estimate of the change in y is
dx

E AM LE E #y U dy=dx
A L IT I
Suppose that the profit P made by dy
selling a certain product at a price of p
Suppose that y D f.x/ and D when x D 3. Estimate the change in y if x changes
dx
per unit is given by P D f .p/ and the from 3 to 3.5.
rate of change of that profit with respect
dP
S We have dy=dx D and #x D 3:5 ! 3 D 0:5. The change in y is given by
to change in price is
dp
D 5 at p D 25. #y, and, from Equation (3),
Estimate the change in the profit P if the dy
price changes from 25 to 25.5. #y $ #x D .0:5/ D 4
dx
We remar that, since #y D f.3:5/ ! f.3/, we have f.3:5/ D f.3/ C #y. For example,
if f.3/ D 5, then f(3.5) can be estimated by 5 C 4 D .
G
 Section .3 e er at e as a ate of C an e 503

E AM LE F R C
A L IT I
The position of an ob ect thrown Find the rate of change of y D x4 with respect to x, and evaluate it when x D 2 and
upward at a speed of 16 feet s from when x D !1. Interpret your results.
a height of 0 feet is given by y.t/ D
16t ! 16t2 . Find the rate of change of S The rate of change is
y with respect to t, and evaluate it when dy
t D 0:5. Use your graphing calculator D 4x3
to graph y.t/. Use the graph to interpret dx
the behavior of the ob ect when t D 0:5. When x D 2; dy=dx D 4.2/3 D 32. This means that if x increases, from 2, by a
small amount, then y increases approximately 32 times as much. ore simply, we
say that, when x D 2, y is increasing 32 times as fast as x does. When x D !1;
dy=dx D 4.!1/3 D !4. The significance of the minus sign on !4 is that, when x D !1,
y is decreasing 4 times as fast as x increases.
Now ork Problem 11 G

E AM LE R C R

et p D 100 ! q2 be the demand function for a manufacturer s product. Find the rate of
change of price, p, per unit with respect to quantity, q. How fast is the price changing
with respect to q when q D 5 Assume that p is in dollars.
S The rate of change of p with respect to q is
dp d
D .100 ! q2 / D !2q
dq dq
Thus,
ˇ
dp ˇˇ
D !2.5/ D !10
dq ˇqD5
This means that when five units are demanded, an increase of one extra unit demanded
corresponds to a decrease of approximately 10 in the price per unit that consumers
are willing to pay.
G

E AM LE R C

A spherical balloon is being filled with air. Find the rate of change of the volume of air
in the balloon with respect to its radius. Evaluate this rate of change when the radius
is 2 ft.
S The formula for the volume of a ball of radius r is D 43 "r3 . The rate of
change of with respect to r is
d 4
D ".3r2 / D 4"r2
dr 3
When r D 2 ft, the rate of change is
ˇ
d ˇˇ 2 ft3
D 4".2/ D 16"
dr ˇrD2 ft

This means that when the radius is 2 ft, changing the radius by 1 ft will change the
volume by approximately 16" ft3 .
G
504 C erent at on

E AM LE R C E

A sociologist is studying various suggested programs that can aid in the education of
preschool-age children in a certain city. The sociologist believes that x years after the
beginning of a particular program, f.x/ thousand preschoolers will be enrolled, where
10
f.x/ D .12x ! x2 / 0 % x % 12

At what rate would enrollment change (a) after three years from the start of this program
and (b) after nine years
S The rate of change of f.x/ is
10
f 0 .x/ D .12 ! 2x/

a After three years, the rate of change is

10 10 20 2
f 0 .3/ D .12 ! 2.3// D "6D D6
3 3
Thus, enrollment would be increasing at the rate of 6 23 thousand preschoolers
per year.
b After nine years, the rate is
10 10 20 2
f 0. / D .12 ! 2. // D .!6/ D ! D !6
3 3
Thus, enrollment would be decreasing at the rate of 6 23 thousand preschoolers
per year.

Now ork Problem 9 G

A R C E
A manufacturer s total cost function, c D f.q/, gives the total cost, c, of producing
and mar eting q units of a product. The rate of change of c with respect to q is called
the marginal cost. Thus,
dc
marginal cost D
dq
For example, suppose c D f.q/ D 0:1q2 C 3 is a cost function, where c is in dollars
and q is in pounds. Then
dc
D 0:2q
dq
The marginal cost when 4 lb are produced is dc dq, evaluated when q D 4:
ˇ
dc ˇˇ
D 0:2.4/ D 0: 0
dq ˇqD4

This means that if production is increased by 1 lb, from 4 lb to 5 lb, then the change in
cost is approximately 0. 0. That is, the additional pound costs about 0. 0. In general,
e interpret margina c st as the appr ximate c st f ne additi na unit f utput.
After all, the difference f.q C 1/ ! f.q/ can be seen as a difference quotient
f.q C 1/ ! f.q/
1
the case where h D 1. Any difference quotient can be regarded as an approximation
of the corresponding derivative and, conversely, any derivative can be regarded as an
approximation of any of its corresponding difference quotients. Thus, for any function
f of q we can always regard f 0 .q/ and f.q C 1/ ! f.q/ as approximations of each other.
 Section .3 e er at e as a ate of C an e 505

In economics, the latter can usually be regarded as the exact value of the cost, or profit
depending upon the function, of the (q C 1)th item when q are produced. The derivative
is often easier to compute than the exact value. In the case at hand, the actual cost of
producing one more pound beyond 4 lb is f.5/ ! f.4/ D 5:5 ! 4:6 D 0: 0.
If c is the total cost of producing q units of a product, then the average cost per
unit, c, is
c
cD
q
For example, if the total cost of 20 units is 100, then the average cost per unit is
c D 100=20 D 5. y multiplying both sides of Equation (4) by q, we have
c D qc
That is, total cost is the product of the number of units produced and the average cost
per unit.

E AM LE M C

If a manufacturer s average-cost equation is

5000
c D 0:0001q2 ! 0:02q C 5 C
q
find the marginal-cost function. What is the marginal cost when 50 units are produced
S

S The marginal-cost function is the derivative of the total-cost function c.

Thus, we first find c by multiplying c by q. We have

c D qc
! "
2 5000
D q 0:0001q ! 0:02q C 5 C
q
c D 0:0001q3 ! 0:02q2 C 5q C 5000
ifferentiating c, we have the marginal-cost function:
dc
D 0:0001.3q2 / ! 0:02.2q/ C 5.1/ C 0
dq
D 0:0003q2 ! 0:04q C 5
The marginal cost when 50 units are produced is
ˇ
dc ˇˇ
D 0:0003.50/2 ! 0:04.50/ C 5 D 3:75
dq ˇ qD50

If c is in dollars and production is increased by one unit, from q D 50 to q D 51,

then the cost of the additional unit is approximately 3.75. If production is increased
by 13 unit, from q D 50, then the cost of the additional output is approximately
%1&
3
.3:75/ D 1:25.
Now ork Problem 21 G
Suppose r D f.q/ is the total revenue function for a manufacturer. The equation
r D f.q/ states that the total dollar value received for selling q units of a product is r.
The marginal revenue is defined as the rate of change of the total dollar value received
with respect to the total number of units sold. Hence, marginal revenue is merely the
derivative of r with respect to q:

dr
marginal revenue D
dq
506 C erent at on

arginal revenue indicates the rate at which revenue changes with respect to units
sold. We interpret it as the appr ximate re enue recei ed fr m se ing ne additi na
unit f utput.

E AM LE M R

Suppose a manufacturer sells a product at 2 per unit. If q units are sold, the total
revenue is given by

r D 2q

The marginal-revenue function is

dr d
D .2q/ D 2
dq dq

which is a constant function. Thus, the marginal revenue is 2 regardless of the number
of units sold. This is what we would expect, because the manufacturer receives 2 for
each unit sold.
Now ork Problem 23 G

R R C
For the total-revenue function in Example , namely r D f.q/ D 2q, we have

dr
D2
dq

This means that revenue is changing at the rate of 2 per unit, regardless of the number
of units sold. Although this is valuable information, it may be more significant when
compared to r itself. For example, if q D 50, then r D 2.50/ D 100. Thus, the rate
of change of revenue is 2=100 D 0:02 f r. n the other hand, if q D 5000, then
r D 2.5000/ D 10; 000, so the rate of change of r is 2=10; 000 D 0:0002 f r.
Although r changes at the same rate at each level, compared to r itself, this rate is
relatively smaller when r D 10; 000 than when r D 100. y considering the ratio

dr=dq
r
we have a means of comparing the rate of change of r with r itself. This ratio is called
the re ati e rate f change of r. We have shown that the relative rate of change when
q D 50 is

dr=dq 2
D D 0:02
r 100
and when q D 5000, it is

dr=dq 2
Percentages can be confusing D D 0:0002
Remember that percent means per r 10; 000
hundred. Thus 100 D 100 100 D 1,
2
y multiplying relative rates by 100 , we obtain the so-called percentage rates f
2 D 100 D 0:02, and so on. change. The percentage rate of change when q D 50 is .0:02/.100 / D 2 when
q D 5000 it is .0:0002/.100 / D 0:02 . For example, if an additional unit beyond
50 is sold, then revenue increases by approximately 2 .
 Section .3 e er at e as a ate of C an e 507

In general, for any function f, we have the following definition:

The relative rate of change of f.x/ is

f 0 .x/
f.x/
The percentage rate of change of f.x/ is
f 0 .x/
" 100
f.x/

A L IT I E AM LE R R C
The volume enclosed by a
etermine the relative and percentage rates of change of
capsule-shaped container with a cylin-
drical height of 4 feet and radius r is y D f.x/ D 3x2 ! 5x C 25
given by
when x D 5.
4 3
.r/ D "r C 4"r2
3 S Here,
etermine the relative and percentage
f 0 .x/ D 6x ! 5
rates of change of volume with respect
to the radius when the radius is 2 feet. Since f 0 .5/ D 6.5/ ! 5 D 25 and f.5/ D 3.5/2 ! 5.5/ C 25 D 75, the relative rate of
change of y when x D 5 is
f 0 .5/ 25
D $ 0:333
f.5/ 75
ultiplying 0.333 by 100 gives the percentage rate of change: .0:333/.100/ D 33:3 .

Now ork Problem 35 G

R BLEMS
Suppose that the position function of an ob ect moving s D 2t2 ! 4tI Œ7; 7:5$I t D 7
along a straight line is s D f .t/ D 2t2 C 3t, where t is in 2
seconds and s is in meters. Find the average velocity #s=#t s D t C 4 Œ3; 3:03$ t D 3
over the interval Œ1; 1 C #t$, where #t is given in the following 3
table: s D 5t3 C 3t C 24 Œ1; 1:01$ t D 1
#t 1 0.5 0.2 0.1 0.01 0.001 s D !3t2 C 2t C 1I Œ1; 1:25$I t D 1
#s=#t s D t4 ! 2t3 C tI Œ2; 2:1$I t D 2
s D 3t4 ! t7=2 I Œ0; 14 $I t D 0
From your results, estimate the velocity when t D 1. Verify your
estimate by using differentiation. Income Education Sociologists studied the relation
p between income and number of years of education for members of
If y D f .x/ D 2x C 5, find the average rate of change of y
a particular urban group. They found that a person with x years
with respect to x over the interval Œ3; 3 C #x$, where #x is given
of education before see ing regular employment can expect to
in the following table:
receive an average yearly income of y dollars per year, where
#x 1 0.5 0.2 0.1 0.01 0.001 =4
y D 6x C 5 00 for 4 % x % 16
#y=#x
Find the rate of change of income with respect to number of years
From your result, estimate the rate of change of y with respect to x of education. Evaluate the function at x D 16.
when x D 3.
Find the rate of change of the volume of a ball, with respect
In each f Pr b ems a p siti n functi n is gi en here t is in to its radius r, when r D 1:5 m. The volume of a ball as a
sec nds and s is in meters function of its radius r is given by
a Find the p siti n at the gi en t a ue.
b Find the a erage e city er the gi en inter a . 4 3
D .r/ D "r
c Find the e city at the gi en t a ue. 3
508 C erent at on

S in emperature The approximate temperature of the Light and Power Plant The total-cost function for an
s in in terms of the temperature e of the environment is given by electric light and power plant is estimated by Nordin5 to be
c D 32:07 ! 0:7 q C 0:02142q2 ! 0:0001q3 20 % q % 0
D 32: C 0:27. e ! 20/
where q is the eight-hour total output (as a percentage of capacity)
where and e are in degrees Celsius.3 Find the rate of change of and c is the total fuel cost in dollars. Find the marginal-cost
with respect to e . function and evaluate it when q D 70.
Biology The volume of a spherical cell is given by rban Concentration Suppose the 100 largest cities in the
D 43 "r3 , where r is the radius. Find the rate of change of United States in 1 20 are ran ed according to area. From ot a,6
volume with respect to the radius when r D 6:3 & 10!4 cm. the following relation holds, approximately:
PR0: 3
D 5; 000; 000
In Pr b ems c st functi ns are gi en here c is the c st f
pr ducing q units f a pr duct In each case nd the where P is the population of the city having ran R. This relation
margina c st functi n hat is the margina c st at the gi en is called the a f urban c ncentrati n for 1 20. etermine P as
a ue s f q a function of R and find how fast the population is changing with
c D 500 C 10qI q D 100 respect to ran .
c D 7500 C 5qI q D 24 Depreciation Under the straight-line method of
2
depreciation, the value, , of a certain machine after t years have
c D 0:2q C 4q C 50 q D 10 elapsed is given by
c D 0:1q2 C 3q C 2I q D 3
D 120;000 ! 15;500t
c D q2 C 50q C 1000I q D 15; q D 16; q D 17
where 0 % t % 6. How fast is changing with respect to t when
c D 0:04q3 ! 0:5q2 C 4:4q C 7500I q D 5; q D 25; q D 1000 t D 2 t D 4 at any time
In Pr b ems c represents a erage c st per unit hich is a inter Moth A study of the winter moth was made in
functi n f the number q f units pr duced Find the Nova Scotia (adapted from Embree).7 The prepupae of the moth
margina c st functi n and the margina c st f r the indicated fall onto the ground from host trees. At a distance of x ft from the
a ues f q base of a host tree, the prepupal density (number of prepupae per
square foot of soil) was y, where
600
c D 0:02q C 3 C q D 40, q D 0 y D 5 :3 ! 1:5x ! 0:5x2 1%x%
q
2000 a At what rate is the prepupal density changing with respect to
cD5C q D 25, q D 250 distance from the base of the tree when x D 6
q
b For what value of x is the prepupal density decreasing at the
20;000 rate of 6 prepupae per square foot per foot
c D 0:00002q2 ! 0:01q C 6 C I q D 100; q D 500
q Cost Function For the cost function
7000
2
c D 0:002q ! 0:5q C 60 C I q D 15; q D 25 c D 0:4q2 C 4q C 5
q
find the rate of change of c with respect to q when q D 2. Also,
In Pr b ems r represents t ta re enue and is a functi n f
what is #c=#q over the interval 2, 3
the number q f units s d Find the margina re enue functi n
and the margina re enue f r the indicated i a ues f q In Pr b ems nd a the rate f change f y ith respect t
x and b the re ati e rate f change f y At the gi en a ue f x
r D 0: qI q D ; q D 300; q D 500 nd c the rate f change f y d the re ati e rate f change f y
r D q.25 ! 1
q/ q D 10, q D 20, q D 100 and e the percentage rate f change f y
20
y D f .x/ D x C 4I x D 5 y D f .x/ D ! 5x x D 3
r D 240q C 40q2 ! 2q3 q D 10 q D 15 q D 20
y D 2x2 C 5 x D 10 y D 5 ! 3x3 I x D 1
r D 2q.30 ! 0:1q/I q D 10; q D 20 3
y D ! x Ix D 1 y D x2 C 3x ! 4I x D !1
osiery Mill The total-cost function for a hosiery mill is Cost Function For the cost function
estimated by ean4 to be
c D 0:4q2 C 3:2q C 11
2
c D !10;4 4:6 C 6:750q ! 0:00032 q
how fast does c change with respect to q when q D 20 etermine
the percentage rate of change of c with respect to q when q D 20.
where q is output in dozens of pairs and c is total cost in dollars.
Find the marginal-cost function and the average cost function and
evaluate each when q D 2000.
5
. A. Nordin, Note on a ight Plant s Cost Curves, Ec n metrica 15
(1 47), 231 35.
3 6 A.
R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or : . ot a, E ements f athematica Bi gy (New or : over
c raw-Hill oo Company, 1 55). Publications, Inc., 1 56).
4 7
. ean, Statistical Cost Functions of a Hosiery ill, Studies in Business . . Embree, The Population ynamics of the Winter oth in Nova Scotia,
Administrati n, I, no. 4 (Chicago: University of Chicago Press, 1 41). 1 54 1 62, em irs f the Ent m gica S ciety f Canada no. 46 (1 65).
 Section .4 e Pro ct e an t e ot ent e 509

Organic Matters Species Diversity In a discussion of and

contemporary waters of shallows seas, dum claims that in such I1:3
waters the total organic matter, y (in milligrams per liter), is a R2 D for 00 % I % 3500
1101:2
function of species diversity, x (in number of species per thousand
individuals). If y D 100=x, at what rate is the total organic matter a For each group, determine the relative rate of change of
changing with respect to species diversity when x D 10 What is response with respect to intensity.
the percentage rate of change when x D 10 b How do these changes compare with each other
c In general, if f1 .x/ D C1 f .x/ and f2 .x/ D C2 f .x/, where C1 and
Revenue For a certain manufacturer, the revenue obtained C2 are constants, how do the relative rates of change of f1 and f2
from the sale of q units of a product is given by compare
Cost A manufacturer of mountain bi es has found that
r D 30q ! 0:3q2 when 20 bi es are produced per day, the average cost is 200 and
the marginal cost is 150. ased on that information, approximate
a How fast does r change with respect to q When q D 10, the total cost of producing 21 bi es per day.
b find the relative rate of change of r, and c to the nearest
Marginal and Average Costs Suppose that the cost
percent, find the percentage rate of change of r.
function for a certain product is c D f .q/. If the relative
Revenue Repeat Problem 41 for the revenue function 1
given by r D 10q ! 0:2q2 and q D 25. rate of change of c (with respect to q) is , prove that the
q
eight of Limb The weight of a limb of a tree is given by marginal-cost function and the average-cost function are equal.
D 2t0:432 , where t is time. Find the relative rate of change of In Pr b ems and use the numerica deri ati e feature f
with respect to t. y ur graphing ca cu at r
Response to Shoc A psychological experiment was If the total-cost function for a manufacturer is given by
conducted to analyze human responses to electrical shoc s
5q2
(stimuli). The sub ects received shoc s of various intensities. The cD p C 5000
response, R, to a shoc of intensity, I (in microamperes), was to be q2 C 3
a number that indicated the perceived magnitude relative to that of where c is in dollars, find the marginal cost when 10 units are
a standard shoc . The standard shoc was assigned a magnitude produced. Round your answer to the nearest cent.
of 10. Two groups of sub ects were tested under slightly different
conditions. The responses R1 and R2 of the first and second groups The population of a city t years from now is given by
to a shoc of intensity I were given by P D 250;000e0:04t
I1:3 Find the rate of change of population with respect to time t three
R1 D for 00 % I % 3500 years from now. Round your answer to the nearest integer.
1 55:24

Objective T R R
o n er at es a n t e The equation F.x/ D .x2 C 3x/.4x C 5/ expresses F.x/ as a product of two functions:
ro ct an ot ent r es an to
e e o t e conce ts of ar na x2 C 3x and 4x C 5. To find F0 .x/ by using only our previous rules, we first multiply
ro ens t to cons e an the functions. Then we differentiate the result, term by term:
ar na ro ens t to sa e
F.x/ D .x2 C 3x/.4x C 5/ D 4x3 C 17x2 C 15x
F0 .x/ D 12x2 C 34x C 15

However, in many problems that involve differentiating a product of functions, the

multiplication is not as simple as it is here. At times, it is not even practical to attempt
it. Fortunately, there is a rule for differentiating a product, and the rule avoids such
multiplications. Since the derivative of a sum of functions is the sum of their derivatives,
one might expect a similar rule for products. There is a rule however, the situation for
products is more subtle than that for sums.

H. T. dum, iological Circuits and the arine Systems of Texas, in P uti n and arine Bi gy
eds T. A. lsen and F. . urgess (New or : Interscience Publishers, 1 67).
H. ab off, agnitude Estimation of Short Electrocutaneous Pulses, Psych gica Research 3 , no. 1
(1 76), 3 4 .
510 C erent at on

C MBINING RULE T R
If f and g are differentiable functions, then the product fg is differentiable, and
Symbolically: . fg/0 D f 0 g C fg0
d d d
. f.x/g.x// D . f.x//g.x/ C f.x/ .g.x//
dx dx dx
That is, the derivative of the product of two functions is the derivative of the first
function times the second, plus the first function times the derivative of the second.
! " ! "
derivative derivative
derivative of product D .second/ C .first/
of first of second

Pr f et F.x/ D f.x/g.x/. We want to show that F0 .x/ D f 0 .x/g.x/ C f.x/g0 .x/. y

the definition of the derivative of F,
F.x C h/ ! F.x/
F0 .x/ D lim
h!0 h
f.x C h/g.x C h/ ! f.x/g.x/
D lim
h!0 h
Now we use a tric . Adding and subtracting f.x/g.x C h/ in the numerator, we have
f.x C h/g.x C h/ ! f.x/g.x/ C f.x/g.x C h/ ! f.x/g.x C h/
F0 .x/ D lim
h!0 h
Regrouping gives
. f.x C h/g.x C h/ ! f.x/g.x C h// C . f.x/g.x C h/ ! f.x/g.x//
F0 .x/ D lim
h!0 h
. f.x C h/ ! f.x//g.x C h/ C f.x/.g.x C h/ ! g.x//
D lim
h!0 h
. f.x C h/ ! f.x//g.x C h/ f.x/.g.x C h/ ! g.x//
D lim C lim
h!0 h h!0 h
f.x C h/ ! f.x/ g.x C h/ ! g.x/
D lim " lim g.x C h/ C lim f.x/ " lim
h!0 h h!0 h!0 h!0 h
Since we assumed that f and g are differentiable,
f.x C h/ ! f.x/
lim D f 0 .x/
h!0 h
and
g.x C h/ ! g.x/
lim D g0 .x/
h!0 h
The differentiability of g implies that g is continuous, so, from Section 10.3,
lim g.x C h/ D g.x/
h!0

Thus,
F0 .x/ D f 0 .x/g.x/ C f.x/g0 .x/

E AM LE A R

If F.x/ D .x2 C 3x/.4x C 5/, find F0 .x/.

S We will consider F as a product of two functions:
F.x/ D .x2 C 3x/.4x C 5/
„ ƒ‚ … „ ƒ‚ …
f.x/ g.x/
 Section .4 e Pro ct e an t e ot ent e 511

Therefore, we can apply the product rule:

It is worth repeating that the derivative of
the product of two functions is somewhat F0 .x/ D f 0 .x/g.x/ C f.x/g0 .x/
subtle. o not be tempted to ma e up a
d d
simpler rule. D .x2 C 3x/ .4x C 5/ C .x2 C 3x/ .4x C 5/
dx ƒ‚ … „ ƒ‚ … „ ƒ‚ … „dx ƒ‚ …
„
Second First
erivative erivative
of first of second
D .2x C 3/.4x C 5/ C .x2 C 3x/.4/
D 12x2 C 34x C 15 simplifying
This agrees with our previous result. See Equation (1). Although there doesn t seem to
be much advantage to using the product rule here, there are times when it is impractical
to avoid it.
Now ork Problem 1 G
E AM LE A R
A L IT I
A taco stand usually sells 225 tacos If y D .x2=3 C 3/.x!1=3 C 5x/, find dy=dx.
per day at 2 each. A business stu-
dent s research tells him that for every S Applying the product rule gives
0.15 decrease in the price, the stand dy d d
will sell 20 more tacos per day. The D .x2=3 C 3/.x!1=3 C 5x/ C .x2=3 C 3/ .x!1=3 C 5x/
dx dx dx
revenue function for the taco stand is ! " ! "
R.x/ D .2 ! 0:15x/.225 C 20x/, where 2 !1=3 !1 !4=3
x is the number of 0.15 reductions in
D x .x!1=3 C 5x/ C .x2=3 C 3/ x C5
3 3
dR
price. Find . 25 2=3 1 !2=3
dx D x C x ! x!4=3 C 15
3 3
Alternatively, we could have found the derivative without the product rule by first find-
ing the product .x2=3 C 3/.x!1=3 C 5x/ and then differentiating the result, term by term.
Now ork Problem 15 G
E AM LE T F

If y D .x C 2/.x C 3/.x C 4/, find y0 .

S

S We would li e to use the product rule, but as given it applies only to t

factors. y treating the first two factors as a single factor, we can consider y to be a
product of two functions:
y D Œ.x C 2/.x C 3/$.x C 4/

The product rule gives

d d
y0 D Œ.x C 2/.x C 3/$.x C 4/ C Œ.x C 2/.x C 3/$ .x C 4/
dx dx
d
D Œ.x C 2/.x C 3/$.x C 4/ C Œ.x C 2/.x C 3/$.1/
dx
Applying the product rule again, we have
! "
0 d d
y D .x C 2/.x C 3/ C .x C 2/ .x C 3/ .x C 4/ C .x C 2/.x C 3/
dx dx
D Œ.1/.x C 3/ C .x C 2/.1/$.x C 4/ C .x C 2/.x C 3/
512 C erent at on

After simplifying, we obtain

y0 D 3x2 C 1 x C 26

Two other ways of finding the derivative are as follows:

ultiply the first two factors of y to obtain

y D .x2 C 5x C 6/.x C 4/

and then apply the product rule.

ultiply all three factors to obtain

y D x3 C x2 C 26x C 24

and then differentiate term by term.

Now ork Problem 19 G
It is sometimes helpful to remember differentiation rules in more streamlined nota-
tion. For example, as noted earlier, in the margin,

. fg/0 D f 0 g C fg0

is a correct equality of functions that expresses the product rule. We can then calculate

. fgh/0 D .. fg/h/0
D . fg/0 h C . fg/h0
D . f 0 g C fg0 /h C . fg/h0
D f 0 gh C fg0 h C fgh0

It is not suggested that you try to commit to memory derived rules li e

. fgh/0 D f 0 gh C fg0 h C fgh0

ecause f 0 g C fg0 D gf 0 C fg0 , using commutativity of the product of functions, we can

express the product rule with the derivatives as second factors:

. fg/0 D gf 0 C fg0

and using commutativity of addition

. fg/0 D fg0 C gf 0

Some people prefer these forms.

A L IT I
E AM LE U R F S
ne hour after x milligrams of a Find the slope of the graph of f.x/ D .7x3 ! 5x C 2/.2x4 C 7/ when x D 1.
particular drug are given to a person,
the change in body temperature .x/, S
in degrees Fahrenheit, is given approx-
% &
imately by .x/ D x2 1 ! 3x . The S We find the slope by evaluating the derivative when x D 1. ecause f
rate at which changes with respect to is a product of two functions, we can find the derivative by using the product rule.
the size of the dosage x, 0 .x/, is called
the sensiti ity of the body to the dosage.
Find the sensitivity when the dosage is 1 We have
milligram. o not use the product rule. d d
f 0 .x/ D .7x3 ! 5x C 2/.2x4 C 7/ C .2x4 C 7/ .7x3 ! 5x C 2/
dx dx
3 3 4 2
D .7x ! 5x C 2/. x / C .2x C 7/.21x ! 5/
 Section .4 e Pro ct e an t e ot ent e 513

Since we must compute f 0 .x/ when x D 1, there is n need t simp ify f 0 .x/ bef re
e a uating it. Substituting into f 0 .x/, we obtain

f 0 .1/ D 4. / C .16/ D 176

Now ork Problem 49 G

Usually, we do not use the product rule when simpler ways are obvious. For exam-
ple, if f.x/ D 2x.x C 3/, then it is quic er to write f.x/ D 2x2 C 6x, from which
f 0 .x/ D 4x C 6. Similarly, we do not usually use the product rule to differentiate
The product rule (and quotient rule that y D 4.x2 ! 3/. Since the 4 is a constant factor, by the constant-factor rule we have
follows) should not be applied when a y0 D 4.2x/ D x.
more direct and e cient method is The next rule is used for differentiating a qu tient of functions.
available.

C MBINING RULE T R
If f and g are differentiable functions and g.x/ ¤ 0, then the quotient f=g is also
differentiable, and
! "
d f.x/ g.x/f 0 .x/ ! f.x/g0 .x/
D
dx g.x/ .g.x//2
With the understanding about the denominator not being zero, we can write
! "0
f gf 0 ! fg0
D
g g2
That is, the derivative of the quotient of two functions is the denominator times
the derivative of the numerator, minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
0 1 0 1
B
derivative C B
derivative C
.denominator/B
@ of C !.numerator/B
A @ of C
A
numerator denominator
derivative of quotient D
(denominator)2

f gf 0 ! fg0
Pr f et F D . We need to show that F0 exists and is given by F0 D but
g g2
here we will ust establish the equation for F0 . Now

Fg D f

ifferentiating both sides of the equation, the product rule gives

Fg0 C gF0 D f 0

Solving for F0 , we have

f 0 ! Fg0
F0 D
g
ut F D f=g. Thus,
fg0
f0 !
g
F0 D
The derivative of the quotient of two g
functions is tric ier still than the product
rule. We must remember where the minus Simplifying, we have
sign goes
gf 0 ! fg0
F0 D
g2
514 C erent at on

E AM LE A R

4x2 C 3
If F.x/ D , find F0 .x/.
2x ! 1
S

S We recognize F as a quotient, so we can apply the quotient rule.

et f.x/ D 4x2 C 3 and g.x/ D 2x ! 1. Then

g.x/f 0 .x/ ! f.x/g0 .x/
F0 .x/ D
.g.x//2
erivative erivative of
of numerator numerator
enominator ‚ …„ ƒ Numerator‚ …„ ƒ ‚ …„ ƒ
‚ …„ ƒ d d
.2x ! 1/ .4x2 C 3/ ! .4x2 C 3/ .2x ! 1/
D dx dx
.2x ! 1/2
„ ƒ‚ …
Square of
denominator
.2x ! 1/. x/ ! .4x2 C 3/.2/
D
.2x ! 1/2
x2 ! x ! 6 2.2x C 1/.2x ! 3/
D 2
D
.2x ! 1/ .2x ! 1/2

Now ork Problem 21 G

E AM LE R
1
ifferentiate y D .
1
xC
xC1
S

S To simplify the differentiation, we will rewrite the function so that no

fraction appears in the denominator.

We have
1 1 xC1
yD D D 2
1 x.x C 1/ C 1 x CxC1
xC
xC1 xC1
2
dy .x C x C 1/.1/ ! .x C 1/.2x C 1/
D quotient rule
dx .x2 C x C 1/2
.x2 C x C 1/ ! .2x2 C 3x C 1/
D
.x2 C x C 1/2
!x2 ! 2x x2 C 2x
D D !
.x2 C x C 1/2 .x2 C x C 1/2

Now ork Problem 45 G

 Section .4 e Pro ct e an t e ot ent e 515

Although a function may have the form of a quotient, this does not necessarily
mean that the quotient rule must be used to find the derivative. The next example illus-
trates some typical situations in which, although the quotient rule can be used, a simpler
and more e cient method is available.

E AM LE U
R

ifferentiate the following functions.

2x3
a f.x/ D
5
S Rewriting, we have f.x/ D 25 x3 . y the constant-factor rule,
2 2 6x2
f 0 .x/ D .3x / D
5 5
4
b f.x/ D
7x3
S Rewriting, we have f.x/ D 47 .x!3 /. Thus,
4 12
f 0 .x/ D .!3x!4 / D ! 4
7 7x
5x2 ! 3x
c f.x/ D
1 4x
To differentiate f .x/ D , we might ! "
x2 ! 2 1 5x2 ! 3x 1
be tempted first to rewrite the quotient as S Rewriting, we have f.x/ D D .5x!3/ for x ¤ 0. Thus,
4 x 4
.x2 ! 2/!1 . Currently, it is not helpful to
do this because we do not yet have a rule 1 5
for differentiating the result. We have no f 0 .x/ D .5/ D for x ¤ 0
4 4
choice now but to use the quotient rule.
However, in the next section we will Since the function f is not defined for x D 0, f 0 is not defined for x D 0 either.
develop a rule that allows us to
differentiate .x2 ! 2/!1 in a direct and
e cient way.
Now ork Problem 17 G

E AM LE M R

If the demand equation for a manufacturer s product is

1000
pD
qC5
where p is in dollars, find the marginal-revenue function and evaluate it when q D 45.
S

S First we must find the revenue function. The revenue, r, received for
selling q units when the price per unit is p is given by
revenue D .price/.quantity/I that is, r D pq
Using the demand equation, we will express r in terms of q only. Then we will
differentiate to find the marginal-revenue function, dr=dq.

The revenue function is

! "
1000 1000q
rD qD
qC5 qC5
516 C erent at on

Thus, the marginal-revenue function is given by

d d
.q C 5/ .1000q/ ! .1000q/ .q C 5/
dr dq dq
D
dq .q C 5/2
.q C 5/.1000/ ! .1000q/.1/ 5000
D 2
D
.q C 5/ .q C 5/2
and
ˇ
dr ˇˇ 5000 5000
D D D2
dq ˇqD45 .45 C 5/2 2500

This means that selling one additional unit beyond 45 results in approximately 2 more
in revenue.
Now ork Problem 59 G

C F
A function that plays an important role in economic analysis, typically, of a country,
is the consumption function. The consumption function C D f.I/ expresses total
national consumption, C, as a function of total national income, I. Usually, both I and
C are expressed in billions of dollars and I is restricted to some interval. In well-run
economies, we should see C.I/ % I, for all I, but, of course, this is not always observed.
If C.I/ > I then the country is in a de cit situation. In any event, C D f.I/ is often
called the pr pensity t c nsume The marginal propensity to consume is defined as
the rate of change of consumption with respect to income. It is the derivative of C with
respect to I:
dC
arginal propensity to consume D
dI

If we assume that the difference between income, I, and consumption, C, is savings, S,

possibly negative, then

SDI!C

ifferentiating both sides with respect to I gives

dS d d dC
D .I/ ! .C/ D 1 !
dI dI dI dI
We define dS=dI as the marginal propensity to save. Thus, the marginal propensity to
save indicates how fast savings change with respect to income, and

arginal propensity arginal propensity

D1!
to save to consume

E AM LE F M C S

If the consumption function is given by

p
5.2 I3 C 3/
CD
I C 10
determine the marginal propensity to consume and the marginal propensity to save
when I D 100.
 Section .4 e Pro ct e an t e ot ent e 517

S
0 d 3=2 p d 1
.I C 10/ .2I C 3/ ! .2 I3 C 3/ .I C 10/
dC B dI dI C
D 5@ A
dI .I C 10/2
p !
.I C 10/.3I1=2 / ! .2 I3 C 3/.1/
D5
.I C 10/2

When I D 100, the marginal propensity to consume is

ˇ ! "
dC ˇˇ 12 7
D 5 $ 0:536
dI ˇID100 12; 100

The marginal propensity to save when I D 100 is 1 ! 0:536 D 0:464. This means
that if a current income of 100 billion increases by 1 billion, the nation consumes
approximately 53.6 .536=1000/ and saves 46.4 .464=1000/ of that increase.

Now ork Problem 69 G

R BLEMS
In Pr b ems di erentiate the functi ns x3 C x2 C 1 x2 ! 4x C 3
f .x/ D yD
f .x/ D .4x C 1/.6x C 3/ f .x/ D .3x ! 1/.7x C 2/ x2 C 1 2x2 ! 3x C 2
3 2
s.t/ D .5 ! 3t/.t ! 2t / .x/ D .x2 C 3x/.7x2 ! 5/ z4 C 4 1
F.z/ D g.x/ D
7 5
f .r/ D .2r ! 3r /.5r ! 2r C 7/ 2
3z x100 C7
C.I/ D .2I2 ! 3/.3I2 ! 4I C 1/ ! 3
C1
yD u. / D
f .x/ D x2 .2x2 ! 5/ 3 2
f .x/ D 3x .x ! 2x C 2/ 7x6
y D .x2 C 5x ! 7/.6x2 ! 5x C 4/ x!5 3x2 ! x ! 1
yD p yD p
%.x/ D .2 C 3x ! 5x2 /.7 C 11x ! 13x2 / x 3
x
2 3
f. / D . C 3 ! 7/.2 ! 4/ x0:3 ! 2 5 2x
2 2 yD yD1! C
f .x/ D .3x ! x /.3 ! x ! x / 2x2:1 C 1 2x C 5 3x C 1
y D .x2 ! 1/.3x3 ! 6x C 5/ ! 4.4x2 C 2x C 1/ 2x C 3 11
q.x/ D x3 C !
h.x/ D 5.x7 C 4/ C 4.5x3 ! 2/.4x2 C 7x/ 5x C 7 x3
p
F.p/ D 57 .2 p ! 3/.11p C 2/ x!5 . x ! 1/.3x C 2/
p p p yD yD
g.x/ D . x C 5x ! 2/. 3 x ! 3 x/ .x C 2/.x ! 4/ 4 ! 5x
yD7" 2 y D .x ! 1/.x ! 2/.x ! 3/ t2 C 3t 17
3
s.t/ D f .s/ D
.t2 ! 1/.t3 C 7/ s.4s3 C 5s ! 23/
y D .5x C 3/.2x ! 5/.7x C /
3x ! 5 5x 3 5 5
yD f .x/ D ! 1!
7x C 11 x!1 x x!1 x2 C 2
y D 2x ! y D 3 ! 12x3 C
x!2 x2 C 5
!5x !13
.x/ D f .x/ D aCx
5!x 3x5 f .x/ D
a!x
3.5x2 ! 7/ ax C b
f .x/ D yD x!1 C a!1
4 cx C d f .x/ D , where a is a constant
x!1 ! a!1
3 2
C5 !1 6 ! 2z
h. / D h.z/ D Find the slope of the curve y D .2x2 ! x C 3/.x3 C x C 1/ at
!3 z2 ! 4
.1; 12/.
2x2 C 5x ! 2 4x2 C 3x C 2 1
zD yD Find the slope of the curve y D 2 at .!1; 12 /.
3x2 C 5x C 3 3x2 ! 2x C 1 x C1
518 C erent at on

In Pr b ems nd an equati n f the tangent ine t the Consumption Function Suppose that a country s
cur e at the gi en p int consumption function is given by
6 xC5 p p
yD (3, 3) yD I .1; 6/ I C 0: I3 ! 0:3I
x!1 x2 CD p
y D .2x C 3/Œ2.x4 ! 5x2 C 4/$ (0, 24) I
x!1 1 where C and I are expressed in billions of dollars.
yD .2; 10 /
x.x2 C 1/ a Find the marginal propensity to save when income is
In Pr b ems and determine the re ati e rate f change f y 25 billion.
ith respect t x f r the gi en a ue f x b etermine the relative rate of change of C with respect to I
x 1!x when income is 25 billion.
yD xD1 yD Ix D 5
xC1 1Cx Marginal Propensities to Consume and to Save Suppose
Motion The position function for an ob ect moving in a that the savings function of a country is
straight line is p
IC I!6
2 SD p
sD 3 IC3
t C1
where the national income .I/ and the national savings .S/ are
where t is in seconds and s is in meters. Find the position and measured in billions of dollars. Find the country s marginal
velocity of the ob ect at t D 1. propensity to consume and its marginal propensity to save when
Motion The position function for an ob ect moving in a the national income is 121 billion. ( int It is helpful to first
straight-line path is factor the numerator of S.)
tC3 Marginal Cost If the total-cost function for a manufacturer
sD 2
t C7 is given by
where t is in seconds and s is in meters. Find the positive value(s) 6q2
of t for which the velocity of the ob ect is 0. cD C 6000
qC2
In Pr b ems each equati n represents a demand functi n
f r a certain pr duct here p den tes the price per unit f r q find the marginal-cost function.
units Find the margina re enue functi n in each case Reca that Marginal and Average Costs iven the cost function
re enue D pq d
c D f .q/, show that if .c/ D 0, then the marginal-cost function
p D 0 ! 0:02q p D 300=q dq
and average-cost function are equal.
10 q C 750
pD !3 pD ost Parasite Relation For a particular host parasite
qC2 q C 50
relationship, it is determined that when the host density (number
Consumption Function For the United States of hosts per unit of area) is x, the number of hosts that are
(1 22 1 42), the consumption function is estimated by10 parasitized is y, where
C D 0:672I C 113:1 00x
yD
Find the marginal propensity to consume. 10 C 45x

Consumption Function Repeat Problem 63 for At what rate is the number of hosts parasitized changing with
C D 0: 36I C 127:2. respect to host density when x D 2
In Pr b ems each equati n represents a c nsumpti n Acoustics The persistence of sound in a room after the
functi n Find the margina pr pensity t c nsume and the source of the sound is turned off is called re erberati n. The
margina pr pensity t sa e f r the gi en a ue f I re erberati n time RT of the room is the time it ta es for the
p p intensity level of the sound to fall 60 decibels. In the acoustical
C D 2 C 3 I C 5 3 I for 40 % I % 70 I D 64 design of an auditorium, the following formula may be used to
p
3I I compute the RT of the room:11
CD6C ! I I D 25
4 3 0:05
p p RT D
16 I C 0: I3 ! 0:2I ACx
CD p I I D 36 Here, is the room volume, A is the total room absorption, and
IC4
p p x is the air absorption coe cient. Assuming that A and x are
20 I C 0:5 I3 ! 0:4I positive constants, show that the rate of change of RT with respect
CD p I I D 100
IC5 to is always positive. If the total room volume increases by one
unit, does the reverberation time increase or decrease

10
T. Haavelmo, ethods of easuring the arginal Propensity to
Consume, urna f the American Statistica Ass ciati n, II (1 47),
105 22.
11 . . oelle, En ir nmenta Ac ustics (New or : c raw-Hill oo
Company, 1 72).
 Section .5 e C an e 519

Predator Prey In a predator-prey experiment,12 it was Verify this. ( int For convenience, let 2 C n D c.) Next, observe
statistically determined that the number of prey consumed, y, that Feldstein s function f is of the form
by an individual predator was a function of the prey density, x A C Bx
(the number of prey per unit of area), where g.x/ D ; where A, B, C, and are constants
CC x
0:7355x
yD Show that g0 .x/ is a constant divided by a nonnegative function
1 C 0:02744x
of x. What does this mean
etermine the rate of change of prey consumed with respect to
prey density. Business The manufacturer of a product has found that
when 20 units are produced per day, the average cost is 150 and
Social Security Benefits In a discussion of social security the marginal cost is 125. What is the relative rate of change of
benefits, Feldstein13 differentiates a function of the form average cost with respect to quantity when q D 20
a.1 C x/ ! b.2 C n/x Use the result . fgh/0 D f 0 gh C fg0 h C fgh0 to find dy=dx if
f .x/ D
a.2 C n/.1 C x/ ! b.2 C n/x y D .3x C 1/.2x ! 1/.x ! 4/
where a, b, and n are constants. He determines that
!1.1 C n/ab
f 0 .x/ D
.a.1 C x/ ! bx/2 .2 C n/

Objective T C R
o ntro ce an a t e c an r e ur next rule, the chain rule, is ultimately the most important rule for finding deriva-
to er e a s ec a case of t e c a n
r e an to e e o t e conce t of t e tives. It involves a situation in which y is a function of the variable u, but u is a function
ar na re en e ro ct as an of x, and we want to find the derivative of y with respect to x. For example, the equations
a cat on of t e c a n r e
y D u2 and u D 2x C 1

define y as a function of u and u as a function of x. If we substitute 2x C 1 for u in the

first equation, we can consider y to be a function of x:

y D .2x C 1/2

To find dy=dx, we first expand .2x C 1/2 :

y D 4x2 C 4x C 1

Then
dy
D xC4
dx
From this example, we can see that finding dy=dx by first performing a substitution
c u d be quite involved. For instance, if originally we had been given y D u100 instead
of y D u2 , we wouldn t even want to try substituting. Fortunately, the chain rule will
allow us to handle such situations with ease.

C MBINING RULE T C R
If y is a differentiable function of u and u is a differentiable function of x, then y is
a differentiable function of x and
dy dy du
D "
dx du dx

12 C. S. Holling,Some Characteristics of Simple Types of Predation and Parasitism, he Canadian Ent m gist
CI, no. 7 (1 5 ), 3 5 .
13 . Feldstein, The ptimal evel of Social Security enefits, he uarter y urna f Ec n mics, C, no. 2
(1 5), 303 20.
520 C erent at on

We can show why the chain rule is reasonable by considering rates of change.
Suppose

y D u C 5 and u D 2x ! 3

et x change by one unit. How does u change To answer this question, we differentiate
and find du=dx D 2. ut for each one-unit change in u, there is a change in y of
dy=du D . Therefore, what is the change in y if x changes by one unit That is, what
dy du dy dy du
is dy=dx The answer is " 2, which is " . Thus, D " .
du dx dx du dx
We will now use the chain rule to redo the problem at the beginning of this sec-
tion. If

y D u2 and u D 2x C 1

then
dy dy du d 2 d
D " D .u / " .2x C 1/
dx du dx du dx
D .2u/2 D 4u

Replacing u by 2x C 1 gives

dy
D 4.2x C 1/ D x C 4
dx
which agrees with our previous result.

A L IT I E AM LE U C R
If an ob ect moves horizontally
according to x D 6t, where t is in a If y D 2u2 ! 3u ! 2 and u D x2 C 4, find dy=dx.
seconds, and vertically according to S y the chain rule,
dy
y D 4x2 , find its vertical velocity .
dt dy dy du d d
D " D .2u2 ! 3u ! 2/ " .x2 C 4/
dx du dx du dx
D .4u ! 3/.2x/

We can write our answer in terms of x alone by replacing u by x2 C 4.

dy
D .4.x2 C 4/ ! 3/.2x/ D .4x2 C 13/.2x/ D x3 C 26x
dx
p
b If y D and D 7 ! t3 , find dy=dt.
S Here, y is a function of and is a function of t, so we can view y as a
function of t. y the chain rule,

dy dy d d p d
D " D . / " .7 ! t3 /
dt d dt d dt
! "
1 !1=2 1
D .!3t2 / D p .!3t2 /
2 2
3t2 3t2
D! p D! p
2 2 7 ! t3

Now ork Problem 1 G

 Section .5 e C an e 521

E AM LE U C R

If y D 4u3 C 10u2 ! 3u ! 7 and u D 4=.3x ! 5/, find dy=dx when x D 1.

S y the chain rule,
! "
dy dy du d d 4
D " D .4u3 C 10u2 ! 3u ! 7/ "
dx du dx du dx 3x ! 5
d d
.3x ! 5/ .4/ ! 4 .3x ! 5/
D .12u2 C 20u ! 3/ " dx dx
.3x ! 5/2
!12
D .12u2 C 20u ! 3/ "
.3x ! 5/2

When x is replaced by a, u D u.x/ must Even though dy=dx is in terms of x s and u s, we can evaluate it when x D 1 if we
be replaced by u.a/. determine the corresponding value of u. When x D 1,

4
u D u.1/ D D !2
3.1/ ! 5

Thus,
ˇ
dy ˇˇ !12
D Œ12.!2/2 C 20.!2/ ! 3$ "
dx ˇxD1 Œ3.1/ ! 5$2
D 5 " .!3/ D !15

Now ork Problem 5 G

The chain rules states that if y D f.u/ and u D g.x/, then

dy dy du
D "
dx du dx
Actually, the chain rule applies to a composite function, because

y D f.u/ D f.g.x// D . f ı g/.x/

Thus y, as a function of x, is f ı g. This means that we can use the chain rule to differen-
tiate a function when we recognize the function as a composition. However, we must
first brea down the function into composite parts.
For example, to differentiate

y D .x3 ! x2 C 6/100

we thin of the function as a composition. et

y D f.u/ D u100 and u D g.x/ D x3 ! x2 C 6

Then y D .x3 ! x2 C 6/100 D .g.x//100 D f.g.x//. Now that we have a composite, we

differentiate. Since y D u100 and u D x3 ! x2 C 6, by the chain rule we have

dy dy du
D "
dx du dx
D .100u /.3x2 ! 2x/
D 100.x3 ! x2 C 6/ .3x2 ! 2x/
522 C erent at on

We have ust used the chain rule to differentiate y D .x3 ! x2 C 6/100 , which is a
power of a functi n of x, not simply a power of x. The following rule, called the power
rule, generalizes our result and is a special case of the chain rule:

d a du
The Power Rule: .u / D aua!1
dx dx

where it is understood that u is a differentiable function of x and a is a real number.

Pr f et y D ua . Since y is a differentiable function of u and u is a differentiable
function of x, the chain rule gives
dy dy du
D "
dx du dx
ut dy=du D aua!1 . Thus,
dy du
D aua!1
dx dx
which is the power rule.

E AM LE U R

If y D .x3 ! 1/7 , find y0 .

S Since y is a power of a functi n of x, the power rule applies. etting
u.x/ D x3 ! 1 and a D 7, we have
y0 D aŒu.x/$a!1 u0 .x/
d
D 7.x3 ! 1/7!1 .x3 ! 1/
dx
D 7.x ! 1/ .3x2 / D 21x2 .x3 ! 1/6
3 6

Now ork Problem 9 G

E AM LE U R
p
If y D 3 .4x2 C 3x ! 2/2 , find dy=dx when x D !2.
S Since y D .4x2 C 3x ! 2/2=3 , we use the power rule with
u D 4x2 C 3x ! 2
and a D 23 . We have
dy 2 d
D .4x2 C 3x ! 2/.2=3/!1 .4x2 C 3x ! 2/
dx 3 dx
2 2
D .4x C 3x ! 2/!1=3 . x C 3/
3
2. x C 3/
D p3
3 4x2 C 3x ! 2
Thus,
ˇ
dy ˇˇ 2.!13/ 13
ˇ D p D!
dx xD!2 3 3
3

Now ork Problem 19 G

 Section .5 e C an e 523

E AM LE U R
1 dy
If y D , find .
The technique used in Example 5 is x2 ! 2 dx
frequently used when the numerator
of a quotient is a constant and the S Although the quotient rule can be used here, a more e cient approach is to
denominator is not. treat the right side as the power .x2 ! 2/!1 and use the power rule. et u D x2 ! 2.
Then y D u!1 , and
dy d
D .!1/.x2 ! 2/!1!1 .x2 ! 2/
dx dx
D .!1/.x2 ! 2/!2 .2x/
2x
D!
.x2 ! 2/2

Now ork Problem 27 G

E AM LE
! "
2s C 5 4 dz
If z D , find .
s2 C 1 ds
The problem here is to recognize the S Since z is a power of a function, we first use the power rule:
form of the function to be differentiated.
! " ! "
In this case it is a power, not a quotient. dz 2s C 5 4!1 d 2s C 5
D4 2
ds s C1 ds s2 C 1

Now we use the quotient rule:

! " ! "
dz 2s C 5 3 .s2 C 1/.2/ ! .2s C 5/.2s/
D4 2
ds s C1 .s2 C 1/2

Simplifying, we have
! "
dz .2s C 5/3 !2s2 ! 10s C 2
D4" 2
ds .s C 1/3 .s2 C 1/2
.s2 C 5s ! 1/.2s C 5/3
D!
.s2 C 1/5

Now ork Problem 41 G

E AM LE

If y D .x2 ! 4/5 .3x C 5/4 , find y0 .

S Since y is a product, we first apply the product rule:
d d
y0 D .x2 ! 4/5 ..3x C 5/4 / C .3x C 5/4 ..x2 ! 4/5 /
dx dx
Now we use the power rule:

y0 D .x2 ! 4/5 .4.3x C 5/3 .3// C .3x C 5/4 .5.x2 ! 4/4 .2x//
D 12.x2 ! 4/5 .3x C 5/3 C 10x.3x C 5/4 .x2 ! 4/4
524 C erent at on

In differentiating a product in which at To simplify, we first remove common factors:

least one factor is a power, simplifying
the derivative usually involves factoring. y0 D 2.x2 ! 4/4 .3x C 5/3 Œ6.x2 ! 4/ C 5x.3x C 5/$
D 2.x2 ! 4/4 .3x C 5/3 .21x2 C 25x ! 24/

Now ork Problem 39 G

Usually, the power rule should be used to differentiate y D Œu.x/$n . Although a
function such as y D .x2 C2/2 can be written y D x4 C4x2 C4 and differentiated easily,
this method is impractical for a function such as y D .x2 C2/1000 . Since y D .x2 C2/1000
is of the form y D Œu.x/$n , we have

y0 D 1000.x2 C 2/ .2x/

M R
et us now use our nowledge of calculus to develop a concept relevant to economic
studies. Suppose a manufacturer hires m employees who produce a total of q units of
a product per day. We can thin of q as a function of m. If r is the total revenue the
manufacturer receives for selling these units, then r can also be considered a function
of m. Thus, we can loo at dr=dm, the rate of change of revenue with respect to the
number of employees. The derivative dr=dm is called the marginal revenue product.
It approximates the change in revenue that results when a manufacturer hires an extra
employee.

E AM LE M R

A manufacturer determines that m employees will produce a total of q units of a product

per day, where

10m2
qD p
m2 C 1

If the demand equation for the product is p D 00=.q C /, determine the marginal-
revenue product when m D .
S We must find dr=dm, where r is revenue. Note that, by the chain rule,
dr dr dq
D "
dm dq dm

Thus, we must find both dr=dq and dq=dm when m D . We begin with dr=dq. The
revenue function is given by
! "
00 00q
r D pq D qD
qC qC

so, by the quotient rule,

dr .q C /. 00/ ! 00q.1/ 100
D D
dq .q C /2 .q C /2

In order top
evaluate this expression when m D , we first use the given equation
2
q D 10m = m2 C 1 to find the corresponding value of q:

10. /2
qD p D 1
2C1
 Section .5 e C an e 525

Hence,
ˇ ˇ
dr ˇˇ dr ˇˇ 100
D D D1
dq ˇmD dq ˇqD 1 . 1 C /2

Now we turn to dq=dm. From the quotient and power rules, we have
! "
dq d 10m2
D p
dm dm m2 C 1
d d
.m2 C 1 /1=2 .10m2 / ! .10m2 / Œ.m2 C 1 /1=2 $
D dm dm
Œ.m2 C 1 /1=2 $2
.m2 C 1 /1=2 .20m/ ! .10m2 /Π12 .m2 C 1 /!1=2 .2m/$
D
m2 C 1
so
ˇ
dq ˇˇ . 1 C 1 /1=2 .20 " / ! .10 " 1/Œ 12 . 1 C 1 /!1=2 .2 " /$
D
dm ˇmD 1C1
D 10:71

Therefore, from the chain rule,

A direct formula for the
ˇ
marginal-revenue product is
! "
dr ˇˇ
D .1/.10:71/ D 10:71
dr
D
dq
pCq
dp dm ˇmD
dm dm dq
This means that if a tenth employee is hired, revenue will increase by approximately
10.71 per day.
Now ork Problem 80 G

R BLEMS
In Pr b ems use the chain ru e y D 2.x2 C 5x ! 2/!5=7 y D 3.5x ! 2x3 /!5=3
3 2
If y D u C 3u and u D x C 1, find dy=dx. 2 p p
y D 5x2 ! x y D 3x2 ! 7
If y D 2u3 ! u and u D 7x ! x3 , find dy=dx. p p3
y D 3 5x C 7 yD x2 ! 1
1 p p
If y D and D 3x ! 5, find dy=dx. y D 4 7 .x2 C 1/3 y D 7 3 .x5 ! 3/5
p
If y D 4 z and z D x5 ! x4 C 3, find dy=dx. 6 2
yD yD
t!1 2x2
!xC1 x3 C5
If D u3 and u D , find d =dt when t D 1.
tC1 1 1
p yD 2 yD
If z D u3 C u C 5 and u D 2s2 C 1, find dz=ds when s D 2. .x ! 3x/2 .3 C 5x/3
2
If y D 3 ! C 4 and D 2x2 C 1, find dy=dx when x D 0. 4 3
yD p yD
If y D 2u3 C 3u2 C 5u ! 1 and u D 3x C 1, find dy=dx when x2 C 1 .3x2 ! x/2=3
x D 1. p p p 1
y D 5 5x C 5 5x yD 2x C p
In Pr b ems 0
nd y 2x
y D .3x C 2/ 6
y D .x2 ! 4/4 y D x3 .2x C 3/7 y D x.x C 4/4
p p
y D .2 C 3x5 /7 y D .x2 C x/4 y D 4x2 5x C 1 y D 2x3 1 ! x5
.2x2 C 1/4 y D .x2 C 2x ! 1/3 .5x/ y D x4 .x4 ! 1/5
y D 5.x3 ! 3x2 C 2x/100 yD
2 y D . x ! 1/3 .2x C 1/4 y D .3x C 2/5 .4x ! 5/2
2 !3 3 2 !10
y D .x ! 2/ y D .3x ! 2x /
526 C erent at on

! "11 ! " Marginal Revenue Product If p D k=q, where k is a

ax C b 2x 4
yD yD constant, is the demand equation for a manufacturer s product and
cx C d xC2
s q D f .m/ defines a function that gives the total number of units
r produced per day by m employees, show that the marginal-
xC1 3 x2 ! 3
yD yD revenue product is always zero.
x!5 x2 C 2
Cost Function The cost c of producing q units of a product
2x ! 5 .2x C 3/5 is given by
yD 2 yD
.x C 4/3 2x4 C
c D 5000 C 10q C 0:1q2
. x ! 1/5 p
yD y D 3 .x ! 3/3 .x C 5/ If the price per unit p is given by the equation
.3x ! 1/3
p q D 1000 ! 2p
y D 6.5x2 C 2/ x4 C 5 y D 6 C 3x ! 4x.7x C 1/2
! " use the chain rule to find the rate of change of cost with respect to
tC1 2t C 3 7
y D 2t C ! price per unit when p D 100.
tC3 5
ospital Discharges A governmental health agency
.2x3 C 6/.7x ! 5/
yD examined the records of a group of individuals who were
.2x C 4/2 hospitalized with a particular illness. It was found that the total
In Pr b ems and use the qu tient ru e and p er ru e t proportion that had been discharged at the end of t days of
nd y0 n t simp ify y ur ans er hospitalization was given by
p ! "3
.3x C 2/3 .x C 1/4 x C 2.4x2 ! 1/2 250
yD y D f .t/ D 1 !
.x2 ! 7/3 x!3 250 C t
If y D .5u C 6/3 and u D .x2 C 1/4 , find dy=dx when x D 0. Find f 0 .100/ and interpret your answer.
If z D 3y3 C 2y2 C y, y D 2x2 C x, and x D t C 1, find dz=dt Marginal Cost If the total-cost function for a manufacturer
when t D 1. is given by
Find the slope of the curve y D .x2 ! 7x ! /3 at the point 4q2
( , 0). cD p C 6000
p q2 C 2
Find the slope of the curve y D x C 2 at the point .7; 3/.
find the marginal-cost function.
In Pr b ems nd an equati n f the tangent ine t the Salary Education For a certain population, if E is the
cur e at the gi en p int number of years of a person s education and S represents average
p
y D 3 .x2 ! /2 I .3; 1/ y D .x C 3/3 I .!1; / annual salary in dollars, then for E # 7,
p
5x C 5 !3 S D 340E2 ! 4360E C 42; 00
yD .4; 1/ yD I .0; !3/
xC1 .3x C 1/3
2
a How fast is salary changing with respect to education when
In Pr b ems and determine the percentage rate f change E D 16
f y ith respect t x f r the gi en a ue f x b At what level of education does the rate of change of salary
1 equal 5000 per year of education
y D .x2 C 1/4 x D 1 yD 2 I xD2
.x ! 1/3 Biology The volume of a spherical cell is given by
In Pr b ems q is the t ta number f units pr duced per D 43 "r3 , where r is the radius. At time t seconds, the radius
day by m emp yees f a manufacturer and p is the price per (in centimeters) is given by
unit at hich the q units are s d In each case nd the r D 10! t2 C 10!7 t
margina re enue pr duct f r the gi en a ue f m
q D 5m, p D !0:4q C 50 m D 6 Use the chain rule to find d =dt when t D 10.

q D .100m ! m2 /=10, p D !0:1q C 50 m D 10 Pressure in Body issue Under certain conditions, the
p pressure, p, developed in body tissue by ultrasonic beams is given
q D 10m2 = m2 C , p D 525=.q C 3/ m D 4 as a function of the beam s intensity, I, via the equation14
p
q D 50m= m2 C 11, p D 100=.q C 10/ m D 5 p D .2& I/1=2
p
Demand Equation Suppose p D 100 ! q2 C 20 is a
demand equation for a manufacturer s product. where & (a ree letter read rho ) is density of the affected tissue
and is the velocity of propagation of the beam. Here & and are
a Find the rate of change of p with respect to q.
constants. a Find the rate of change of p with respect to I. b
b Find the relative rate of change of p with respect to q.
Find the relative rate of change of p with respect to I.
c Find the marginal-revenue function.

14
R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or :
c raw-Hill oo Company, 1 55).
 Chapter e e 527

Demography Suppose that, for a certain group of 20,000 Suppose y D f .x/, where x D g.t/. iven that g.2/ D 3,
births, the number of people surviving to age x years is g0 .2/ D 4, f .2/ D 5, f 0 .2/ D 6, g.3/ D 7, g0 .3/ D , f .3/ D ,
dy ˇ
x D !0:000354x4 C 0:00452x3 C 0: 4 x2 ! 34: x C 20;000 and ˇtD2 D 40 determine f 0 .3/.
dt
0 % x % 5:2
Business A manufacturer has determined that, for his
a Find the rate of change of x with respect to x, and evaluate product, the daily average cost (in hundreds of dollars) is given by
your answer for x D 65.
b Find the relative rate of change and the percentage rate of 324 5 1
cD p C C
change of x when x D 65. Round your answers to three decimal q2 C 35 q 1
places.
a As daily production increases, the average cost approaches a
Muscle Contraction A muscle has the ability to shorten constant dollar amount. What is this amount
when a load, such as a weight, is imposed on it. The equation b etermine the manufacturer s marginal cost when 17 units are
.P C a/. C b/ D k produced per day.
is called the fundamental equation of muscle contraction. 15 c The manufacturer determines that if production (and sales)
Here P is the load imposed on the muscle, is the velocity of were increased to 1 units per day, revenue would increase by
the shortening of the muscle fibers, and a, b, and k are positive 275. Should this move be made Why
constants. Express as a function of P. Use your result to find If
d =dP. p
y D .u C 2/ u C 3
Economics Suppose pq D 100 is the demand equation for
a manufacturer s product. et c be the total cost, and assume that and
the marginal cost is 0.01 when q D 200. Use the chain rule to find
dc=dp when q D 200. u D x.x2 C 3/3

Marginal Revenue Product A monopolist who employs find dy=dx when x D 0:1. Round your answer to two decimal
m wor ers finds that they produce places.
q D 2m.2m C 1/3=2 If
units of product per day. The total revenue, r (in dollars), is 2u C 3
yD
given by u3 ! 2
50q and
rD p
1000 C 3q xC4
a What is the price per unit (to the nearest cent) when there are uD
.2x C 3/3
12 wor ers
b etermine the marginal revenue when there are 12 wor ers. find dy=dx when x D !1. Round your answer to two decimal
c etermine the marginal-revenue product when m D 12. places.

Chapter 11 Review
I T S E
S he Derivative
secant line tangent line slope of a curve Ex. 1, p. 4 5
f.x C h/ ! f.x/ f.z/ ! f.x/
derivative lim lim Ex. 2, p. 4 6
h!0 h z!x z!x
0 0 d dy
difference quotient f .x/ y . f.x// Ex. 4, p. 4 7
dx dx
S Rules for Differentiation
power function constant factor rule sum or difference rule Ex. 5, p. 4 6
S he Derivative as a Rate of Change
position function #x velocity rate of change Ex. 1, p. 501
total-cost function marginal cost average cost per unit Ex. 7, p. 505
total-revenue function marginal revenue Ex. , p. 506
relative rate of change percentage rate of change Ex. , p. 507

15
Ibid.
528 C erent at on

S he Product Rule and the Quotient Rule

product rule quotient rule Ex. 5, p. 514
consumption function marginal propensity to consume and to save Ex. , p. 516
S he Chain Rule
chain rule power rule marginal-revenue product Ex. , p. 524

S
The tangent line (or tangent) to a curve at point P is the lim- In particular, if s D f.t/ is a position function, where s is
iting position of secant lines P as approaches P along the position at time t, then
curve. The slope of the tangent at P is called the slope of the
ds
curve at P. D velocity at time t
If y D f.x/, the derivative of f at x is f 0 .x/ defined by dt
In economics, the term margina is used to describe deriva-
f.x C h/ ! f.x/ tives of specific types of functions. If c D f.q/ is a total-cost
f 0 .x/ D lim
h!0 h function (c is the total cost of q units of a product), then the
rate of change
eometrically, the derivative gives the slope of the curve
y D f.x/ at the point .x; f.x//. At a particular point .a; f.a// dc
is called marginal cost
the slope of the tangent line is f 0 .a/, thus the point-slope form dq
of the tangent line at .a; f.a// is y ! f.a/ D f 0 .a/.x ! a/. Any
function that is differentiable at a point is also continuous at We interpret marginal cost as the approximate cost of
that point. one additional unit of output. (Average cost per unit, c, is
The rules for finding derivatives, discussed so far, are as related to total cost c by c D c=q equivalently, c D cq.)
follows, where all functions are assumed to be differentiable: A total-revenue function r D f.q/ gives a manufacturer s
revenue r for selling q units of product. (Revenue r and price
p are related by r D pq.) The rate of change
d
.c/ D 0; where c is any constant dr
dx is called marginal revenue
d a dq
.x / D axa!1 ; where a is any real number
dx which is interpreted as the approximate revenue obtained
d from selling one additional unit of output.
.cf.x// D cf 0 .x/; where c is a constant If r is the revenue that a manufacturer receives when the
dx
total output of m employees is sold, then the derivative
d
. f.x/ C g.x// D f 0 .x/ C g0 .x/ dr=dm D dr=dq"dq=dm is called the marginal-revenue prod-
dx uct and gives the approximate change in revenue that results
d when the manufacturer hires an extra employee.
. f.x/ ! g.x// D f 0 .x/ ! g0 .x/
dx If C D f.I/ is a consumption function, where I is national
d income and C is national consumption, then
. f.x/g.x// D f 0 .x/g.x/ C f.x/g0 .x/
dx dC
! " is marginal propensity to consume
d f.x/ g.x/f 0 .x/ ! f.x/g0 .x/ dI
D
dx g.x/ .g.x//2 and
dy dy du dC
D " ; where y is a function of u and u is a 1! is marginal propensity to save
dx du dx dI
function of x
d a du For any function, the relative rate of change of f.x/ is
.u / D aua!1 ; where u is a function of x and a
dx dx f 0 .x/
is any real number
f.x/
The derivative dy=dx can also be interpreted as giving which compares the rate of change of f.x/ with f.x/ itself.
the (instantaneous) rate of change of y with respect to x: The percentage rate of change is
dy #y change in y f 0 .x/
D lim D lim " 100
dx !x!0 #x !x!0 change in x f.x/
 Chapter e e 529

R
In Pr b ems use the de niti n f the deri ati e t nd f 0 .x/ Marginal Revenue If r D q.20 ! 0:1q/ is a total-revenue
f .x/ D 2 ! x 2 4
f .x/ D 7x C 5x C 3 2 function, find the marginal-revenue function.
p 2 Marginal Cost If
f .x/ D 3x f .x/ D
1 C 4x c D 0:0001q3 ! 0:02q2 C 3q C 6000
In Pr b ems di erentiate is a total-cost function, find the marginal cost when q D 100.
p
y D 74 y D ex y D ex3 C 3 3x2 C 7x2 C 5x C 2 Consumption Function If
p
2
y D 4.x C 5/ ! 7x C D 5 C 0:6I ! 0:4 I
p is a consumption function, find the marginal propensity to
f .s/ D s2 .s2 C 2/ yD xC3 consume and the marginal propensity to save when I D 25.
x2 C 1 1 q C 12
yD yD Demand Equation If p D is a demand equation,
5 xn qC5
find the rate of change of price, p, with respect to quantity, q.
y D .x3 C 7x2 /.x3 ! x2 C 5/
Demand Equation If p D !0:1q C 500 is a demand
y D .x2 C 1/100 .x ! 6/ equation, find the marginal-revenue function.
p 3
f .x/ D .2x2 C 4x/100 f. / D C 2
Average Cost If c D 0:03q C 1:2 C is an average-cost
q
.ax C b/2 5x2 ! x function, find the marginal cost when q D 100.
yD yD
.cx C d/2 2x Power Plant Cost Function The total-cost function of an
y D . C 2x/.x2 C 1/4 g.z/ D .2z/3=5 C 5 electric light and power plant is estimated by16
z2 ! 1 p c D 16:6 C 0:125q C 0:0043 q2 20 % q % 0
f .z/ D y D a2 ! x2
z2 C 4
p where q is the eight-hour total output (as a percentage of capacity)
yD 3
4x ! 1 f .x/ D .1 C 23 /12 and c is the total fuel cost in dollars. Find the marginal-cost
1 x.x C 1/ function and evaluate it when q D 70.
yD p yD 2
1 ! x2 2x C 3 Marginal Revenue Product A manufacturer has
determined that m employees will produce a total of q units of
y D .x C a/m .x C b/n .x C c/p product per day, where

.x C 3/5 q D m.50 ! m/
yD
x If the demand function is given by
5x ! 4 p 3
yD f .x/ D 5x 3 C 2x4 p D !0:05q C 10
xC6
x a find the marginal-revenue product when m D 2.
y D 2x!3= C .2x/!3= yD C
a x inter Moth In a study of the winter moth in Nova
x2 C 6 p Scotia,17 it was determined that the average number of eggs, y, in
yD p y D 3 .7 ! 3x2 /2
x2 C 5 a female moth was a function of the female s abdominal width, x
(in millimeters), where
y D .x3 C 6x2 C /3=5 z D 0:4x2 .x C 1/!3 C 0:5
.2z C 3/2 !3 y D f .x/ D 14x3 ! 17x2 ! 16x C 34
g.z/ D g.z/ D
.5z C 7/!3 4.z C 2z ! 5/4
5
and 1:5 % x % 3:5. At what rate does the number of eggs change
In Pr b ems nd an equati n f the tangent ine t the with respect to abdominal width when x D 2
cur e at the p int c rresp nding t the gi en a ue f x ost Parasite Relation For a particular host parasite
2
y D x ! 6x C 4, x D 1 3
y D !2x C 6x C 1, x D 2 relationship, it is found that when the host density (number of
hosts per unit of area) is x, the number of hosts that are
p x3 parasitized is
y D 3 x, x D yD 2 ,xD2
x !3 ! "
1
If f .x/ D 4x2 C 2x C , find the relative and percentage rates y D 12 1 ! x# 0
1 C 3x
of change of f.x/ when x D 1.
1
If f .x/ D x=.x C 4/, find the relative and percentage rates of For what value of x does dy=dx equal 3
change of f.x/ when x D 1.
16
. A. Nordin, Note on a ight Plant s Cost Curves, Ec n metrica
15 (1 47), 231 55.
17 . . Embree, The Population ynamics of the Winter oth in Nova
Scotia, 1 54 1 62, em irs f the Ent m gica S ciety f Canada no. 46
(1 65).
530 C erent at on

Bacteria Growth acteria are growing in a culture. The A manufacturer has found that when m employees are
time, t (in hours), for the number of bacteria to double in number wor ing, the number of units of product produced per day is
(the generation time) is a function of the temperature, (in p
degrees Celsius), of the culture and is given by q D 10 m2 C 4 00 ! 700
(1
C 11 if 30 % % 36 The demand equation for the product is
t D f . / D 24 4
4
175
3
! 4 if 36 < % 3 q C p2 ! 1 ;300 D 0
Find dt=d when (a) D 3 and (b) D 35. where p is the selling price when the demand for the product is
Motion The position function of a particle moving in a q units per day.
straight line is a etermine the manufacturer s marginal-revenue product when
m D 240.
b Find the relative rate of change of revenue with respect to the
sD number of employees when m D 240.
2t2 C 3
c Suppose it would cost the manufacturer 400 more per day to
where t is in seconds and s is in meters. Find the velocity of the hire an additional employee. Would you advise the manufacturer
particle at t D 1. to hire the 241st employee Why or why not
Rate of Change The volume of an in atable ball is given If f .x/ D ex , use the definition of the derivative ( limit of a
by D 43 "r3 , where r is the radius. Find the rate of change of difference quotient ) to estimate f 0 .0/ correct to three decimal
with respect to r when r D 10 cm. places.
p
Motion The position function for a ball thrown vertically If f .x/ D 3 x2 C 3x ! 4, use the numerical derivative feature
upward from the ground is of your graphing calculator to estimate the derivative when
x D 10. Round your answer to three decimal places.
s D 21 t ! 16t2
The total-cost function for a manufacturer is given by
where s is the height in feet above the ground after t seconds. For 5q2 C 4
what value(s) of t is the velocity 64 ft s cD p C 2500
q2 C 6
Find the marginal-cost function if the average-cost function is
where c is in dollars. Use the numerical derivative feature of your
10;000 graphing calculator to estimate the marginal cost when 15 units
c D 2q C
q2 are produced. Round your answer to the nearest cent.
Find an equation of the tangent line to the curve Show that asic Rule 1 is actually a consequence of
p Combining Rule 1 and the a D 0 case of asic Rule 2.
.x3 C 2/ x C 1
yD Show that asic Rule 2 f r p siti e integers is a consequence
x4 C 2x of Combining Rule 3 (the Product Rule) and the a D 1 case of
at the point on the curve where x D 1. asic Rule 2.
 t ona
fferent at on
o cs
A
fter an uncomfortable trip in a vehicle, passengers sometimes describe the
12.1 er at es of ride as er y although, if pressed for what they mean, they may be unable
o ar t c nct ons to define er or er iness . In fact, jerk admits a precise definition that
12.2 er at es of uses ideas introduced in this chapter.
onent a nct ons Travel in a straight line at a constant speed is called unif rm m ti n, and there is
nothing er y about it. ut if the speed changes, the ride may become er y. Change in
12.3 ast c t of e an speed over time is the derivative of speed, called acce erati n. Since speed is itself the
12.4 ct erent at on derivative of position with respect to time, acceleration is the derivative of the derivative
of position and also called, naturally enough, the sec nd deri ati e of position with
12.5 o ar t c respect to time.
erent at on However, nonzero acceleration is not necessarily er y. For example, when people
ump off high diving boards, their downward speed increases as they fall, so they are
12.6 e ton s et o
accelerating, but there is nothing er y about their fall until the moment of impact.
12.7 er r er The acceleration in that case is due to gravity, and it is well nown that gravitational
er at es acceleration due to the earth is essentially constant within the earth s atmosphere. To
say it more carefully, constantly accelerated motion is not er y.
C er 12 e e Recall that the derivative of a constant is zero. If acceleration is constant, the deriva-
tive of acceleration is zero. In fact, the derivative of acceleration, the third deri ati e
of position with respect to time, is called jerk, and when it is nonzero, motion feels, as
people say, er y. When riding in a very powerful car with the gas pedal pressed all
the way to the oor, the passengers will feel themselves pushed bac in their seats but
without any er iness until the driver releases and depresses the gas pedal, producing
changes in acceleration.
Second derivatives, third derivatives, and so on are all called higher-order deriva-
tives and constitute an important topic in this chapter. er , for example, has impli-
cations not only for passenger comfort in vehicles but also for equipment reliability.
Engineers designing equipment for spacecraft, for example, follow guidelines about
the er the equipment must be able to withstand without damage to its components.

531
532 C t ona erent at on o cs

Objective L F
o e eo a erent at on for a for So far, the only derivatives we have been able to calculate are those of functions that
y D ln u to a t e for a an to
se t to erent ate a o ar t c are constructed from power functions using multiplication by a constant, arithmetic
f nct on to a ase ot er t an e operations, and composition. (As pointed out in Review Problem 65 of Chapter 11, we
can calculate the derivative of a constant function, c, by writing c D cx0 then
d d ! 0" d ! 0"
.c/ D cx D c x D c ! 0x!1 D 0
dx dx dx
Thus, we really have only one basic differentiation formula so far. The logarithmic
functions logb x and the exponential functions bx cann t be constructed from power
functions using multiplication by a constant, arithmetic operations, and composition.
It follows that we will need at least another truly basic differentiation formula.
In this section, we develop formulas for differentiating logarithmic functions. We
begin with the derivative of ln x, commenting further on the numbered steps at the end
of the calculation.
d .1/ ln.x C h/ " ln x
.ln x/ D lim definition of derivative
dx h!0 h
# $
xCh
ln
.2/ x
D lim since ln m " ln n D ln.m=n/
h!0 h
# # $$
.3/ 1 h
D lim ln 1 C algebra
h!0 h x
# # $$
.4/ 1 x h 1 1 x
D lim ! ln 1 C writing D !
h!0 x h x h x h
# $ !
.5/ 1 h x=h
D lim ln 1 C since r ln m D ln mr
h!0 x x
# $ !
.6/ 1 h x=h
D ! lim ln 1 C by limit Property 1 in Section 10.1
x h!0 x
# $ !
.7/ 1 h x=h
D ! ln lim 1 C ln is continuous
x h!0 x
# $ !
. / 1 h x=h
D ! ln lim 1 C for fixed x > 0
x h=x!0 x
# $
. / 1
D ! ln lim .1 C k/1=k setting k D h=x
x k!0

.10/ 1
D ! ln.e/ as shown in Section 10.1
x
.11/1
D since ln e D 1
x
The calculation is long, but following it step by step allows for review of many impor-
tant ideas. Step (1) is the ey definition introduced in Section 11.1. Steps (2), (5), and
(11) involve properties found in 4.3. In step (3), labeled simply a gebra, we use proper-
ties of fractions first given in 0.2. Step (4) is admittedly a trick that requires experience
to discover. Note that, necessarily, x ¤ 0 since x is in the domain of ln, which is .0; 1/.
To understand the ustification for step (6), we must observe that x, and hence 1=x, is
constant with respect to the limit variable h. We have already remar ed in Section 10.3
 Section 2. er at es of o ar t c nct ons 533

that logarithmic functions are continuous and this is what allows us to interchange the
processes of applying the ln function and ta ing a limit in (7). In ( ) the point is that,
for fixed x > 0, h=x goes to 0 when h goes to 0 and, conversely, h goes to 0 when h=x
goes to 0. Thus, we can regard h=x as a new limit variable, k, and this we do in step ( ).
In conclusion, we have derived the following:

BASIC RULE ln x
d 1
.ln x/ D for x > 0
dx x

Some care is required with this rule because while the left-hand side is defined
only for x > 0, the right-hand side is defined for all x ¤ 0. For x < 0, ln."x/ is defined
and by the chain rule we have
d 1 d "1 1
.ln."x// D ."x/ D D for x < 0
dx "x dx "x x

We can combine the last two equations by using the absolute function to get

d 1
.ln jxj/ D for x ¤ 0
dx x

E AM LE F I ln x

a ifferentiate f.x/ D 5 ln x.
S Here f is a constant (5) times a function (ln x), so by asic Rule 3, we have

d 1 5
f 0 .x/ D 5 .ln x/ D 5 ! D for x > 0
dx x x
ln x
b ifferentiate y D .
x2
S y the quotient rule and asic Rule 3,

d d
x2 .ln x/ " .ln x/ .x2 /
y0 D dx dx
.x2 /2
# $
1
x2 " .ln x/.2x/
x x " 2x ln x 1 " 2 ln x
D 4
D 4
D for x > 0
x x x3

Now ork Problem 1 G

We will now extend Equation (1) to cover a broader class of functions. et y D ln juj,
The chain rule is used to develop the where u is a differentiable function of x. y the chain rule,
differentiation formula for ln juj.
d dy du d du 1 du
.ln juj/ D ! D .ln juj/ ! D ! for u ¤ 0
dx du dx du dx u dx
Thus,
d 1 du
.ln juj/ D ! for u ¤ 0
du u dx

d 1 du
f course, Equation (2) gives us .ln u/ D ! for u > 0.
du u dx
534 C t ona erent at on o cs

E AM LE F I ln u

a ifferentiate y D ln.x2 C 1/.

S This function has the form ln u with u D x2 C 1, and since x2 C 1 > 0, for
all x, y D ln.x2 C 1/ is defined for all x. Using Equation (2), we have
A L IT I
The supply of q units of a product dy 1 d 2 1 2x
D 2 .x C 1/ D 2 .2x/ D 2
at a price of p dollars per unit is given dx x C 1 dx x C1 x C1
by q.p/ D 25 C 2 ln.3p2 C 4/. Find the
rate of change of supply with respect to b ifferentiate y D x2 ln.4x C 2/.
dq
price, . S Using the product rule gives
dp

dy d d
D x2 .ln.4x C 2// C .ln.4x C 2// .x2 /
dx dx dx
y Equation (2) with u D 4x C 2,
# $
dy 1
D x2 .4/ C .ln.4x C 2//.2x/
dx 4x C 2
2x2
D C 2x ln.4x C 2/ for 4x C 2 > 0
2x C 1

Since 4x C 2 > 0 exactly when x > "1=2, we have

d 2 2x2
.x ln.4x C 2// D C 2x ln.4x C 2/ for x > "1=2
dx 2x C 1

c ifferentiate y D ln j ln jxjj.
S This has the form y D ln juj with u D ln jxj. Using Equation (2), we obtain
# $
0 1 d 1 1 1
y D .ln jxj/ D D for x; u ¤ 0
ln jxj dx ln jxj x x ln jxj

Since ln jxj D 0 when x D "1; 1, we have

d 1
.ln j ln jxjj/ D for x ¤ "1; 0; 1
dx x ln jxj

Now ork Problem 9 G

Frequently, we can reduce the wor involved in differentiating the logarithm of a
product, quotient, or power by using properties of logarithms to rewrite the logarithm
bef re differentiating. The next example will illustrate.

E AM LE R L F
dy
a Find if y D ln.2x C 5/3 .
dx
S Here we have the logarithm of a power. First, we simplify the right side by
using properties of logarithms. Then we differentiate. We have

y D ln.2x C 5/3 D 3 ln.2x C 5/ for 2x C 5 > 0

Comparing both methods, we note that # $
the easier one is to simplify first and then dy 1 6
differentiate. D3 .2/ D for x > "5=2
dx 2x C 5 2x C 5
 Section 2. er at es of o ar t c nct ons 535

Alternatively, if the simplification were not performed first, we would write

dy 1 d
D ..2x C 5/3 /
dx .2x C 5/3 dx
1 6
D 3
.3/.2x C 5/2 .2/ D
.2x C 5/ 2x C 5
b Find f 0 .p/ if f.p/ D ln..p C 1/2 .p C 2/3 .p C 3/4 /.
S We simplify the right side and then differentiate:
f.p/ D 2 ln.p C 1/ C 3 ln.p C 2/ C 4 ln.p C 3/
# $ # $ # $
0 1 1 1
f .p/ D 2 .1/ C 3 .1/ C 4 .1/
pC1 pC2 pC3
2 3 4
D C C
pC1 pC2 pC3
Now ork Problem 5 G
E AM LE F I L
r
1C 2
a Find f 0 . / if f. / D ln 2"1
.

S We simplify by using properties of logarithms and then differentiate:

1
f. / D .ln.1 C 2 / " ln. 2
" 1//
2
# $
0 1 1 1
f . /D .2 / " .2 /
2 1C 2 2"1

2
D 2
" 2"1
D" 4
1C "1
b Find f 0 .x/ if f.x/ D ln3 .2x C 5/.
S The exponent 3 refers to the cubing of ln.2x C 5/. That is,
o not confuse ln3 .2x C 5/ with
ln.2x C 5/3 , which occurred in f.x/ D ln3 .2x C 5/ D .ln.2x C 5//3
Example 3(a). It is advisable to write
ln3 .2x C 5/ explicitly as .ln.2x C 5//3 . y the power rule,
d
f 0 .x/ D 3.ln.2x C 5//2 .ln.2x C 5//
dx
# $
2 1
D 3.ln.2x C 5// .2/
2x C 5
6
D .ln.2x C 5//2
2x C 5
Now ork Problem 39 G

L F B b
To differentiate a logarithmic function to a base different from e, we can first convert the
logarithm to natural logarithms via the change-of-base formula and then differentiate
the resulting expression. For example, consider y D logb u, where u is a differentiable
function of x. y the change-of-base formula,
ln u
y D logb u D for u > 0
ln b
536 C t ona erent at on o cs

ifferentiating, we have
# $
d d ln u 1 d 1 1 du
Note that ln b is ust a constant .logb u/ D D .ln u/ D !
dx dx ln b ln b dx ln b u dx
Summarizing,
d 1 du
.logb u/ D ! for u > 0
dx .ln b/u dx

Rather than memorizing this rule, we suggest remembering the procedure used to
obtain it.

logb u
ln u
Convert logb u to natural logarithms to obtain , and then differentiate.
ln b

E AM LE L F B 2

ifferentiate y D log2 x.
S Following the foregoing procedure, we have
# $
d d ln x 1 d 1
.log2 x/ D D .ln x/ D
dx dx ln 2 ln 2 dx .ln 2/x
It is worth mentioning that we can write our answer in terms of the original base.
ecause
1 1 logb e
D D D logb e
ln b logb b 1
logb e
1 log2 e d logb e du
we can express as . ore generally, .logb u/ D ! .
.ln 2/x x dx u dx
Now ork Problem 15 G
A L IT I E AM LE L F B 10
The intensity of an earthqua e is If y D log.2x C 1/, find the rate of change of y with respect to x.
measured on the Richter scale. The
I
reading is given by R D log , where I S The rate of change is dy=dx, and the base involved is 10. Therefore, we have
I0 # $
dy d d ln.2x C 1/
is the intensity and I0 is a standard min- D .log.2x C 1// D
dR dx dx dx ln 10
imum intensity. If I0 D 1, find , the
dI 1 1 2
rate of change of the Richter-scale read- D ! .2/ D
ing with respect to the intensity.
ln 10 2x C 1 ln 10.2x C 1/
G
R BLEMS
In Pr b ems di erentiate the functi ns If p ssib e rst use f .t/ D t ln t " t y D x2 ln x
pr perties f garithms t simp ify the gi en functi n
y D x2 ln.ax C b/ y D .ax C b/3 ln.ax C b/
5 ln x
y D a ln x yD y D ln.ax C b/ y D log3 . x " 1/ f . / D log. 2
C 2 C 1/
a
y D ln.5x " 6/ y D ln x 2 2
y D x C log2 .x C 4/ 2 y D x logb x
3 2
y D ln.5x C 3x C 2x C 1/ y D ln.1 " x /2 ln z x2
f .z/ D yD
z ln x
y D ln.ax2 C bx C c/ f .X / D ln.4X 6 C 2X 3 /
x4 C 3x2 C x
yD y D ln x100
f .r/ D ln.2r4 " 3r2 C 2r C 1/ ln x
 Section 2.2 er at es of onent a nct ons 537
p
y D ln.ax2 C bx C c/d y D 6 ln 3
x Marginal Cost A manufacturer s average-cost function, in
# $ dollars, is given by
p t4
y D ln 1 C x2 f .t/ D ln 500
1 C 6t C t2 cN D
# $ # $ ln.q C 20/
1C ax C b
f . / D ln y D ln Find the marginal cost (rounded to two decimal places) when
1" cx C d
q D 50.
r s
4 1 C x2 3
3 x " 1 Supply Change The supply of q units of a product at a
y D ln y D ln
1 " x2 3
x C1 price of p dollars per unit is given by q.p/ D 27 C 11 ln.2p C 1/.
dq
y D lnŒ.ax2 C bx C c/p .hx2 C kx C /q ! Find the rate of change of supply with respect to price, .
dp
y D lnŒ.5x C 2/4 . x " 3/6 ! y D ln. f .x/g.x// Sound Perception The loudness of sound, , measured in
x decibels, perceived by the human ear depends upon intensity
y D 6 ln p y D .x2 C 1/ ln.2x C 1/ I
2x C 1 levels, I, according to D 10 log , where I0 is the standard
I0
y D .ax2 C bx C c/ ln.hx2 C kx C / d
threshold of audibility. If I0 D 17, find , the rate of change of
y D ln x3 C ln3 x y D xlog3 5 dI
the loudness with respect to the intensity.
y D ln4 .ax/ y D ln2 .2x C 11/
p ! p " Biology In a certain experiment with bacteria, it is
y D ln f .x/ y D ln x3 4 2x C 1 observed that the relative activeness of a given bacteria colony
p ! p " is described by
y D f .x/ C ln x y D ln x C 1 C x2 # $
Find an equation of the tangent line to the curve A D 6 ln "a
a"
y D ln.x2 " 3x " 3/ where a is a constant and is the surrounding temperature. Find
when x D 4. the rate of change of A with respect to .
Find an equation of the tangent line to the curve Show that the relative rate of change of y D f .x/ with respect
to x is equal to the derivative of y D ln f .x/.
y D x ln x " x d 1 du
Show that .logb u/ D .logb e/ .
at the point where x D 1. dx u dx
x
Find the slope of the curve y D when x D 3. In Pr b ems and use di erentiati n ru es t nd f 0 .x/ hen
ln x
use y ur graphing ca cu at r t nd a r ts f f 0 .x/ D 0 R und
Marginal Revenue Find the marginal-revenue function if
y ur ans ers t t decima p aces
the demand function is p D 50= ln.q C 3/.
ln.x2 /
Marginal Cost A total-cost function is given by f .x/ D x3 ln x f .x/ D
x2
c D 25 ln.q C 1/ C 12
Find the marginal cost when q D 6.

Objective E F
o e eo a erent at on for a for As we pointed out in Section 12.1, the exponential functions cannot be constructed
y D eu to a t e for a an to
se t to erent ate an e onent a from power functions using multiplication by a constant, arithmetic operations, and
f nct on t a ase ot er t an e composition. However, the functions bx , for b > 0 and b ¤ 1, are inverse to the
functions logb .x/, and if an invertible function f is differentiable, it is fairly easy to see
that its inverse is differentiable. The ey idea is that the graph of the inverse of a function
is obtained by re ecting the graph of the original function in the line y D x. This
re ection process preserves smoothness so that if the graph of an invertible function is
smooth, then so is the graph of its inverse. ifferentiating f. f !1 .x// D x, we have
d d
. f. f !1 .x/// D .x/
dx dx
d !1
f 0 . f !1 .x// . f .x// D 1 Chain Rule
dx
d !1 1
. f .x// D 0 !1
dx f . f .x//
538 C t ona erent at on o cs

Thus we have

C MBINING RULE I F R
If f is an invertible, differentiable function, then f !1 is differentiable and
d !1 1
. f .x// D 0 !1
dx f . f .x//

As with the chain rule, eibniz notation is well suited for inverse functions. Indeed,
dy d !1 dx
if y D f !1 .x/, then D . f .x// and since f.y/ D x, f 0 .y/ D . When we
dx dx dy
substitute these in Combining Rule 6, we get
dy d 1 1 1
D . f !1 .x// D 0 !1 D 0 D
dx dx f . f .x// f .y/ dx
dy
so that Combining Rule 6 can be rewritten as

dy 1
D
dx dx
dy
In the immediate case of interest, with y D ex so that x D ln y and dx=dy D 1=y D
x
1=e , we have
d x 1
.e / D D ex
dx 1
ex
The power rule does n t apply to ex and which we record as
other exponential functions, bx The
power rule applies to power functions, xa . d x
Note the location of the variable.
.e / D ex
dx
For u a differentiable function of x, an application of the Chain Rule gives
d u du
.e / D eu
dx dx

E AM LE F I ex

d
a Find .3ex /. Since 3 is a constant factor,
dx
d d
.3ex / D 3 .ex /
dx dx
D 3ex by Equation (2)
x dy
If a quotient can be easily rewritten b If y D , find .
as a product, then we can use the ex dx
somewhat simpler product rule rather S We could use first the quotient rule and then Equation (2), but it is a little
than the quotient rule. easier first to rewrite the function as y D xe!x and use the product rule and Equation (3):
dy d d 1"x
D .x/e!x C x .e!x / D .1/e!x C x.e!x /."1/ D e!x .1 " x/ D x
dx dx dx e

c If y D e2 C ex C ln 3, find y0 .
S Since e2 and ln 3 are constants, y0 D 0 C ex C 0 D ex .
Now ork Problem 1 G
 Section 2.2 er at es of onent a nct ons 539

A L IT I
E AM LE F I eu
When an ob ect is moved from
one environment to another, the change d ! x3 C3x "
in temperature of the ob ect is given a Find e .
dx
by D Cekt , where C is the tem-
perature difference between the two S The function has the form eu with u D x3 C 3x. From Equation (2),
environments, t is the time in the new d ! x3 C3x " 3 d 3
environment, and k is a constant. Find e D ex C3x .x3 C 3x/ D ex C3x .3x2 C 3/
the rate of change of temperature with dx dx
3 C3x
respect to time. D 3.x2 C 1/ex

d u du du d xC1
.e / D eu . on t forget the . b Find .e ln.x2 C 1//.
dx dx dx dx
S y the product rule,
d xC1 d d
.e ln.x2 C 1// D exC1 .ln.x2 C 1// C .ln.x2 C 1// .exC1 /
dx dx dx
# $
1
D exC1 2 .2x/ C .ln.x2 C 1//exC1 .1/
x C1
# $
xC1 2x 2
De C ln.x C 1/
x2 C 1
Now ork Problem 3 G

y E AM LE T N F

An important function used in the social sciences is the normal distribution density
function
1 2
x y D f.x/ D p e!.1=2/..x!!/="/
m " 2#
FIGURE The normal-distribution where " (a ree letter read sigma ) and $ (a ree letter read mu ) are constants.
density function. The graph of this function, called the normal curve, is bell shaped. (See Figure 12.1.)
etermine the rate of change of y with respect to x when x D $ C ".
S The rate of change of y with respect to x is dy=dx. We note that the factor
1
p is a constant and the second factor has the form eu , where
" 2#
1 % x " $ &2
uD"
2 "
Thus,
# % x " $ & # 1 $$
dy 1 ! !.1=2/..x!!/=" /2 " 1
D p e " .2/
dx " 2# 2 " "
Evaluating dy=dx when x D $ C ", we obtain
ˇ # $# $
dy ˇˇ 1 ! !.1=2/..!C"!!/=" /2 " $C" "$ 1
D p e "
dx ˇxD!C" " 2# " "
# $
1 ! !.1=2/ " 1
D p e "
" 2# "
"e!.1=2/ "1
D p D p
2
" 2# 2
" 2#e
G
540 C t ona erent at on o cs

E F B b
Now that we are familiar with the derivative of eu , we consider the derivative of the
more general exponential function bu . ecause b D eln b , we can express bu as an
exponential function with the base e, a form we can differentiate. We have
d u d d
.b / D ..eln b /u / D .e.ln b/u /
dx dx dx
d
D e.ln b/u ..ln b/u/
dx
# $
.ln b/u du
De .ln b/
dx
du
D bu .ln b/ since e.ln b/u D bu
dx
Summarizing,
d u du
.b / D bu .ln b/
dx dx

Note that if b D e, then the factor ln b in Equation (4) is 1. Thus, if exponential func-
tions to the base e are used, we have a simpler differentiation formula with which to
wor . This is the reason natural exponential functions are used extensively in calcu-
lus. Rather than memorizing Equation (4), we advocate remembering the procedure
for obtaining it.

bu
Convert bu to a natural exponential function by using the property that b D eln b ,
and then differentiate.

The next example will illustrate this procedure.

E AM LE E F B 4
d x
Find .4 /.
dx
S Using the preceding procedure, we have
d x d
.4 / D ..eln 4 /x /
dx dx
d ! .ln 4/x " d u
D e form W .e /
dx dx
D e.ln 4/x .ln 4/ by Equation (2)
x
D 4 .ln 4/
Verify the result by using Equation (4) Now ork Problem 15 G
directly.

E AM LE F
d! 2 p "
Find e C xe C 2 x .
dx
S Here we must differentiate three different forms do not confuse them The
first .e2 / is a constant base to a constant power, so it is a constant itself. Thus, its deriva-
tive is zero. The second .xe / is a variable base to a constant power, so the power rule
 Section 2.2 er at es of onent a nct ons 541
p
applies. The third .2 x / is a constant base to a variable power, so we must differentiate
an exponential function. Ta en all together, we have
d! 2 p " d ' .ln 2/px (
e C xe C 2 x D 0 C exe!1 C e
dx dx
# $
e!1
' .ln 2/px ( 1
D ex C e .ln 2/ p
2 x
p
x
e!1 2 ln 2
D ex C p
2 x

Now ork Problem 17 G

E AM LE F A

We have often used the rule d=dx.xa / D axa!1 , but we have only pr ed it for a a
positive integer and a few other special cases. At least for x > 0, we can now improve
our understanding of power functions, using Equation (2).
For x > 0, we can write xa D ea ln x . So we have
d a d d
.x / D ea ln x D ea ln x .a ln x/ D xa .ax!1 / D axa!1
dx dx dx
Now ork Problem 19 G

R BLEMS
In Pr b ems di erentiate the functi ns F r each f the demand equati ns in Pr b ems and nd
x aex the rate f change f price p ith respect t quantity q hat is
y D 5e yD the rate f change f r the indicated a ue f q
b
2
y D e2x C3
3 2
y D e3x C5x C7xC11 p D 15e!0:001q I q D 500 p D 5e!q=100 q D 100
y D e !5x f .q/ D e!q
3 C6q!1
In Pr b ems and cN is the a erage c st f pr ducing q units
f .r/ D e4r
3 C5r2 C2rC6
y D ex
2 C6x3 C1
f a pr duct Find the margina c st functi n and the margina
c st f r the gi en a ues f q
y D xe ex y D 3x4 e!x
!x2
7000eq=700
y D x2 e y D xeax cN D I q D 350; q D 700
ex C e!x ex C e!x q
yD yD x
3 e " e!x 50 e.2qC6/= 00

2x3 cN D C 4000 I q D 7; q D 1 7
yD5 y D 2x x2 q q
ea p
f. / D 2 y D ex! x 2 tC1 d
C C1 If D ex and x D , find when t D 2.
p t"1 dt
y D e1! x y D .e2x C 1/3 If f 0 .x/ D x3 and u D ex , show that
y D x5 " 5x f .z/ D e1=z
ex " 1 d
yD x y D eax .bx C c/ Œf .u/! D e4x
e C1 dx
y D ln ex y D e!x ln x
If c is a positive constant and
y D xx y D ln e4xC1 ˇ
2 d x ˇ
c ˇ
If f .x/ D ec ebx eax , find f 0 .1/. .c " x /ˇ D0
2 dx
If f .x/ D 5x ln x
, find f 0 .1/. xD1

Find an equation of the tangent line to the curve y D ex when prove that c D e.
x D "2. Calculate the relative rate of change of
Find an equation of the tangent line to the curve y D ex at the
point .1; e/. Show that this tangent line passes through .0; 0/ and f .x/ D 10!x C ln. C x/ C 0:01ex!2
show that it is the only tangent line to y D ex that passes through
.0; 0/. when x D 2. Round your answer to four decimal places.
542 C t ona erent at on o cs

Production Run For a firm, the daily output on the tth day sub ect was signaled by a light to recite the three-letter consonant
of a production run is given by syllable. The time between the experimenter s completion of the
last consonant to the onset of the light was called the reca
q D 500.1 " e!0:2t /
inter a . The time between the onset of the light and the
Find the rate of change of output q with respect to t on the completion of a response was referred to as atency. After many
tenth day. trials, it was determined that, for a recall interval of t seconds, the
approximate proportion of correct recalls with latency below 2. 3
ormal Density Function For the normal-density
seconds was
function
1 2 p D 0: Œ0:01 C 0: .0: 5/t !
f .x/ D p e!x =2
2# a Find dp=dt and interpret your result.
find f 0 ."1/. b Evaluate dp=dt when t D 2. Round your answer to two
decimal places.
Population The population, in millions, of the greater
Seattle area t years from 1 70 is estimated by P D 1: 2e0:0176t . Medicine Suppose a tracer, such as a colored dye, is
Show that dP=dt D kP, where k is a constant. This means that the in ected instantly into the heart at time t D 0 and mixes uniformly
rate of change of population at any time is proportional to the with blood inside the heart. et the initial concentration of the
population at that time. tracer in the heart be C0 , and assume that the heart has constant
volume . Also assume that, as fresh blood ows into the heart,
Mar et Penetration In a discussion of diffusion of a new
the diluted mixture of blood and tracer ows out at the constant
process into a mar et, Hurter and Rubenstein1 refer to an equation
positive rate r. Then the concentration of the tracer in the heart at
of the form
time t is given by
t
D k˛ ˇ
C.t/ D C0 e!.r= /t

where is the cumulative level of diffusion of the new process at

Show that dC=dt D ."r= /C.t/.
time t and k, ˛, and ˇ are positive constants. Verify their claim that
Medicine In Problem 4 , suppose the tracer is in ected at a
d t constant rate R. Then the concentration at time t is
D k˛ ˇ .ˇ t ln ˛/ ln ˇ
dt R' (
C.t/ D 1 " e!.r= /t
Finance After t years, the value S of a principal of P r
dollars invested at the annual rate of r compounded continuously a Find C(0).
is given by S D Pert . Show that the relative rate of change of S dC R r
with respect to t is r. b Show that D " C.t/.
dt
Predator Prey Relationship In an article concerning Schi ophrenia Several models have been used to analyze
predators and prey, Holling2 refers to an equation of the form the length of stay in a hospital. For a particular group of
schizophrenics, one such model is5
y D .1 " e!ax / f .t/ D 1 " e!0:00 t

where x is the prey density, y is the number of prey attac ed, and where f .t/ is the proportion of the group that was discharged at the
and a are constants. Verify his statement that end of t days of hospitalization. Find the rate of discharge (the
proportion discharged per day) at the end of 100 days. Round your
dy
D a. " y/ answer to four decimal places.
dx
Savings and Consumption A country s savings S (in
Earthqua es According to Richter,3 the number of billions of dollars) is related to its national income I (in billions
earthqua es of magnitude or greater per unit of time is given by of dollars) by the equation
D 10A 10!b , where A and b are constants. Find d =d . 3
Psychology Short-term retention was studied by Peterson S D ln
2 C e!I
and Peterson.4 The two researchers analyzed a procedure in which a Find the marginal propensity to consume as a function of
an experimenter verbally gave a sub ect a three-letter consonant income.
syllable, such as CH , followed by a three-digit number, such as b To the nearest million dollars, what is the national income
30 . The sub ect then repeated the number and counted bac ward 1
by 3 s, such as 30 , 306, 303, : : : : After a period of time, the when the marginal propensity to save is
7
In Pr b ems and use di erentiati n ru es t nd f 0 .x/ hen
1
A. P. Hurter, r., A. H. Rubenstein, et al., ar et Penetration by New use y ur graphing ca cu at r t nd a rea zer s f f 0 .x/ R und
Innovations: The Technological iterature, echn gica F recasting and y ur ans ers t t decima p aces
S cia Change 11 (1 7 ), 1 7 221. 3 2
f .x/ D e2x Cx !3x f .x/ D xe!x
2 C. S. Holling, Some Characteristics of Simple Types of Predation and

Parasitism, he Canadian Ent m gist, CI, no. 7 (1 5 ), 3 5 .

3 C. F. Richter, E ementary Seism gy (San Francisco: W. H. Freeman and
Company, Publishers, 1 5 ). 5
Adapted from W. W. Eaton and . A. Whitmore, ength of Stay as a
4
. R. Peterson and . . Peterson, Short-Term Retention of Individual Verbal Stochastic Process: A eneral Approach and Application to Hospitalization
Items, urna f Experimenta Psych gy 5 (1 5 ), 1 3 . for Schizophrenia, urna f athematica S ci gy 5 (1 77), 273 2.
 Section 2.3 ast c t of e an 543

Objective E
o e a at e at ca ana s s of t e efore embar ing on this topic, of considerable importance in Economics, it is advis-
econo c conce t of e ast c t
able to re ect on the convention of graphing equations relating prices, p, and quantities,
q, so that the vertical axis is the p-axis and the horizontal axis is the q-axis. This was
touched upon, very brie y in the sub-section emand and Supply Curves of Section
3.2. In particular, a demand equati n relating p and q, is typically (uniquely) solv-
able for p in terms of q and the result, p D f.q/, is called the demand functi n. Some
people find it hard to thin about quantity, q, as the independent variable upon which
price, p, depends. Why not solve the demand equation for q in terms of p getting, say,
q D g.p/, so that price determines quantity sold In fact, typically, this is equally possi-
ble. It should be noted that a demand cur e, be it p D f.q/ or q D g.p/, falls from left
to right as q increases, p decreases, and as p increases q decreases. In the terminol-
ogy of Section 2.4, the functions f and g are functions that have inverses and when they
come from the same demand equation they are inverses of each other, so that g D f !1
and f D g!1 . If we need demand derivatives (and we will), the results of the previous
section inform us that g0 .p/ D 1=f 0 .g.p/ equivalently,
dq 1
D
dp dp
dq
E asticity f demand is a means by which economists measure how a change in
the price of a product will affect the quantity demanded. That is, it measures consumer
response to price changes. ore precisely, it can be defined as the ratio of the resulting
percentage change in quantity demanded to a given percentage change in price:
percentage change in quantity
p elasticity of demand D
percentage change in price
For example, if, for a price increase of 5 , quantity demanded were to decrease by 2 ,
Demand function we would say that elasticity of demand is "2=5 the minus sign being used to signal
f (q ) that the 2 decrease is a "2 increase.
p = f (q ) Continuing, suppose p D f.q/ is the demand function for a product. Consumers
f (q + h)
will demand q units at a price of f.q/ per unit and will demand q C h units at a price
of f.q C h/ per unit (Figure 12.2). The percentage change in quantity demanded from
q to q C h is
q
q q+h .q C h/ " q h
! 100 D ! 100
q q
FIGURE Change in demand.
The corresponding percentage change in price per unit is
f.q C h/ " f.q/
! 100
f.q/
The ratio of these percentage changes is
h
! 100
q h f.q/
D !
f.q C h/ " f.q/ q f.q C h/ " f.q/
! 100
f.q/
f.q/ h
D !
q f.q C h/ " f.q/
f.q/
q
D
f.q C h/ " f.q/
h
544 C t ona erent at on o cs

If f is differentiable, then as h ! 0, the limit of . f.q C h/ " f.q//=h is f 0 .q/ D dp=dq.

Thus, the limit of (1) is
f.q/ p
q q
0
D since p D f.q/
f .q/ dp
dq
which is called the p int e asticity f demand.

If p D f.q/ is a differentiable demand function, the point elasticity of demand,

Since p is a function of q, dp=dq is denoted by the ree letter % (eta), at (q, p) is given by
a function of q, and, thus, the ratio that p
defines % is a function of q. That is why
we write % D %.q/. q
% D %.q/ D
dp
dq

To illustrate, let us find the point elasticity of demand for the demand function
p D 1200 " q2 . We have
p 1200 " q2
# $
q q 1200 " q2 600 1
%D D D" D" "
dp "2q 2q2 q2 2
dq
! "
For example, if q D 10, then % D " .600=102 / " 12 D "5 12 . Since
change in demand
%#
change in price
we have
. change in price/.%/ # change in demand
Thus, if price were increased by 1 when q D 10, then quantity demanded would
change by approximately
# $
1 1
.1 / "5 D "5
2 2
That is, demand would decrease 5 12 . Similarly, decreasing price by 1
2
when q D 10
results in a change in demand of approximately
# $# $
1 1 3
" "5 D2
2 2 4
Hence, demand increases by 2 34 .
Note that when elasticity is evaluated, no units are attached to it it is nothing
more than a real number. In fact, the 100 s arising from the word percentage cancel,
so that elasticity is really an approximation of the ratio
relative change in quantity
relative change in price
and each of the relative changes is no more than a real number. For usual behavior of
demand, an increase (decrease) in price corresponds to a decrease (increase) in quantity.
This means that if price is plotted as a function of quantity, then the graph will have a
negative slope at each point. Thus, dp=dq will typically be negative, and since p and
q are positive, % will typically be negative too. Some economists disregard the minus
sign in the preceding situation, they would consider the elasticity to be 5 12 . We will
not adopt this practice.
 Section 2.3 ast c t of e an 545

There are three categories of elasticity:

When j%j > 1, demand is elastic.
When j%j D 1, demand has unit elasticity.
When j%j < 1, demand is inelastic.
For example, in Equation (2), since j%j D 5 12 when q D 10, demand is elastic. If
ˇ ' (ˇ
q D 20, then j%j D ˇ" .600=202 / " 12 ˇ D 1 so demand has unit elasticity. If q D 25,
ˇ ˇ
then j%j D ˇ" 23
50
ˇ, and demand is inelastic.
If demand is inelastic, then for a given percentage change in price there is a greater
percentage change in quantity demanded. If demand is inelastic, then for a given per-
centage change in price there is a smaller percentage change in quantity demanded.
Unit elasticity means that for a given percentage change in price there is an equal per-
centage change in quantity demanded. To better understand elasticity, it is helpful to
thin of typical examples. emand for an essential utility such as electricity tends to be
inelastic through a wide range of prices. If electricity prices are increased by 10 , con-
sumers can be expected to reduce their consumption somewhat, but a full 10 decrease
may not be possible if most of their electricity usage is for essentials of life, such as
heating and food preparation. n the other hand, demand for luxury goods tends to
be elastic. A 10 increase in the price of ewelry, for example, may result in a 50
decrease in demand.

E AM LE F E

etermine the point elasticity of the demand equation

k
pD ; where k > 0 and q > 0
q
S From the definition, we have
p k
q q2
%D D D "1
dp "k
dq q2
Thus, the demand has unit elasticity for all q > 0. The graph of p D k=q is called an
equi atera hyperb a and is often found in economics texts in discussions of elasticity.
(See Figure 2.11 for a graph of such a curve.)
Now ork Problem 1 G
If we are given p D f.q/ for our demand equation, as in our discussion thus far,
then it is usually straightforward to calculate dp=dq D f 0 .q/. However, if instead we
are given q as a function of p, then we will have q D f !1 .p/ and, from Section 12.2,
dp 1
D
dq dq
dp
It follows that
p
q p dq
%D D !
dp q dp
dq

provides another useful expression for %. Notice too that if q D g.p/, then
p dq p g0 .p/
% D %.p/ D ! D ! g0 .p/ D p !
q dp g.p/ g.p/
546 C t ona erent at on o cs

and, thus,
elasticity D price ! relative rate of change of quantity as a function of price

E AM LE F E

etermine the point elasticity of the demand equation

q D p2 " 40p C 400 D .p " 20/2
S Here we have q given as a function of p, and it is easy to see that
dq=dp D 2p " 40. Thus,
p dq p
%.p/ D ! D .2p " 40/
q dp q.p/
For example, if p D 15, then q D q.15/ D 25 hence, %.15/ D .15."10//=25 D "6,
so demand is elastic for p D 15.
Now ork Problem 13 G
Here we analyze elasticity for linear Point elasticity for a inear demand equation is interesting. Suppose the equation
demand. has the form
p D mq C b; where m < 0 and b > 0
(See Figure 12.3.) We assume that q > 0 thus, p < b. The point elasticity of demand is
p
p p
q q p p
%D D D D
b h 7 1, elastic dp m mq p"b
h = 1, unit elasticity dq
b
2
h 6 1, inelastic y considering d%=dp, we will show that % is a decreasing function of p. y the quo-
tient rule,
p = mq + b
q d% .p " b/ " p b
D 2
D"
dp .p " b/ .p " b/2
FIGURE Elasticity for linear
demand. Since b > 0 and .p " b/2 > 0, it follows that d%=dp < 0, meaning that the graph
of % D %.p/ has a negative slope. This means that as price p increases, elasticity %
decreases. However, p ranges between 0 and b, and at the midpoint of this range, b=2,
b b
2 2
% D %.b/ D D D "1
b b
"b "
2 2
Therefore, if p < b=2, then % > "1 if p > b=2, then % < "1. ecause we typi-
cally have % < 0, we can state these facts another way: When p < b=2; j%j < 1, and
demand is inelastic when p D b=2; j%j D 1, and demand has unit elasticity when
p > b=2; j%j > 1 and demand is elastic. This shows that the slope of a demand curve
is not a measure of elasticity. The slope of the line in Figure 12.3 is m everywhere, but
elasticity varies with the point on the line. f course, this is in accord with Equation (4).

E R
Here we analyze the relationship between Turning to a different situation, we can relate how elasticity of demand affects changes
elasticity and the rate of change of in revenue (marginal revenue). If p D f.q/ is a manufacturer s demand function, the
revenue. total revenue is given by
r D pq
 Section 2.3 ast c t of e an 547

To find the marginal revenue, dr=dq, we differentiate r by using the product rule:
dr dp
DpCq :
dq dq
Factoring the right side of Equation (5), we have
# $
dr q dp
Dp 1C
dq p dq
ut
dp
q dp dq 1
D p D
p dq %
q
Thus,
# $
dr 1
Dp 1C
dq %
1
If demand is elastic, then % < "1, so 1 C > 0. If demand is inelastic, then % > "1,
%
1
so 1 C < 0. We can assume that p > 0. From Equation (6) we can conclude that
%
dr=dq > 0 on intervals for which demand is elastic. As we will soon see, a function is
increasing on intervals for which its derivative is positive, and a function is decreasing
on intervals for which its derivative is negative. Hence, total revenue r is increasing on
intervals for which demand is elastic, and total revenue is decreasing on intervals for
which demand is inelastic.
Thus, we conclude from the preceding argument that as more units are sold, a
manufacturer s total revenue increases if demand is elastic but decreases if demand is
inelastic. That is, if demand is elastic, a lower price will increase revenue. This means
that a lower price will cause a large enough increase in demand to actually increase
revenue. If demand is inelastic, a lower price will decrease revenue. For unit elasticity,
a lower price leaves total revenue unchanged.
If we solve the demand equation to obtain the form q D g.p/, rather than p D f.q/,
then a similar analysis gives
dr
D q.1 C %/
dp
and the conclusions of the last paragraph follow even more directly.

R BLEMS
In Pr b ems nd the p int e asticity f the demand q D p2 " 50p C 50 p D 20
equati ns f r the indicated a ues f q r p and determine
For the linear demand equation p D 15 " q, verify that
hether demand is e astic is ine astic r has unit e asticity
demand is elastic when p D 10, is inelastic when p D 5, and has
p D 40 " 2q q D 5 p D 10 " 0:04q q D 100 unit elasticity when p D 7:5.
3000 500 For what value (or values) of q do the following demand
pD q D 300 pD q D 52
q q2 equations have unit elasticity
a p D 36 " 0:25q
100 00 b p D 300 " q2
pD q D 100 pD q D 24
qC1 2q C 1
The demand equation for a product is
q=100 !q=50
p D 150 " e q D 100 p D 250e q D 50
q D 500 " 40p C p2
q D 1200 " 150p p D 4 p D 100 " q p D 50
p p where p is the price per unit (in dollars) and q is the quantity of
q D 500 " p p D 400 q D 2500 " p2 p D 20 units demanded (in thousands). Find the point elasticity of
demand when p D 15. If this price of 15 is increased by 12 ,
q D .p " 50/2 p D 10 what is the approximate change in demand
548 C t ona erent at on o cs

The demand equation for a certain product is a etermine the point elasticity of demand when p D 4, and
p classify the demand as elastic, inelastic, or of unit elasticity at this
q D 3000 " p2 price level.
b If the price is lowered by 2 (from 4.00 to 3. 2), use the
where p is in dollars. Find the point elasticity of demand when answer to part (a) to estimate the corresponding percentage
p D 40, and use this value to compute the percentage change in change in quantity sold.
demand if the price of 40 is increased by 7 . c Will the changes in part (b) result in an increase or decrease
For the demand equation p D 500 " 2q, verify that demand is in revenue Explain.
elastic and total revenue is increasing for 0 < q < 125. Verify that The demand equation for a manufacturer s product is
demand is inelastic and total revenue is decreasing for p
125 < q < 250. p D 50.151 " q/0:02 qC1
Show that if the demand equation can be written as q D g.p/ a Find the value of dp=dq when 150 units are demanded.
dr b Using the result in part (a), determine the point elasticity of
then D q.1 C %/.
dp demand when 150 units are demanded. At this level, is demand
1000 elastic, inelastic, or of unit elasticity
Repeat Problem 20 for p D 2 . c Use the result in part (b) to approximate the price per unit if
q
demand decreases from 150 to 140 units.
Suppose p D mq C b is a linear demand equation, where
d If the current demand is 150 units, should the manufacturer
m ¤ 0 and b > 0.
increase or decrease price in order to increase revenue ( ustify
a Show that limp!b! % D "1.
your answer.)
b Show that % D 0 when p D 0.
A manufacturer of aluminum doors currently is able to sell
The demand equation for a manufacturer s product has
500 doors per wee at a price of 0 each. If the price were
the form
lowered to 75 each, an additional 50 doors per wee could be
p
q D a b " cp2 sold. Estimate the current elasticity of demand for the doors,
and also estimate the current value of the manufacturer s
where a, b, and c are positive constants. marginal-revenue function.
a Show that elasticity does not depend on a. iven the demand equation
b etermine the interval of prices for which demand is elastic. p D 2000 " q2
c For which price is there unit elasticity
where 5 $ q $ 40, for what value of q is j%j a maximum For
iven the demand equation q2 .1 C p/2 D p, determine the what value is it a minimum
point elasticity of demand when p D .
Repeat Problem 2 for
The demand equation for a product is 200
pD
60 qC5
qD C ln.65 " p3 /
p such that 5 $ q $ 5.

Objective I
o sc ss t e not on of a f nct on Implicit differentiation is a technique for differentiating functions that are not given
e ne c t an to eter ne
er at es eans of ct in the form y D f.x/ nor in the form x D g.y/. To introduce this technique, we will find
erent at on the slope of a tangent line to a circ e. Circles are smooth curves, and it is clear that at
any point on any circle there is a tangent line. ut, for any circle with positive radius,
there will be some vertical lines that intersect the circle at more than one point. So we
now that any such circle cannot be described as the graph of a sing e function. For
definiteness in our discussion, let us ta e the circle of radius 2 whose center is at the
origin (Figure 12.4). Its equation is
y x2 C y2 D 4
x2 + y2 = 4 equivalently x2 C y2 " 4 D 0
p p
( 2, 2) The point . 2; 2/ lies on the circle. To find the slope at this point, we need to
find dy=dx there. Until now, we have always had y given explicitly (directly) in terms
x
-2 2 2 of x before determining y0 that is, our equation was always in the form y D f.x/ or
in the form x D g.y/. In Equation (1), this is not so. We say that Equation (1) has the
form F.x; y/ D 0, where F(x, y) denotes a function of two variables as introduced in
Section 2. . The obvious thing to do is solve Equation (1) for y in terms of x:
FIGURE The circle x2 C y2 D 4. x2 C y2 " 4 D 0
y2 D 4 " x2
p
y D ˙ 4 " x2
 Section 2.4 ct erent at on 549

y y

y = 4 - x2
( 2, 2)

x x

y = - 4 - x2

(a) (b)

FIGURE x2 C y2 D 4 gives rise to two different functions.

A problem now occurs: Equation (2) may give two values of y for a value of x. It
does not define y explicitly as a function of x. We can, however, suppose that Equa-
tion (1) defines y as ne f t di erent functi ns f x,
p p
y D C 4 " x2 or y D " 4 " x2
p p
whose graphs are given in Figure 12.5. Since the point . 2; 2/ lies on the graph of
p
y D 4 " x2 , we should differentiate that function:
p
y D 4 " x2
dy 1
D .4 " x2 /!1=2 ."2x/
dx 2
x
D "p
4 " x2
So
ˇ p
dy ˇˇ 2
ˇ p
D "p D "1
dx xD 2 4"2
p p
Thus, the slope of the circle x2 C y2 " 4 D 0 at the point . 2; 2/ is "1.
et us summarize the di culties we had. First, y was not originally given explicitly
in terms of x. Second, after we tried to find such a description, we ended up with more
than one function of x. In fact, depending on the equation given, it may be very compli-
cated or even impossible to find an explicit expression for y. For example, it would be
di cult, perhaps impossible, to solve yex C ln.x C y/ D 0 for y. We will now consider
a method that avoids such di culties.
An equation of the form F.x; y/ D 0, such as we had originally, is said to express y
imp icit y as a function of x. The word imp icit y is used because y is not given exp icit y
as a function of x. However, we will assume that the equation defines y as at least one
differentiable function of x. Thus, we assume that Equation (1), x2 Cy2 "4 D 0, defines
s me differentiable function of x, say, y D f.x/.
bserve that if y is a function of x, then y2 is also a function of x and then x2 Cy2 "4
is yet another function of x. f course, 0 can be regarded as the function of x that
is constantly 0. And now the left side of Equation (1) is the differentiable function
x2 C. f.x//2 "4 of x while the right side is the differentiable function 0 of x. Equation (1)
says these functions are equal and so their deri ati es must be equa This observation
gives us
d 2 d
.x C . f.x//2 " 4/ D .0/
dx dx
from which we get
d 2 d d d
.x / C .. f.x//2 / " .4/ D .0/
dx dx dx dx
and, using the chain rule for the second term,
2x C 2f.x/f 0 .x/ " 0 D 0
550 C t ona erent at on o cs

While we don t now explicitly what f.x/ is, the equation above tells us that its deriva-
tive, f 0 .x/, satisfies

2f.x/f 0 .x/ D "2x

and, hence,
x
f 0 .x/ D "
f.x/
Finally, noting that y D f.x/, we can write
dy x
D"
dx y
and
ˇ ˇ p
dy ˇˇ dy ˇˇ2 p 3
2
D ˇ D " p D "1
dx ˇ.p2;p2/ dx 4 x D p2 5 2
yD 2
p p
showing again that the slope of the circle x2 C y2 D 4 at . 2; 2/ D "1.
Notice that we didn t have to solve Equation (1) and we didn t end up with two
functions competing
p p for our attention. We didn tp needp to choose an equation appropriate
for the point . 2; 2/. It su ced to now that . 2; 2/ is a point on the curve that is,
a point whose coordinates satisfy Equation (1). Implicit differentiation, as the technique
above is called, can seem mysterious, and this is why our treatment was at first very
careful and somewhat labored. It is actually easier to carry out the calculation above
ith ut intr ducing the name f.x/ for y and using eibniz notation throughout. We start
again.
First, treat y as a differentiable function of x, and differentiate both sides of
Equation (1) with respect to x. Second, solve the resulting equation for dy=dx. Applying
this procedure, we obtain
d 2 d
.x C y2 " 4/ D .0/
dx dx
d 2 d d d
.x / C .y2 / " .4/ D .0/
dx dx dx dx
d 2 dy
2x C .y / " 0 D 0
dy dx
dy
2x C 2y D0
dx
dy x
D" for y ¤ 0
dx y
d 2
The only point above that requires special care is the treatment of .y / in
dx
Equation (3). ifferentiation is with respect to x, so we use the chain rule as shown.
Notice that the expression for dy=dx involves the variable y as well as x. This means
that to find dy=dx at a point, both coordinates of the point must be substituted into
dy=dx. Thus,
ˇ p
dy ˇˇ 2
ˇ p p
D " p D "1
dx . 2; 2/ 2
as before. We note that Equation (4) is not defined when y D 0. eometrically, this is
clear, since the tangent line to the circle at either .2; 0/ or ."2; 0/ is vertical and the
slope is not defined.
 Section 2.4 ct erent at on 551

Here again are the steps to follow when differentiating implicitly:

I
For an equation that we assume defines y implicitly as a differentiable function of
dy
x, the derivative can be found as follows:
dx
ifferentiate both sides of the equation with respect to x.
dy
Solve for , noting any restrictions.
dx

E AM LE I
dy
Find by implicit differentiation if y C y3 " x D 7.
dx
S Here y is not given as an explicit function of x. It is not at all clear if the
equation can be rewritten in the form y D f.x/. Thus, we assume that y is an implicit
differentiable function of x and apply the preceding two-step procedure:
ifferentiating both sides with respect to x, we have
d d
.y C y3 " x/ D .7/
dx dx
The derivative of y3 ith respect t x is d d d d
dy .y/ C .y3 / " .x/ D .7/
3y2 . dx dx dx dx
dx
dy dy
C 3y2 " 1 D 0
dx dx
dy
Solving for side, gives
dx
dy
.1 C 3y2 / D 1
In an implicit-differentiation problem, dx
we are able to find the derivative of a dy 1
function without nowing the function. D
dx 1 C 3y2
ecause step 2 of the process often involves division by an expression involv-
ing the variables, the answer obtained must often be restricted to exclude those values
of the variables that would ma e the denominator zero. Here the denominator is always
greater than or equal to 1, so there is no restriction.
Now ork Problem 3 G

A L IT I E AM LE I
Suppose that P, the proportion of dy
people affected by#a certain Find if x3 C 4xy2 " 27 D y4 .
$ disease, dx
P
is described by ln D 0:5t, S Since y is not given explicitly in terms of x, we will use the method of
1"P
dP implicit differentiation:
where t is the time in months. Find , Assuming that y is a function of x and differentiating both sides with respect to x,
dt
the rate at which P grows with respect we get
to time.
d 3 d
.x C 4xy2 " 27/ D .y4 /
dx dx
d 3 d d d
.x / C 4 .xy2 / " .27/ D .y4 /
dx dx dx dx
552 C t ona erent at on o cs

d 2 d dy
3x2 C 4.x .y / C y2 .x// " 0 D 4y3
dx dx dx
dy dy
3x2 C 4.x2y C y2 .1/ D 4y3
dx dx
dy dy
3x2 C xy C 4y2 D 4y3
dx dx

dy
Solving for , we have
dx

dy
. xy " 4y3 / D "3x2 " 4y2
dx
dy "3x2 " 4y2
D for xy " 4y3 ¤ 0
dx xy " 4y3
Note the use of the product rule in line dy 3x2 C 4y2
three of step one. D 3 for y3 " 2xy ¤ 0
dx 4y " xy

Now ork Problem 11 G

A L IT I
E AM LE I
The volume enclosed by a spher- Find the slope of the curve x3 D .y " x2 /2 at .1; 2/.
ical balloon of radius r is given by
4 S The slope at .1; 2/ is the value of dy=dx at that point. Finding dy=dx by
the equation D #r3 . If the radius is
3 implicit differentiation, we have
increasing at a rate of 5 inches minute
ˇ
dr d ˇˇ d 3 d
.that is; D 5/, then find ,
dt dt ˇ .x / D ..y " x2 /2 /
rD12 dx dx
the rate of increase of the volume, when # $
the radius is 12 inches. 2 2 d 2
3x D 2.y " x / .y " x /
dx
# $
dy
3x2 D 2.y " x2 / " 2x
dx
dy dy
3x2 D 2y " 4xy " 2x2 C 4x3
dx dx
dy dy
2y " 2x2 D 3x2 C 4xy " 4x3
dx dx
dy
2.y " x2 / D 3x2 C 4xy " 4x3
dx
dy 3x2 C 4xy " 4x3
D for y " x2 ¤ 0
dx 2.y " x2 /

For the point .1; 2/, y " x2 D 2 " 12 D 1 ¤ 0. Thus, the slope of the curve at (1, 2) is
ˇ
dy ˇˇ 3.1/2 C 4.1/.2/ " 4.1/3 7
D D
dx ˇ.1;2/ 2.2 " .1/2 / 2

Now ork Problem 25 G

 Section 2.4 ct erent at on 553

E AM LE I
A L IT I
A 10-foot ladder is placed against If q " p D ln q C ln p, find dq=dp.
a vertical wall. Suppose the bottom of
the ladder slides away from the wall at a S We assume that q is a function of p and differentiate both sides with respect
dx to p:
constant rate of 3 ft s. (That is, D 3.)
dt d d
How fast is the top of the ladder sliding .q " p/ D .ln q C ln p/
dp dp
down the wall when the top of the ladder
is feet from the ground (that is, when dq dp d d
dy " D .ln q/ C .ln p/
y D ) (That is, what is ) (Use the dp dp dp dp
dt
Pythagorean theorem for right triangles, dq 1 dq 1
x2 C y2 D z2 , where x and y are the legs "1D C
of the triangle and z is the hypotenuse.)
dp q dp p
# $
dq 1 1
1" D C1
dp q p
# $
dq q " 1 1Cp
D
dp q p
dq .1 C p/q
D for p.q " 1/ ¤ 0
dp p.q " 1/
Now ork Problem 19 G
R BLEMS
In Pr b ems nd dy=dx by imp icit di erentiati n Repeat Problem 2 for the curve
2 2 2 2
x "y D1 3x C 6y D 1 y2 C xy " x2 D 5
3 2 2 2
2y " 7x D 5 5y " 2x D 10
p p p p at the point (4, 3).
3
xC 3 yD3 x" yD1
F r the demand equati ns in Pr b ems nd the rate f
x3=4 C y3=4 D 5 y3 D 4x change f q ith respect t p
p
xy D 36 x2 C xy " 2y2 D 0 p D 100 " q3 p D 400 " q
x C xy C y D 1 x3 " y3 D 3x2 y " 3xy2 20 3
pD pD 2
.q C 5/2 q C1
2x3 C y3 " 12xy D 0 5x3 C 6xy C 7y3 D 0
p p Radioactivity The relative activity I=I0 of a radioactive
xD yC 4 y x2 y2 D 1
element varies with elapsed time according to the equation
5x3 y4 " x C y2 D 25 y2 C y D ln x # $
I
ln.xy/ D exy ln.xy/ C x D 4 ln D "& t
I0
xey C yex D 1 4x2 C y2 D 16
.1 C e3x /2 D 3 C ln.x C y/ ex!y D ln.x " y/ where & (a ree letter read lambda ) is the disintegration
constant and I0 is the initial intensity (a constant). Find the rate of
If x C xy C y2 D 7, find dy=dx at (1, 2). change of the intensity, I, with respect to the elapsed time, t.
p p
If .x C 1/ y D .y C 1/ x, find dy=dx at .2; 2/. Earthqua es The magnitude, , of an earthqua e and its
! "
Find the slope of the curve 4x2 C y2 D 1 at the point 0; 13 energy, E, are related by the equation6
at the point .x0 ; y0 /. # $
E
Find the slope of the curve .x2 C y2 /2 D 4y2 at the point 1:5 D log
.0; 2/. 2:5 % 1011

Find equations of the tangent lines to the curve Here is given in terms of Richter s preferred scale of 1 5 and
3 3 E is in ergs. etermine the rate of change of energy with respect
x C xy C y D "1
to magnitude and the rate of change of magnitude with respect to
at the points ."1; "1/, ."1; 0/, and ."1; 1/. energy.

6
. E. ullen, An Intr ducti n t the he ry f Seism gy (Cambridge, U. .:
Cambridge at the University Press, 1 63).
554 C t ona erent at on o cs

Physical Scale The relationship among the speed . /, echnological Substitution New products or technologies
frequency . f/, and wavelength (&) of any wave is given by often tend to replace old ones. For example, today most
commercial airlines use et engines rather than prop engines. In
D f&
discussing the forecasting of technological substitution, Hurter
Find df=d& by differentiating implicitly. (Treat as a constant.) and Rubenstein refer to the equation
Then show that the same result is obtained if you first solve the f .t/ 1
equation for f and then differentiate with respect to &. ln C" D C1 C C2 t
1 " f .t/ 1 " f .t/
Biology The equation .P C a/. C b/ D k is called the
where f t is the mar et share of the substitute over time t and
fundamental equation of muscle contraction. 7 Here P is the load
C1 ; C2 , and " (a ree letter read sigma ) are constants. Verify
imposed on the muscle is the velocity of the shortening of the
their claim that the rate of substitution is
muscle fibers and a, b, and k are positive constants. Use implicit
differentiation to show that, in terms of P, C2 f .t/Œ1 " f .t/!2
f 0 .t/ D
d k "f .t/ C Œ1 " f .t/!
D"
dP .P C a/2
Marginal Propensity to Consume A country s savings, S,
is defined implicitly in terms of its national income, I, by the
equation
1
S2 C I2 D SI C I
4
where both S and I are in billions of dollars. Find the marginal
propensity to consume when I D 16 and S D 12.

Objective L
o escr e t e et o of o ar t c A technique called logarithmic differentiation often simplifies the differentiation
erent at on an to s o o to
erent ate a f nct on of t e for u of y D f.x/ when f.x/ involves products, quotients, or powers. The procedure is as
follows:

L
To differentiate y D f.x/,
Ta e the natural logarithm of both sides. This results in
ln y D ln. f.x//
Simplify ln. f.x// by using properties of logarithms.
ifferentiate both sides with respect to x.
dy
Solve for .
dx
Express the answer in terms of x only. This requires substituting f(x) for y.

There are a couple of points worth noting. First, irrespective of any simplification,
the procedure produces
y0 d
D .ln. f.x//
y dx
so that
dy d
D y .ln. f.x//
dx dx

7
R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or : c raw-Hill oo Company,
1 55).
A. P. Hurter, r., A. H. Rubenstein et al., ar et Penetration by New Innovations: The Technological itera-
ture, echn gica F recasting and S cia Change 11 (1 7 ), 1 7 221.
 Section 2.5 o ar t c erent at on 555

f 0 .x/
is a formula that you can memorize, if you prefer. Second, the quantity , which
f.x/
results from differentiating ln. f.x//, is what was called the relative rate of change of
f.x/ in Section 11.3.
The next example illustrates the procedure.

E AM LE L

.2x " 5/3

Find y0 if y D p :
x2 4 x2 C 1
S ifferentiating this function in the usual way is messy because it involves
the quotient, power, and product rules. ogarithmic differentiation lessens the wor .

We ta e the natural logarithm of both sides:

.2x " 5/3
ln y D ln p
x2 4 x2 C 1
Simplifying by using properties of logarithms, we have
! p 4 "
ln y D ln.2x " 5/3 " ln x2 x2 C 1
D 3 ln.2x " 5/ " .ln x2 C ln.x2 C 1/1=4 /
1
D 3 ln.2x " 5/ " 2 ln x " ln.x2 C 1/
4
Since y is a function of x, differentiating ifferentiating with respect to x gives
y0 # $ # $ # $
ln y with respect to x gives .
y
y0 1 1 1 1
D3 .2/ " 2 " .2x/
y 2x " 5 x 4 x2 C 1
6 2 x
D " " 2
2x " 5 x 2.x C 1/
Solving for y0 yields
# $
0 6 2 x
y Dy " " 2
2x " 5 x 2.x C 1/
Substituting the original expression for y gives y0 in terms of x only:
) *
0 .2x " 5/3 6 2 x
y D p " "
x2 4 x2 C 1 2x " 5 x 2.x2 C 1/

Now ork Problem 1 G

ogarithmic differentiation can also be used to differentiate a function of the form
y D u , where both u and are differentiable functions of x. ecause neither the base
nor the exponent is necessarily a constant, the differentiation techniques for ua and b
do not apply here.

E AM LE F u

ifferentiate y D xx by using logarithmic differentiation.

S This example is a good candidate for the f rmu a approach to logarithmic
differentiation.
# # $$
d d 1
y0 D y .ln xx / D xx .x ln x/ D xx .1/.ln x/ C .x/ D xx .ln x C 1/
dx dx x
556 C t ona erent at on o cs

It is worthwhile mentioning that an alternative technique for differentiating a func-

tion of the form y D u is to convert it to an exponential function to the base e. To
illustrate, for the function in this example, we have

y D xx D .eln x /x D ex ln x
# $
1
y0 D ex ln x 1 ln x C x D xx .ln x C 1/
x

Now ork Problem 15 G

E AM LE R R C

Show that the relative rate of change of a product is the sum of the relative rates of
change of its factors. Use this result to express the percentage rate of change in revenue
in terms of the percentage rate of change in price.
S For the moment, suppose that r is a function of an unspecified variable x
dr
and that r0 denotes , differentiation with respect to x. Recall that the relative rate of
dx
r0
change of r with respect to x is . We are to show that if r D pq, where also p and q
r
are functions of x, then

r0 p0 q0
D C
r p q

From r D pq we have

ln r D ln p C ln q

which, when both sides are differentiated with respect to x, gives

r0 p0 q0
D C
r p q

as required. ultiplying both sides by 100 gives an expression for the percentage
rate of change of r in terms of those of p and q:

r0 p0 q0
100 D 100 C 100
r p q

If p is price per item and q is quantity sold, then r D pq is total re enue. In this case we
ta e x D p so that differentiation is with respect to p, and note that now, with p0 D 1,
q0 p0
Equation (3) of Section 12.3 gives D % , where % is the elasticity of demand. It
q p
follows that in this case we have
r0 p0
100 D .1 C %/ 100
r p

expressing the percentage rate of change in revenue in terms of the percentage rate
of change in price. For example, if at a given price and quantity, % D "5, then a 1
increase in price will result in a .1 " 5/ D "4 increase in revenue, which is to say
a 4 decrease in revenue, while a 3 decrease in price that is, a "3 increase in
price will result in a .1 " 5/."3/ D 12 increase in revenue. It is also clear that
at points at which there is unit elasticity (% D "1), any percentage change in price
produces no percentage change in revenue.
Now ork Problem 29 G
 Section 2.5 o ar t c erent at on 557

E AM LE F u

Find the derivative of y D .1 C ex /ln x .

S This has the form y D u , where u D 1Cex and D ln x. Using logarithmic
differentiation, we have
ln y D ln..1 C ex /ln x /
ln y D .ln x/ ln.1 C ex /
# $ # $
y0 1 x 1 x
D .ln.1 C e // C .ln x/ !e
y x 1 C ex
y0 ln.1 C ex / ex ln x
D C
y x 1 C ex
# $
0 ln.1 C ex / ex ln x
y Dy C
x 1 C ex
# $
ln.1 C ex / ex ln x
y0 D .1 C ex /ln x C
x 1 C ex
Now ork Problem 17 G
Alternatively, we can differentiate even a general function of the form y D u.x/ .x/

with u.x/ > 0 by using the equation

ln u
u De
Indeed, if y D u.x/ .x/ D e .x/ ln u.x/ for u.x/ > 0, then
dy d % .x/ ln u.x/ & d
D e D e .x/ ln u.x/ . .x/ ln u.x//
dx dx dx
# $
0 u0 .x/
Du .x/ ln u.x/ C .x/
u.x/
which could be summarized as
# $
0 0 u0
.u / D u ln u C
u
As is often the case, there is no suggestion that the preceding formula should be mem-
orized. The point here is that we have shown that any function of the form u can be
differentiated using the equation u D e ln u . The same result will be obtained from
logarithmic differentiation:
ln y D ln.u /
ln y D ln u
y0 0 u0
D ln u C
y u
# $
u0
y0 D y 0
ln u C
u
# $
0 u0 0
.u / D u ln u C
u
After completing this section, we understand how to differentiate each of the following
forms:
8̂
ˆ . f.x//a (a)
<
g.x/
yD b (b)
ˆ
:̂ . f.x//g.x/ (c)
558 C t ona erent at on o cs

For type (a), use the power rule. For type (b), use the differentiation formula for expo-
nential functions. (If b ¤ e, first convert bg.x/ to an eu function.) For type (c), use loga-
rithmic differentiation or first convert to an eu function. o not apply a rule in a situation
where the rule does not apply. For example, the power rule does not apply to xx .

R BLEMS
In Pr b ems nd y0 by using garithmic di erentiati n Find an equation of the tangent line to
2 2
y D .x C 1/ .x " 2/.x C 3/ y D .x C 1/.x C 2/2 .x C 3/2
y D .2x " 3/.5x " 7/2 .11x " 13/3 at the point where x D 0.
p
y D .3x3 " 1/2 .2x C 5/3 y D .2x2 C 1/ x2 " 1 Find an equation of the tangent line to the graph of
p p p y D xx
y D x C 1 x " 1 x2 C 1
p p at the point where x D 1.
y D .2x C 1/ x3 C 2 3 2x C 5
s Find an equation of the tangent line to the graph of
p3 2
1 C x2 x C5 y D xx
yD yD
1Cx xC
at the point where x D e.
.2x2 C 2/2 x2 .1 C x2 / If y D xx , find the relative rate of change of y with respect to x
yD yD p
.x C 1/2 .3x C 2/ x2 C 4 when x D 1.
r r
.x C 3/.x " 2/ 2
5 .x C 1/
2 If y D xx , find the value of x for which the percentage rate of
yD yD change of y with respect to x is 50 .
2x " 1 x2 e!x
Suppose f .x/ is a positive differentiable function and g is a
In Pr b ems nd y0 differentiable function and y D . f .x//g.x/ . Use logarithmic
2 C1
p
y D xx y D .2x/ x differentiation to show that
# $x # $
p 3 dy g.x/
yDx x
yD D . f .x//g.x/ f 0 .x/ C g0 .x/ ln. f .x//
x2 dx f .x/
The demand equation for a V is
y D .2x C 3/5x y D .x2 C 1/xC1
p q D 500 " 40p C p2
y D 4ex x3x y D . x/x
If the price of 15 is increased by 1 2 , find the corresponding
If y D .4x " 3/2xC1 , find dy=dx when x D 1. percentage change in revenue.
x
If y D .ex /.e / , find dy=dx when x D 0. Repeat Problem 2 with the same information except for a
5 decrease in price.

Objective N M
o a ro ate rea roots of an It is easy to solve equations of the form f.x/ D 0 when f is a linear or quadratic function.
e at on s n ca c s e
et o s o n s s ta e for For example, we can solve x2 C3x"2 D 0 by the quadratic formula. However, if f.x/ has
ca c ators a degree greater than 2 or is not a polynomial, it may be di cult, or even impossible, to
find solutions of f.x/ D 0, in terms of nown functions, even when it is proveable that
at least one solution exists. For this reason, we may settle for approximate solutions,
which can be obtained in a variety of e cient ways. For example, a graphing calculator
can be used to estimate the real roots of f.x/ D 0. In this section, we will study how
the derivative can be so used (provided that f is differentiable). The procedure we will
develop, called e t n s meth d is well suited to a calculator or computer.
Newton s method requires an initial estimate for a root of f.x/ D 0. ne way of
obtaining this estimate is by ma ing a rough s etch of the graph of y D f.x/. A point
on the graph where y D 0 is an x-intercept, and the x-value of this point is a solution
of f.x/ D 0. Another way of estimating a root is based on the following fact:

If f is continuous on the interval Œa; b! and f.a/ and f.b/ have opposite signs, then
the equation f.x/ D 0 has at least one real root between a and b.
 Section 2.6 e ton s et o 559

y = f(x)

y f(x1)

y = f(x)
f(a) 7 0
root of f(x) = 0
b Tangent
a x line

x
f(b) 6 0 r x3 x2 x1

FIGURE Root of f .x/ D 0 between a FIGURE Improving approximation of root

and b, where f .a/ and f .b/ have opposite signs. via tangent line.

Figure 12.6 depicts this situation. The x-intercept between a and b corresponds to
a root of f.x/ D 0, and we can use either a or b to approximate this root.
Assuming that we have an estimated (but incorrect) value for a root, we turn to a
way of getting a better approximation. In Figure 12.7, we see that f.r/ D 0, so r is
a root of the equation f.x/ D 0. Suppose x1 is an initial approximation to r (and one
that is close to r). bserve that the tangent line to the curve at .x1 ; f.x1 // intersects the
x-axis at the point .x2 ; 0/, and x2 is a better approximation to r than is x1 .
We can find x2 from the equation of the tangent line. The slope of the tangent line
is f 0 .x1 /, so a point-slope form for this line is
y " f.x1 / D f 0 .x1 /.x " x1 /
Since .x2 ; 0/ is on the tangent line, its coordinates must satisfy Equation (1). This gives
0 " f.x1 / D f 0 .x1 /.x2 " x1 /
f.x1 /
" D x2 " x1 if f 0 .x1 / ¤ 0
f 0 .x1 /
Thus,
f.x1 /
x2 D x1 "
f 0 .x1 /
To get a better approximation to r, we again perform the procedure described, but this
time we use x2 as our starting point. This gives the approximation
f.x2 /
x3 D x2 "
f 0 .x2 /
Repeating this computation over and over, we hope to obtain better and better approx-
imations, in the sense that the sequence of values obtained
x1 ; x2 ; x3 ; : : :
approaches r. In practice, we terminate the process when we have reached a desired
degree of accuracy.
Analyzing Equations (2) and (3), we see how x2 is obtained from x1 and how x3
is obtained from x2 . In general, xnC1 is obtained from xn by means of the following
general formula, called ewton’s method:

N M
f.xn /
xnC1 D xn " n D 1; 2; 3; : : :
f 0 .xn /
560 C t ona erent at on o cs

To review recursively defined sequences, A formula, li e Equation (4), that indicates how one number in a sequence is obtained
see Section 1.6. from the preceding one is called a recursion formula.

E AM LE A R N M
A L IT I
If the total profit (in dollars) from Approximate the root of x4 "4xC1 D 0 that lies between 0 and 1. Continue the approx-
the sale of x televisions is P.x/ D 20x " imation procedure until two successive approximations differ by less than 0.0001.
0:01x2 " 50 C 3 ln.x/, use Newton s
method to approximate the brea -even S etting f.x/ D x4 " 4x C 1, we have
quantities. ( te There are two brea -
even quantities one is between 10 and f.0/ D 0 " 0 C 1 D 1
50, and the other is between 1 00 and
2000.) ive the x-value to the nearest and
integer.
f.1/ D 1 " 4 C 1 D "2

In the event that a root lies between a and (Note the change in sign.) Since f(0) is closer to 0 than is f(1), we choose 0 to be our
b, and f .a/ and f .b/ are equally close to first approximation, x1 . Now,
0, choose either a or b as the first
approximation.
f 0 .x/ D 4x3 " 4

so

f.xn / D x4n " 4xn C 1 and f 0 .xn / D 4x3n " 4

Substituting into Equation (4) gives the recursion formula

f.xn / x4n " 4xn C 1 4x4n " 4xn " x4n C 4xn " 1
xnC1 D xn " D xn " D
f 0 .xn / 4x3n " 4 4x3n " 4
so
3x4n " 1
xnC1 D
4x3n " 4
Since x1 D 0, letting n D 1 above gives

3x41 " 1 3.0/4 " 1

x2 D 3
D D 0:25
4x1 " 4 4.0/3 " 4
etting n D 2 gives

3x42 " 1 3.0:25/4 " 1

x3 D D # 0:250 206
4x32 " 4 4.0:25/3 " 4
Table 1 .1 etting n D 3 gives
n xn xnC1
3x43 " 1 3.0:250 206/4 " 1
x4 D D # 0:250 216
1 0.00000 0.25000000 3
4x3 " 4 4.0:250 206/3 " 4
2 0.25000 0.250 206
3 0.250 0.250 216
The data obtained thus far are displayed in Table 12.1. Since the values of x3 and x4 dif-
fer by less than 0.0001, we ta e our approximation of the root to be x4 # 0:250 216.
Now ork Problem 1 G
Notice that in simplifying xnC1 the appearance of the n s on the right side is an
unnecessary distraction. It saves writing to simplify
f.x/ xf 0 .x/ " f.x/
.x/ D x " D
f 0 .x/ f 0 .x/
so that the recursion formula becomes xnC1 D .xn /. The next example will illustrate.
 Section 2.6 e ton s et o 561

E AM LE A R N M

Approximate the root of x3 D 3x"1 that lies between "1 and "2. Continue the approx-
imation procedure until two successive approximations differ by less than 0.0001.
S etting f.x/ D x3 " 3x C 1, so that the equation becomes f.x/ D 0, we
find that

f."1/ D ."1/3 " 3."1/ C 1 D 3

and

f."2/ D ."2/3 " 3."2/ C 1 D "1

(Note the change in sign.) Since f."2/ is closer to 0 than is f."1/, we choose "2 to be
our first approximation, x1 . Now,
f 0 .x/ D 3x2 " 3
so
x3 " 3x C 1 3x3 " 3x " x3 C 3x " 1 2x3 " 1
.x/ D x " D D
3x2 " 3 3x2 " 3 3x2 " 3
and
2x3n " 1
xnC1 D
3x2n " 3
Table 1 . Since x1 D "2, letting n D 1 gives
n xn xn C 1
2x31 " 1 2."2/3 " 1
1
x2 D D # 1:
"2:00000 "1: 3x21 " 3 3."2/2 " 3
2 "1: "1: 7 45
3 "1: 7 45 "1: 7 3
Continuing in this way, we obtain Table 12.2. ecause the values of x3 and x4 differ by
less than 0.0001, we ta e our approximation of the root to be x4 # "1: 7 3 .
Now ork Problem 3 G
If our choice of x1 has f 0 .x1 / D 0, then Newton s method will fail to produce
a value for x2 . hen this happens the ch ice f x1 must be rejected and a di erent
number c se t the desired r t must be ch sen f r x1 . A graph of f can be helpful in
this situation. Finally, there are times when the sequence of approximations does not
approach the root. A discussion of such situations is beyond the scope of this boo .

R BLEMS
In Pr b ems use e t n s meth d t appr ximate the x4 " x3 C x " 2 D 0 root between 1 and 2
indicated r t f the gi en equati n C ntinue the appr ximati n
Estimate, to three-decimal-place accuracy, the cube root of
pr cedure unti the di erence f t successi e appr ximati ns
73. int Show that the problem is equivalent to finding a root of
is ess than
f .x/ D x3 " 73 D 0. Choose 4 as the initial estimate. Continue
x3 " 5x C 1 D 0 root between 0 and 1 the iteration until two successive approximations, rounded to three
3 2
x C 2x " 1 D 0 root between 0 and 1 decimal places, are the same.
p
x3 C x C 1 D 0 root between "1 and 2. Estimate 4 1 , to two-decimal-place accuracy. Use 2 as your
initial estimate.
x3 " x C 6 D 0 root between 2 and 3
3
Find, to three-decimal-place accuracy, all p siti e real
x CxC1D0 root between "1 and 0 solutions of the equation ex D x C 3. ( int A rough s etch of the
x3 D 2x C 6 root between 2 and 3 graphs of y D ex and y D x C 3 in the same plane ma es it clear
4 how many such solutions there are. Use the integer values
x D 3x " 1 root between 0 and 1
suggested by the graph to choose the initial values.)
4 3
x Cx "1D0 root between 0 and 1
Find, to three-decimal-place accuracy, all real solutions of the
x4 " 2x3 C x2 " 3 D 0 root between 1 and 2 equation ln x D 5 " x.
562 C t ona erent at on o cs

Brea Even Quantity The cost of manufacturing q tons of Equilibrium iven the supply equation p D 2q C 5 and
a certain product is given by 100
the demand equation p D 2 , use Newton s method to
c D 250 C 2q " 0:1q3 q C1
estimate the mar et equilibrium quantity. ive your answer to
and the revenue obtained by selling the q tons is given by three-decimal-place accuracy.
r D 3q Equilibrium iven the supply equation
Approximate, to two-decimal-place accuracy, the brea -even p D 0:2q3 C 0:6q C 2
quantity. ( int Approximate a root of r " c D 0 by choosing 13
as your initial estimate.) and the demand equation p D " q, use Newton s method to
estimate the mar et equilibrium quantity, and find the
Brea Even Quantity The total cost of manufacturing q corresponding equilibrium price. Use 2 as an initial estimate for
hundred pencils is c dollars, where the required value of q, and give the answer to two-decimal-place
q2 1 accuracy.
c D 50 C 4q C C
1000 q Use Newton s method to approximate (to two-decimal-place
accuracy) a critical value of the function
Pencils are sold for per hundred.
a Show that the brea -even quantity is a solution of the x3
f .x/ D " x2 " 5x C 1
equation 3
q3 on the interval 3, 4 .
f .q/ D " 4q2 C 50q C 1 D 0
1000
b Use Newton s method to approximate the solution of
f .q/ D 0, where f .q/ is given in part (a). Use 10 as your initial
approximation, and give your answer to two-decimal-place
accuracy.

Objective H
o n er or er er at es ot We now that the derivative of a function y D f.x/ is itself a function, f 0 .x/. If we
rect an ct
differentiate f 0 .x/, the resulting function . f 0 /0 .x/ is called the second derivative of f
at x. It is denoted f 00 .x/, which is read f double prime of x. Similarly, the derivative of
the second derivative is called the third derivative, written f 000 .x/. Continuing in this
way, we get higher order derivatives. Some notations for higher-order derivatives are
given in Table 12.3. To avoid clumsy notation, primes are not used beyond the third
derivative.

Table 1 .
dy d
First derivative: y0 f 0 .x/ . f .x// xy
dx dx
d2 y d2
Second derivative: y00 f 00 .x/ . f .x// 2y
x
dx2 dx2
d3 y d3
The symbol d 2 y=dx2 represents the Third derivative: y000 f 000 .x/ . f .x// 3y
x
dx3 dx3
second derivative of y. It is not the same
as .dy=dx/2 , the square of the first d4 y d4 4y
Fourth derivative: y.4/ f.4/ .x/ . f .x// x
derivative of y. dx4 dx4
dn y d n
ny
nth derivative: y.n/ f.n/ .x/ . f .x// x
dxn dxn

The eibniz notation for higher derivatives is a little less mysterious when we note
# $n
dn y d
that n is a convention for .y/, which is to say, differentiation with respect to x,
dx dx
d
, applied n times to y.
dx
 Section 2.7 er r er er at es 563

E AM LE F H

If f.x/ D 6x3 " 12x2 C 6x " 2, find all higher-order derivatives.

S ifferentiating f.x/ gives

f 0 .x/ D 1 x2 " 24x C 6

ifferentiating f 0 .x/ yields

f 00 .x/ D 36x " 24

Similarly,

f 000 .x/ D 36

f.4/ .x/ D 0

and for n & 5, f.n/ .x/ D 0

Now ork Problem 1 G

E AM LE F S
A L IT I
The height h.t/ of a roc dropped 2 d 2y
off of a 200-foot building is given by
If y D ex , find .
dx2
h.t/ D 200 " 16t2 , where t is the
d2 h
S
time measured in seconds. Find 2 ,
dt dy 2 2
the acceleration of the roc at time t. D ex .2x/ D 2xex
dx
y the product rule,

d 2y 2 2 2
D 2.x.ex /.2x/ C ex .1// D 2ex .2x2 C 1/
dx2

Now ork Problem 5 G

E AM LE E S
A L IT I
If the cost to produce q units of a 16 d 2y
product is If y D f.x/ D , find 2 and evaluate it when x D 4.
xC4 dx
c.q/ D 7q2 C 11q C 1 S Since y D 16.x C 4/!1 , the power rule gives
and the marginal-cost function is c0 .q/,
dy
find the rate of change of the marginal D "16.x C 4/!2
cost function with respect to q when dx
q D 3.
d 2y 32
2
D 32.x C 4/!3 D
dx .x C 4/3
Evaluating when x D 4, we obtain
ˇ
d 2 y ˇˇ 32 1
2 ˇ D 3 D
dx xD4 16

The second derivative evaluated at x D 4 is also denoted by f 00 .4/ and by y00 .4/.
Now ork Problem 21 G
564 C t ona erent at on o cs

E AM LE F R C f 00 .x/

If f.x/ D x ln x, find the rate of change of f 00 .x/.

S To find the rate of change of any function, we must find its derivative. Thus,
The rate of change of f 00 .x/ is f 000 .x/. we want the derivative of f 00 .x/, which is f 000 .x/. Accordingly,
# $
0 1
f .x/ D x C .ln x/.1/ D 1 C ln x
x
1 1
f 00 .x/ D 0 C D
x x
d !1 1
f 000 .x/ D .x / D ."1/x!2 D " 2
dx x

Now ork Problem 17 G

H I
We will now find a higher-order derivative by means of implicit differentiation. eep
in mind that we will assume y to be a function of x.

E AM LE H I

d2 y
Find if x2 C 4y2 D 4.
dx2
S ifferentiating both sides with respect to x, we obtain
dy
2x C y D0
dx
dy "x
D
dx 4y
d d
4y
."x/ " ."x/ .4y/
d2 y dx dx
D
dx2 .4y/2
# $
dy
4y."1/ " ."x/ 4
dx
D 2
16y
dy
"4y C 4x
dx
D
16y2
dy
"y C x
d2 y dx
D
dx2 4y2

Although we have found an expression for d2 y=dx2 , our answer involves the derivative
dy=dx. It is customary to express the answer without the derivative that is, in terms
dy "x
of x and y only. This is easy to do. From Equation (1), D , so by substituting
dx 4y
into Equation (2), we have
# $
"x
2
"y C x
dy 4y "4y2 " x2 4y2 C x2
D D D "
dx2 4y2 16y3 16y3
 Section 2.7 er r er er at es 565

In Example 5, the simplification of We can further simplify the answer. Since x2 C 4y2 D 4 (the original equation),
d2 y=dx2 by ma ing use of the original
equation is not unusual. d 2y 4 1
2
D" 3
D" 3
dx 16y 4y

Now ork Problem 23 G

E AM LE H I

d 2y
Find if y2 D exCy .
dx2
S ifferentiating both sides with respect to x gives
# $
dy xCy dy
2y D e 1C
dx dx
Solving for dy=dx, we obtain
dy dy
2y D exCy C exCy
dx dx
dy dy
2y " exCy D exCy
dx dx
dy
.2y " exCy / D exCy
dx
dy exCy
D
dx 2y " exCy
Since y2 D exCy (the original equation),
dy y2 y
D 2
D
dx 2y " y 2"y
# $
dy dy dy
.2 " y/ " y " 2
d2 y dx dx dx
D D
dx2 .2 " y/2 .2 " y/2
dy y
Now we express our answer without dy=dx. Since D ,
dx 2"y
# $
y
2
d2 y 2"y 2y
2
D D
dx .2 " y/ 2 .2 " y/3

Now ork Problem 31 G

R BLEMS
In Pr b ems nd the indicated deri ati es 1 p 00
f .q/ D , f 000 .q/ f .x/ D x, f .x/
3 2
y D 4x " 12x C 6x C 2, y 000 3q3
p
" r, f 00 .r/
2
y D x5 C x4 C x3 C x2 C x C 1, y000 f .r/ D y D eax , y00
d2 y d3 y 1 d2 y
y D " x, 2 y D ax2 C bx C c, 3 yD , 2 y D .ax C b/6 , y000
dx dx 2x C 3 dx
d3 F x C 1 00
y D x3 C ex , y.4/ F.q/ D ln.q C 1/, 3 yD ,y y D 2x1=2 C .2x/1=2 , y00
dq x"1
1 000 .2x C 5/.5x " 2/ 00
f .x/ D x3 ln x, f 000 .x/ yD ,y y D lnŒx.x C a/!, y00 y D ln ,y
x xC1
566 C t ona erent at on o cs

x d2 y Find the rate of change of f 0 .x/ if f .x/ D .5x " 3/4 .

f .z/ D z3 ez , f 000 .z/ yD ,
ex dx2 Find the rate of change of f 00 .x/ if
ˇ
2x 3x d5 y ˇˇ p 1
If y D e C e , find 5 ˇ . f .x/ D 6 x C p
dx xD0 6 x
2 C1/
If y D e2 ln.x , find y00 when x D 1. Marginal Cost If c D 0:2q2 C 2q C 500 is a cost function,
how fast is marginal cost changing when q D 7:357
In Pr b ems nd y00
Marginal Revenue If p D 400 " 40q " q2 is a demand
x2 C 4y2 " 16 D 0 x2 C y2 D 1 equation, how fast is marginal revenue changing when q D 4
y2 D 4x x2 C 16y2 D 25 1 4 1 3
p p If f .x/ D x " x C x2 " 4, determine the values of x for
a xCb yDc y2 " 6xy D 4 12 2
which f 00 .x/ D 0.
x C xy C y D 1 x2 C 2xy C y2 D 1 Suppose that ey D y2 ex . a etermine dy=dx, and express
your answer in terms of y only. b etermine d2 y=dx2 , and
y D exCy ex C ey D x2 C y2
express your answer in terms of y only.
If x2 C 3x C y2 D 4y, find d2 y=dx2 when x D 0 and y D 0. In Pr b ems and determine f 00 .x/ hen use y ur graphing
Show that the equation ca cu at r t nd a rea r ts f f 00 .x/ D 0 R und y ur ans ers
t t decima p aces
f 00 .x/ C 2f 0 .x/ C f .x/ D 0 f .x/ D 6ex " x3 " 15x2

is satisfied if f .x/ D .x C 1/e!x . x5 x4 5x3 x2

f .x/ D C C C
20 12 6 2

Chapter 12 Review
I T S E
S Derivatives of Logarithmic Functions
derivative of ln x and of logb u Ex. 5, p. 536
S Derivatives of Exponential Functions
derivative of ex and of bu Ex. 4, p. 540
S Elasticity of Demand
point elasticity of demand, % elastic unit elasticity inelastic Ex. 2, p. 546
S Implicit Differentiation
implicit differentiation Ex. 1, p. 551
S Logarithmic Differentiation
logarithmic differentiation relative rate of change of f.x/ Ex. 3, p. 556
S ewton’s Method
f.xn / xn f 0 .xn / " f.xn /
recursion formula, xnC1 D xn " 0 D Ex. 1, p. 560
f .xn / f 0 .xn /
S igher Order Derivatives
d3 y d4
higher-order derivatives, f 00 .x/, , Œf.x/!; : : : Ex. 1, p. 563
dx3 dx4

S
The derivative formulas for natural logarithmic and exponen- To differentiate logarithmic and exponential functions in
tial functions are bases other than e, first transform the function to base e and
then differentiate the result. Alternatively, differentiation for-
d 1 du mulas can be applied:
.ln u/ D
dx u dx
d 1 du
and .logb u/ D !
dx .ln b/u dx
d u du d u du
.e / D eu .b / D bu .ln b/ !
dx dx dx dx
 Chapter 2 e e 567

Point elasticity of demand is a function that measures ferentiate both sides of the equation with respect to x. When
how consumer demand is affected by a change in price. It is doing this, remember that
given by
d n dy
.y / D nyn!1
dx dx
p dq and, more generally, that
%D
q dp
d dy
. f.y// D f 0 .y/
dx dx
where p is the price per unit at which q units are demanded.
Finally, we solve the resulting equation for dy=dx.
The three categories of elasticity are as follows:
To differentiate y D f.x/, where f.x/ consists of prod-
ucts, quotients, or powers, the method of logarithmic differ-
j%.p/j > 1 demand is elastic entiation may be used. In that method, we ta e the natural
j%.p/j D 1 unit elasticity logarithm of both sides of y D f.x/ to obtain ln y D ln. f.x//.
After simplifying ln. f.x// by using properties of logarithms,
j%.p/j < 1 demand is inelastic we differentiate both sides of ln y D ln. f.x// with respect to
x and then solve for y0 . ogarithmic differentiation can also
For a given percentage change in price, if there is a be used to differentiate y D u , where both u and are dif-
greater (respectively, lesser) percentage change in quantity ferentiable functions of x.
demanded, then demand is elastic (respectively, inelastic). Newton s method is the name given to the following for-
Two relationships between elasticity and the rate of mula, which is used to approximate the roots of the equation
change of revenue are given by f.x/ D 0, provided that f is differentiable:

# $ xn f 0 .xn / " f.xn /

dr 1 dr xnC1 D ; n D 1; 2; 3; : : :
Dp 1C D q.1 C %/ f 0 .xn /
dq % dp
In many cases encountered, the approximation improves as
n increases.
If an equation implicitly defines y as a function of x ecause the derivative f 0 .x/ of a function y D f.x/ is
(rather than defining it explicitly in the form y D f.x/), itself a function, it can be successively differentiated to obtain
then dy=dx can be found by implicit differentiation. With this the second derivative f 00 .x/, the third derivative f 000 .x/, and
method, we treat y as a differentiable function of x and dif- other higher-order derivatives.

R
In Pr b ems di erentiate 1 C ex
y D .x2 C 1/xC1 yD
y D 3ex C e2 C e C x x2 e2
f. / D e C 2 1 " ex
p
f .r/ D ln.7r2 C 4r C 5/ y D eln x '.t/ D ln.t2 5 " t3 / y D .x C 3/ln x
p
y D e3x
2 C5xC7
f .t/ D log6 t2 C 1 .x2 C 1/1=2 .x2 C 2/1=3 p
yD y D .ln x/ x
.2x3 C 6x/2=5
y D ex .x2 C 2/ y D 23x
2

p p y D .xx /x y D x.x /
x

y D .x " 6/.x C 5/. " x/ t

f .t/ D e
ln x ex C e!x In Pr b ems e a uate y0 at the gi en a ue f x
yD yD
ex x2 ex C1
2

y D .x C 1/ ln x2 ; x D 1 yD p ;x D 1
f .q/ D lnŒ.q C a/m .q C b/n ! x2 C 1
y D .x C 2/3 .x C 1/4 .x " 2/2 y D .1=x/x , x D e
y D 35x
2 C3xC2
y D .e C e2 /0 " #!1
25x .x2 " 3x C 5/1=3
4e3x ln x yD ;x D 0
yD yD .x2 " 3x C 7/3
xex!1 ex
# $
3x " 7 In Pr b ems and nd an equati n f the tangent ine t the
y D log2 . x C 5/2 y D ln cur e at the p int c rresp nding t the gi en a ue f x
x2 C 5x " 2
f . / D ln.1 C C 2
C 3/ 2 y D 2ex , x D ln 2 y D x C x2 ln x; x D 1
y D .x2 /x
568 C t ona erent at on o cs
4 3 2
Find the y-intercept of the tangent line to the graph of If f .x/ D ex !10x C36x !2x , find all real roots of f 0 .x/ D 0.
2
y D x.22!x / at the point where x D 1. Round your answers to two decimal places.
If D 2x C ln.1 C x2 / and x D ln.1 C t2 /, find and d =dt x5 x4 2x3
If f .x/ D C C C x2 C 1, find all roots of
when t D 0. 10 6 3
In Pr b ems nd the indicated deri ati e at the gi en p int f 00 .x/ D 0. Round your answers to two decimal places.
yDe x2 !2xC1 00
; y , .1; 1/ y D x3 ex , y000 , .1; e/ F r the demand equati ns in Pr b ems determine hether
demand is e astic is ine astic r has unit e asticity f r the
y D ln.2x/; y000 , (1, ln 2) y D x ln x; y00 , (1, 0) indicated a ue f q
100 p D 00 " q2 q D 10
In Pr b ems nd dy=dx pD q D 100
q
x2 C 2xy C y2 D 4 x3 y3 D 3
p D 1 " 0:02q q D 600
ln.xy/ D xy y2 ey ln x D e2
The demand equation for a product is
2 2 # $
In Pr b ems and nd d y=dx at the gi en p int 20 " p 2
qD for 0 $ p $ 20
x C xy C y D 5, (2, 1) x2 C xy C y2 D 1, .0; "1/ 2

If y is defined implicitly by ey D .y C 1/ex , determine both a Find the point elasticity of demand when p D .
dy=dx and d2 y=dx2 as explicit functions of y only. b Find all values of p for which demand is elastic.
d 2y The demand equation of a product is
If ex C ey D 1, find 2 . p
dx q D 2500 " p2
Schi ophrenia Several models have been used to analyze
Find the point elasticity of demand when p D 30. If the price of 30
the length of stay in a hospital. For a particular group of
decreases 23 , what is the approximate change in demand
schizophrenics, one such model is
The demand equation for a product is
f .t/ D 1 " .0: e!0:01t C 0:2e!0:0002t / p
q D 144 " p; where 0 < p < 144
where f .t/ is the proportion of the group that was discharged at the a Find all prices that correspond to elastic demand.
end of t days of hospitalization. etermine the discharge rate
b Compute the point elasticity of demand when p D 100. Use
(proportion discharged per day) at the end of t days.
the answer to estimate the percentage increase or decrease in
Earthqua es According to Richter,10 the number of demand when price is increased by 5 to p D 105.
earthqua es of magnitude or greater per unit of time is given by
The equation x3 " 2x " 2 D 0 has a root between 1 and 2. Use
log D A " b , where A and b are constants. Richter claims that
Newton s method to estimate the root. Continue the approximation
# $ # $ procedure until the difference of two successive approximations is
d b
log " D A C log "b less than 0.0001. Round your answer to four decimal places.
d q
Find, to an accuracy of three decimal places, all real solutions
where q D log e. Verify this statement. of the equation ex D 3x.

Adapted from W. W. Eaton and . A. Whitmore, ength of Stay as a

Stochastic Process: A eneral Approach and Application to Hospitalization
for Schizophrenia, urna f athematica S ci gy 5 (1 77) 273 2.
10
C. F. Richter, E ementary Seism gy (San Francisco: W. H. Freeman and
Company, Publishers, 1 5 ).
 C r e etc n

I
n the mid-1 70s, economist Arthur affer was explaining his views on taxes to a
13.1 e at e tre a politician. To illustrate his argument, affer grabbed a paper nap in and s etched
13.2 so te tre a on a the graph that now bears his name: the affer curve.
C ose nter a The affer curve describes total government tax revenue as a function of the
tax rate. bviously, if the tax rate is zero, the government gets nothing. ut if the tax
13.3 Conca t rate is 100 , revenue would again equal zero, because there is no incentive to earn
13.4 e econ er at e money if it will all be ta en away. Since tax rates between 0 and 100 do generate
est revenue, affer reasoned, the curve relating revenue to tax rate must loo , qualitatively,
more or less as shown in the figure below.
13.5 s totes affer s argument was not meant to show that the optimal tax rate was 50 . It was
13.6 e a a an meant to show that under some circumstances, namely, when the tax rate is to the right
n a of the pea of the curve, it is possible to raise g ernment re enue by ering taxes.
This was a ey argument made for the tax cuts passed by Congress during the first term
C er 13 e e of the Reagan presidency.
ecause the affer curve is only a qualitative picture, it does not actually give an
optimal tax rate. Revenue-based arguments for tax cuts involve the claim that the point
of pea revenue lies to the left of the current taxation scheme on the horizontal axis. y
the same to en, those who urge raising taxes to raise government income are assuming
either a different relationship between rates and revenues or a different location of the
curve s pea .
y itself, then, the affer curve is too abstract to be of much help in determining the
optimal tax rate. ut even very simple s etched curves, li e supply and demand curves
and the affer curve, can help economists describe the causal factors that drive an econ-
omy. In this chapter, we will discuss techniques for s etching and interpreting curves.
Tax revenue

0 100%
Tax rate

569
570 C C r e etc n

Objective R E
o n en a f nct on s ncreas n or
ecreas n to n cr t ca a es to I N F
ocate re at e a a an re at e
n a an to state t e rst er at e Examining the graphical behavior of functions is a basic part of mathematics and has
test so to s etc t e ra of a applications to many areas of study. When we s etch a curve, ust plotting points may
f nct on s n t e nfor at on not give enough information about its shape. For example, the points .!1; 0/; .0; !1/,
o ta ne fro t e rst er at e
and (1, 0) satisfy the equation given by y D .xC1/3 .x!1/. n the basis of these points,
we might hastily conclude that the graph should appear as in Figure 13.1(a), but in fact
the true shape is given in Figure 13.1(b). In this chapter we will explore the powerful
role that differentiation plays in analyzing a function so that we can determine the true
shape and behavior of its graph.

y y

x x
-1 1 -1 1

-1
-1

(a) (b)

FIGURE Curves passing through .!1; 0/, .0; !1/, and .1; 0/.

We begin by analyzing the graph of the function y D f.x/ in Figure 13.2. Notice
that as x increases (goes from left to right) on the interval I1 , between a and b, the values
of f.x/ increase and the curve is rising. athematically, this observation means that if
x1 and x2 are any two points in I1 such that x1 < x2 , then f.x1 / < f.x2 /. Here f is said
to be an increasing functi n on I1 . n the other hand, as x increases on the interval I2
between c and d, the curve is falling. n this interval, x3 < x4 implies that f.x3 / > f.x4 /,
and f is said to be a decreasing functi n on I2 . We summarize these observations in the
following definition.

A function f is said to be increasing on an interval I when, for any two numbers

x1 ; x2 in I, if x1 < x2 , then f.x1 / < f.x2 /. A function f is decreasing on an interval I
when, for any two numbers x1 ; x2 in I, if x1 < x2 , then f.x1 / > f.x2 /.

y
y = f (x)
Positive slope Negative slope
f ¿ (x) 7 0 f¿ (x) 6 0

f (x1) 6 f (x2) f (x3) 7 f (x4)

x
a x1 x2 b c x3 x4 d

I1 I2
f increasing f decreasing

FIGURE Increasing or decreasing nature of function.

 Section 3. e at e tre a 571

In terms of the graph of the function, f is increasing on I if the curve rises to the
right and f is decreasing on I if the curve falls to the right. Recall that a straight line
with positive slope rises to the right, while a straight line with negative slope falls to
the right.
Turning again to Figure 13.2, we note that over the interval I1 , tangent lines to
the curve have positive slopes, so f 0 .x/ must be positive for all x in I1 . A positive
derivative implies that the curve is rising. ver the interval I2 , the tangent lines have neg-
ative slopes, so f 0 .x/ < 0 for all x in I2 . The curve is falling where the derivative is
negative. We, thus, have the following rule, which allows us to use the derivative to
determine when a function is increasing or decreasing:

R C I F
et f be differentiable on the interval .a; b/. If f 0 .x/ > 0 for all x in .a; b/, then f is
increasing on .a; b/. If f 0 .x/ < 0 for all x in .a; b/, then f is decreasing on .a; b/.

To illustrate these ideas, we will use Rule 1 to find the intervals on which
y D 1 x! 23 x3 is increasing and the intervals on which y is decreasing. etting y D f.x/,
we must determine when f 0 .x/ is positive and when f 0 .x/ is negative. We have
f 0 .x/ D 1 ! 2x2 D 2. ! x2 / D 2.3 C x/.3 ! x/
Using the technique of Section 10.4, we can find the sign of f 0 .x/ by testing the intervals
determined by the roots of 2.3 C x/.3 ! x/ D 0, namely, !3 and 3. These should be
arranged in increasing order on the top of a sign chart for f 0 so as to divide the domain
of f into intervals. (See Figure 13.3.) In each interval, the sign of f 0 .x/ is determined by
the signs of its factors:
-q -3 3 q
3+x - 0 + +

3-x + + 0 -

f¿(x) - 0 + 0 -
-
f(x)
-

FIGURE Sign chart for f 0 .x/ D 1 ! x2 and its interpretation for f .x/.
y
y = 18x - 23 x3 If x < !3; then sign. f 0 .x// D 2.!/.C/ D !; so f is decreasing:
36
If !3 < x < 3; then sign. f 0 .x// D 2.C/.C/ D C; so f is increasing:
If x > 3; then sign. f 0 .x// D 2.C/.!/ D !; so f is decreasing:
These results are indicated in the sign chart given by Figure 13.3, where the bottom line
-3 3
x
is a schematic version of what the signs of f 0 say about f itself. Notice that the horizontal
line segments in the bottom row indicate horizontal tangents for f at !3 and at 3. Thus,
f is decreasing on .!1; !3/ and .3; 1/ and is increasing on .!3; 3/. This corresponds
to the rising and falling nature of the graph of f shown in Figure 13.4. Indeed, the point
-36 of a well-constructed sign chart is to provide a schematic for subsequent construction
of the graph itself.
Decreasing Increasing Decreasing
E
FIGURE Increasing decreasing for
y D 1 x ! 23 x3 .
oo now at the graph of y D f.x/ in Figure 13.5. Some observations can be
made. First, there is something special about the points P, , and R. Notice that P
is higher than any other nearby point on the curve and li ewise for R. The point
is er than any other nearby point on the curve. Since P, , and R may not
necessarily be the highest or lowest points on the entire curve, we say that the graph
572 C C r e etc n

y
f¿(c)
f¿(a) = 0
does not exist
sign (f¿(x)) = + P R
Relative
sign (f¿(x)) = - sign (f¿(x)) = + maximum
Relative
maximum Q sign (f ¿(x)) = -

Relative f¿ (b) = 0
minimum
x
a b c

FIGURE Relative maxima and relative minima.

of f has re ati e maxima at a and at c and has a re ati e minimum at b. The function
e sure to note the difference between f has re ati e maximum a ues f f.a/ at a and f.c/ at c and has a re ati e minimum
relative extreme a ues and here they a ue f f.b/ at b. We also say that .a; f.a// and .c; f.c// are re ati e maximum p ints
occur. and .b; f.b// is a re ati e minimum p int n the graph f f.
Turning bac to the graph, we see that there is an abs ute maximum (highest point
on the entire curve) at a, but there is no abs ute minimum (lowest point on the entire
curve) because the curve is assumed to extend downward indefinitely. ore precisely,
we define these new terms as follows:

A function f has a relative maximum at a if there is an open interval containing a

on which f.a/ " f.x/ for all x in the interval. The relative maximum value is f.a/. A
function f has a relative minimum at a if there is an open interval containing a on
which f.a/ # f.x/ for all x in the interval. The relative minimum value is f.a/.

If it exists, an absolute maximum value is

unique however, it may occur at more
than one value of x. A similar statement A function f has an a solute maximum at a if f.a/ " f.x/ for all x in the domain of
is true for an absolute minimum. f. The absolute maximum value is f.a/. A function f has an a solute minimum at a
if f.a/ # f.x/ for all x in the domain of f. The absolute minimum value is f.a/.

We refer to either a relative maximum or a relative minimum as a relative extremum

(plural: re ati e extrema). Similarly, we spea of absolute extrema.
When dealing with relative extrema, we compare the function value at a point with
values of nearby points however, when dealing with absolute extrema, we compare the
function value at a point with all other values determined by the domain. Thus, relative
extrema are ca in nature, whereas absolute extrema are g ba in nature.
Referring to Figure 13.5, we notice that at a relative extremum the derivative may
not be defined (as when x D c). ut whenever it is defined at a relative extremum, it is
0 (as when x D a and when x D b), and hence, the tangent line is horizontal. We can
state the following:

R AN C R E
If f has a relative extremum at a, then f .a/ D 0 or f 0 .a/ does not exist.
0

The implication in Rule 2 goes in only one direction:

8
! < f 0 .a/ D 0
relative extremum
implies or
at a :f 0 .a/ does not exist
 Section 3. e at e tre a 573
y y

f¿(a) does not exist

y = f(x) but no relative
extremum at a
f¿(a) = 0
but no relative
extremum at a

x x
a a
(a) (b)

FIGURE No relative extremum at a.

Rule 2 does n t say that if f 0 .a/ is 0 or f 0 .a/ does not exist, then there must be a relative
extremum at a. In fact, there may not be one at all. For example, in Figure 13.6(a), f 0 .a/
is 0 because the tangent line is horizontal at a, but there is no relative extremum there.
In Figure 13.6(b), f 0 .a/ does not exist because the tangent line is vertical at a, but again,
there is no relative extremum there.
ut if we want to find all relative extrema of a function and this is an important
tas what Rule 2 d es tell us is that we can limit our search to those values of x in the
domain of f for which either f 0 .x/ D 0 r f 0 .x/ does not exist. Typically, in applications,
this cuts down our search for relative extrema from the infinitely many x for which f
is defined to a small finite number of p ssibi ities. ecause these values of x are so
important for locating the relative extrema of f, they are called the critica a ues for f,
and if a is a critical value for f, then we also say that .a; f.a// is a critica p int on the
graph of f. Thus, in Figure 13.5, the numbers a, b, and c are critical values, and P, ,
and R are critical points.

For a in the domain of f, if either f 0 .a/ D 0 or f 0 .a/ does not exist, then a is called a
critical value for f. If a is a critical value, then the point .a; f.a// is called a critical
point for f.

At a critical point, there may be a relative maximum, a relative minimum, or neither.

oreover, from Figure 13.5, we observe that each relative extremum occurs at a point
around which the sign of f 0 .x/ is changing. For the relative maximum at a, the sign of
f 0 .x/ goes from C for x < a to ! for x > a, as ng as x is near a. For the relative
minimum at b, the sign of f 0 .x/ goes from ! to C, and for the relative maximum at
c, it again goes from C to !. Thus, ar und re ati e maxima f is increasing and then
decreasing and the re erse h ds f r re ati e minima. ore precisely, we have the
following rule:

R C R E
Suppose f is continuous on an open interval I that contains the critical value a and f
is differentiable on I, except possibly at a.
If f 0 .x/ changes from positive to negative as x increases through a, then f has a
relative maximum at a.
If f 0 .x/ changes from negative to positive as x increases through a, then f has a
relative minimum at a.
574 C C r e etc n

y = f (x) = 12
x
-q 0 q
1
- +
x3
f¿(x) + -
f¿(x) 7 0 f¿ (x) 6 0
f(x)
x

We point out again that not every critical

value corresponds to a relative extremum.
For example, if y D f .x/ D x3 , then (a) (b)
f 0 .x/ D 3x2 . Since f 0 .0/ D 0, 0 is a
critical value. ut if x < 0, then 3x2 > 0, FIGURE f 0 .0/ is not defined, but 0 is not a critical value because 0 is not in the domain of f.
and if x > 0, then 3x2 > 0. Since f 0 .x/
does not change sign at 0, there is no
relative extremum at 0. Indeed, since
f 0 .x/ " 0 for all x, the graph of f never To illustrate Rule 3 with a concrete example, refer again to Figure 13.3, the
falls, and f is said to be n ndecreasing. sign chart for f 0 .x/ D 1 ! 2x2 . The row labeled by f 0 .x/ shows clearly that
(See Figure 13. .) 2
f.x/ D 1 x ! x2 has a relative minimum at !3 and a relative maximum at 3. The
3
row providing the interpretation of the chart for f, labeled f.x/, is immediately deduced
y from the row above it. The significance of the f.x/ row is that it provides an interme-
diate step in actually s etching the graph of f. In this row it stands out, visually, that f
has a relative minimum at !3 and a relative maximum at 3.
y = f (x) = x3 When searching for extrema of a function f, care must be paid to those a that are
not in the domain of f but that are near values in the domain of f. Consider the following
f¿(x) 7 0 example. If
x
1 2
f¿(x) 7 0
y D f.x/ D ; then f 0 .x/ D !
f¿(x) = 0 x2 x3
Although f 0 .x/ does not exist at 0, 0 is not a critical value, because 0 is not in the
domain of f. Thus, a relative extremum cannot occur at 0. Nevertheless, the derivative
FIGURE ero is a critical value,
may change sign around any x-value where f 0 .x/ is not defined, so such values are
but does not give a relative extremum. important in determining intervals over which f is increasing or decreasing. In partic-
ular, such values should be included in a sign chart for f 0 . See Figure 13.7(a) and the
accompanying graph in Figure 13.7(b).
bserve that the thic vertical rule at 0 on the chart serves to indicate that 0 is not
in the domain of f. Here there are no extrema of any ind.
In Rule 3 the hypotheses must be satisfied, or the conclusion need not hold. For
example, consider the case-defined function
8
<1
if x ¤ 0
f.x/ D x2
: 0 if x D 0

Here, 0 is explicitly in the domain of f but f is not continuous at 0. We recall from

Section 11.1 that if a function f is not continuous at a, then f is not differentiable at a,
meaning that f 0 .a/ does not exist. Thus, f 0 .0/ does not exist, and 0 is a critical value
that must be included in the sign chart for f 0 shown in Figure 13. (a). We extend our
sign chart conventions by indicating with a $ symbol those values for which f 0 does not
exist. We see in this example that f 0 .x/ changes from positive to negative as x increases
through 0, but f does n t have a relative maximum at 0. Here Rule 3 does not apply
because its continuity hypothesis is not met. In Figure 13. (b), 0 is displayed in the
domain of f. It is clear that f has an absolute minimum at 0 because f.0/ D 0 and, for
all x ¤ 0, f.x/ > 0.
Summarizing the results of this section, we have the first derivative test for the
relative extrema of y D f.x/:
 Section 3. e at e tre a 575

-q 0 q
1
- * + 1/x2 if x Z 0
x3 y = f (x) =
0 if x = 0
f¿(x) + * -

f(x)
0 x

(a) (b)

FIGURE ero is a critical value, but Rule 3 does not apply.

F T R E
Step Find f 0 .x/.
Step etermine all critical values of f (those a where f 0 .a/ D 0 or f 0 .a/ does
not exist) and any a that are not in the domain of f but that are near values
in the domain of f, and construct a sign chart that shows for each of the
intervals determined by these values whether f is increasing (f 0 .x/ > 0) or
decreasing (f 0 .x/ < 0).
Step For each critical value a at which f is continuous, determine whether
f 0 .x/ changes sign as x increases through a. There is a relative maximum
at a if f 0 .x/ changes from C to ! going from left to right and a relative
minimum if f 0 .x/ changes from ! to C going from left to right. If f 0 .x/
does not change sign, there is no relative extremum at a.
Step For critical values a at which f is not continuous, analyze the situation by
using the definitions of extrema directly.

A L IT I E AM LE F T
The cost equation for a hot dog 4
stand is given by
If y D f.x/ D x C , for x ¤ !1 use the first-derivative test to find where relative
xC1
extrema occur.
c.q/ D 2q3 ! 21q2 C 60q C 500
S
where q is the number of hot dogs sold,
and c.q/ is the cost in dollars. Use the Step f.x/ D x C 4.x C 1/!1 , so
first-derivative test to find where relative
extrema occur. 4 .x C 1/2 ! 4
f 0 .x/ D 1 C 4.!1/.x C 1/!2 D 1 ! D
.x C 1/2 .x C 1/2
x2 C 2x ! 3 .x C 3/.x ! 1/
D D for x ¤ !1
.x C 1/2 .x C 1/2

Note that we expressed f 0 .x/ as a quotient with numerator and denominator

fully factored. This enables us in Step 2 to determine easily where f 0 .x/ is 0
or does not exist and the signs of f 0 .
Step Setting f 0 .x/ D 0 gives x D !3; 1. The denominator of f 0 .x/ is 0 when x is
!1. We note that !1 is not in the domain of f but all values near !1 are in the
domain of f. We construct a sign chart, headed by the values !3, !1, and 1
(which we have placed in increasing order). See Figure 13.10.
576 C C r e etc n

-q -3 -1 1 q
x+3 - 0 + + +

(x + 1)-2 + + + +

x-1 - - - 0 +

f¿(x) + 0 - - 0 +
-
f(x)
-
.x C 3/.x ! 1/
FIGURE Sign chart for f 0 .x/ D .
.x C 1/2
The three values lead us to test four intervals as shown in our sign chart.
n each of these intervals, f is differentiable and is not zero. We determine the
sign of f 0 on each interval by first determining the sign of each of its factors on
each interval. For example, considering first the interval .!1; !3/, it is not
easy to see immediately that f 0 .x/ > 0 there but it is easy to see that x C 3 < 0
for x < !3, while .xC1/!2 > 0 for all x ¤ !1, and x!1 < 0 for x < 1. These
y
observations account for the signs of the factors in the .!1; !3/ column of
the chart. The sign of f 0 .x/ in that column is obtained by multiplying signs
(downward): .!/.C/.!/ D C. We repeat these considerations for the other
y=x+ 4 three intervals. Note that the thic vertical line at !1 in the chart indicates that
x+1 !1 is not in the domain of f and, hence, cannot give rise to any extrema. In the
3 bottom row of the sign chart we record, graphically, the nature of tangent lines
to f.x/ in each interval and at the values where f 0 is 0.
x
-3 -1 1 Step From the sign chart alone we conclude that at !3 there is a relative maximum
(since f 0 .x/ changes from C to ! at !3). oing beyond the chart, we compute
f.!3/ D !3 C .4= ! 2/ D !5, and this gives the relative maximum value of
-5
!5 at !3. We also conclude from the chart that there is a relative minimum
at 1 (because f 0 .x/ changes from ! to C at 1). From f.1/ D 1 C 4=2 D 3 we
see that at 1 the relative minimum value is 3.
FIGURE raph of Step There are no critical values at which f is not continuous, so our considerations
4 above provide the whole story about the relative extrema of f.x/, whose graph
yDxC : is given in Figure 13.11. Note that the general shape of the graph was indeed
xC1
forecast by the bottom row of the sign chart (Figure 13.10).
-q 0 q
(x)-1/3 - * + Now ork Problem 37 G
f¿(x) - * +
E AM LE AR E f 0 .x/ N E
f(x)
Test y D f.x/ D x2=3 for relative extrema.
FIGURE Sign chart for
S We have
2 2 !1=3
f 0 .x/ D p .
33x f 0 .x/ Dx
3
2
D p
y 33x
and the sign chart is given in Figure 13.12. Again, we use the symbol $ on the vertical
line at 0 to indicate that the factor x!1=3 does not exist at 0. Hence, f 0 .0/ does not exist.
Since f is continuous at 0, we conclude from Rule 3 that f has a relative minimum at 0 of
y = x2/3 f.0/ D 0, and there are no other relative extrema. We note further, by inspection of the
x
sign chart, that f has an abs ute minimum at 0. The graph of f follows as Figure 13.13.
Note that we could have predicted its shape from the bottom line of the sign chart in
Figure 13.12, which shows there can be no tangent with a slope at 0. ( f course, the
tangent does exist at 0, but it is a vertical line.)
FIGURE erivative does not
exist at 0, and there is a minimum at 0. Now ork Problem 41 G
 Section 3. e at e tre a 577

E AM LE F R E

Test y D f.x/ D x2 ex for relative extrema.

S y the product rule,

f 0 .x/ D x2 ex C ex .2x/ D xex .x C 2/

A L IT I
Noting that ex is always positive, we obtain the critical values 0 and !2. From the
A drug is in ected into a patient s
sign chart of f 0 .x/ given in Figure 13.14, we conclude that there is a relative maximum
bloodstream. The concentration of the
drug in the bloodstream t hours after the
when x D !2 and a relative minimum when x D 0.
in ection is approximated by
0:14t -q -2 0 q
C.t/ D
t2 C 4t C 4 x+2 - 0 + +
Find the relative extrema for t > 0, and
use them to determine when the drug is x - - 0 +
at its greatest concentration.
ex + + +

f¿(x) + 0 - 0 +
-
f(x)
-

FIGURE Sign chart for f 0 .x/ D x.x C 2/ex .

Now ork Problem 49 G

C S
In the next example we show how the first-derivative test, in con unction with the
notions of intercepts and symmetry, can be used as an aid in s etching the graph of
a function.

E AM LE C S

S etch the graph of y D f.x/ D 2x2 ! x4 with the aid of intercepts, symmetry, and the
first-derivative test.
S nter epts If x D 0, then f.x/ D 0 so that the y-intercept is .0; 0/. Next
note that
p p
f.x/ D 2x2 ! x4 D x2 .2 ! x2 / D x2 . 2 C x/. 2 ! x/
p p p
So if y D 0, then x D 0, ˙ 2 and the x-intercepts are .! 2; 0/, .0; 0/, and . 2; 0/.
We have the sign chart f r f itse f (Figure 13.15), which shows the intervals over which
the graph of y D f.x/ is above the x-axis .C/ and the intervals over which the graph of
y D f.x/ is below the x-axis .!/.

-q - 2 0 2 q
2+x - 0 + + +

x2 + + 0 + +

2-x + + + 0 -

f(x) - 0 + 0 + 0 -

p p
FIGURE Sign chart for f .x/ D . 2 C x/x2 . 2 ! x/.
578 C C r e etc n

-q -1 0 1 q
1+x - 0 + + +

4x - - 0 + +

1-x + + + 0 -

f¿(x) + 0 - 0 + 0 -
- -
f(x)
-

FIGURE Sign chart of y0 D .1 C x/4x.1 ! x/.

ymmetry Testing for y-axis symmetry, we have

f.!x/ D 2.!x/2 ! .!x/4 D 2x2 ! x4 D f.x/
So the graph is symmetric with respect to the y-axis. ecause y is a function (and not the
zero function), there is no x-axis symmetry and, hence, no symmetry about the origin.

irst eri ati e est

Step y0 D 4x ! 4x3 D 4x.1 ! x2 / D 4x.1 C x/.1 ! x/
Step Setting y0 D 0 gives the critical values x D 0, ˙1. Since f is a polynomial, it
is defined and differentiable for all x. Thus, the only values to head the sign
chart for f 0 are !1, 0, 1 (in increasing order) and the sign chart is given in
Figure 13.16. Since we are interested in the graph, the critical p ints are impor-
tant to us. y substituting the critical values into the rigina equation,
y D 2x2 ! x4 , we obtain the y-coordinates of these points. We find the critical
points to be .!1; 1/, .0; 0/, and .1; 1/.
Step From the sign chart and evaluations in step 2, it is clear that f has relative
maxima .!1; 1/ and .1; 1/ and relative minimum .0; 0/. (Step 4 does not apply
here.)
is ussion In Figure 13.17(a), we have indicated the horizontal tangents at the rel-
ative maximum and minimum points. We now the curve rises from the left, has a
relative maximum, then falls, has a relative minimum, then rises to a relative maxi-
mum, and falls thereafter. y symmetry, it su ces to s etch the graph on one side
of the y-axis and construct a mirror image on the other side. We also now, from the
sign chart for f, where the graph crosses and touches the x-axis, and this adds further
precision to our s etch, which is shown in Figure 13.17(b).
As a passing comment, we note that abs ute maxima occur at x D ˙1. See
Figure 13.17(b). There is no absolute minimum.

y y
Relative Relative
y = 2x2 - x4
maximum maximum (-1, 1)
1 1 (1, 1)

x x
-1 1 - 2 -1 (0, 0) 1 2
Relative
minimum

(a) (b)

FIGURE Putting together the graph of y D 2x2 ! x4 .

Now ork Problem 59 G

 Section 3. e at e tre a 579

R BLEMS
In Pr b ems the graph f a functi n is gi en In Pr b ems the deri ati e f a di erentiab e functi n f is
Figures Find the pen inter a s n hich the gi en Find the pen inter a s n hich f is a increasing
functi n is increasing the pen inter a s n hich the functi n is b decreasing and c nd the x a ues f a re ati e extrema
decreasing and the c rdinates f a re ati e extrema f 0 .x/ D .x C 3/.x ! 1/.x ! 2/
y f 0 .x/ D x2 .x ! 2/3
x.x C 2/
f 0 .x/ D .x C 1/.x ! 3/2 f 0 .x/ D
x2 C 1
4
y = f(x) In Pr b ems determine here the functi n is a increasing
3 b decreasing and c determine here re ati e extrema ccur
2 n t sketch the graph
1 y D !x3 ! 1 y D x2 C 4x C 3
x
-2 -1 1 2 3 4 5 5
-1 y D 5 ! 2x ! x2 y D x3 ! x2 ! 2x C 6
2
x3 x4
yD! ! 2x2 C 5x ! 2 yD! ! x3
FIGURE 3 4
3
y y D x4 ! 2x2 y D x3 ! x2 ! 36x
2
7
y D x3 ! x2 C 2x ! 5 y D x3 ! 6x2 C 12x ! 6
1 2
1 2
x y D 2x3 ! x C 10x C 2 y D !5x3 C x2 C x ! 7
-2 -1 1 2 2
y = f(x) 47 3
-1 y D 1 ! 3x C 3x2 ! x3 yD x5 ! x C 10x
5 3
FIGURE y D 3x5 ! 5x3
y x6
y D 3x ! (Remar : x4 C x3 C x2 C x C 1 D 0 has no
2
real roots.)
x4 5x3 7x2
y D !x5 ! 5x4 C 200 yD ! C ! 3x
4 3 2
4 5 13 3
y D x4 ! x yD x ! x C 3x C 4
5 3
p
y = f (x) y D .x2 ! 4/4 y D 3 x.x ! 2/
1 3 3
x yD yD
-4 -2 4 xC2 x
10 ax C b
yD p yD
FIGURE x cx C d
a for ad ! bc > 0
y b for ad ! bc < 0
3 x2 27x2 1
yD yD C
2 y = f (x)
2!x 2 x
x2 ! 3 2x2
1 yD yD
xC2 4x2 ! 25
-3 -1
x ax2 C b p
3 3
1 2 3 yD for d=c < 0 yD x ! x
cx2 C d
a for ad ! bc > 0
-1

-2 b for ad ! bc < 0

-3 y D .x C 1/2=3 y D x2 .x C 3/4

FIGURE y D x3 .x ! 6/4 y D .1 ! x/2=3

580 C C r e etc n

y D e!!x C ! y D x3 ln x Labor Mar ets Eswaran and otwal1 consider agrarian

economies in which there are two types of wor ers, permanent
y D x2 ! ln x y D x!1 ex and casual. Permanent wor ers are employed on long-term
y D ex ! e!x 2 =2 contracts and may receive benefits such as holiday gifts and
y D e!x
emergency aid. Casual wor ers are hired on a daily basis and
y D x2 ln x y D .x2 C 1/e!x perform routine and menial tas s such as weeding, harvesting,
and threshing. The difference, z, in the present-value cost of hiring
In Pr b ems determine inter a s n hich the functi n is a permanent wor er over that of hiring a casual wor er is given by
increasing inter a s n hich the functi n is decreasing re ati e
extrema symmetry and th se intercepts that can be btained z D .1 C b/ p !b c

c n enient y hen sketch the graph where p and c are wage rates for permanent labor and casual
y D x2 ! 3x ! 10 y D 2x2 C x ! 10 labor, respectively, b is a positive constant, and p is a function
of c .
y D 3x ! x3 y D x4 ! 1 a Show that
y D 2x3 ! x2 C 12x y D 2x3 ! x2 ! 4x C 4 ! "
dz d p b
D .1 C b/ !
6 d c d c 1Cb
y D x4 ! 2x2 y D x6 ! x5
5 b If d p =d c < b=.1 C b/, show that z is a decreasing function
p 2
y D .x ! 1/3 .x ! 2/2 y D x.x ! x ! 2/ of c .
p hermal Pollution In Shonle s discussion of thermal
yD2 x!x y D x5=3 ! 2x2=3
pollution,2 the e ciency of a power plant is given by
# $
S etch the graph of a continuous function f such that E D 0:71 1 !
c
f .2/ D 2, f .4/ D 6, f 0 .2/ D f 0 .4/ D 0, f 0 .x/ < 0 for x < 2, h
f 0 .x/ > 0 for 2 < x < 4, f has a relative maximum at 4, and
limx!1 f .x/ D 0. where h and c are the respective absolute temperatures of the
hotter and colder reservoirs. Assume that c is a positive constant
S etch the graph of a continuous function f such that and that h is positive. Using calculus, show that as h increases,
f .0/ D 0, f .1/ D 1, f .2/ D 2, f .3/ D 1, f 0 .0/ D 0 D f 0 .2/, there the e ciency increases.
is a vertical tangent line when x D 1 and when x D 3, f 0 .x/ < 0
for x in .!1; 0/ and x in .2; 3/, f 0 .x/ > 0 for x in .0; 1/ and x in elephone Service In a discussion of the pricing of local
.1; 2/ and x in .3; 1/. telephone service, Renshaw3 determines that total revenue r is
given by
Average Cost If cf D 25;000 is a fixed-cost function, show %
that the average fixed-cost function cf D cf =q is a decreasing a& a2
r D 2F C 1 ! p ! p2 C
function for q > 0. Thus, as output q increases, each unit s portion b b
of fixed cost declines. where p is an indexed price per call, and a, b, and F are constants.
Marginal Cost If c D 3q ! 3q2 C q3 is a cost function, etermine the value of p that maximizes revenue.
when is marginal cost increasing Storage and Shipping Costs In his model for storage and
Marginal Revenue iven the demand function shipping costs of materials for a manufacturing process,
ancaster4 derives the cost function
# $
p D 500 ! 5q 144
C.k/ D 100 100 C k C 1 " k " 100
k
find when marginal revenue is increasing. where C.k/ is the total cost (in dollars) of storage and
p transportation for 100 days of operation if a load of k tons of
Cost Function For the cost function c D q, show that
marginal and average costs are always decreasing for q > 0. material is moved every k days.
Revenue For a manufacturer s product, the revenue a Find C(1).
function is given by r D 1 0q C 7q2 ! 2q3 . etermine the output b For what value of k does C.k/ have a minimum
for maximum revenue. c What is the minimum value

1
. Eswaran and A. otwal, A Theory of Two-Tier abor ar ets in
Agrarian Economics, he American Ec n mic Re ie , 75, no. 1 (1 5),
162 77.
2
. I. Shonle, En ir nmenta App icati ns f Genera Physics (Reading, A:
Addison-Wesley Publishing Company, Inc., 1 75).
3 E. Renshaw, A Note on Equity and E ciency in the Pricing of ocal

Telephone Services, he American Ec n mic Re ie , 75, no. 3 (1 5),

515 1 .
4 P. ancaster, athematics de s f the Rea r d (Englewood Cliffs, N :
Prentice-Hall, Inc., 1 76).
 Section 3.2 so te tre a on a C ose nter a 581

Physiology he Bends When a deep-sea diver raph the function

undergoes decompression or a pilot climbs to a high altitude,
f .x/ D .x.x ! 2/.2x ! 3//2
nitrogen may bubble out of the blood, causing what is
commonly called the bends. Suppose the percentage P of in a calculator window with !1 " x " 3, !1 " y " 3. At first
people who suffer effects of the bends at an altitude of h thousand glance, it may appear that this function has two relative minimum
feet is given by5 points and one relative maximum point. However, in reality, it has
three relative minimum points and two relative maximum points.
100 etermine the x-values of all these points. Round answers to two
PD decimal places.
1 C 100;000e!0:36h
If f .x/ D 3x3 ! 7x2 C 4x C 2, display the graphs of f and f 0
Is P an increasing function of h on the same screen. Notice that f 0 .x/ D 0 where relative extrema
of f occur.
In Pr b ems fr m the graph f the functi n nd the
c rdinates f a re ati e extrema R und y ur ans ers t t et f .x/ D 6 C 4x ! 3x2 ! x3 . a Find f 0 .x/. b raph f 0 .x/.
decima p aces c bserve where f 0 .x/ is positive and where it is negative. ive
the intervals (rounded to two decimal places) where f is increasing
y D 0:3x2 C 2:3x C 5:1 y D 3x4 ! 4x3 ! 5x C 1 and where f is decreasing. d raph f and f 0 on the same screen,
:2x ex .3 ! x/ and verify your results to part (c).
yD yD
0:4x2 C 3 7x2 C 1 If f .x/ D x4 ! x2 ! .x C 2/2 , find f 0 .x/. etermine the critical
values of f. Round your answers to two decimal places.

Objective A E C I
o n e tre e a es on a c ose If a function f is c ntinu us on a c sed interval Œa; b", it can be shown that of a
nter a
the function values f.x/ for x in Œa; b", there must be an absolute maximum value and
an absolute minimum value. These two values are called extreme values of f on that
interval. This important property of continuous functions is called the extreme value
theorem.

E T
If a function is continuous on a closed interval, then the function has b th a maxi-
mum value and a minimum value on that interval.

For example, each function in Figure 13.22 is continuous on the closed interval Œ1; 3".
eometrically, the extreme-value theorem assures us that over this interval each graph
has a highest point and a lowest point.

y y
Highest Highest
point point

Lowest Lowest
point point

x x
1 3 1 3

FIGURE Illustrating the extreme-value theorem.

In the extreme-value theorem, it is important that we are dealing with

a closed interval, and
a function continuous on that interval.

5
Adapted from . E. Fol , r., extb k f En ir nmenta Physi gy 2nd ed. (Philadelphia: ea Febiger, 1 74).
582 C C r e etc n

y y

f (x) = 12
x
f (x) = x2
1 1

x x
-1 1 -1 1

Open interval (-1, 1) Not continuous at 0

No maximum, minimum = 0 No maximum, minimum = 1
(a) (b)

FIGURE Extreme-value theorem does not apply.

If either condition (1) or condition (2) is not met, then extreme values are not guaran-
teed. For example, Figure 13.23(a) shows the graph of the continuous function f.x/ D x2
on the pen interval .!1; 1/. ou can see that f has no maximum value on the inter-
val (although f has a minimum value there). Now consider the function f.x/ D 1=x2
on the closed interval Œ!1; 1". Here f is n t c ntinu us at 0. From the graph of f in
Figure 13.23(b), it can be seen that f has no maximum value (although there is a
minimum value).
In the previous section, our emphasis was on relative extrema. Now we will focus
our attention on absolute extrema and ma e use of the extreme-value theorem where
possible. If the domain of a function is a closed interval, to determine abs ute extrema
we must examine the function not only at critical values, but also at the endpoints. For
example, Figure 13.24 shows the graph of the continuous function y D f.x/ over Œa; b".
The extreme-value theorem guarantees absolute extrema over the interval. Clearly, the
important points on the graph occur at x D a, b, c, and d, which correspond to endpoints
or critical values. Notice that the absolute maximum occurs at the critical value c and
the absolute minimum occurs at the endpoint a. These results suggest the following
procedure:

F A E F fT I C
Œa; b"
Step Find the critical values of f.
Step Evaluate f.x/ at the endpoints a and b and at the critical values in .a; b/.
Step The maximum value of f is the greatest of the values found in step 2. The
minimum value of f is the least of the values found in step 2.

y = f (x)

f (c) Absolute
maximum, f (c)

f (a) Absolute minimum, f (a)

x
a c d b

Endpoint Critical values Endpoint

FIGURE Absolute extrema.

 Section 3.3 Conca t 583

E AM LE F E C I

Find absolute extrema for f.x/ D x2 ! 4x C 5 over the closed interval 1, 4 .

S Since f is continuous on 1, 4 , the foregoing procedure applies.
Step To find the critical values of f, we first find f 0 :
y f 0 .x/ D 2x ! 4 D 2.x ! 2/
y = x 2 - 4x + 5, 1 … x … 4
This gives the critical value x D 2.
5
Absolute Step Evaluating f.x/ at the endpoints 1 and 4 and at the critical value 2, we have
maximum, f (4)

f.1/ D 2
2 Absolute values of f at endpoints
1 minimum, f (2) f.4/ D 5
x
1 2 4
and
FIGURE Extreme values for
Example 1. f.2/ D 1 value of f at critical value 2 in .1; 4/

Step From the function values in Step 2, we conclude that the maximum is f.4/ D 5
and the minimum is f.2/ D 1. (See Figure 13.25.)

Now ork Problem 1 G

R BLEMS
In Pr b ems nd the abs ute extrema f the gi en functi n f .x/ D x4 ! x2 C 2, Œ!1; 3"
n the gi en inter a x
f .x/ D 2 , Œ2; 3"
f .x/ D x2 ! 2x C 3, Œ0; 3" x !1

f .x/ D !3x2 C 12x C 1, Œ1; 3" f .x/ D .x ! 1/2=3 , Œ!26; 2 "

f .x/ D 13 x3 C 12 x2 ! 2x C 1, Œ!1; 0" f .x/ D 0:2x3 ! 3:6x2 C 2x C 1, Œ!1; 2"

f .x/ D 14 x4 ! 23 x2 , Œ0; 1" Consider the function

f .x/ D x4 C x3 C 21x2 C 20x C
f .x/ D x3 ! 5x2 ! x C 50, Œ0; 5"
over the interval Œ!4; ".
f .x/ D x2=3 , Œ! ; "
a etermine the value(s) (rounded to two decimal places) of x
f .x/ D .1=6/x6 ! .3=4/x4 ! 2x2 , Œ!1; 1" at which f attains a minimum value.
b What is the minimum value (rounded to two decimal places)
f .x/ D 73 x3 C 2x2 ! 3x C 1, Œ0; 3" of f
f .x/ D 3x4 ! x6 , Œ!1; 2" c etermine the value(s) of x at which f attains a maximum
value.
f .x/ D x4 ! x3 C 22x2 ! 24x C 2, Œ0; 4" d What is the maximum value of f

Objective C
o test a f nct on for conca t an The first derivative provides a lot of information for s etching curves. It is used to
n ect on o nts o s etc c r es t
t e a of t e nfor at on o ta ne determine where a function is increasing, is decreasing, has relative maxima, and has
fro ot rst an secon er at es relative minima. However, to be sure we now the true shape of a curve, we may need
more information. For example, consider the curve y D f.x/ D x2 . Since f 0 .x/ D 2x,
x D 0 is a critical value. If x < 0, then f 0 .x/ < 0, and f is decreasing if x > 0,
then f 0 .x/ > 0, and f is increasing. Thus, there is a relative minimum at x D 0. In
Figure 13.26, both curves meet the preceding conditions. ut which one truly describes
the curve y D x2 This question will be settled easily by using the second derivative
and the notion of c nca ity.
584 C C r e etc n

y y

x x

(a) (b)

FIGURE Two functions with f 0 .x/ < 0 for x < 0 and f 0 .x/ > 0 for x > 0.

y y y
y = f (x)

y = f (x) y = f (x)
Slope
Slope Slope
increasing
increasing increasing
x x x

(a) (b) (c)

FIGURE Each curve is concave up.

In Figure 13.27, note that each curve y D f.x/ bends (that is opens) upward.
This means that if tangent lines are drawn to each curve, the curves lie ab e them.
oreover, the slopes of the tangent lines increase in value as x increases: In part (a),
the slopes go from small positive values to larger values in part (b), they are negative
and approaching zero (and thus increasing) in part (c), they pass from negative values
to positive values. Since f 0 .x/ gives the slope at a point, an increasing slope means that
f 0 must be an increasing function. To describe this property, each curve in Figure 13.27
is said to be c nca e up.
In Figure 13.2 , it can be seen that each curve lies be the tangent lines and
the curves are bending downward. As x increases, the slopes of the tangent lines are
decreasing. Thus, f 0 must be a decreasing function here, and we say that f is c nca e
d n.

y y y

y = f (x) y = f (x)
Slope
decreasing
Slope Slope
y = f (x) decreasing decreasing

x x x

(a) (b) (c)

FIGURE Each curve is concave down.

Concavity relates to whether f 0 , not f,

is increasing or decreasing. In et f be differentiable on the interval .a; b/. Then f is said to be concave up
Figure 13.27(b), note that f is concave
up and decreasing however, in concave down on .a; b/ if f 0 is increasing decreasing on .a; b/.
Figure 13.2 (a), f is concave down
and decreasing.
Remember If f is concave up on an interval, then geometrically its graph is bend-
ing upward there. If f is concave down, then its graph is bending downward.
Since f 0 is increasing when its derivative f 00 .x/ is positive, and f 0 is decreasing
when f 00 .x/ is negative, we can state the following rule:
 Section 3.3 Conca t 585

R C C
et f be differentiable on the interval .a; b/. If f 00 .x/ > 0 for all x in .a; b/, then f
0

is concave up on .a; b/. If f 00 .x/ < 0 for all x in .a; b/, then f is concave down on
.a; b/.

A function f is also said to be concave up at a point c if there exists an open interval

around c on which f is concave up. In fact, for the functions that we will consider, if
f 00 .c/ > 0, then f is concave up at c. Similarly, f is concave down at c if f 00 .c/ < 0.

E AM LE T C

etermine where the given function is concave up and where it is concave down.
a y D f.x/ D .x ! 1/3 C 1.
S To apply Rule 1, we must examine the signs of y00 . Now, y0 D 3.x ! 1/2 , so
y00 D 6.x ! 1/
Thus, f is concave up when 6.x ! 1/ > 0 that is, when x > 1. And f is concave down
when 6.x ! 1/ < 0 that is, when x < 1. We now use a sign chart for f 00 (together with
an interpretation line for f) to organize our findings. (See Figure 13.2 .)

-q 1 q
Concave
x-1 - 0 + down 1 Concave
up
f¿(x) - 0 +
x
f(x) 1
y = f (x ) = (x - 1)3 + 1

FIGURE Sign chart for f 00 and concavity for f .x/ D .x ! 1/3 C 1.

b y D x2 .
S We have y0 D 2x and y00 D 2. ecause y00 is always positive, the graph of
2
y D x must always be concave up, as in Figure 13.26(a). The graph cannot appear as
in Figure 13.26(b), for that curve is sometimes concave down.
Now ork Problem 1 G
A point on a graph where concavity changes from concave down to concave up,
or vice versa, such as .1; 1/ in Figure 13.2 , is called an in ecti n p int, equivalently
a p int f in ecti n. Around such a point, the sign of f 00 .x/ goes from ! to C or from
C to !. ore precisely, we have the following definition:

The definition of an in ection point A function f has an in ection point at a if and only if f is continuous at a and f
implies that a is in the domain of f. changes concavity at a.

To test a function for concavity and in ection points, first find the values of x where
either f 00 .x/ D 0 or f 00 .x/ is not defined. These values of x determine intervals. n each
interval, determine whether f 00 .x/ > 0 (f is concave up) or f 00 .x/ < 0 (f is concave
down). If concavity changes around one of these x-values and f is continuous there,
then f has an in ection point at this x-value. The continuity requirement implies that
586 C C r e etc n

the x-value must be in the domain of the function. In brief, a candidate for an in ection
point must satisfy two conditions:
f 00 must be 0 or fail to exist at that point.
f must be continuous at that point.
y

f (x) = x1/3
f–(x) 7 0
f concave up Inflection point
x

f–(x) 6 0
f concave down

FIGURE In ection point for f .x/ D x1=3 .

The candidate i be an in ection point if concavity changes around it. For example,
if f.x/ D x1=3 , then f 0 .x/ D 13 x!2=3 and
2 2
f 00 .x/ D ! x!5=3 D ! 5=3
x
ecause f 00 does not exist at 0, but f is continuous at 0, there is a candidate for an
in ection point at 0. If x > 0, then f 00 .x/ < 0, so f is concave down for x > 0 if x < 0,
then f 00 .x/ > 0, so f is concave up for x < 0. ecause concavity changes at 0, there is
an in ection point there. (See Figure 13.30.)

E AM LE C I

Test y D 6x4 ! x3 C 1 for concavity and in ection points.

S We have
y0 D 24x3 ! 24x2
y00 D 72x2 ! 4 x D 24x.3x ! 2/

-q 0 2/3 q
x - 0 + +
y
y = 6x4 - 8x3 + 1 3x - 2 - - 0 +

y– + 0 - 0 +

Inflection y
points

1 FIGURE Sign chart of y00 D 24x.3x ! 2/ for y D 6x4 ! x3 C 1.

2
3 To find where y00 D 0, we set each factor in y00 equal to 0. This gives x D 0, 23 . We also
x
-5 note that y00 is never undefined. Thus, there are three intervals to consider, as recorded
27
on the top of the sign chart in Figure 13.31. Since y is continuous at 0 and 23 , these
points are candidates for in ection points. Having completed the sign chart, we see
that concavity changes at 0 and at 23 . Thus, these candidates are indeed in ection points.
Concave Concave Concave
up down up (See Figure 13.32.) In summary, the curve is concave up on .!1; 0/ and on . 23 ; 1/
FIGURE raph of
and is concave down on .0; 23 /. In ection points occur at 0 and at 23 . These points are
y D 6x4 ! x3 C 1. .0; y.0// D .0; 1/ and . 23 ; y. 23 // D . 23 ; ! 27
5
/.
Now ork Problem 13 G
 Section 3.3 Conca t 587

As we did in the analysis of increasing and decreasing, so we must in concavity

analysis consider also those points a that are not in the domain of f but that are near
points in the domain of f. The next example will illustrate.

E AM LE AC C N I
-q 0 q
1
- + 1
x3 iscuss concavity and find all in ection points for f.x/ D .
x
f–(x) - +
S Since f.x/ D x!1 for x ¤ 0,
f(x)
f 0 .x/ D !x!2 for x ¤ 0
FIGURE Sign chart for f 00 .x/. 2
f 00 .x/ D 2x!3 D 3 for x ¤ 0
x
y We see that f 00 .x/ is never 0 but it is not defined when x D 0. Since f is not continuous at
0, we conclude that 0 is not a candidate for an in ection point. Thus, the given function
y = x1 has no in ection point. However, 0 must be considered in an analysis of concavity. See
Concave
down the sign chart in Figure 13.33 note that we have a thic vertical line at 0 to indicate
that 0 is not in the domain of f and cannot correspond to an in ection point. If x > 0,
x
then f 00 .x/ > 0 if x < 0, then f 00 .x/ < 0. Hence, f is concave up on .0; 1/ and concave
Concave down on .!1; 0/. (See Figure 13.34.) Although concavity changes around x D 0,
up
there is no in ection point there because f is not continuous at 0 (nor is it even defined
there).

1 Now ork Problem 23 G

FIGURE raph of y D .
x
A candidate for an in ection point may C S
not necessarily be an in ection point. For
example, if f .x/ D x4 , then f 00 .x/ D 12x2 E AM LE C S
and f 00 .0/ D 0. ut f 00 .x/ > 0 both when
x < 0 and when x > 0. Thus, concavity
does not change, and there are no S etch the graph of y D 2x3 ! x2 C 12x.
in ection points. (See Figure 13.35.) S
nter epts If x D 0, then y D 0. Setting y D 0 gives 0 D x.2x2 ! x C 12/. Clearly,
x D 0 is a solution, and using the quadratic formula on 2x2 ! x C 12 D 0 gives no
real roots. Thus, the only intercept is .0; 0/. In fact, since 2x2 ! x C 12 is a contin-
y
uous function whose value at 0 is 2 # 02 ! # 0 C 12 D 12 > 0, we conclude that
y = f (x) = x4 2x2 ! x C 12 > 0 for all x, which gives the sign chart in Figure 13.36 for y.
Note that the sign chart for y itself tells us the graph of y D 2x3 ! x2 C 12x is
confined to the third and first quadrants of the xy-plane.
ymmetry None.
x
axima an inima We have
0
FIGURE raph of y D 6x ! 1 x C 12 D 6.x2 ! 3x C 2/ D 6.x ! 1/.x ! 2/
2
f .x/ D x4 .
The critical values are x D 1 and x D 2, so these and the factors x ! 1 and x ! 2
determine the sign chart of y0 (Figure 13.37).

-q 1 2 q
x-1 - 0 + +
-q 0 q
x-2 - - 0 +
x - 0 +
y¿ + 0 - 0 +
2x2 - 9x + 12 + +
-
0 y
y - +
-

FIGURE Sign chart for y D 2x3 ! x2 C 12x. FIGURE Sign chart for y0 D 6.x ! 1/.x ! 2/.
588 C C r e etc n

-q 3/2 q
2x - 3 - 0 +

y– - 0 +

FIGURE Sign chart for y00 D 6.2x ! 3/.

From the sign chart for y0 we see that there is a relative maximum at 1 and a relative
minimum at 2. Note, too, that the bottom line of Figure 13.37, together with that of
y
Figure 13.36, comes close to determining a precise graph of y D 2x3 ! x2 C 12x.
f course, it will help to now the relative maximum y.1/ D 5, which occurs at 1, and
the relative minimum y.2/ D 4, which occurs at 2, so that in addition to the intercept
.0; 0/ we will actually plot also .1; 5/ and .2; 4/.

5
on a ity

4 y00 D 12x ! 1 D 6.2x ! 3/

y = 2x3 - 9x2 + 12x Setting y00 D 0 gives a possible in ection point at x D 32 , from which we construct the
simple sign chart for y00 in Figure 13.3 .
Since concavity changes at x D 32 , at which point f is certainly continuous, there
is an in ection point at 32 .

1 3 2
x is ussion We now the coordinates of three of the important points on the graph.
2 The only other important point from our perspective is the in ection point, and since
y.3=2/ D 2.3=2/3 ! .3=2/2 C 12.3=2/ D =2 the in ection point is .3=2; =2/.
FIGURE raph of We plot the four points noted above and observe from all three sign charts ointly
y D 2x3 ! x2 C 12x. that the curve increases through the third quadrant and passes through .0; 0/, all the
while concave down until a relative maximum is attained at .1; 5/. The curve then falls
until it reaches a relative minimum at .2; 4/. However, along the way the concavity
20 changes at .3=2; =2/ from concave down to concave up and remains so for the rest
of the curve. After .2; 4/ the curve increases through the first quadrant. The curve is
shown in Figure 13.3 .
-2 8 Now ork Problem 39 G
Suppose that we need to find the in ection points for
-20
1 5 17 4 273 3 4225 2 750
FIGURE raph of f 00 roots of f.x/ D x ! x C x ! x C
00
f D 0 are approximately 3.25 and 6.25. 20 16 32 12 4

The second derivative of f is given by

300
51 2 1 4225
f 00 .x/ D x3 ! x C x!
4 16 64
-5 10
Here the roots of f 00 D 0 are not obvious. Thus, we will graph f 00 . (See Figure 13.40.)
We find that the roots of f 00 D 0 are approximately 3:25 and 6:25. Around x D 6:25,
f 00 .x/ goes from negative to positive values. Therefore, at x D 6:25, there is an in ection
point. Around x D 3:25, f 00 .x/ does not change sign, so no in ection point exists at
-300

FIGURE raph of f in ection x D 3:25. Comparing our results with the graph of f in Figure 13.41, we see that
point at x D 6:25, but not at x D 3:25. everything chec s out.
 Section 3.3 Conca t 589

R BLEMS
In Pr b ems a functi n and its sec nd deri ati e are gi en In Pr b ems determine inter a s n hich the functi n
etermine the c nca ity f f and nd x a ues here p ints f is increasing decreasing c nca e up and c nca e d n re ati e
in ecti n ccur maxima and minima in ecti n p ints symmetry and th se inter
f .x/ D x4 ! 3x3 ! 6x2 C 6x C 1 f 00 .x/ D 6.2x C 1/.x ! 2/ cepts that can be btained c n enient y hen sketch the graph

x5 x4 y D x2 ! x ! 6 y D x2 C a for a > 0
f .x/ D C ! 2x2 f 00 .x/ D .x ! 1/.x C 2/2
20 4
y D 5x ! 2x2 y D !1 ! x2 C 2x
2
x C 3x C 1 00 2x ! 4
f .x/ D f .x/ D
x2 C 2x C 1 .x C 1/4 y D x3 ! x2 C 24x ! 1 y D x3 ! 25x2

x2 2.2x C 1/ x3
f .x/ D f 00 .x/ D yD ! 5x y D x3 ! 6x2 C x
.x ! 1/2 .x ! 1/4 3
x2 C 1 00 6.3x2 C 2/ 5
f .x/ D f .x/ D y D x3 C 3x2 C 3x C 1 y D 2x3 C x2 C 2x
x2 ! 2 .x2 ! 2/3 2
p x.2x2 ! 3a2 / y D 4x3 ! 3x4 y D !x3 C x2 ! 5x C 3
f .x/ D x a2 ! x2 f 00 .x/ D 2
.a ! x2 /3=2
y D !2 C 12x ! x3 y D !.3x C 2/3
In Pr b ems determine c nca ity and the x a ues here
x5 x4
p ints f in ecti n ccur n t sketch the graphs y D 2x3 ! 6x2 C 6x ! 2 yD !
100 20
y D !2x2 C 4x y D 4x2 ! 375x C 47
y D 16x ! x5 y D x2 .x ! 1/2
y D 4x3 C 12x2 ! 12x y D x3 ! 6x2 C x C 1
y D 6x4 ! x3 C 3 y D 3x5 ! 5x3
y D ax3 C bx2 C cx C d y D x4 ! x2 ! 6

110 3 y D 4x2 ! x4 y D x2 ex
y D x5 ! 10x4 C 3
x ! 60x2

x4 x2 y D x1=3 .x ! / y D .x C 1/2 .x ! 2/2

yD! C C 2x
4 2 p
y D 4x1=3 C x4=3 y D .x C 1/ x C 4
a
y D 2x1=5 yD 3
x y D 2x2=3 ! x y D 5x2=3 ! x5=3
x4 1 x3 7x2
yD C ! CxC5
2 6 2 S etch the graph of a continuous function f such that
2 4 11 3 3 2 7 3 f .0/ D 0 D f .3/, f 0 .1/ D 0 D f 0 .3/, f 00 .x/ < 0 for x < 2, and
yD x C x C x C xC f 00 .x/ > 0 for x > 2.
4 6 2 5 5
1 5 1 4 1 3 1 2 S etch the graph of a continuous function f such that
yD x ! x C x ! x! f .4/ D 4, f 0 .4/ D 0, f 00 .x/ < 0 for x < 4, and f 00 .x/ > 0 for
20 4 6 2 3
x > 4.
1 5
yD x ! 3x3 C 17x C 43
10 S etch the graph of a continuous function f such that
1 6 7 f .1/ D 1, f 0 .1/ D 0, and f 00 .x/ < 0 for all x.
yD x ! x4 C 6x2 C 5x ! 4
30 12 S etch the graph of a continuous function f such that
6 4 x!1 f .1/ D 1, both f 0 .x/ < 0 and f 00 .x/ < 0 for x < 1, and both
y D x ! 3x yD
xC1 f .x/ > 0 and f 00 .x/ < 0 for x > 1.
1 x2 Demand Equation Show that the graph of the demand
yD1! yD
x2 x2 C 1 100
equation p D is decreasing and concave up for q > 0.
21x C 40 qC2
ax2
yD yD
xCb 6.x C 3/2 Average Cost For the cost function
y D .x2 ! 12/2 y D 5ex
c D q2 C 3q C 2
y D ex ! e!x y D axex
2 x2 C 1
y D xex y D ln x yD show that the graph of the average-cost function c is concave up
3ex for all q > 0.
590 C C r e etc n

Species of Plants The number of species of plants on a Entomology In a study of the effects of food deprivation
plot may depend on the size of the plot. For example, in on hunger, an insect was fed until its appetite was completely
Figure 13.42, we see that on 1-m2 plots there are three species satisfied. Then it was deprived of food for t hours (the deprivation
(A, , and C on the left plot, A, , and on the right plot), and period). At the end of this period, the insect was re-fed until its
on a 2-m2 plot there are four species (A, , C, and ). appetite was again completely satisfied. The weight (in grams)
of the food that was consumed at this time was statistically found
2 sq meters to be a function of t, where

A D 1:00Œ1 ! e!.0:0464tC0:0670/ "

A
B
Here is a measure of hunger. Show that is increasing with
C
respect to t and is concave down.
B
D
Insect Dispersal In an experiment on the dispersal of a
particular insect,10 a large number of insects are placed at a release
point in an open field. Surrounding this point are traps that are
1 sq meter 1 sq meter placed in a concentric circular arrangement at a distance of 1 m,
2 m, 3 m, and so on from the release point. Twenty-four hours
FIGURE
after the insects are released, the number of insects in each trap is
In a study of rooted plants in a certain geographic region,6 it was counted. It is determined that at a distance of r meters from the
determined that the average number of species, S, occurring on release point, the average number of insects contained in a trap is
plots of size A (in square meters) is given by 7
n D f .r/ D 0:1 ln.r/ C ! 0: 1 " r " 10
p r
4
S D f .A/ D 12 A 0 " A " 625 a Show that the graph of f is always falling and concave up.
b S etch the graph of f. (c) When r D 5, at what rate is the
S etch the graph of f. ( te our graph should be rising and average number of insects in a trap decreasing with respect to
concave down. Thus, the number of species is increasing with distance
respect to area, but at a decreasing rate.) raph y D !0:35x3 C 4:1x2 C :3x ! 7:4, and from the graph
Inferior Good In a discussion of an inferior good, Pers y7 determine the number of a relative maximum points, b relative
considers a function of the form minimum points, and c in ection points.
raph y D x5 .x ! 2:3/, and from the graph determine the
!x2 =.2A/ number of in ection points. Now, prove that for any a ¤ 0, the
g.x/ D e. 0 =A/
e
curve y D x5 .x ! a/ has two points of in ection.
where x is a quantity of a good, 0 is a constant that represents raph y D xe!x and determine the number of in ection
utility, and A is a positive constant. Pers y claims that the graph of points, first using a graphing calculator and then using the
p p
g is concave down for x < A and concave up for x > A. techniques of this chapter. If a demand equation has the form
Verify this. q D q.p/ D e!Rp for constants and R, relate the graph of the
resulting revenue function to that of the function graphed above,
Psychology In a psychological experiment involving by ta ing D 1 D R.
conditioned response, sub ects listened to four tones, denoted 0,
1, 2, and 3. Initially, the sub ects were conditioned to tone 0 by raph the curve y D x3 ! 2x2 C x C 3, and also graph the
receiving a shoc whenever this tone was heard. ater, when each tangent line to the curve at x D 2. Around x D 2, does the curve
of the four tones (stimuli) was heard without shoc s, the sub ects lie above or below the tangent line From your observation
responses were recorded by means of a trac ing device that determine the concavity at x D 2.
measures galvanic s in reaction. The average response to each et f be a function for which both f 0 .x/ and f 00 .x/ exist.
stimulus (without shoc ) was determined, and the results were Suppose that f 0 has a a relative minimum at a. Show that f
plotted on a coordinate plane where the x- and y-axes represent changes its direction of bending at a. This means that the
the stimuli (0, 1, 2, 3) and the average galvanic responses, concavity of f changes at x D a which means that the direction
respectively. It was determined that the points fit a curve that is of bending of the graph of f changes at x D a.
approximated by the graph of
If f .x/ D x6 C 3x5 ! 4x4 C 2x2 C 1, find the x-values
(rounded to two decimal places) of the in ection points of f.
y D 12:5 C 5: .0:42/x xC1
If f .x/ D 2 , find the x-values (rounded to two decimal
x C1
Show that this function is decreasing and concave up. places) of the in ection points of f.

6
Adapted from R. W. Poole, An Intr ducti n t uantitati e Ec gy (New
or : c raw-Hill oo Company, 1 74).
7
A. . Pers y, An Inferior ood and a Novel Indifference ap, he American
Ec n mist I , no. 1 (1 5), 67 6 .
Adapted from C. I. Hovland, The eneralization of Conditioned Responses: C. S. Holling, The Functional Response of Invertebrate Predators to Prey
I. The Sensory eneralization of Conditioned Responses with Varying ensity, em irs f the Ent m gica S ciety f Canada no. 4 (1 66).
10 Adapted from Poole, op. cit.
Frequencies of Tone, urna f Genera Psych gy 17 (1 37), 125 4 .
 Section 3.4 e econ er at e est 591

Objective T S T
o ocate re at e e tre a a n The second derivative can be used to test certain critical values for relative extrema.
t e secon er at e test
bserve in Figure 13.43 that at a there is a horizontal tangent that is, f 0 .a/ D 0.
Furthermore, around a the function is concave up (that is, f 00 .a/ > 0). This leads us to
y Concave up and
conclude that there is a relative minimum at a. n the other hand, around b the function
relative minimum is concave down (that is, f 00 .b/ < 0). ecause the tangent line is horizontal at b, we
Concave down and
relative maximum conclude that a relative maximum exists there. This technique of examining the second
derivative at points where the first derivative is 0 is called the second derivative test
for relative extrema.
y = f(x)

a b
x S T R E
0
Suppose f .a/ D 0.
FIGURE Relating concavity to
relative extrema.
If f 00 .a/ < 0, then f has a relative maximum at a.
If f 00 .a/ > 0, then f has a relative minimum at a.

We want to emphasize that the sec nd deri ati e test d es not app y hen
f 00 .a/ D 0. If both f 0 .a/ D 0 and f 00 .a/ D 0, then there may be a relative maxi-
mum, a relative minimum, or neither at a. In such cases, the first-derivative test should
be used to analyze what is happening at a. (Also, the second-derivative test does not
apply when f 0 .a/ does not exist.)

E AM LE S T

Test the following for relative maxima and minima. Use the second-derivative test,
if possible.

a y D 1 x ! 23 x3 .
S

y0 D 1 ! 2x2 D 2. ! x2 / D 2.3 C x/.3 ! x/

y00 D !4x

Solving y0 D 0 gives the critical values x D ˙3.

If x D 3; then y00 D !4.3/ D !12 < 0:

There is a relative maximum when x D 3.

If x D !3; then y00 D !4.!3/ D 12 > 0:

There is a relative minimum when x D !3. (Refer to Figure 13.4.)

Although the second-derivative test can
be very useful, do not depend entirely on b y D 6x4 ! x3 C 1.
it. Not only may the test fail to apply, but
also it may be aw ward to find the S
second derivative.
y0 D 24x3 ! 24x2 D 24x2 .x ! 1/
y00 D 72x2 ! 4 x

Solving y0 D 0 gives the critical values x D 0, 1. We see that

if x D 0; then y00 D 0
592 C C r e etc n

and

if x D 1; then y00 > 0

y the second-derivative test, there is a relative minimum when x D 1. We cannot

apply the test when x D 0 because y00 D 0 there. To analyze what is happening at 0,
we turn to the first-derivative test:

If x < 0; then y0 < 0:

y
If 0 < x < 1; then y0 < 0:

y = x2 Thus, no maximum or minimum exists when x D 0. (Refer to Figure 13.35.)

x Now ork Problem 5 G

Relative and
absolute extremum
when x = 0 If a continuous function has exact y ne relative extremum on an interval, it can be
shown that the relative extremum must also be an abs ute extremum on the interval. To
illustrate, in Figure 13.44 the function y D x2 has a relative minimum when x D 0, and
FIGURE Exactly one relative there are no other relative extrema. Since y D x2 is continuous, this relative minimum
extremum implies an absolute extremum. is also an absolute minimum for the function.

E AM LE A E

If y D f.x/ D x3 ! 3x2 ! xC 5, determine when absolute extrema occur on the interval

.0; 1/.
S We have

f 0 .x/ D 3x2 ! 6x ! D 3.x2 ! 2x ! 3/

D 3.x C 1/.x ! 3/
y
y = x 3 - 3x 2 - 9x + 5 The only critical value on the interval .0; 1/ is 3. Applying the second-derivative test
5 at this point gives
x
3
f 00 .x/ D 6x ! 6
f 00 .3/ D 6.3/ ! 6 D 12 > 0
-22

FIGURE n .0; 1/, there is Thus, there is a relative minimum at 3. Since this is the only relative extremum on
an absolute minimum at 3. .0; 1/ and f is continuous there, we conclude by our previous discussion that there is
an abs ute minimum value at 3 this value is f.3/ D !22. (See Figure 13.45.)
Now ork Problem 3 G

R BLEMS
In Pr b ems test f r re ati e maxima and minima se the y D 2x3 ! 3x2 ! 36x C 17 y D x4 ! 2x2 C 4
sec nd deri ati e test if p ssib e In Pr b ems state hether
the re ati e extrema are a s abs ute extrema y D 3 C 5x4 y D !2x7
y D x2 ! 5x C 6 y D 3x2 C 12x C 14 y D 1x5 ! 5x y D 15x3 C x2 ! 15x C 2
y D !4x2 C 2x ! y D !5x2 C 11x ! 7 y D .x2 C 7x C 10/2 y D 2x3 ! x2 ! 60x C 42
y D 13 x3 C 2x2 ! 5x C 1 y D x3 ! 12x C 1
 Section 3.5 s totes 593

Objective A
o eter ne or onta an ert ca
as totes for a c r e an to s etc
t e ra s of f nct ons a n
A
as totes In this section, we conclude our discussion of curve-s etching techniques by investi-
gating functions having asympt tes. An asymptote is a line that a curve approaches
arbitrarily closely. For example, in each part of Figure 13.46, the dashed line x D a
is an asymptote. ut to be precise about it, we need to ma e use of infinite limits.
In Figure 13.46(a), notice that as x ! aC , f.x/ becomes positively infinite:

lim f.x/ D 1
x!aC

In Figure 13.46(b), as x ! aC , f.x/ becomes negatively infinite:

lim f.x/ D !1
x!aC

In Figures 13.46(c) and (d), we have

lim f.x/ D 1 and lim f.x/ D !1

x!a! x!a!

respectively.

f(x) f(x) f (x) f(x)

x=a x=a x=a

a x x x x
a a a

x=a
(a) (b) (c) (d)

FIGURE Vertical asymptotes x D a.

oosely spea ing, we can say that each graph in Figure 13.46 blows up around
the dashed vertical line x D a, in the sense that a one-sided limit of f.x/ at a is either
1 or !1. The line x D a is called a ertica asympt te for the graph. A vertical
asymptote is not part of the graph but is a useful aid in s etching it because part of
the graph approaches the asymptote. ecause of the explosion around x D a, the
function is n t continuous at a.

The line x D a is a vertical asymptote for the graph of the function f if and only if
at least one of the following is true:
lim f.x/ D ˙1
x!aC
or
lim f.x/ D ˙1
x!a!

To determine vertical asymptotes, we must find values of x around which f.x/

increases or decreases without bound. For a rational function (a quotient of two poly-
nomials) expressed in est terms, these x-values are precisely those for which the
594 C C r e etc n

To see that the proviso about est terms denominator is zero but the numerator is not zero. For example, consider the rational
is necessary, observe that function
3x ! 5 .3x ! 5/.x ! 2/
f .x/ D D so that 3x ! 5
x!2 .x ! 2/2 f.x/ D
x D 2 is a vertical asymptote of x!2
.3x ! 5/.x ! 2/
, and here 2 ma es both When x is 2, the denominator is 0, but the numerator is not. If x is slightly larger than 2,
.x ! 2/2
then x ! 2 is both close to 0 and positive, and 3x ! 5 is close to 1. Thus, .3x ! 5/=.x ! 2/
the denominator and the numerator 0.
is very large, so
3x ! 5
y lim D1
x!2 C x!2
3x - 5
Vertical y=
x-2 This limit is su cient to conclude that the line x D 2 is a vertical asymptote. ecause
asymptote
we are ultimately interested in the behavior of a function around a vertical asymptote,
3 it is worthwhile to examine what happens to this function as x approaches 2 from the
left. If x is slightly less than 2, then x ! 2 is very close to 0 but negative, and 3x ! 5 is
close to 1. Hence, .3x ! 5/=.x ! 2/ is very negative, so
x
2 3x ! 5
lim D !1
x!2! x!2

3x ! 5 We conclude that the function increases without bound as x ! 2C and decreases with-
FIGURE raph of y D
x!2
. out bound as x ! 2! . The graph appears in Figure 13.47.
In summary, we have a rule for vertical asymptotes.

A R R F
Suppose that
P.x/
f.x/ D
.x/
where P and are polynomial functions and the quotient is in lowest terms. The
line x D a is a vertical asymptote for the graph of f if and only if .a/ D 0 and
P.a/ ¤ 0.

Although the vertical-asymptote rule

guarantees that the lines x D 3 and x D 1 It might be thought here that lowest terms rules out the possibility of a value a
are vertical asymptotes, it does not ma ing b th denominator and numerator 0, but consider the rational function
indicate the precise nature of the .3x ! 5/.x ! 2/
blow-up around these lines. A precise . This rational function is in lowest terms. Here we cannot divide numer-
analysis requires the use of one-sided .x ! 2/
limits. ator and denominator by x ! 2, to obtain the polynomial 3x ! 5, because the d main
.3x ! 5/.x ! 2/
of the latter is not equal to the domain of the former. The graph of is
.x ! 2/
f (x)
a straight line with a hole in it and it does not have a vertical asymptote.

E AM LE F A

etermine vertical asymptotes for the graph of

x=1 x=3
x2 ! 4x
f.x/ D
x2 ! 4x C 3
1 S Since f is a rational function, the vertical-asymptote rule applies. Writing
x x.x ! 4/
1 3 f.x/ D factoring
.x ! 3/.x ! 1/
ma es it clear that the denominator is 0 if x is either 3 or 1. Neither of these values
FIGURE raph of
ma es the numerator 0. Thus, the lines x D 3 and x D 1 are vertical asymptotes. (See
Figure 13.4 .)
x2 ! 4x
f .x/ D
x2 ! 4x C 3
.
Now ork Problem 1 G
 Section 3.5 s totes 595

H A
A curve y D f.x/ may have other inds of asymptote. In Figure 13.4 (a), as x increases
without bound .x ! 1/, the graph approaches the horizontal line y D b. That is,

lim f.x/ D b
x!1

In Figure 13.4 (b), as x becomes negatively infinite, the graph approaches the horizon-
tal line y D b. That is,

lim f.x/ D b
x!!1

In each case, the dashed line y D b is called a h riz nta asympt te for the graph. It is
a horizontal line around which the graph settles either as x ! 1 or as x ! !1.

f (x) f (x)

b y=b b
y=b

x x

(a) (b)

FIGURE Horizontal asymptotes y D b.

In summary, we have the following definition:

et f be a function. The line y D b is a hori ontal asymptote for the graph of f if

and only if at least one of the following is true:
lim f.x/ D b or lim f.x/ D b
x!1 x!!1

To test for horizontal asymptotes, we must find the limits of f.x/ as x ! 1 and as
x ! !1. To illustrate, we again consider

3x ! 5
f.x/ D
y x!2

3x - 5 Since this is a rational function, we can use the procedures of Section 10.2 to find the
y=
x-2 limits. ecause the dominant term in the numerator is 3x and the dominant term in the
3 denominator is x, we have
Horizontal 3x ! 5 3x
asymptote lim D lim D lim 3 D 3
x!1 x!2 x!1 x x!1
x
2 Thus, the line y D 3 is a horizontal asymptote. See Figure 13.50. Also,

3x ! 5 3x
lim D lim D lim 3 D 3
x!!1 x ! 2 x!!1 x x!!1
FIGURE raph of
3x ! 5 Hence, the graph settles down near the horizontal line y D 3 both as x ! 1 and as
f .x/ D .
x!2 x ! !1.
596 C C r e etc n

E AM LE F H A

Find horizontal asymptotes for the graph of

x2 ! 4x
f.x/ D
x2 ! 4x C 3

S We have

x2 ! 4x x2
lim D lim D lim 1 D 1
x!1 x2 ! 4x C 3 x!1 x2 x!1

Therefore, the line y D 1 is a horizontal asymptote. The same result is obtained for
x ! !1. Refer to Figure 13.4 .
Now ork Problem 11 G
Horizontal asymptotes arising from limits such as limt!1 f.t/ D b, where t is
thought of as time can be important in business applications as expressions of long-
term behavior. For example, in Section .3 we discussed steady states that can be used
to determine long-term mar et shares.
If we rewrite limx!1 f.x/ D b as limx!1 . f.x/ ! b/ D 0, then another possibility
is suggested. For it might be that the long-term behavior of f, while not constant, is
linear. This leads us to the following:

et f be a function. The line y D mx C b is a non erti al asymptote for the graph of

f if and only if at least one of the following is true:
lim . f.x/ ! .mx C b// D 0 or lim . f.x/ ! .mx C b// D 0
x!1 x!!1

f course, if m D 0, then we have ust repeated the definition of horizontal asymp-

tote. ut if m ¤ 0, then y D mx C b is the equation of a nonhorizontal (and non-
vertical) line with slope m that is sometimes described as b ique. Thus to say that
limx!1 . f.x/ ! .mx C b// D 0 is to say that for large values of x, the graph settles
down near the line y D mx C b, often called an oblique asympote for the graph.
P.x/
If f.x/ D , where the degree of P is one more than the degree of , then long
.x/
P.x/ R.x/
divison allows us to write D .mx C b/ C , where m ¤ 0 and where either
.x/ .x/
R.x/ is the zero polynomial or the degree of R is strictly less than the degree of . In
this case, y D mx C b will be an oblique asymptote for the graph of f. Example 3 will
illustrate.

E AM LE F A

Find the oblique asymptote for the graph of the rational function

10x2 C x C 5
y D f.x/ D
5x C 2

S Since the degree of the numerator is 2, one greater than the degree of the
denominator, we use long division to express

10x2 C x C 5 3
f.x/ D D 2x C 1 C
5x C 2 5x C 2
 Section 3.5 s totes 597

Thus
3
lim . f.x/ ! .2x C 1// D lim D0
x!˙1 x!˙1 5x C 2

which shows that y D 2x C 1 is an oblique asymptote, in fact the only nonvertical

2
asymptote, as we explain below. n the other hand, it is clear that x D ! is a vertical
5
asymptote and the only one. (See Figure 13.51.)

-2
x=
5

y = 2x + 1

2
f (x) = 10x + 9x + 5
5x + 2

10x2 C x C 5
FIGURE raph of f .x/ D has an oblique asymptote.
5x C 2

Now ork Problem 35 G

A few remar s about asymptotes are appropriate now. With vertical asymptotes,
we are examining the behavior of a graph around specific x-values. However, with non-
vertical asymptotes we are examining the graph as x becomes unbounded. Although a
graph may have numerous vertical asymptotes, it can have at most two different non-
vertical asymptotes possibly one for x ! 1 and possibly one for x ! !1. If,
for example, the graph has two horizontal asymptotes, then there can be no oblique
asymptotes.
598 C C r e etc n

From Section 10.2, when the numerator of a rational function has degree greater
than that of the denominator, no limit exists as x ! 1 or x ! !1. From this obser-
vation, we conclude that hene er the degree f the numerat r f a rati na functi n
is greater than the degree f the den minat r the graph f the functi n cann t ha e
a h riz nta asympt te. Similarly, it can be shown that if the degree of the numerator
of a rational function is more than one greater than the degree of the denominator, the
function cannot have an oblique asymptote.

E AM LE F H A

Find vertical and horizontal asymptotes for the graph of the polynomial function
y D f.x/ D x3 C 2x

y S We begin with vertical asymptotes. This is a rational function with denomi-

nator 1, which is never zero. y the vertical-asymptote rule, there are no vertical asymp-
totes. ecause the degree of the numerator (3) is greater than the degree of the denom-
3 inator (0), there are no horizontal asymptotes. However, let us examine the behavior of
the graph of f as x ! 1 and x ! !1. We have
-1
x
1
-3 lim .x3 C 2x/ D lim x3 D 1
y = f (x) x!1 x!1
= x3 + 2x
and
FIGURE raph of y D x3 C 2x lim .x3 C 2x/ D lim x3 D !1
has neither horizontal nor vertical x!!1 x!!1
asymptotes. Thus, as x ! 1, the graph must extend indefinitely upward, and as x ! !1, the
graph must extend indefinitely downward. See Figure 13.52.
Now ork Problem 9 G
The results in Example 4 can be generalized to any polynomial function:

A polynomial function of degree greater than 1 has no asymptotes.

E AM LE F H A

Find horizontal and vertical asymptotes for the graph of y D ex ! 1.

y S Testing for horizontal asymptotes, we let x ! 1. Then ex increases without
bound, so
y = e x -1
lim .ex ! 1/ D 1
x!1

Thus, the graph does not settle down as x ! 1. However, as x ! !1, we have
x
ex ! 0, so
y = -1
lim .ex ! 1/ D lim ex ! lim 1 D 0 ! 1 D !1
-1
x!!1 x!!1 x!!1

FIGURE raph of y D ex ! 1
Therefore, the line y D !1 is a horizontal asymptote. The graph has no vertical asymp-
has a horizontal asymptote. totes because ex ! 1 neither increases nor decreases without bound around any fixed
value of x. See Figure 13.53.

Now ork Problem 23 G

C S
In this section we show how to graph a function by ma ing use of all the curve-
s etching tools that we have developed.
 Section 3.5 s totes 599

E AM LE C S
1
S etch the graph of y D .
4 ! x2
S
nter epts When x D 0, y D 14 . If y D 0, then 0 D 1=.4 ! x2 /, which has no solution.
Thus .0; 14 / is the only intercept. However, the factorization
1 1
yD 2
D
4!x .2 C x/.2 ! x/
allows us to construct the following sign chart, Figure 13.54, for y, showing where the
graph lies below the x-axis .!/ and where it lies above the the x-axis .C/.

-q -2 2 q
1
2+x - + +
1
+ + -
2-x
y - + -

1
FIGURE Sign chart for y D .
4 ! x2
ymmetry There is symmetry about the y-axis:
1 1
y.!x/ D 2
D D y.x/
4 ! .!x/ 4 ! x2
Since y is a function of x (and not the constant function 0), there can be no symmetry
about the x-axis and, hence, no symmetry about the origin. Since x is not a function of
y (and y is a function of x), there can be no symmetry about y D x either.

symptotes From the factorization of y above, we see that x D !2 and x D 2 are

vertical asymptotes. Testing for horizontal asymptotes, we have
1 1 1
lim 2
D lim 2
D ! lim 2 D 0
x!˙1 4 ! x x!˙1 !x x!˙1 x

Thus, y D 0 (the x-axis) is the only nonvertical asymptote.

axima an inima Since y D .4 ! x2 /!1 ,

2x
y0 D !1.4 ! x2 /!2 .!2x/ D
.4 ! x2 /2
We see that y0 is 0 when x D 0 and y0 is undefined when x D ˙2. However, only 0
is a critical value, because y is not defined at ˙2. The sign chart for y0 follows. (See
Figure 13.55.)

-q -2 0 2 q
2x - - 0 + +
1
+ + + +
(4 - x2)2
y¿ - - 0 + +

y
-
2x
FIGURE Sign chart for y0 D .
.4 ! x2 /2

The sign chart shows clearly that the function is decreasing on .!1; !2/ and
.!2; 0/, increasing on .0; 2/ and .2; 1/, and that there is a relative minimum at 0.
600 C C r e etc n

on a ity
.4 ! x2 /2 .2/ ! .2x/.2/.4 ! x2 /.!2x/
y00 D
.4 ! x2 /4
2.4 ! x2 /..4 ! x2 / ! .2x/.!2x// 2.4 C 3x2 /
D D
.4 ! x2 /4 .4 ! x2 /3
Setting y00 D 0, we get no real roots. However, y00 is undefined when x D ˙2. Although
concavity may change around these values of x, the values cannot correspond to in ec-
tion points because they are not in the domain of the function. There are three intervals
to test for concavity. See the sign chart in Figure 13.56.
The sign chart shows that the graph is concave up on .!2; 2/ and concave down
on .!1; !2/ and on .2; 1/.

Concave up Concave up
decreasing increasing

x 0 1 3 1
y=
4 - x2
y 1 1 1
4 3
-
5 1
3
x
-2 2
-q -2 2 q
-1
4 + 3x2 + + +
1
- + -
(4 - x2)3
Concave down Concave down
y– - + - decreasing increasing

y ¨ ´ ¨

1
FIGURE Concavity analysis. FIGURE raph of y D .
4 ! x2

is ussion nly one point on the curve, .0; 1=4/, has arisen as a special point that
must be plotted (both because it is an intercept and a local minimum). We might wish
to plot a few more points as in the table in Figure 13.57, but note that any such extra
points are of value only if they are on the same side of the y-axis (because of symmetry).
Ta ing account of all the information gathered, we obtain the graph in Figure 13.57.
Now ork Problem 31 G
E AM LE C S
4x
S etch the graph of y D .
x2 C1
S
nter epts When x D 0, y D 0 when y D 0, x D 0. Thus, .0; 0/ is the only intercept.
Since the denominator of y is always positive, we see that the sign of y is that of x. So
here we dispense with a sign chart for y. From the observations so far it follows that
the graph proceeds from the third quadrant (negative x and negative y) through .0; 0/
to the positive quadrant (positive x and positive y).
ymmetry There is symmetry about the origin:
4.!x/ !4x
y.!x/ D 2
D 2 D !y.x/
.!x/ C 1 x C1
No other symmetry exists.
 Section 3.5 s totes 601

symptotes The denominator of this rational function is never 0, so there are no ver-
tical asymptotes. Testing for horizontal asymptotes, we have
4x 4x 4
lim D lim 2 D lim D0
x!˙1x2 C 1 x!˙1 x x!˙1 x

Thus, y D 0 (the x-axis) is a horizontal asymptote and the only nonvertical asymptote.
axima an inima We have
2
.x C 1/.4/ ! 4x.2x/ 4 ! 4x2 4.1 C x/.1 ! x/
y0 D 2 2
D 2 2
D
.x C 1/ .x C 1/ .x2 C 1/2
The critical values are x D ˙1, so there are three intervals to consider in the sign chart
for y0 . See Figure 13.5 .
We see that y is decreasing on .!1; !1/ and on .1; 1/, increasing on .!1; 1/,
with relative minimum at !1 and relative maximum at 1. The relative minimum is
.!1; y.!1// D .!1; !2/ the relative maximum is .1; y.1// D .1; 2/.

-q -1 1 q
1+x - 0 + +

1-x + + 0 -
1
+ + +
(x2 + 1)2
y¿ - 0 + 0 -
-
y
-

FIGURE Sign chart for y0 .

4 ! 4x2
on a ity Since y0 D ,
.x2 C 1/2
.x2 C 1/2 .! x/ ! .4 ! 4x2 /.2/.x2 C 1/.2x/
y00 D
.x2 C 1/4
p p
x.x2 C 1/.x2 ! 3/ x.x C 3/.x ! 3/
D D
.x2 C 1/4 .x2 C 1/3
p
Setting y00 D 0, we conclude that the possible points of in ection are when x D ˙ 3,
and x D 0. There are four intervals to consider in the sign chart. See Figure 13.5 .

-q - 3 0 3 q
x+ 3 - 0 + + +

x - - 0 + +

x- 3 - - - 0 +
1
+ + + +
(x2 + 1)3
y– - 0 + 0 - 0 +

4x
FIGURE Concavity analysis for y D .
x2 C 1

p
In ection points occur at x D 0 and at x D ˙ 3. The in ection points are
p p p p p p p p
.! 3; y. 3// D .! 3; ! 3/ .0; y.0// D .0; 0/ . 3; y. 3// D . 3; 3/
602 C C r e etc n

y= 4x
x2 + 1
x 0 1 3 -1 - 3
2
y 0 2 3 -2 - 3

x
- 3 -1 1 3

-2

4x
FIGURE raph of y D .
x2 C 1

is ussion After consideration of all of the preceding information, the graph of

y D 4x=.x2 C 1/ is given in Figure 13.60, together with a table of important points.
Now ork Problem 39 G

R BLEMS
In Pr b ems nd the ertica asympt tes and the n n ertica x2 ! 16 2 2x
asympt tes f r the graphs f the functi ns n t sketch the yD yD C
2.3x C 4/2 5 2
12x C 5x ! 2
graphs
x xC1 y D 5ex!3 ! 2 f .x/ D 12e!x
yD yD
x!1 x
xC5 2x C 1 In Pr b ems determine inter a s n hich the functi n is
f .x/ D yD increasing decreasing c nca e up and c nca e d n re ati e
2x C 7 2x C 1
maxima and minima in ecti n p ints symmetry ertica and
3 2 n n ertica asympt tes and th se intercepts that can be btained
yD yD1!
x3 x2 c n enient y hen sketch the cur e
1 x 1 2
yD yD yD yD
x2 ! 1 x2 ! x3 2x ! 3
x3 x 50
y D x2 ! 5x C 5 yD yD yD p
x2 !1 x!1 3x
2x2 x3 1 x2 C x C 1
f .x/ D f .x/ D y D x2 C yD
x2 C x ! 6 5 x2 x!2
15x2 C 31x C 1 2x3 C 1 1 1
yD yD yD yD
x2 ! 7 3x.2x ! 1/.4x ! 3/ x2 ! 1 x2 C 1
3 x2 ! 1 2Cx 1Cx
yD C7 f .x/ D yD yD
x!5 2x2 ! x C 4 3!x x2
3 ! x4 5x2 C 7x3 C x4 x2 x3 C 1
f .x/ D yD yD yD
x3 C x2 3x2 x!1 x
x2 ! 3x ! 4 x3 C 1 4x2 C 2x C 1
yD yD yD yD
1 C 4x C 4x2 1 ! x3 x2 ! 6x ! 2x2
 Section 3.6 e a a an n a 603

3x C 1 3x C 5 S etch the graphs of y D 6 ! 3e!x and y D 6 C 3e!x . Show

yD yD
.3x ! 2/2 .7x C 11/2 that they are asymptotic to the same line. What is the equation of
this line
x2 ! 1 3x
yD yD Mar et for Product For a new product, the yearly number
x3 .x ! 2/2 of thousand pac ages sold, y, t years after its introduction is
1 3x4 C 1 predicted to be given by
y D 2x C 1 C yD
x!1 x3 y D f .t/ D 250 ! 3e!t
1 ! x2 1
yD y D 3x C 2 C Show that y D 250 is a horizontal asymptote for the graph. This
x2 ! 1 3x C 2
reveals that after the product is established with consumers, the
S etch the graph of a function f such that f .0/ D 0, there mar et tends to be constant.
is a horizontal asymptote y D 1 for x ! ˙1, there is a vertical x2 ! 2
asymptote x D 2, both f 0 .x/ < 0 and f 00 .x/ < 0 for x < 2, and raph y D 3 7 2 . From the graph, locate any
x C 2 x C 12x C 1
both f 0 .x/ < 0 and f 00 .x/ > 0 for x > 2.
horizontal or vertical asymptotes.
S etch the graph of a function f such that f .0/ D !4 and
f .4/ D !2, there is a horizontal asymptote y D !3 for x ! ˙1, 2x3 ! 2x2 C 6x ! 1
With a graphing utility, graph y D . From
there is a vertical asymptote x D 2, both f 0 .x/ < 0 and f 00 .x/ < 0 x3 ! 6x2 C 11x ! 6
for x < 2, and both f 0 .x/ < 0 and f 00 .x/ > 0 for x > 2. the graph, locate any horizontal or vertical asymptotes.
S etch the graph of a function f such that f .0/ D 0, there ln.x C 4/
is a horizontal asymptote y D 0 for x ! ˙1, there are vertical raph y D in the standard window. The graph
x2 ! x C 5
asymptotes x D !1 and x D 2; f 0 .x/ < 0 for x < !1 and suggests that there are two vertical asymptotes of the form x D k,
!1 < x < 2, and f 00 .x/ < 0 for x > 2. where k > 0. Also, it appears that the graph begins near x D !4.
S etch the graph of a function f such that f .0/ D 0, there are As x ! !4C ; ln.x C 4/ ! !1 and x2 ! x C 5 ! 53. Thus,
vertical asymptotes x D !1 and x D 1, there is a horizontal limx!4C y D !1. This gives the vertical asymptote x D !4. So,
asymptote y D 0 for x ! ˙1. f 0 .x/ < 0 for x in .!1; !1/, in in reality, there are three vertical asymptotes. Use the zoom
.!1; 1/, and in .1; 1/. f 00 .0/ D 0 f 00 .x/ < 0 for x in .!1; !1/ feature to ma e the asymptote x D !4 apparent from the display.
and in .0; 1/ f 00 .x/ > 0 for x in .!1; 0/ and in .1; 1/. 0:34e0:7x
Purchasing Power In discussing the time pattern of raph y D , where x > 0. From the graph,
4:2 C 0:71e0:7x
purchasing, antell and Sing11 use the curve determine an equation of the horizontal asymptote by examining
x the y-values as x ! 1. To confirm this equation algebraically,
yD
a C bx find limx!1 y by first dividing both the numerator and
as a mathematical model. Find the asymptotes for their model. denominator by e0:7x .

Objective A M M
o o e s t at ons n o n y using techniques from this chapter, we can solve problems that involve maximizing
a n or n n a ant t
or minimizing a quantity. For example, we might want to maximize profit or minimize
cost. he crucia part is expressing the quantity t be maximized r minimized as a
functi n f s me ariab e in the pr b em Then we differentiate and test the resulting
critical values. For this, the first-derivative test or the second-derivative test can be used,
although it may be obvious from the nature of the problem whether or not a critical
value represents an appropriate answer. ecause our interest is in abs ute maxima and
minima, sometimes we must examine endpoints of the domain of the function. Very
often the function used to model the situation of a problem will be the restriction to a
closed interval of a function that has a large natural domain. Such real-world limitations
tend to generate endpoints.

The aim of this example is to set up a cost E AM LE M C F

function from which cost is minimized.
For insurance purposes, a manufacturer plans to fence in a 10, 00-ft2 rectangular stor-
age area ad acent to a building by using the building as one side of the enclosed area.
The fencing parallel to the building faces a highway and will cost 3 per foot, installed,
whereas the fencing for the other two sides costs 2 per foot, installed. Find the amount

11
. H. antell and F. P. Sing, Ec n mics f r Business ecisi ns (New or : c raw-Hill oo Company,
1 72), p. 107.
604 C C r e etc n

of each type of fence so that the total cost of the fence will be a minimum. What is the
Building minimum cost
y y S As a first step in a problem li e this, it is a good idea to draw a diagram that
re ects the situation. In Figure 13.61, we have labeled the length of the side parallel to
x the building as x and the lengths of the other two sides as y, where x and y are in feet.
Highway Since we want to minimize cost, our next step is to determine a function that gives
cost. The cost obviously depends on how much fencing is along the highway and how
FIGURE Fencing problem of
Example 1. much is along the other two sides. Along the highway the cost per foot is 3 (dollars),
so the total cost of that fencing is 3x. Similarly, along each of the other two sides, the
cost is 2y. Thus, the total cost of the fencing is given by the cost function
C D 3x C 2y C 2y
that is,
C D 3x C 4y
We need to find the absolute minimum value of C. To do this, we use the techniques
discussed in this chapter that is, we examine C at critical values (and any endpoints)
in the domain. ut in order to differentiate, we need to first express C as a function of
one variable only. Equation (1) gives C as a function of t variables, x and y. We can
accomplish this by first finding a relationship between x and y. From the statement of
the problem, we are told that the storage area, which is xy, must be 10, 00:
xy D 10; 00
With this equation, we can express one variable (say, y) in terms of the other (x). Then,
substitution into Equation (1) will give C as a function of one variable only. Solving
Equation (2) for y gives
10; 00
yD
x
Substituting into Equation (1), we have
# $
10; 00
C D C.x/ D 3x C 4
x
43;200
C.x/ D 3x C
x
From the physical nature of the problem, the domain of C is x > 0.
We now find dC=dx, set it equal to 0, and solve for x. We have
dC 43;200 d
D3! 2
.43;200x!1 / D !43;200x!2
dx x dx
43;200
3! D0
x2
43;200
3D
x2
from which it follows that
43;200
x2 D D 14;400
3
x D 120 since x > 0
Thus, 120 is the n y critical value, and there are no endpoints to consider. To test this
value, we will use the second-derivative test.
d2 C 6;400
D
dx2 x3
When x D 120, d2 C=dx2 > 0, so we conclude that x D 120 gives a relative minimum.
However, since 120 is the only critical value on the open interval .0; 1/ and C is con-
tinuous on that interval, this relative minimum must also be an absolute minimum.
We are not done yet. The questions posed in the problem must be answered. For
minimum cost, the number of feet of fencing along the highway is 120. When x D 120,
 Section 3.6 e a a an n a 605

we have, from Equation (3), y D 10; 00=120 D 0. Therefore, the number of feet of
fencing for the other two sides is 2y D 1 0. It follows that 120 ft of the 3 fencing and
1 0 ft of the 2 fencing are needed. The minimum cost can be obtained from the cost
function, Equation (4), and is
ˇ
43;200 ˇˇ 43;200
C.120/ D 3x C D 3.120/ C D 720
x ˇ xD120 120
Now ork Problem 3 G
ased on Example 1, the following guide may be helpful in solving an applied
maximum or minimum problem:

G S A M M
Step When appropriate, draw a diagram that re ects the information in the
problem.
Step Set up an expression for the quantity that you want to maximize or
minimize.
Step Write the expression in step 2 as a function of one variable, and note the
domain of this function. The domain may be implied by the nature of the
problem itself.
Step Find the critical values of the function. After testing each critical value,
determine which one gives the absolute extreme value you are see ing. If
the domain of the function includes endpoints, be sure to examine function
values at the endpoints too.
Step ased on the results of step 4, answer the question(s) posed in the problem.

E AM LE M R

The demand equation for a manufacturer s product is

This example involves maximizing
revenue when a demand equation is 0!q
nown. pD 0"q" 0
4
where q is the number of units and p is the price per unit. At what value of q will there
be maximum revenue What is the maximum revenue
S et r represent total revenue, which is the quantity to be maximized. Since
revenue D .price/.quantity/
we have
0!q 0q ! q2
r D pq D #qD D r.q/
4 4
where 0 " q " 0. Setting dr=dq D 0, we obtain
dr 0 ! 2q
D D0
dq 4
0 ! 2q D 0
q D 40
Thus, 40 is the only critical value. Now we determine whether this gives a maximum.
Examining the first derivative for 0 " q < 40, we have dr=dq > 0, so r is increasing.
If q > 40, then dr=dq < 0, so r is decreasing. ecause to the left of 40 we have r
increasing, and to the right r is decreasing, we conclude that q D 40 gives the abs ute
maximum revenue, namely,
r.40/ D . 0.40/ ! .40/2 /=4 D 400

Now ork Problem 7 G

606 C C r e etc n

E AM LE M A C

A manufacturer s total-cost function is given by

This example involves minimizing
average cost when the cost function is q2
nown. c D c.q/ D C 3q C 400
4
where c is the total cost of producing q units. At what level of output will average cost
per unit be a minimum What is this minimum
S The quantity to be minimized is the average cost cN . The average-cost
function is
q2
C 3q C 400
c 4 q 400
cN D cN .q/ D D D C3C
q q 4 q
Here q must be positive. To minimize cN , we differentiate:
dNc 1 400 q2 ! 1600
D ! 2 D
dq 4 q 4q2
To get the critical values, we solve dNc=dq D 0:
q2 ! 1600 D 0
.q ! 40/.q C 40/ D 0
q D 40 since q > 0
To determine whether this level of output gives a relative minimum, we will use the
second-derivative test. We have
d2 cN 00
2
D 3
dq q
which is positive for q D 40. Thus, cN has a relative minimum when q D 40. We note that
cN is continuous for q > 0. Since q D 40 is the only relative extremum, we conclude
that this relative minimum is indeed an absolute minimum. Substituting q D 40 in
40 400
Equation (5) gives the minimum average cost cN .40/ D C3C D 23.
4 40
Now ork Problem 5 G
E AM LE M A E

This example is a biological application An enzyme is a protein that acts as a catalyst for increasing the rate of a chemical
involving maximizing the rate at which reaction that occurs in cells. In a certain reaction, an enzyme is converted to another
an enzyme is formed. The equation enzyme called the product. The product acts as a catalyst for its own formation. The
involved is a literal equation.
rate R at which the product is formed (with respect to time) is given by
R D kp. ! p/
where is the total initial amount of both enzymes, p is the amount of the product
enzyme, and k is a positive constant. For what value of p will R be a maximum
S We can write R D k.p ! p2 /. Setting dR=dp D 0 and solving for p gives
dR
D k. ! 2p/ D 0
dp
pD
2
Now, d2 R=dp2 D !2k. Since k > 0, the second derivative is always negative. Hence,
p D =2 gives a relative maximum. oreover, since R is a continuous function of p,
we conclude that we indeed have an absolute maximum when p D =2.
G
Calculus can be applied to inventory decisions, as the following example shows.
 Section 3.6 e a a an n a 607

E AM LE E L S

A company annually produces and sells 10,000 units of a product. The 10,000 units are
This example involves determining the
number of units in a production run in produced in several production runs of equal sizes. The number of units in a produc-
order to minimize certain costs. tion run is called the lot size. Sales are uniformly distributed throughout the year. The
company wishes to determine the lot size that will minimize total annual setup costs
and carrying costs. This number is referred to as the economic lot si e. The produc-
tion cost of each unit is 20, and carrying costs (insurance, interest, storage, etc.) are
estimated to be 10 of the value of the average inventory. Setup costs per production
run are 40. Find the economic lot size.
S et q be the lot size. Since sales are distributed at a uniform rate, we will
assume that inventory varies uniformly from q to 0 between production runs. Thus, we
ta e the average inventory to be q=2 units. The production costs are 20 per unit, so
the value of the average inventory is 20.q=2/. Carrying costs are 10 of this value:
%q&
0:10.20/ Dq
2
The number of production runs per year is 10;000=q. Hence, total setup costs are
# $
10;000 400;000
40 D
q q
Therefore, the total of the annual carrying costs and setup costs, call it C, is given by
400;000
CDqC q>0
q
dC 400;000 q2 ! 400;000
D1! 2
D
dq q q2
Setting dC=dq D 0, we get
q2 D 400;000
Since q > 0,
p p
qD 400;000 D 200 10 $ 632:5
To determine whether this value of q minimizes C, we will examine the first derivative.
p p
If 0 < q < 400;000, then dC=dq < 0. If q > 400;000, then dC=dq > 0. We
conclude that there is an abs ute minimum at q D 632:5. The number of production
runs is 10;000=632:5 $ 15: . For practical purposes, there would be 16 lots, each
having the economic lot size of 625 units.
Now ork Problem 29 G
E AM LE M T C C R

The Vista TV Cable Co. currently has 100,000 subscribers who are each paying a
The aim of this example is to set up a
revenue function from which revenue is monthly rate of 40. A survey reveals that there will be 1000 more subscribers for
maximized over a closed interval. each 0.25 decrease in the rate. At what rate will maximum revenue be obtained, and
how many subscribers will there be at this rate
S et x be the number of 0.25 decreases. The monthly rate is then 40!0:25x.
We have x % 0 but, because the rate cannot be negative, we also have x " 160. With
x 0.25 decreases, the number of ne subscribers is 1000x so that the total number of
subscribers is 100;000 C 1000x. We want to maximize the revenue, which is given by
r D .number of subscribers/.rate per subscriber/
D .100;000 C 1000x/.40 ! 0:25x/
D 1000.100 C x/.40 ! 0:25x/
r D 1000.4000 C 15x ! 0:25x2 / for x in Œ0; 160"
608 C C r e etc n

Setting r0 D 0 and solving for x, we have

r0 D 1000.15 ! 0:5x/ D 0
x D 30
Since the domain of r is the closed interval 0, 160 , the absolute maximum value of r
must occur at x D 30 or at one of the endpoints of the interval. We now compute r at
these three points:
r.0/ D 1000.4000 C 15.0/ ! 0:25.0/2 / D 4;000;000
r.30/ D 1000.4000 C 15.30/ ! 0:25.30/2 / D 4;225;000
r.160/ D 1000.4000 C 15.160/ ! 0:25.160/2 / D 0
Accordingly, the maximum revenue occurs when x D 30. This corresponds to thirty
0.25 decreases, for a total decrease of 7.50 ma ing the monthly rate 40 ! 7:50 D
32:50. The number of subscribers at that rate is 100;000 C 30.1000/ D 130;000.
Now ork Problem 19 G

E AM LE M R H C B

An article in a sociology ournal stated that if a particular health-care program for

Here we maximize a function over a
closed interval. the elderly were initiated, then t years after its start, n thousand elderly people would
receive direct benefits, where
t3
nD ! 6t2 C 32t 0 " t " 12
3
For what value of t does the maximum number receive benefits
S Setting dn=dt D 0, we have
dn
D t2 ! 12t C 32 D 0
dt
n .t ! 4/.t ! / D 0
t D 4 or tD
t3 Since the domain of n is the closed interval Œ0; 12", the absolute maximum value of n
n= - 6t 2 + 32t
3
96
must occur at t D 0; 4; , or 12:
03
n.0/ D ! 6.02 / C 32.0/ D 0
3
43 160
t n.4/ D ! 6.42 / C 32.4/ D
4 8 12 3 3
3
FIGURE raph of 12
n. / D ! 6. 2 / C 32. / D
t3 2 3 3
n D ! 6t C 32t on Œ0; 12".
3 123 2
n.12/ D ! 6.122 / C 32.12/ D D 6
This example illustrates that endpoints 3 3
must not be ignored when finding Thus, an absolute maximum occurs when t D 12. A graph of the function is given in
absolute extrema on a closed interval. Figure 13.62.
Now ork Problem 15 G
In the next example, we use the word m n p ist. Under a situation of monopoly,
there is only one seller of a product for which there are no similar substitutes, and
the seller that is the monopolist controls the mar et. y considering the demand
equation for the product, the monopolist may set the price (or volume of output) so that
maximum profit will be obtained.
 Section 3.6 e a a an n a 609

This example involves maximizing E AM LE M M

profit for a monopolist when the demand
and average-cost functions are nown. Suppose that the demand equation for a monopolist s product is p D 400 ! 2q and the
In the last part, a tax is imposed on the
monopolist, and a new profit function is average-cost function is cN D 0:2q C 4 C .400=q/, where q is number of units, and both
analyzed. p and cN are expressed in dollars per unit.
a etermine the level of output at which profit is maximized.
b etermine the price at which maximum profit occurs.
c etermine the maximum profit.
d If, as a regulatory device, the government imposes a tax of 22 per unit on the
monopolist, what is the new price for profit maximization
S We now that
profit D total revenue ! total cost
Since total revenue, r, and total cost, c, are given by
r D pq D 400q ! 2q2
and
c D qNc D 0:2q2 C 4q C 400
the profit is
P D r ! c D 400q ! 2q2 ! .0:2q2 C 4q C 400/
so that
P.q/ D 3 6q ! 2:2q2 ! 400 for q > 0
a To maximize profit, we set dP=dq D 0:
dP
D 3 6 ! 4:4q D 0
dq
qD 0
Now, d2 P=dq2 D !4:4 is always negative, so it is negative at the critical value
q D 0. y the second-derivative test, then, there is a relative maximum there.
Since q D 0 is the only critical value on .0; 1/, we must have an absolute maxi-
mum at q D 0.
b The price at which maximum profit occurs is obtained by setting q D 0 in the
demand equation:
p D 400 ! 2. 0/ D 220

c The maximum profit is obtained by evaluating P. 0/. We have

P. 0/ D 3 6. 0/ ! 2:2. 0/2 ! 400 D 17; 420
d The tax of 22 per unit means that for q units, the total cost increases by 22q. The
new cost function is c1 D 0:2q2 C 4q C 400 C 22q, and the new profit is given by
P1 D 400q ! 2q2 ! .0:2q2 C 4q C 400 C 22q/
D 374q ! 2:2q2 ! 400
Setting dP1 =dq D 0 gives
dP1
D 374 ! 4:4q D 0
dq
qD 5
Since d2 P1 =dq2 D !4:4 < 0, we conclude that, to maximize profit, the monopolist
must restrict output to 5 units at the higher price of p1 D 400 ! 2. 5/ D 230.
Since this price is only 10 more than before, only part of the tax has been shifted
610 C C r e etc n

to the consumer, and the monopolist must bear the cost of the balance. The profit
now is 15,4 5, which is less than the former profit.

Now ork Problem 13 G

This discussion leads to the economic
principle that when profit is maximum, We conclude this section by using calculus to develop an important principle in
marginal revenue is equal to economics. Suppose p D f.q/ is the demand function for a firm s product, where p is
marginal cost. price per unit and q is the number of units produced and sold. Then the total revenue
is given by r D qp D qf.q/, which is a function of q. et the total cost of producing
q units be given by the cost function c D g.q/. Thus, the total profit, which is total
revenue minus total cost, is also a function of q, namely,
P.q/ D r ! c D qf.q/ ! g.q/

et us consider the most profitable output for the firm. Ignoring special cases, we now
that profit is maximized when dP=dq D 0 and d2 P=dq2 < 0. We have
dP d dr dc
$ D .r ! c/ D !
dq dq dq dq
Consequently, dP=dq D 0 when
Total revenue
dr dc
D
Total cost
dq dq
That is, at the level of maximum profit, the slope of the tangent to the total-revenue
curve must equal the slope of the tangent to the total-cost curve (Figure 13.63). ut
dr=dq is the marginal revenue R, and dc=dq is the marginal cost C. Thus, under
q typical conditions, to maximize profit, it is necessary that
q1
RD C
FIGURE At maximum profit,
marginal revenue equals marginal cost. For this to indeed correspond to a maximum, it is necessary that d2 P=dq2 < 0:
$ d2 P d2 d2 r d2 c d2 r d2 c
D .r ! c/ D ! < 0 equivalently <
dq2 dq2 dq2 dq2 dq2 dq2
That is, when R D C, in order to ensure maximum profit, the slope of the marginal-
MC
revenue curve must be less than the slope of the marginal-cost curve.
The condition that d2 P=dq2 < 0 when dP=dq D 0 can be viewed another way.
Equivalently, to have R D C correspond to a maximum, dP=dq must go from
C to ! that is, it must go from dr=dq ! dc=dq > 0 to dr=dq ! dc=dq < 0. Hence, as
MR output increases, we must have R > C and then R < C. This means that at
q1
q the point q1 of maximum profit, the margina re enue cur e must cut the margina c st
cur e fr m ab e (Figure 13.64). For production up to q1 , the revenue from additional
FIGURE At maximum profit, output would be greater than the cost of such output, and the total profit would increase.
the marginal-cost curve cuts the For output beyond q1 ; C > R, and each unit of output would add more to total costs
marginal-revenue curve from below. than to total revenue. Hence, total profits would decline.

R BLEMS
In this set f pr b ems un ess ther ise speci ed p is price per Fencing A company has set aside 000 to fence in a
unit in d ars and q is utput per unit f time Fixed c sts refer rectangular portion of land ad acent to a stream by using the
t c sts that remain c nstant at a e e s f pr ducti n during a stream for one side of the enclosed area. The cost of the fencing
gi en time peri d An examp e is rent parallel to the stream is 15 per foot, installed, and the fencing for
Find two numbers whose sum is 6 and whose product is as the remaining two sides costs per foot, installed. Find the
big as possible. dimensions of the maximum enclosed area.

Find two nonnegative numbers whose sum is 20 and for which

the product of twice one number and the square of the other
number will be a maximum.
 Section 3.6 e a a an n a 611

Fencing The owner of the aurel Nursery arden Center Drug Dose The severity of the reaction of the human body
wants to fence in 1400 ft2 of land in a rectangular plot to be used to an initial dose, , of a drug is given by13
for different types of shrubs. The plot is to be divided into six # $
equal plots with five fences parallel to the same pair of sides, as 2 C
R D f. / D !
shown in Figure 13.65. What is the least number of feet of fence 2 3
needed
where the constant C denotes the maximum amount of the drug
that may be given. Show that R has a maximum rate f change
when D C=2.
Profit For a monopolist s product, the demand function is

FIGURE p D 75 ! 0:05q
Average Cost A manufacturer finds that the total cost, c, of
producing a product is given by the cost function and the cost function is

c D 0:05q2 C 5q C 500 c D 500 C 40q

At what level of output will average cost per unit be a minimum
At what level of output will profit be maximized At what price
Automobile Expense The cost per hour (in dollars) of does this occur, and what is the profit
operating an automobile is given by
Profit For a monopolist, the cost per unit of producing a
C D 0:0015s2 ! 0:24s C 1 0 " s " 100 product is 3, and the demand equation is
10
pD p
q

What price will give the greatest profit

Profit For a monopolist s product, the demand equation is

p D 42 ! 4q

and the average-cost function is

where s is the speed in ilometers per hour. At what speed is the
cost per hour a minimum 0
cN D 2 C
Revenue The demand equation for a monopolist s product is q

p D !5q C 30 Find the profit-maximizing price.

At what price will revenue be maximized Profit For a monopolist s product, the demand function is
Revenue Suppose that the demand function for a 50
monopolist s product is of the form pD p
q
q D Ae!Bp
and the average-cost function is
for positive constants A and B. In terms of A and B, find the value
of p for which maximum revenue is obtained. Can you explain 1 2500
cN D C
why your answer does not depend on A 4 q
eight Gain A group of biologists studied the nutritional Find the profit-maximizing price and output.
effects on rats that were fed a diet containing 10 protein.12 The
protein consisted of yeast and cottonseed our. y varying the Profit A manufacturer can produce at most 120 units of a
percent, p, of yeast in the protein mix, the group found that the certain product each year. The demand equation for the product is
(average) weight gain (in grams) of a rat over a period of time was
p D q2 ! 100q C 3200
1600
f .p/ D 170 ! p ! 0 " p " 100
p C 15 and the manufacturer s average-cost function is
Find a the maximum weight gain and b the minimum
2 2 10;000
weight gain. cN D q ! 40q C
3 q
etermine the profit-maximizing output q and the corresponding
maximum profit.

12 13
Adapted from R. ressani, The Use of east in Human Foods, in R. . Thrall, . A. ortimer. . R. Rebman, and R. F. aum, eds., S me
Sing e Ce Pr tein eds. R. I. ateles and S. R. Tannenbaum (Cambridge, athematica de s in Bi gy rev. ed., Report No. 40241-R-7. Prepared at
A: IT Press, 1 6 ). University of ichigan, 1 67.
612 C C r e etc n

Cost A manufacturer has determined that, for a certain Poster Design A rectangular cardboard poster is to have
product, the average cost (in dollars per unit) is given by 720 in2 for printed matter. It is to have a 5-in. margin on each side
200 and a 4-in. margin at the top and bottom. Find the dimensions of
cN D 2q2 ! 4 q C 210 C the poster so that the amount of cardboard used is minimized.
q
where 2 " q " 7. (See Figure 13.67.)
a At what level within the interval Œ2; 7" should production be
fixed in order to minimize total cost What is the minimum total
cost 4–
b If production were required to lie in the interval Œ3; 7", what
value of q would minimize total cost
Profit For anufacturing Co., total fixed costs are
5– 5–
1200, material and labor costs combined are 2 per unit, and the y
demand equation is
100
pD p x
q
What level of output will maximize profit Show that this occurs 4–
when marginal revenue is equal to marginal cost. What is the
price at profit maximization
FIGURE
Revenue A real-estate firm owns 100 garden-type
apartments. At 400 per month, each apartment can be rented. Container Design A cylindrical can, open at the top, is to
However, for each 10-per-month increase, there will be two have a fixed volume of . Show that if the least amount of
vacancies with no possibility of filling them. What rent per material is to be used, then both the radius and height are equal to
p
apartment will maximize monthly revenue 3
=!. (See Figure 13.6 .)
Revenue A TV cable company has 6400 subscribers who
are each paying 24 per month. It can get 160 more subscribers
for each 0.50 decrease in the monthly fee. What rate will yield Volume = rr 2h
maximum revenue, and what will this revenue be Surface area = 2rrh + rr2

Profit A manufacturer of a product finds that, for the first

600 units that are produced and sold, the profit is 40 per unit.
The profit on each of the units beyond 600 is decreased by 0.05
h
times the number of additional units produced. For example, the
total profit when 602 units are produced and sold is 600.40/C
2.3 : 0/. What level of output will maximize profit
Container Design A container manufacturer is designing a
r
rectangular box, open at the top and with a square base, that is to
have a volume of 13.5 ft3 . If the box is to require the least amount Open at top
of material, what must be its dimensions FIGURE
Container Design An open-top box with a square base
is to be constructed from 1 2 ft2 of material. What should be the Container Design A cylindrical can, including both top
dimensions of the box if the volume is to be a maximum What is and bottom, is to be made from a fixed amount of material, S. If
the maximum volume the volume is to be a maximum, show that the radius is equal to
r
Container Design An open box is to be made by cutting S
equal squares from each corner of a -inch-square piece of . Try also to show that h D 2r. (See Figure 13.6 .)
6!
cardboard and then folding up the sides. Find the length of the
side of the square (in terms of ) that must be cut out if the Profit The demand equation for a monopolist s product is
volume of the box is to be maximized. What is the maximum
volume (See Figure 13.66.) p D 600 ! 2q
L and the total-cost function is

c D 0:2q2 C 2 q C 200
Fold
Find the profit-maximizing output and price, and determine the
corresponding profit. If the government were to impose a tax of
Fold

Fold

L 22 per unit on the manufacturer, what would be the new

profit-maximizing output and price What is the profit now
Fold Profit Use the rigina data in Problem 27, and assume
that the government imposes a license fee of 1000 on the
manufacturer. This is a lump-sum amount without regard to
output. Show that the profit-maximizing price and output remain
FIGURE the same. Show, however, that there will be less profit.
 Section 3.6 e a a an n a 613

Economic Lot Si e A manufacturer has to produce Profit s. ones owns a small insurance agency that sells
3000 units annually of a product that is sold at a uniform rate policies for a large insurance company. For each policy sold,
during the year. The production cost of each unit is 12, and s. ones, who does not sell policies herself, is paid a commission
carrying costs (insurance, interest, storage, etcetera) are estimated of 50 by the insurance company. From previous experience,
to be 1 .2 of the value of average inventory. Setup costs per s. ones has determined that, when she employs m salespeople,
production run are 54. Find the economic lot size.
q D m3 ! 15m2 C 2m
Profit For a monopolist s product, the cost function is
policies can be sold per wee . She pays each of the m salespeople
c D 0:004q3 C 20q C 5000 a salary of 1000 per wee , and her wee ly fixed cost is 3000.
Current o ce facilities can accommodate at most eight
and the demand function is salespeople. etermine the number of salespeople that s. ones
should hire to maximize her wee ly profit. What is the
p D 450 ! 4q corresponding maximum profit
Profit A manufacturing company sells high-quality ac ets
Find the profit-maximizing output. through a chain of specialty shops. The demand equation for these
ac ets is
or shop Attendance Imperial Educational Services
(I.E.S.) is considering offering a wor shop in resource allocation p D 1000 ! 50q
to ey personnel at Acme Corp. To ma e the offering
economically feasible, I.E.S. says that at least 40 persons must where p is the selling price (in dollars per ac et) and q is the
attend at a cost of 200 each. oreover, I.E.S. will agree to demand (in thousands of ac ets). If this company s marginal-cost
reduce the charge for e eryb dy by 2.50, for each person over function is given by
the committed 40, who attends. How many people should be in
the group for I.E.S. to maximize revenue Assume that the dc 1000
maximum allowable number in the group is 70. D
dq qC5
Cost of Leasing Motor The iddie Toy Company
show that there is a maximum profit, and determine the number of
plans to lease an electric motor that will be used 0,000
ac ets that must be sold to obtain this maximum profit.
horsepower-hours per year in manufacturing. ne
horsepower-hour is the wor done in 1 hour by a 1-horsepower Chemical Production Each day, a firm ma es x tons of
motor. The annual cost to lease a suitable motor is 200, plus chemical A .x " 4/ and
0.40 per horsepower. The cost per horsepower-hour of operating
the motor is 0.00 = , where is the horsepower. What size 24 ! 6x
yD
motor, in horsepower, should be leased in order to minimize cost 5!x
ransportation Cost The cost of operating a truc on a tons of chemical . The profit on chemical A is 2000 per ton,
thruway (excluding the salary of the driver) is and on it is 1000 per ton. How much of chemical A should be
produced per day to maximize profit Answer the same question
s
0:165 C if the profit on A is P per ton and that on is P=2 per ton.
200
Rate of Return To erect an o ce building, fixed costs are
dollars per mile, where s is the (steady) speed of the truc in miles 1.44 million and include land, architect s fees, a basement, a
per hour. The truc driver s salary is 1 per hour. At what speed foundation, and so on. If x oors are constructed, the cost
should the truc driver operate the truc to ma e a 700-mile trip (excluding fixed costs) is
most economical
c D 10xŒ120;000 C 3000.x ! 1/"

The revenue per month is 60,000 per oor. How many oors will
yield a maximum rate of return on investment (Rate of return D
total revenue=total cost.)
Gait and Power Output of an Animal In a model by
Smith,14 the power output of an animal at a given speed as a
function of its movement or gait j, is found to be
4 3 2

Cost For a manufacturer, the cost of ma ing a part is P.j/ D Aj CB

1Cj
30 per unit for labor and 10 per unit for materials overhead is
fixed at 20,000 per wee . If more than 5000 units are made each where A and B are constants, j is a measure of the umpiness of
wee , labor is 45 per unit for those units in excess of 5000. At the gait, is a constant representing linear dimension, and is a
what level of production will average cost per unit be a minimum constant forward speed.

14
. . Smith, athematica Ideas in Bi gy ( ondon: Cambridge University
Press, 1 6 ).
614 C C r e etc n

a Find the value of that minimizes the denominator.

b Evaluate your answer in part (a) when a D !1 :6 .ft s2 /,
D 20 (ft), and tr D 0:5 (s). our answer will be in feet per
second.
c Find the corresponding value of to one decimal place. our
answer will be in cars per second. Convert your answer to cars
Assume that P is a minimum when dP=dj D 0. Show that when per hour.
this occurs, d Find the relative change in that results when is reduced
B 4 from 20 ft to 15 ft, for the maximizing value of .
.1 C j/2 D Average Cost uring the Christmas season, a promotional
A 2
As a passing comment, Smith indicates that at top speed, j is zero company purchases cheap red felt stoc ings, glues fa e white fur
for an elephant, 0.3 for a horse, and 1 for a greyhound, and sequins onto them, and pac ages them for distribution. The
approximately. total cost of producing q cases of stoc ings is given by
ra c Flow In a model of tra c ow on a lane of a c D 3q2 C 50q ! 1 q ln q C 120
freeway, the number of cars the lane can carry per unit time is
Find the number of cases that should be processed in order to
given by15
minimize the average cost per case. etermine (to two decimal
!2a places) this minimum average cost.
D
2a Profit A monopolist s demand equation is given by
!2atr C !
p D q2 ! 20q C 160
where a is the acceleration of a car when stopping .a < 0/; tr is
the reaction time to begin bra ing, is the average speed of the where p is the selling price (in thousands of dollars) per ton when
cars, and is the length of a car. Assume that a, tr , and are q tons of product are sold. Suppose that fixed cost is 50,000 and
constant. To find how many cars a lane can carry at most, we want that each ton costs 30,000 to produce. If current equipment has
to find the speed that maximizes . To maximize , it su ces to a maximum production capacity of 12 tons, use the graph of
minimize the denominator the profit function to determine at what production level the
2a maximum profit occurs. Find the corresponding maximum profit
!2atr C ! and selling price per ton.

Chapter 13 Review
I T S E
S Relative Extrema
increasing function decreasing function Ex. 1, p. 575
relative maximum relative minimum Ex. 2, p. 576
relative extrema absolute extrema Ex. 3, p. 577
critical value critical point first-derivative test Ex. 4, p. 577
S Absolute Extrema on a Closed Interval
extreme-value theorem Ex. 1, p. 5 3
S Concavity
concave up concave down in ection point Ex. 1, p. 5 5
S he Second Derivative est
second-derivative test Ex. 1, p. 5 1
S Asymptotes
vertical asymptote horizontal asymptote Ex. 1, p. 5 4
oblique asymptote Ex. 3, p. 5 6
S Applied Maxima and Minima
economic lot size Ex. 5, p. 607

15
. I. Shonle, En ir nmenta App icati ns f Genera Physics (Reading, A: Addison-Wesley Publishing Co., 1 75).
 Chapter 3 e e 615

S
Calculus provides the best way of understanding the graphs is concave up over that interval, meaning that its graph bends
of functions. Even the best electronic computational aids upward. If f 00 .x/ < 0 over an interval, then f is concave down
need the udgement added by calculus to tell the user here throughout that interval, and its graph bends downward. A
to loo at a graph. point on the graph where f is continuous and its concav-
The first derivative is used to determine where a function ity changes is an in ection point. The point .a; f.a// on the
is increasing or decreasing and to locate relative maxima and graph is a possible in ection point if either f 00 .a/ D 0 or
minima. If f 0 .x/ is positive throughout an interval, then over f 00 .a/ is not defined and f is continuous at a.
that interval, f is increasing and its graph rises (from left to The second derivative also provides a means for testing
right). If f 0 .x/ is negative throughout an interval, then over certain critical values for relative extrema:
that interval, f is decreasing and its graph is falling.
A point .a; f.a// on the graph at which f 0 .a/ is 0 or is S T R E
not defined is a candidate for a relative extremum, and a is Suppose f 0 .a/ D 0. Then
called a critical value. For a relative extremum to occur at a, If f 00 .a/ < 0, then f has a relative maximum at a.
the first derivative must change sign around a. The following
If f 00 .a/ > 0, then f has a relative minimum at a.
procedure is the first-derivative test for the relative extrema
of y D f.x/:
Asymptotes are also aids in curve s etching. raphs
F T R E
blow up near vertical asymptotes, and they settle near
Step Find f 0 .x/. horizontal asymptotes and oblique asymptotes. The line
Step etermine all values a where f 0 .a/ D 0 or f 0 .a/ x D a is a vertical asymptote for the graph of a function f if
is not defined. either limx!aC f.x/ D ˙1 or limx!a! f.x/ D ˙1. For the
case of a rational function, f.x/ D P.x/= .x/ in lowest terms,
Step n the intervals defined by the values in Step 2,
we can find vertical asymptotes without evaluating limits. If
determine whether f is increasing (f 0 .x/ > 0) or
.a/ D 0 but P.a/ ¤ 0, then the line x D a is a vertical
decreasing (f 0 .x/ < 0).
asymptote.
Step For each critical value a at which f is continu- The line y D b is a horizontal asymptote for the graph of
ous, determine whether f 0 .x/ changes sign as x a function f if at least one of the following is true:
increases through a. There is a relative maximum
at a if f 0 .x/ changes from C to !, and a rela- lim f.x/ D b or lim f.x/ D b
x!1 x!!1
tive minimum if f 0 .x/ changes from ! to C. If
The line y D mx C b is an oblique asymptote for the graph of
f 0 .x/ does not change sign, there is no relative
a function f if at least one (note ˙) of the following is true:
extremum at a.
lim . f.x/ ! .mx C b// D 0
x!˙1

Under certain conditions, a function is guaranteed to In particular, a polynomial function of degree greater
have absolute extrema. The extreme-value theorem states than 1 has no asymptotes. oreover, a rational function
that if f is continuous on a closed interval, then f has an abso- whose numerator has degree greater than that of the denom-
lute maximum value and an absolute minimum value over the inator does not have a horizontal asymptote, and a ratio-
interval. To locate absolute extrema, the following procedure nal function whose numerator has degree more than one
can be used: greater than that of the denominator does not have an oblique
asymptote.
F A E F
fC Œa; b" A M M
Step Find the critical values of f. In applied wor , calculus is very important in maximization
Step Evaluate f.x/ at the endpoints a and b and at the and minimization problems. For example, in the area of eco-
critical values in .a; b/. nomics, we can use it to maximize profit or minimize cost.
Step The maximum value of f is the greatest of the val- Some important relationships that are used in economics
ues found in Step 2. The minimum value of f is problems are the following:
the least of the values found in Step 2. c total cost
cN D average cost per unit D
q quantity
r D pq revenue D (price)(quantity)
The second derivative is used to determine concavity and
in ection points. If f 00 .x/ > 0 throughout an interval, then f PDr!c profit D total revenue ! total cost
616 C C r e etc n

R
In Pr b ems nd h riz nta and ertica asympt tes et f .x/ D x ln x.
3x2 xC2 a etermine the values of x at which relative maxima and
yD 2 yD
x ! 16 5x ! x2 relative minima, if any, occur.
b etermine the interval(s) on which the graph of f is concave
5x2 ! 3 4x C 1 3x C 1 up, and find the coordinates of all points of in ection, if any.
yD yD !
.3x C 2/2 3x ! 5 2x ! 11
x
et f .x/ D .
In Pr b ems nd critica a ues x2 ! 1
3x2 a etermine whether the graph of f is symmetric about the
f .x/ D f .x/ D .x ! 1/2 .x C 6/4 x-axis, y-axis, or origin.
! x2
p b Find the interval(s) on which f is increasing.
3
x!1 13xe!5x=6 c Find the coordinates of all relative extrema of f.
f .x/ D f .x/ D d etermine limx!!1 f .x/ and limx!1 f .x/.
2 ! 3x 6x C 5
e S etch the graph of f.
In Pr b ems nd inter a s n hich the functi n is f State the absolute minimum and absolute maximum values of
increasing r decreasing f.x/ (if they exist).
f .x/ D ! 53 x3 C 15x2 C 35x C 10 In Pr b ems indicate inter a s n hich the functi n is
3x2 6x4 increasing decreasing c nca e up r c nca e d n indicate
f .x/ D f .x/ D 2
.x C 2/2 x !3 re ati e maximum p ints re ati e minimum p ints in ecti n
p ints h riz nta asympt tes ertica asympt tes symmetry and
p
3
f .x/ D 5 2x3 ! 3x th se intercepts that can be btained c n enient y hen sketch
the graph
In Pr b ems nd inter a s n hich the functi n is y D x2 ! 4x ! 21 y D 2x3 C 15x2 C 36x C
c nca e up r c nca e d n
y D x3 ! 12x C 20 y D e1=x
x!2
f .x/ D x4 ! x3 ! 14 f .x/ D
xC2 xC1
y D x3 ! x yD
1 x!1
f .x/ D f .x/ D x3 C 2x2 ! 5x C 2
3x C 2
100.x C 5/ x2 ! 4
3 2 2 f .x/ D yD
f .x/ D .3x ! 1/.2x ! 5/ f .x/ D .x ! x ! 1/ x2 x2 ! 1
x
In Pr b ems test f r re ati e extrema yD y D 6x1=3 .2x ! 1/
.x ! 1/3
3 2
f .x/ D 2x ! x C 12x C 7
ax C b ex ! e!x
f .x/ D for a > 0 and b > 0 f .x/ D f .x/ D 1 ! ln.x3 /
x2 2
x10 x5 2x2
f .x/ D C f .x/ D 2 Are the following statements true or false
10 5 x !1
a If f 0 .x0 / D 0, then f must have a relative extremum at x0 .
f .x/ D x2=3 .x C 1/ f .x/ D x3 .x ! 2/4 b Since the function f .x/ D 1=x is decreasing on the intervals
.!1; 0/ and .0; 1/, it is impossible to find x1 and x2 in the
domain of f such that x1 < x2 and f .x1 / < f .x2 /.
In Pr b ems nd the x a ues here in ecti n p ints ccur
c n the interval .!1; 1", the function f .x/ D x4 has an absolute
y D 3x5 C 20x4 ! 30x3 ! 540x2 C 2x C 3 maximum and an absolute minimum.
x2 C 2 d If f 00 .x0 / D 0, then .x0 ; f .x0 // must be a point of in ection.
yD y D 2.x ! 3/.x4 C 1/ e A function f defined on the interval .!2; 2/ with exactly one
5x
relative maximum must have an absolute maximum.
x3
y D x2 C 2 ln.!x/ yD An important function in probability theory is the standard
ex
normal-density function
y D .x2 ! 5/3
1 2
f .x/ D p e!x =2
In Pr b ems test f r abs ute extrema n the gi en inter a 2!
f .x/ D 3x ! 4x3 I Œ0; 2"
4
a etermine whether the graph of f is symmetric about the
f .x/ D x3 ! . =2/x2 ! 12x C 2 Œ0; 5" x-axis, y-axis, or origin.
x b Find the intervals on which f is increasing and those on which
f .x/ D I Œ!2; 0" it is decreasing.
.5x ! 6/2
c Find the coordinates of all relative extrema of f.
f .x/ D .x C 1/2 .x ! 1/2=3 Œ2; 3" d Find limx!!1 f .x/ and limx!1 f .x/.
 Chapter 3 e e 617

e Find the intervals on which the graph of f is concave up and The graph of f has a shape similar to that in Figure 13.6 and is
those on which it is concave down. best fit by a third-degree polynomial, such as
f Find the coordinates of all points of in ection.
g S etch the graph of f. f .t/ D A t3 C Bt2 C Ct C
h Find all absolute extrema.
Marginal Cost If c D q3 ! 6q2 C 12q C 1 is a total-cost f (t)
function, for what values of q is marginal cost increasing
Marginal Revenue If r D 200q3=2 ! 3q2 is the revenue
P
function for a manufacturer s product, determine the intervals on
which the marginal-revenue function is increasing. (a, f (a))

Revenue Function The demand equation for a

t
manufacturer s product is a
p FIGURE
q
p D 200 ! where q > 0
5
The point P has special meaning. It is such that the value a
Show that the graph of the revenue function is concave down separates interresponse times ithin chun s from those bet een
wherever it is defined. chun s. athematically, P is a critical point that is also a point of
in ection. Assume these two conditions, and show that a
Contraception In a model of the effect of contraception on a D !B=.3A/ and b B2 D 3AC.
birthrate,16 the equation
Mar et Penetration In a model for the mar et penetration
x of a new product, sales S of the product at time t are given by1
R D f .x/ D 0"x" 1
4:4 ! 3:4x 2 3
gives the proportional reduction R in the birthrate as a function of
m.p C q/2 6 e!.pCq/t 7
6 7
the e ciency x of a contraception method. An e ciency of 0.2 (or S D g.t/ D 6# $2 7
20 ) means that the probability of becoming pregnant is 0 of p 4 q !.pCq/t 5
e C1
the probability of becoming pregnant without the contraceptive. p
Find the reduction (as a percentage) when e ciency is a 0, b
0.5, and c 1. Find dR=dx and d2 R=dx2 , and s etch the graph of where p, q, and m are nonzero constants.
the equation. a Show that
Learning and Memory If you were to recite members of a ! "
category, such as four-legged animals, the words that you utter m 3 !.pCq/t q !.pCq/t
.p C q/ e e !1
would probably occur in chun s, with distinct pauses between dS p p
such chun s. For example, you might say the following for the D # $3
dt q !.pCq/t
category of four-legged animals: e C1
p
dog, cat, mouse, rat,
.pause/ b etermine the value of t for which maximum sales occur. ou
horse, don ey, mule, may assume that S attains a maximum when dS=dt D 0.
.pause/
cow, pig, goat, lamb, In Pr b ems here appr priate r und the ans ers t t
etc. decima p aces
From the graph of y D 3: x3 C 5:2x2 ! 7x C 3, using a
The pauses may occur because you must mentally search for graphing utility, find the coordinates of all relative extrema.
subcategories (animals around the house, beasts of burden, farm
animals, etc.). From the graph of f .x/ D x4 ! 2x3 C 3x ! 1, determine the
The elapsed time between onsets of successive words is absolute extrema of f over the interval !1; 1 .
called interresp nse time. A function has been used to analyze the The graph of a function f has exactly one in ection point. If
length of time for pauses and the chun size (number of words in
a chun ).17 This function f is such that x3 C 3x C 2
f 00 .x/ D
8 5x2 ! 2x C 4
<the average number of words
f .t/ D that occur in succession with use the graph of f 00 to determine the x-value of the in ection point
:interresponse times less than t of f.

16
R. . ei and . F. ee er, athematica S ci gy (Englewood Cliffs, N :
Prentice-Hall, Inc., 1 75).
17 1
A. raesser and . andler, imited Processing Capacity Constrains the A. P. Hurter, r., A. H. Rubenstein et al., ar et Penetration by New
Storage of Unrelated Sets of Words and Retrieval from Natural Categories, Innovations: The Technological iterature, echn gica F recasting and
uman earning and em ry 4, no. 1 (1 7 ), 6 100. S cia Change vol. 11 (1 7 ), 1 7 221.
618 C C r e etc n

5x2 C 2x Container Design A rectangular box is to be made by

raph y D . From the graph, locate any cutting out equal squares from each corner of a piece of cardboard
x3 C 2x C 1
horizontal or vertical asymptotes. 10 in. by 16 in. and then folding up the sides. What must be the
length of the side of the square cut out if the volume of the box is
Maximi ation of Production A manufacturer determined to be maximum
that m employees on a certain production line will produce q units
per month, where Fencing A rectangular portion of a field is to be enclosed
by a fence and divided equally into four parts by three fences
q D 0m2 ! 0:1m4 parallel to one pair of the sides. If a total of meter of fencing is
to be used, find the dimensions (in terms of ) that will maximize
To obtain maximum monthly production, how many employees the fenced area.
should be assigned to the production line
Poster Design A rectangular poster having an area of
Revenue The demand function for a manufacturer s
500 in2 is to have a 4-in. margin at each side and at the bottom
product is given by p D 0e!0:05q . For what value of q does the
and a 6-in. margin at the top. The remainder of the poster is for
manufacturer maximize total revenue
printed matter. Find the dimensions of the poster so that the area
Revenue The demand function for a monopolist s for the printed matter is maximized.
product is
Cost A furniture company ma es personal-computer
p stands. For a certain model, the total cost (in thousands of dollars)
pD 500 ! q when q hundred stands are produced is given by

If the monopolist wants to produce at least 100 units, but not more c D 2q3 ! q2 C 12q C 20
than 200 units, how many units should be produced to maximize a The company is currently capable of manufacturing between
total revenue 75 and 600 stands (inclusive) per wee . etermine the number of
Average Cost If c D 0:01q2 C 5q C 100 is a cost function, stands that should be produced per wee to minimize the total
find the average-cost function. At what level of production q is cost, and find the corresponding average cost per stand.
there a minimum average cost b Suppose that between 300 and 600 stands must be produced.
How many should the company now produce in order to minimize
Profit The demand function for a monopolist s product is
total cost

p D 700 ! 2q Bacteria In a laboratory, an experimental antibacterial

agent is applied to a population of 100 bacteria. ata indicate that
the number of bacteria t hours after the agent is introduced is
and the average cost per unit for producing q units is given by
1000 12;100 C 110t C 100t2
cN D q C 100 C D
q 121 C t2
where p and cN are in dollars per unit. Find the maximum profit For what value of t does the maximum number of bacteria in the
that the monopolist can achieve. population occur What is this maximum number
 nte rat on

A
nyone who runs a business nows the need for accurate cost estimates.
14.1 erent a s
When obs are individually contracted, determining how much a ob will
14.2 e n e n te nte ra cost is generally the first step in deciding how much to bid.
For example, a painter must determine how much paint a ob will ta e.
14.3 nte rat on t nta Since a gallon of paint will cover a certain number of square feet, the ey is to deter-
Con t ons
mine the area of the surfaces to be painted. Normally, even this requires only simple
14.4 ore nte rat on arithmetic walls and ceilings are rectangular, and so total area is a sum of products
or as of base and height.
ut not all area calculations are as simple. Suppose, for instance, that the bridge
14.5 ec n es of
nte rat on shown below must be sandblasted to remove accumulated soot. How would the con-
tractor who charges for sandblasting by the square foot calculate the area of the vertical
14.6 e e n te nte ra face on each side of the bridge
14.7 e n a enta A B
eore of Ca c s
C er 14 e e

D C

The area could be estimated as perhaps three-quarters of the area of the trapezoid
formed by points A, B, C, and . ut a more accurate calculation which might be
desirable if the bid were for dozens of bridges of the same dimensions (as along a
stretch of railroad) would require a more refined approach.
If the shape of the bridge s arch can be described mathematically by a function, the
contractor could use the method introduced in this chapter: integration. Integration has
many applications, the simplest of which is finding areas of regions bounded by curves.
ther applications include calculating the total de ection of a beam due to bending
stress, calculating the distance traveled underwater by a submarine, and calculating the
electricity bill for a company that consumes power at differing rates over the course of
a month.
Chapters 11 13 dealt with differential calculus. We differentiated a function and
obtained another function, its derivative. Rather surprisingly, integra ca cu us, involv-
ing area considerations as mentioned above, is deeply connected with the reverse pro-
cess of differentiation: We are given the derivative of a function and must find the orig-
inal function. The need for solving this reverse problem also arises in a natural way.
For example, we might have a marginal-revenue function and want to find the revenue
function from it.

619
620 C nte rat on

Objective
o e ne t e erent a nter ret We will soon give a reason for using the symbol dy=dx to denote the derivative of y
t eo etr ca an se t n
a ro at ons so to restate t e with respect to x. To do this, we introduce the notion of the di erentia of a function.
rec roca re at ons et een dx=dy
an dy=dx

et y D f.x/ be a differentiable function of x, and let !x denote a change in x,

where !x can be any real number. Then the differential of y, denoted by either dy
or d. f.x// is given by
dy D f 0 .x/!x

Note that dy depends on two variables, namely, x and !x. In fact, dy is a function of
two variables.
To review functions of several variables,
see Section 2. .
E AM LE C

Find the differential of y D x3 !2x2 C3x!4, and evaluate it when x D 1 and !x D 0:04.
S The differential is
d 3
dy D .x ! 2x2 C 3x ! 4/!x
dx
D .3x2 ! 4x C 3/!x

When x D 1 and !x D 0:04,

dy D dy.1; 0:04/ D .3.1/2 ! 4.1/ C 3/.0:04/ D 0:0

Now ork Problem 1 G

If f is the identity functi n, then f.x/ D x. Following the notation above applied to
y D f.x/ D x we have dy D d.x/ D 1!x D !x. Said otherwise, the differential of x is
!x. We abbreviate d.x/ by dx. Thus, dx D !x. From now on, we will write dx for !x
when finding a differential. For example,
d 2
d.x2 C 5/ D .x C 5/dx D 2xdx
dx
Summarizing, we say that if y D f.x/ defines a differentiable function of x, then

dy D f 0 .x/dx

where dx is any real number. Provided that dx ¤ 0, we can divide both sides of the last
equation by dx:
dy
D f 0 .x/
dx
That is, dy=dx can be viewed either as the quotient of two differentials, namely, dy
divided by dx, or as one symbol for the derivative of f at x. It is for this reason that the
symbol dy=dx is widely used to denote the derivative.
dy
In Section 11.1 we noted marginally that for y D f.x/, is eibniz notation for
dx
the derivative of f. The notation f 0 for the derivative could equally be called Newton
notation (although Pf is more closely related to Newton s original writings). eibniz and
Newton independently discovered calculus in the middle of the 17th century.
 Section 4. erent a s 621

E AM LE F T dx
p
a If f.x/ D x, then
p d p 1 1
d. x/ D . x/dx D x!1=2 dx D p dx
dx 2 2 x
b If u D .x2 C 3/5 , then du D 5.x2 C 3/4 .2x/dx D 10x.x2 C 3/4 dx.
Now ork Problem 3 G
The differential can be interpreted geometrically. In Figure 14.1, the point P.x; f.x//
is on the curve y D f.x/. Suppose x changes by dx, a real number, to the new value
x C dx. Then the new function value is f.x C dx/, and the corresponding point on the
curve is .x C dx; f.x C dx//. Passing through P and are horizontal and vertical lines,
respectively, that intersect at S. A line tangent to the curve at P intersects segment S
at R, forming the right triangle PRS. bserve that the graph of f near P is approximated
by the tangent line at P. The slope of is f 0 .x/ but it is also given by SR=PS so that
SR
f 0 .x/ D
PS
0
Since dy D f .x/ dx and dx D PS,
SR
dy D f 0 .x/ dx D
" PS D SR
PS
Thus, if dx is a change in x at P, then dy is the corresponding vertical change along
the tangent line at P. Note that for the same dx, the vertical change along the curve is
!y D S D f.x C dx/ ! f.x/. o not confuse !y with dy. However, from Figure 14.1,
the following is apparent:

When dx is close to 0, dy is an approximation to !y. Therefore,

!y # dy

This fact is useful in estimating !y, a change in y, as Example 3 shows.

y = f(x )

Q
f (x + dx )
L
f (x + dx ) - f(x ) ¢y
R
dy
P
f(x ) S

dx
x
x x + dx

FIGURE eometric interpretation of dy and !x.

E AM LE U E C

A governmental health agency examined the records of a group of individuals who

were hospitalized with a particular illness. It was found that the total proportion P that
622 C nte rat on

are discharged at the end of t days of hospitalization is given by

! "3
300
P D P.t/ D 1 !
300 C t
Use differentials to approximate the change in the proportion discharged if t changes
from 300 to 305.
S The change in t from 300 to 305 is !t D dt D 305 ! 300 D 5. The change
in P is !P D P.305/ ! P.300/. We approximate !P by dP:
! "2 ! "
0 300 300 3003
!P # dP D P .t/dt D !3 ! dt D 3 dt
300 C t .300 C t/2 .300 C t/4
When t D 300 and dt D 5,
3003 15 1 1
dP D 3 5D 3 D 3 D # 0:0031
6004 2 600 2 40 320
For a comparison, the true value of !P is
P.305/ ! P.300/ D 0: 7 07 ! 0: 7500 D 0:00307
(to five decimal places).
Now ork Problem 11 G
We said that if y D f.x/, then !y # dy if dx is close to zero. Thus,
!y D f.x C dx/ ! f.x/ # dy
Formula (1) is used to approximate a so that
function value, whereas the formula
!y # dy is used to approximate a change f.x C dx/ # f.x/ C dy
in function values.
This formula gives us a way of estimating a function value f.x C dx/. For example,
suppose we estimate ln.1:06/. etting y D f.x/ D ln x, we need to estimate f.1:06/.
Since d.ln x/ D .1=x/dx, we have, from Formula (1),
f.x C dx/ # f.x/ C dy
1
ln.x C dx/ # ln x C dx
x
We now the exact value of ln.1/, so we will let x D 1 and dx D 0:06. Then
x C dx D 1:06, and dx is close to 0. Therefore,
1
ln.1 C 0:06/ # ln.1/ C .0:06/
1
ln.1:06/ # 0 C 0:06 D 0:06
The true value of ln.1:06/ to five decimal places is 0:05 27.

E AM LE U E F

The demand function for a product is given by

p
p D f.q/ D 20 ! q
where p is the price per unit in dollars for q units. y using differentials, approximate
the price when units are demanded.
S We want to approximate f. /. y Formula (1),
f.q C dq/ # f.q/ C dp
where
1 dp 1
dp D ! p dq since D ! q!1=2
2 q dq 2
 Section 4. erent a s 623

We choose q D 100 and dq D !1 because q C dq D , dq is small, and it is easy to

p
compute f.100/ D 20 ! 100 D 10. We, thus, have
1
f. / D f.100 C .!1// # f.100/ ! p .!1/
2 100
f. / # 10 C 0:05 D 10:05
Hence, the price per unit when units are demanded is approximately 10.05.
Now ork Problem 17 G
The equation y D x3 C 4x C 5 defines y as a function of x. We could write
f.x/ D x3 C4xC5. However, the equation also defines x implicitly as a function of y. In
fact, if we restrict the domain of f to some set of real numbers x so that y D f.x/ is a one-
to-one function, then in principle we could solve for x in terms of y and get x D f !1 .y/.
Actually, no restriction of the domain is necessary here. Since f 0 .x/ D 3x2 C 4 > 0,
for all x, we see that f is strictly increasing on .!1; 1/ and is thus one-to-one on
.!1; 1/. As we did in Section 12.2, we can loo at the derivative of x with respect
to y, dx=dy, and we have seen that it is given by

dx 1
D provided that dy=dx ¤ 0
dy dy
dx
Since dx=dy can be considered a quotient of differentials, we now see that it is the
reciprocal of the quotient of differentials dy=dx. Thus,
dx 1
D 2
dy 3x C 4
It is important to understand that it is not necessary to be able to solve y D x3 C 4x C 5
dx 1
for x in terms of y, and the equation D 2 holds for all x.
dy 3x C 4

E AM LE F dp=dq dq=dp
dp p
Find if q D 2500 ! p2 .
dq
S

S There are a number of ways to find dp=dq. ne approach is to solve the

given equation for p explicitly in terms of q and then differentiate directly. Another
approach to find dp=dq is to use implicit differentiation. However, since q is given
explicitly as a function of p, we can easily find dq=dp and then use the preceding
reciprocal relation to find dp=dq. We will ta e this approach.

We have
dq 1 p
D .2500 ! p2 /!1=2 .!2p/ D ! p
dp 2 2500 ! p2
Hence,
p
dp 1 2500 ! p2
D D!
dq dq p
dp

Now ork Problem 27 G

624 C nte rat on

R BLEMS
In Pr b ems nd the di erentia f the functi n in terms f x Profit Suppose that the profit (in dollars) of producing
and dx q units of a product is
y D ax C b yD2
p P D 3 7q ! 2:3q2 ! 400
2 3
f .x/ D x3 ! 27 f .x/ D .4x ! 5x C 2/
Using differentials, find the approximate change in profit if the
1 p level of production changes from q D 0 to q D 1. Find the true
uD 2 uD x
x change.
p D ln.x2 C 7/ p D ex
4 C3x2 C1
Revenue iven the revenue function
p
2x2 C3 y D ln x2 C 12
y D . x C 3/e r D 200q C 40q2 ! q3
In Pr b ems nd !y and dy f r the gi en a ues f x use differentials to find the approximate change in revenue if the
and dx number of units increases from q D 10 q D 11. Find the true
y D ax C b for any x and any dx change.
y D 5x2 x D !1, dx D !0:02 Demand The demand equation for a product is
2
y D x C 3x C 5 x D 2, dx D 0:01 10
pD p
2
y D .3x C 2/ x D !1, dx D !0:03 q
p
y D 32 ! x2 x D 4, dx D !0:05 Round your answer to Using differentials, approximate the price when 24 units are
three decimal places. demanded.
y D ln x x D 1, dx D 0:01 Demand iven the demand function
xC5 200
et f .x/ D . pD p
xC1 qC
a Evaluate f 0 .1/.
b Use differentials to estimate the value of f(1.1). use differentials to estimate the price per unit when 40 units are
demanded.
et f .x/ D xx .
If y D f .x/, then the pr p rti na change in y is defined to be
a Evaluate f 0 .1/. !y=y, which can be approximated with differentials by dy=y. Use
b Use differentials to estimate the value of f .1:001/. this last form to approximate the proportional change in the cost
function
In Pr b ems appr ximate each expressi n by using
di erentia s q2
p p c D f .q/ D C 5q C 300
2 ( int 172 D 2 .) 122 2
p3
p
4
16:3 when q D 10 and dq D 2. Round your answer to one decimal
place.
ln.0: / ln 1.01
Status Income Suppose that S is a numerical value of
e 0:001
e!0:002 status based on a person s annual income I (in thousands of
p
dollars). For a certain population, suppose S D 20 I. Use
In Pr b ems nd dx=dy r dp=dq as makes sense differentials to approximate the change in S if annual income
y D 2x ! 1 y D 2x3 C 2x C 3 decreases from 45,000 to 44,500.
p Biology The volume of a spherical cell is given by
q D . p2 C 5/3 qD pC5
4
1 D "r3 , where r is the radius. Estimate the change in volume
qD q D e4!2p 3
p2
when the radius changes from 5:40 $ 10!4 cm to 5:45 $ 10!4 cm.
ˇ
7 2 ˇ Muscle Contraction The equation
If y D 5x ! x C 3, find dx=dyˇˇ
3
.
2 xD1=3
.P C a/. C b/ D k
If y D ln x2 , find the value of dx=dy when x D 3.
In Pr b ems and nd the rate f change f q ith respect t is called the fundamental equation of muscle contraction. 1 Here
p f r the indicated a ue f q P is the load imposed on the muscle, is the velocity of the
500 p shortening of the muscle fibers, and a, b, and k are positive
pD Iq D 1 p D 60 ! 2qI q D 50 constants. Find P in terms of , and then use the differential to
qC2
approximate the change in P due to a small change in .

1
R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or :
c raw-Hill, 1 55).
 Section 4.2 e n e n te nte ra 625

Demand The demand, q, for a monopolist s product is Profit The demand equation for a monopolist s product is
related to the price per unit, p, according to the equation
1 2
q2 4000 pD q ! 66q C 7000
2C D 2 2
200 p
and the average-cost function is
a Verify that 40 units will be demanded when the price per unit
is 20. 0;000
dq cN D 500 ! q C
b Show that D !2:5 when the price per unit is 20. 2q
dp
c Use differentials and the results of parts (a) and (b) to a Find the profit when 100 units are demanded.
approximate the number of units that will be demanded if the b Use differentials and the result of part (a) to estimate the
price per unit is reduced to 1 .20. profit when 101 units are demanded.

Objective T I I
o e ne t e ant er at e an t e iven a function f, if F is a function such that
n e n te nte ra an to a as c
nte rat on for as
F0 .x/ D f.x/
then F is called an antideri ati e of f. Thus,

An antideri ati e of f is simply a function whose derivative is f.

ultiplying both sides of Equation (1) by the differential dx gives F0 .x/dx D f.x/dx.
However, because F0 .x/dx is the differential of F, we have dF D f.x/dx. Hence, we can
also thin of an antiderivative of f as a function whose differential is f.x/dx.

An antiderivative of a function f is a function F such that

F0 .x/ D f.x/
Equivalently, in differential notation,
dF D f.x/dx

For example, because the derivative of x2 is 2x, x2 is an antiderivative of 2x. How-

ever, it is not the only antiderivative of 2x: Since
d 2 d 2
.x C 1/ D 2x and .x ! 5/ D 2x
dx dx
both x2 C1 and x2 !5 are also antiderivatives of 2x. In fact, it is obvious that because the
derivative of a constant is zero, x2 C C is also an antiderivative of 2x for any constant C.
Thus, 2x has infinitely many antiderivatives. ore importantly, although not obviously,
e ery antiderivative of 2x is a function of the form x2 CC, for some constant C. It can be
shown that if a continuous function has a derivative of 0 on an interval then the function
is constant on that interval. We note:

Any two antiderivatives of a function differ only by a constant.

Since x2 C C describes all antiderivatives ofR 2x, we refer to it as being the m st

genera antideri ati e of 2x, and denote it by 2xdx, which is read the inde nite
integra of 2x with respect to x. In fact, we write
Z
2xdx D x2 C C
626 C nte rat on
R
The symbol is called the integral sign, 2x is the integrand, and C is the constant of
integration. The dx is part of the integral notation and indicates the variable involved.
Here x is the variable of integration.
R ore generally, the indefinite integral of any function f with respect to x is written
f.x/dx and denotes the most general antiderivative of f. Since all antiderivatives of f
differ only by a constant, if F is any antiderivative of f, then
Z
f.x/dx D F.x/ C C; where C is a constant
R
To integrate f means to find f.x/dx. In summary,
Z
f.x/dx D F.x/ C C if and only if F0 .x/ D f.x/

It follows that both

!Z " Z
d d
f.x/dx D f.x/ and .F.x//dx D F.x/ C C
dx dx
which show the extent to which differentiation and indefinite integration are inverse
operations. Z
It is good to begin by thin ing of . /dx as an operation that is applied to a
d
function, in the same way that we initially regarded . /. In the previous section
dx
d
we have introduced differentials that somewhat rationalize . /, but we suggest not
Z dx
trying to rationalize . /dx at this time. It should be simply accepted as a (somewhat
odd) notation for now.

E AM LE F I I
A L IT I
Z
If the marginal cost
R for a company Find 5 dx.
is f .q/ D 2 :3, find 2 :3 dq, which
gives the form of the cost function.
S

S First we must find a function whose derivative is 5. Then we add the

constant of integration.
A common mista e when finding
integrals is to omit C, the constant of
integration. Since we now that the derivative of 5x is 5, 5x is an antiderivative of 5. Therefore,
Z
5dx D 5x C C

Now ork Problem 1 G

Using differentiation formulas from Chapters 11 and 12, we have compiled a list of
elementary integration formulas in Table 14.1. These formulas are easily verified. For
example, Formula (2) is true because the derivative of xaC1 =.a C 1/ is xa for a ¤ !1.
(We must have a ¤ !1 because the denominator is 0 when a D !1.) Formula (2) states
that the indefinite integral of a power of x, other than x!1 , is obtained by increasing the
exponent of x by 1, dividing by the new exponent, and adding a constant of integration.
The indefinite integral of x!1 will be discussed in Section 14.4.R
To verify Formula
R (5), we must show that the derivative ofRk f.x/dx is kf.x/. Since
the derivative
R of k f.x/ dx is simply k times the derivative of f.x/ dx, and the deriva-
tive of f.x/dx is f.x/, Formula (5) is verified. The reader should verify the other for-
mulas. Formula (6) can be extended to any number of terms.
 Section 4.2 e n e n te nte ra 627

Table 1 .1 E I F
Z
1. k dx D kx C C k is a constant
Z
xaC1
2. xa dx D CC a ¤ !1
aC1
Z Z Z
1 dx
3. x!1 dx D dx D D ln x C C for x > 0
x x
Z
4. ex dx D ex C C
Z Z
5. kf .x/ dx D k f .x/ dx k is a constant
Z Z Z
6. . f .x/ ˙ g.x// dx D f .x/ dx ˙ g.x/ dx

E AM LE I I C x
Z
a Find 1dx.

S y Formula (1) with k D 1

Z
1dx D 1x C C D x C C
R R R
Usually, we write 1dx as dx. Thus, dx D x C C.
Z
b Find x5 dx.

S y Formula (2) with a D 5,

Z
x5C1 x6
x5 dx D CCD CC
5C1 6

Now ork Problem 3 G

E AM LE I I C T F
A L IT I
Z
If the rate of change of a com-
pany s revenues can R be modeled by Find 7xdx.
dR=dt D 0:12t2 , find 0:12t2 dt, which
gives the company s revenue function to S y Formula (5) with k D 7 and f.x/ D x,
within a constant. Z Z
7x dx D 7 x dx

Since x is x1 , by Formula (2) we have

Z
x1C1 x2
x1 dx D C C1 D C C1
1C1 2
nly a c nstant factor of the integrand where C1 is the constant of integration. Therefore,
can be passed through an integral sign. Z Z ! 2 "
x 7
7x dx D 7 x dx D 7 C C1 D x2 C 7C1
2 2
Since 7C1 is ust an arbitrary constant, we will replace it by C for simplicity. Thus,
Z
7
7x dx D x2 C C
2
628 C nte rat on

It is not necessary to write all intermediate steps when integrating. ore simply,
we write
Z
x2 7
7xdx D .7/ C C D x2 C C
2 2

Now ork Problem 5 G

E AM LE I I C T F
Z
3
Find ! ex dx
5
S
Z Z
3 x 3
! e dx D ! ex dx Formula (5)
5 5
3
D ! ex C C Formula (4)
5

Now ork Problem 21 G

A L IT I
ue to new competition, the num- E AM LE F I I
ber of subscriptions to a certain mag- Z
dS 1
azine is declining at a rate of D a Find p dt.
dt t
4 0
! 3 subscriptions per month, where t S Here t is the variable of integration. We rewrite the integrand so that a basic
t p
is the number of months since the com- formula can be used. Since 1= t D t!1=2 , applying Formula (2) gives
petition entered the mar et. Find the
Z Z
form of the equation for the number of 1 t.!1=2/C1 t1=2 p
subscribers to the magazine. p dt D t!1=2 dt D CCD CCD2 tCC
t 1 1
! C1
2 2
Z
1
b Find dx
6x3 Z Z ! " !3C1
1 1 !3 1 x
S dx D x dx D CC
6x 3 6 6 !3 C 1
x!2 1
D! CCD! CC
12 12x2

Now ork Problem 9 G

E AM LE I I S
A L IT I
Z
The rate of growth of the popula-
tion of a new city ispestimated to be Find .x2 C 2x/dx.
d =dt D 500R C 300 t,pwhere t is in
years. Find .500 C 300 t/dt. S y Formula (6),
Z Z Z
2 2
.x C 2x/ dx D x dx C 2x dx

Now,
Z
x2C1 x3
x2 dx D C C1 D C C1
2C1 3
 Section 4.2 e n e n te nte ra 629

and
Z Z
x1C1
2x dx D 2 x dx D .2/ C C2 D x2 C C2
1C1
Thus,
Z
x3
.x2 C 2x/ dx D C x2 C C1 C C2
3
When integrating an expression involving For convenience, we will replace the constant C1 C C2 by C. We then have
more than one term, only one constant of Z
integration is needed. x3
.x2 C 2x/ dx D C x2 C C
3
mitting intermediate steps, we simply integrate term by term and write
Z
x3 x2 x3
.x2 C 2x/dx D C .2/ C C D C x2 C C
3 2 3

Now ork Problem 11 G

E AM LE I I S
A L IT I
Z p
Suppose the rate of savings 5
in the United States is given by Find .2 x4 ! 7x3 C 10ex ! 1/dx.
dS
D 2:1t2 ! 65:4t C 4 1:6, where t is S
dt Z p
the time in years and S is the amount of 5
money saved in billions of dollars. Find .2 x4 ! 7x3 C 10ex ! 1/dx
the form of the equation for the amount Z Z Z Z
of money saved. D 2 x4=5 dx ! 7 x3 dx C 10 ex dx ! 1dx Formulas (5) and (6)

x =5
x4
D .2/ ! .7/ C 10ex ! x C C Formulas (1), (2), and (4)
4
5
10 7
D x =5
! x4 C 10ex ! x C C
4
Now ork Problem 15 G
Sometimes, in order to apply the basic integration formulas, it is necessary first to
perform algebraic manipulations on the integrand, as Example shows.

E AM LE U A M F
I I
Z ! "
2
Find y2 y C dy
3
In Example , we first multiplied the S The integrand does not fit a familiar integration form. However, by multi-
factors in the integrand. The answer plying the integrand we get
could not have been found simply in Z ! " Z ! "
R R 2 2 2 3 2 2
terms of y2 dy and .y C /dy. There is y yC dy D y C y dy
3 3 3
not a formula for the integral of a genera
4 ! " 3
product of functions. y 2 y y4 2y3
D C CCD C CC
4 3 3 4
Now ork Problem 41 G
630 C nte rat on

E AM LE U A M F
I I
Z
.2x ! 1/.x C 3/
a Find dx.
6
S y factoring out the constant 16 and multiplying the binomials, we get
Z Z
.2x ! 1/.x C 3/ 1
dx D .2x2 C 5x ! 3/ dx
6 6
! "
1 x3 x2
D .2/ C .5/ ! 3x C C
6 3 2
x3 5x2 x
D C ! CC
12 2
Z
x3 ! 1
b Find dx.
Another algebraic approach to part (b) is x2
Z 3 Z
x !1 S We can brea up the integrand into fractions by dividing each term in the
dx D .x3 ! 1/x!2 dx
x2 numerator by the denominator:
Z Z 3 Z ! 3 " Z
D .x ! x!2 / dx x !1 x 1
dx D ! dx D .x ! x!2 / dx
x2 x2 x2
and so on.
x2 x!1 x2 1
D ! CCD C CC
2 !1 2 x
Now ork Problem 49 G

R BLEMS
Z
In Pr b ems nd the inde nite integra s
Z Z .x :3
! x6 C 3x!4 C x!3 / dx
1
7 dx dx Z
x
.0:3y4 C 2y!2 /dy
Z Z
Z p Z
x dx 3x37 dx !2 x
dx dz
Z Z 3
z!3 Z Z
5x!7 dx dz 5 !4
3 p dx dx
Z Z
3 2
3 x .3x/3
5 7 Z ! 4 " Z ! "
dx dx x 4 1 1
x7 x4 ! 4 dx ! dx
Z Z 4 x 2x3 x4
1 7 Z ! 2 " Z
dt dx 3 2
t5=2 2x =4 ! d 7e!s ds
Z Z 2 3 2
.4 C t/ dt .7r5 C 4r2 C 1/ dr Z Z ! "
3u ! 4 1 2 x
du e dx
Z Z 5 e 3
.y5 ! 5y/ dy .2 ! 3 ! 5 2
/d Z Z ! "
ey
Z Z .ue C eu / du 3y3 ! 2y2 C dy
6
2 2 4 6
.3t ! 4t C 5/ dt .1 C t C t C t / dt Z ! " Z
3 p
Z p Z p ! 12 3 x dx 0 dt
!1 x
. 2 C e/ dx .5 ! 2 /dx
Z p !
! " Z ! "
5 3
x 2
x 2 5 2x2 ! p C 711x dx
! x ! x4 dx 5 5 x
7 3 7 3
Z Z Z ! " Z
p 1
"ex dx .ex C 3x2 C 2x/ dx
3
uC p du .x2 C 5/.x ! 3/ dx
u
 Section 4.3 nte rat on t n t a Con t ons 631
Z Z Z
p .x2 C 1/3
x3 .x2 C 5x C 2/ dx x.x C 3/ dx dx
x
Z Z
.z ! 3/3 dz .3u C 2/3 du If F.x/ and G.x/ are such that F0 .x/ D G0 .x/, is it true that
F.x/ ! G.x/ must be zero
Z ! "2 Z R
2 a Find a function F such that F.x/dx D x2 ex C C.
p ! 1 dx x!2 .3x4 C 4x2 ! 5/ dx
5
x b How many functions F are there which satisfy the equation
Z Z given in part (a)
p z5 C 7z2
.6eu ! u3 . u C 1// du dz Z ! "
3z d 1
Find p dx.
Z Z dx x2 C 1
x4 ! 5x2 C 2x ex C e2x
dx dx
5x2 ex

Objective I I C
o n a art c ar ant er at e If we now the rate of change, f 0 , of the function f, then the function f itself is an
of a f nct on t at sat s es certa n
con t ons s n o es e a at n antiderivative of f 0 (since the derivative of f is f 0 ). f course, there are many antideriva-
constants of nte rat on tives of f 0 , and the most general one is denoted by the indefinite integral. For example, if
f 0 .x/ D 2x
then Z Z
f.x/ D f 0 .x/dx D 2xdx D x2 C:

That is, any function of the form f.x/ D x2 C C has its derivative equal to 2x. ecause
of the constant of integration, notice that we do not now f.x/ specifically. However, if f
must assume a certain function value for a particular value of x, then we can determine
the value of C and thus determine f.x/ specifically. For instance, if f.1/ D 4, then, from
Equation (1),
f.1/ D 12 C C
4D1CC
CD3
Thus,
f.x/ D x2 C 3
That is, we now now the particular function f.x/ for which f 0 .x/ D 2x and f.1/ D 4.
The condition f.1/ D 4, which gives a function value of f for a specific value of x, is
called an initial condition.

E AM LE I C
A L IT I
The rate of growth of a species of If y is a function of x such that y0 D x ! 4 and y.2/ D 5, find y. (Note: y.2/ D 5 means
d that y D 5 when x D 2.) Also, find y.4/.
bacteria is estimated by D 00 C
dt
t
200e , where is the number of bacteria
S Here y.2/ D 5 is the initial condition. Since y0 D x!4, y is an antiderivative
(in thousands) after t hours. If .5/ D of x ! 4,
Z
40;000, find .t/. x2
y D . x ! 4/dx D " ! 4x C C D 4x2 ! 4x C C
2
We can determine the value of C by using the initial condition. ecause y D 5 when
x D 2, from Equation (2), we have
5 D 4.2/2 ! 4.2/ C C
5 D 16 ! C C
C D !3
632 C nte rat on

Replacing C by !3 in Equation (2) gives the function that we see :

y D 4x2 ! 4x ! 3
To find y.4/, we let x D 4 in Equation (3):
y.4/ D 4.4/2 ! 4.4/ ! 3 D 64 ! 16 ! 3 D 45

Now ork Problem 1 G

E AM LE I C I y00
A L IT I
The acceleration of an ob ect after iven that y00 D x2 ! 6, y0 .0/ D 2, and y.1/ D !1, find y.
t seconds is given by y00 D 4t C 24,
the velocity at seconds is given by S
y0 . / D 2 1 ft s, and the position at
2 seconds is given by y.2/ D 1 5 ft. S To go from y00 to y, two integrations are needed: the first to ta e us from
Find y.t/. y to y and the other to ta e us from y0 to y. Hence, there will be two constants of
00 0

integration, which we will denote by C1 and C2 .

d 0
Since y00 D .y / D x2 ! 6, y0 is an antiderivative of x2 ! 6. Thus,
dx
Z
x3
y0 D .x2 ! 6/dx D ! 6x C C1
3
Now, y0 .0/ D 2 means that y0 D 2 when x D 0 therefore, from Equation (4), we have
03
2D ! 6.0/ C C1
3
Hence, C1 D 2, so
x3
y0 D ! 6x C 2
3
y integration, we can find y:
Z ! "
x3
yD ! 6x C 2 dx
3
! " 4
1 x x2
D ! .6/ C 2x C C2
3 4 2
so
x4
yD ! 3x2 C 2x C C2
12
Now, since y D !1 when x D 1, we have, from Equation (5),
14
!1 D ! 3.1/2 C 2.1/ C C2
12
1
Thus, C2 D ! , so
12
x4 1
yD ! 3x2 C 2x !
12 12
Now ork Problem 5 G
Integration with initial conditions is applicable to many applied situations, as the
next three examples illustrate.
 Section 4.3 nte rat on t n t a Con t ons 633

E AM LE I E

For a particular urban group, sociologists studied the current average yearly income, y
(in dollars), that a person can expect to receive with x years of education before see ing
regular employment. They estimated that the rate at which income changes with respect
to education is given by
dy
D 100x3=2 4 % x % 16
dx
where y D 2 ;720 when x D . Find y.
S Here y is an antiderivative of 100x3=2 . Thus,
Z Z
y D 100x3=2 dx D 100 x3=2 dx

x5=2
D .100/ CC
5
2
y D 40x5=2 C C
The initial condition is that y D 2 ;720 when x D . y putting these values into
Equation (6), we can determine the value of C:
2 ;720 D 40. /5=2 C C
D 40.243/ C C
2 ;720 D 720 C C
Therefore, C D 1 ;000, and
y D 40x5=2 C 1 ;000

Now ork Problem 17 G

E AM LE F F M R

If the marginal-revenue function for a manufacturer s product is

dr
D 2000 ! 20q ! 3q2
dq
find the demand function.
S

S y integrating dr=dq and using an initial condition, we can find the

revenue function, r. ut revenue is also given by the general relationship r D pq,
where p is the price per unit. Thus, p D r=q. Replacing r in this equation by the
revenue function yields the demand function.

Since dr=dq is the derivative of total revenue, r,

Z
r D .2000 ! 20q ! 3q2 / dq

q2 q3
D 2000q ! .20/ ! .3/ C C
2 3
so that
r D 2000q ! 10q2 ! q3 C C
634 C nte rat on

Revenue is 0 when q is 0. We assume that hen n units are s d there is n re enue that is, r D 0 when q D 0.
This is our initial condition. Putting these values into Equation (7) gives
0 D 2000.0/ ! 10.0/2 ! 03 C C
Although q D 0 gives C D 0, this is not Hence, C D 0, and
true in general. It occurs in this section
because the revenue functions are r D 2000q ! 10q2 ! q3
polynomials. In later sections, evaluating
at q D 0 may produce a nonzero value To find the demand function, we use the fact that p D r=q and substitute for r:
for C. r 2000q ! 10q2 ! q3
pD D
q q
p D 2000 ! 10q ! q2
Now ork Problem 11 G
E AM LE F C M C

In the manufacture of a product, fixed costs per wee are 4000. Fixed costs are costs,
such as rent and insurance, that remain constant at all levels of production during a
given time period. If the marginal-cost function is
dc
D 0:000001.0:002q2 ! 25q/ C 0:2
dq
where c is the total cost (in dollars) of producing q ilograms of product per wee , find
the cost of producing 10,000 g in 1 wee .
S Since dc=dq is the derivative of the total cost c,
Z
c.q/ D .0:000001.0:002q2 ! 25q/ C 0:2/dq
Z Z
D 0:000001 .0:002q2 ! 25q/dq C 0:2dq
! "
0:002q3 25q2
c.q/ D 0:000001 ! C 0:2q C C
3 2
When q is 0, total cost is equal to Fixed costs are constant regardless of output. Therefore, when q D 0, c D 4000,
fixed cost. which is our initial condition. Putting c.0/ D 4000 in the last equation, we find that
C D 4000, so
! "
0:002q3 25q2
Although q D 0 gives C a value equal to c.q/ D 0:000001 ! C 0:2q C 4000
fixed costs, this is not true in general. It 3 2
occurs in this section because the cost 2
functions are polynomials. In later From Equation ( ), we have c.10; 000/ D 5416 . Thus, the total cost for producing
sections, evaluating at q D 0 may 3
produce a value for C that is different 10,000 pounds of product in 1 wee is 5416.67.
from fixed cost.
Now ork Problem 15 G

R BLEMS
In Pr b ems and nd y subject t the gi en c nditi ns In Pr b ems nd y subject t the gi en c nditi ns
dy=dx D 3x ! 4I y.!1/ D 13
2
y00 D !5x2 C 2xI y0 .1/ D 0; y.0/ D 3
2 1
dy=dx D x ! xI y.3/ D 2 y00 D x C 1I y0 .0/ D 0; y.0/ D 5
In Pr b ems and if y satis es the gi en c nditi ns nd y.x/ y000 D 2xI y00 .!1/ D 3; y0 .3/ D 10; y.0/ D 13
f r the gi en a ue f x
y000 D 2e!x C 3I y00 .0/ D 7; y0 .0/ D 5; y.0/ D 1
0
y D p , y.16/ D 10 x D
x
0 2
y D ! x C 2x; y.2/ D 1I x D 1
 Section 4.4 ore nte rat on or as 635

In Pr b ems dr=dq is a margina re enue functi n Find the Fluid Flow In the study of the ow of uid in a tube of
demand functi n constant radius R, such as blood ow in portions of the body, one
1 can thin of the tube as consisting of concentric tubes of radius r,
dr=dq D 0:7 dr=dq D 12 ! q where 0 % r % R. The velocity, , of the uid is a function of r
15
dr=dq D 275 ! q ! 0:3q2 dr=dq D 5;000 ! 3.2q C 2q3 / and is given by4
Z
.P1 ! P2 /r
In Pr b ems dc=dq is a margina c st functi n and xed D ! dr
c sts are indicated in braces F r Pr b ems and nd the 2#
t ta c st functi n F r Pr b ems and nd the t ta c st f r where P1 and P2 are pressures at the ends of the tube, # (a ree
the indicated a ue f q letter read eta ) is uid viscosity, and is the length of the tube.
If D 0 when r D R, show that
dc=dq D 2:47I f15 g dc=dq D 2q C 75I f2000g
2
.P1 ! P2 /.R2 ! r2 /
dc=dq D 0:0 q ! 1:4q C 6:7I f 500gI q D 20 D
4#
dc=dq D 0:000204q2 ! 0:046q C 6I f15;000gI q D 200
Elasticity of Demand The sole producer of a product has
Diet for Rats A group of biologists studied the nutritional determined that the marginal-revenue function is
effects on rats that were fed a diet containing 10 protein.2 The
dr
protein consisted of yeast and corn our. D 00 ! 6q2
dq
etermine the point elasticity of demand for the product when
q D 5. ( int First find the demand function.)
ver a period of time, the group found that the (approximate) rate Average Cost A manufacturer has determined that the
of change of the average weight gain G (in grams) of a rat with marginal-cost function is
respect to the percentage P of yeast in the protein mix was dc
D 0:003q2 ! 0:4q C 40
dG P dq
D! C2 0 % P % 100
dP 25 where q is the number of units produced. If marginal cost is
If G D 3 when P D 10, find G. 27.50 when q D 50 and fixed costs are 5000, what is the
a erage cost of producing 100 units
inter Moth A study of the winter moth was made in
Nova Scotia.3 The prepupae of the moth fall onto the ground from If f 00 .x/ D 30x4 C 12x and f 0 .1/ D 10, evaluate
host trees. It was found that the (approximate) rate at which f . 65:335245/ ! f .! 65:335245/
prepupal density, y (the number of prepupae per square foot of
soil), changes with respect to distance, x (in feet), from the base
of a host tree is
dy
D !1:5 ! x 1 % x %
dx
If y D 5 :6 when x D 1, find y.

Objective M I F
o earn an a t e for as for
Z
a
Z
u
Z
1 R I
u du e xdu an du
u The formula
Z
xaC1
xa dx D CC if a ¤ !1
nC1
which applies to a power of x, can be generalized to handle a power of a functi n
of x. et u be a differentiable function of x. y the power rule for differentiation, if
a ¤ !1, then
! "
d .u.x//aC1 .a C 1/.u.x//a " u0 .x/
D D .u.x//a " u0 .x/
dx aC1 aC1

2 Adapted from R. ressani, The Use of east in Human Foods, in Sing e Ce Pr tein eds. R. I. ateles and
S. R. Tannenbaum (Cambridge, A: IT Press, 1 6 ).
3 Adapted from . . Embree, The Population ynamics of the Winter oth in Nova Scotia, 1 54 1 62,
em irs f the Ent m gica S ciety f Canada no. 46 (1 65).
4 R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or : c raw-Hill, 1 55).
636 C nte rat on

Thus,
Z
.u.x//aC1
.u.x//a " u0 .x/ dx D C C a ¤ !1
aC1
We call this the power rule for integration. Note that u0 .x/dx is the differential of u,
namely, du. If we abbreviate u.x/ by u and replace u0 .x/dx by du, we get

R I If u is di erentiab e then
Z
uaC1
ua du D CC if a ¤ !1
aC1

It is important toR appreciate the difference between the power rule for integration and
the formula for xa dx. In the power rule, u represents a function of x.

E AM LE A R I

Z
a Find .x C 1/20 dx.

S Since theR integrand is a power of Rthe function x C 1, we will set u D x C 1.

Then du D dx, and .xC1/20 dx has the form u20 du. y the power rule for integration,
Z Z
u21 .x C 1/21
.x C 1/20 dx D u20 du D CCD CC
21 21
Note that we give our answer not in terms of u, but explicitly in terms of x.
Z
b Find 3x2 .x3 C 7/3 dx.

S We observe that the integrand contains a power of the function x3 C 7. et

u D x C 7. Then du D 3x2 dx. Fortunately, 3x2 appears as a factor in the integrand and
3

we have
Z Z Z
3x2 .x3 C 7/3 dx D .x3 C 7/3 .3x2 dx/ D u3 du

This example is more typical than u4 .x3 C 7/4

Example 1(a). Note again that D CCD CC
du D 3x2 dx. 4 4
Now ork Problem 3 G
In order to apply the power rule for integration, sometimes an ad ustment must be
made to obtain du in the integrand, as Example 2 illustrates.

E AM LE A du
Z p
Find x x2 C 5dx.
R
S We can write this as x.x2 C 5/1=2 dx. Notice that the integrand contains a
power of the function x2 C 5. If u D x2 C 5, then du D 2xdx. Since the c nstantR factor
2 in du does n t appear in the integrand, this integral does not have the form un du.
du
However, from du D 2xdx we can write xdx D so that the integral becomes
2
Z Z Z
du
x.x C 5/ dx D .x C 5/ .xdx/ D u1=2
2 1=2 2 1=2
2
 Section 4.4 ore nte rat on or as 637
1
oving the c nstant factor 2
in front of the integral sign, we have
0 1
Z Z 3=2
1 1 Bu C 1 3=2
x.x2 C 5/1=2 dx D u1=2 du D @ ACCD u CC
2 2 3 3
2
The answer to an integration problem which, in terms of x, as is required, gives
must be expressed in terms of the
original variable. Z p
.x2 C 5/3=2
x x2 C 5dx D CC
3
Now ork Problem 15 G
p
In Example 2, the integrand x x2 C 5 missed being of the form .u.x//1=2 u0 .x/
Z
u0 .x/
by the c nstant fact r of 2. In general, if we have .u.x//a dx, for k a nonzero
k
It must be stressed strongly that the k in constant, then we can write
this displayed equation cannot be Z Z Z
u0 .x/ du 1
variable. The equation applies only for .u.x//a dx D ua D ua du
nonzero constants. k k k
to simplify the integral, but such adjustments of the integrand are n t p ssib e f r ari
ab e fact rs. R
When using the form ua du, do not neglect du. For example,
Z
.4x C 1/3
.4x C 1/2 dx ¤ CC
3
The correct way to do this problem is as follows. et u D 4x C 1, from which it follows
du
that du D 4dx. Thus dx D and
4
Z Z ! " Z
du 1 1 u3 .4x C 1/3
.4x C 1/2 dx D u2 D u2 du D " CCD CC
4 4 4 3 12

E AM LE A R I
Z p
a Find 3
6ydy.

S The integrand is .6y/1=3 , a power of a function. However, in this case the

obvious substitution u D 6y can be avoided. ore simply, we have
Z p Z p Z p
p
3 1=3 1=3 3 1=3 3 y4=3 3 3 6 4=3
6ydy D 6 y dy D 6 y dy D 6 CCD y CC
4 4
3
Z 3
2x C 3x
b Find dx.
.x4 C 3x2 C 7/4
R
S We can write this as .x4 C 3x2 C 7/!4 .2x3 C 3x/dx. et us try to use the
power rule for integration. If u D x4 C 3x2 C 7, then du D .4x3 C 6x/dx, which is two
du
times the quantity .2x3 C 3x/dx in the integral. Thus, .2x3 C 3x/dx D and we again
2
illustrate the adjustment technique:
Z Z ! " Z
du 1
.x4 C 3x2 C 7/!4 ..2x3 C 3x/dx/ D u!4 D u!4 du
2 2
1 u!3 1 1
D " CCD! 3 CCD! 4 CC
2 !3 6u 6.x C 3x2 C 7/3

Now ork Problem 5 G

638 C nte rat on

In using the power rule for integration, ta e care when ma ing a choice for u.
In Example 3(b), letting u D 2x3 C 3x does not lead very far. At times it may be
necessary to try many different choices. Sometimes a wrong choice will provide a hint
as to what does wor . S ill at integration comes only after many hours of practice
and conscientious study

E AM LE A I R N A
Z
Find 4x2 .x4 C 1/2 dx.

S If we set u D x4 C 1, then du D 4x3 dx. To get du in the integral, we need an

additional factor of the ariab e x. However, we can ad ust only for constant factors.
Thus, we cannot use the power rule. Instead, to find the integral, we will first expand
.x4 C 1/2 :
Z Z
4x .x C 1/ dx D 4 x2 .x C 2x4 C 1/dx
2 4 2

Z
D4 .x10 C 2x6 C x2 /dx
! "
x11 2x7 x3
D4 C C CC
11 7 3

Now ork Problem 67 G

I N E F
We now turn our attention to integrating exponential functions. If u is a differentiable
function of x, then
d u du
.e / D eu
dx dx
Corresponding to this differentiation formula is the integration formula
Z
du
eu dx D eu C C
dx
du
ut dx is the differential of u, namely, du. Thus,
dx
Z
eu du D eu C C

A L IT I
When an ob ect is moved from
one environment to another, its tem- E AM LE I I E F
perature, , changes at a rate given by
d Z
D kCekt , where t is the time (in 2
dt a Find 2xex dx.
hours) after changing environments, C
is the temperature difference (original
minus new) between the environments,
S et u D x2 . Then du D 2xdx, and, by Equation (2),
and k is a constant. If the original envi- Z Z Z
x2 x2
ronment is 70ı , the new environment 2xe dx D e .2xdx/ D eu du
is 60ı , and k D !0:5, find the general
form of .t/. D eu C C D ex C C
2
 Section 4.4 ore nte rat on or as 639
Z
3 C3x
b Find .x2 C 1/ex dx.

S If u D x3 C 3x, then du D .3x2 C 3/dx D 3.x 2

R u C 1/dx. If the integrand
contained a factor of 3, the integral would have the form e du. Thus, we write
Z Z
x3 C3x 3
2
.x C 1/e dx D ex C3x Œ.x2 C 1/ dx$
Z
1 1
eu du D eu C C
D
3 3
1 3
D ex C3x C C
3
1 1
where in the second step we replaced .x2 C1/dx by du but wrote outside the integral.
3 3

Now ork Problem 41 G

I I L F
R a aC1
As we now, the power-rule formula u du D u =.a C 1/ C C does not apply when
Z Z
1
a D !1. To handle that situation, namely, u!1 du D du, we first recall from
u
Section 12.1 that
d 1 du
.ln juj/ D for u ¤ 0
dx u dx
which gives us the integration formula
Z
1
du D ln juj C C for u ¤ 0
u

In particular, if u D x, then du D dx, and

Z
1
dx D ln jxj C C for x ¤ 0
x

1
E AM LE I I du
A L IT I Z u
7
If the rate of vocabulary memoriza- a Find dx.
tion of the average student in a for- x
d 35 S From Equation (4),
eign language is given by D , Z Z
dt tC1
7 1
where is the number of vocabulary dx D 7 dx D 7 ln jxj C C
words memorized in t hours of study, x x
find the general form of .t/. Using properties of logarithms, we can write this answer another way:
Z
7
dx D ln jx7 j C C
x
Z
2x
b Find 2
dx.
x C5
S et u D x2 C 5. Then du D 2xdx. From Equation (3),
Z Z Z
2x 1 1
2
dx D 2
.2xdx/ D du
x C5 x C5 u
D ln juj C C D ln jx2 C 5j C C
640 C nte rat on

Since x2 C 5 is always positive, we can omit the absolute-value bars:

Z
2x
dx D ln.x2 C 5/ C C
x2 C 5
Now ork Problem 31 G
1
E AM LE A I I du
u
Z
.2x3 C 3x/dx
Find .
x4 C 3x2 C 7
S If u D x4 C 3x2 C 7, then du D .4x3 C 6x/dx, which is two times the
du
numerator giving .2x3 C 3x/dx D . To apply Equation (3), we write
2
Z Z
2x3 C 3x 1 1
4 2
dx D du
x C 3x C 7 2 u
1
D ln juj C C
2
1
D ln jx4 C 3x2 C 7j C C Rewrite u in terms of x.
2
1
D ln.x4 C 3x2 C 7/ C C x4 C 3x2 C 7 > 0 for all x
2
Now ork Problem 51 G
E AM LE A I I T F
Z ! "
1 1
Find C d .
.1 ! /2 !1
S
Z ! " Z Z
1 1 !2 1
2
C d D .1 ! / d C d
.1 ! / !1 !1
Z Z
1
D !1 .1 ! /!2 .!d / C d
!1
Z
R !2 1
The first integral has the form u du, and the second has the form d . Thus,
Z ! "
1 1 .1 ! /!1
C d D ! C ln j ! 1j C C
.1 ! /2 !1 !1
1
D C ln j ! 1j C C
1!
G
R BLEMS
Z Z
In Pr b ems nd the inde nite integra s 5 4x
Z Z dx dx
5 .3x ! 1/3 .2x2 ! 7/10
.x C 3/ dx 15.x C 2/4 dx Z Z
p 1
Z Z 7x C 3 dx p dx
2x.x2 C 3/5 dx .4x C 3/.2x2 C 3x C 1/ dx x!5
Z Z
Z 5
.5x ! 2/ dx x2 .3x3 C 7/3 dx
2 3 2 2=3
.3y C 6y/.y C 3y C 1/ dy Z Z p
Z u.5u2 ! /14 du x 3 C 5x2 dx
2 3 2
.12t ! 4t C 3/.4t ! 2t C 3t/ dt
 Section 4.4 ore nte rat on or as 641
Z Z Z
4x4 .27 C x5 /1=3 dx .3 ! 2x/7 dx !.x2 ! 2x5 /.x3 ! x6 /!10 dx
Z Z Z Z
2 C 2
3x
3e dx 5e 3tC7
dt . C 4/e2C d .2x3 C x/.x4 C x2 / dx
7
Z Z Z Z
3:1 2 C1 x
.3t C 1/e 3t2 C2tC1
dt !3 2 !
e
3
d .e / dx dx
.5 ! x ! x2 /4
Z Z Z Z ! "
.ex ! e!x /2 dx
3 2
3xe 5x2
dx x e 3 4x4
dx x C 5x e3x C5x C2 dx
3
2
Z Z Z Z p
6 C7 2
4e!3x dx 24x5 e!2x dx .u3 ! ue6!3u / du x . ! 5x2 /3 dx
Z Z Z Z !p "
1 302 C x C 6 1
dx dx eax dx 2x ! p dx
xC5 3x C 2x2 C 5x3 2x
Z Z Z Z
3x2 C 4x3 6x2 ! 6x x7
dx dx 4 dx .x2 C 1/2 dx
x3 C x4 1 ! 3x2 C 2x3 ex
Z Z Z # $
z 3 1
dz d 2
x.x ! 16/ ! 2
dx
.z ! 5/7
2 .5 ! 1/4 2x C 5
Z Z Z ! " Z # $
7 3 x x 3 1
dx dy C dx C dx
x 1 C 2y x2 C 1 .x2 C 1/2 x!1 .x ! 1/2
Z Z
s2 32x3 Z ! "
ds dx 3
3
s C5 4x4 C ! .5x2 C 10x5 /.x3 C x6 /!5 dx
Z Z 5x C 2
5 4t Z Z # $
dx dt p x
4 ! 2x 3t2C1 .r3 C 5/2 dr 3x C 1 ! dx
x2 C 3
Z p Z
1 Z ! "
5x dx dx x x2
.3x/6 ! 3 dx
7x C 2 .x C 2/4
2
Z Z
x Z p Z
p dx dx e x
ax2 C b 1 ! 3x p dx .e7 ! 7e /dx
x
Z Z p
4 C1 Z Z r
2y3 ey dy 2 2x ! 1 dx 1 C e2x 2 1
dx C dt
Z Z 4ex t2 t
2 !2 3 C1 x2 C x C 1 Z Z
e d q dx 4x C 3 p p
3
x4
3
x3 C 32 x2 C 3x ln.2x2 C 3x/dx 3
xe dx
2x2 C 3x
Z Z p
.e!5x C 2ex / dx 7 5 y C 3dy In Pr b ems nd y subject t the gi en c nditi ns
x
Z y0 D .5 ! 7x/3 I y.0/ D 2 y0 D 2 I y.1/ D 0
x C6
. x C 10/.7 ! 2x2 ! 5x/3 dx 1
y00 D 2 I y0 .!2/ D 3; y.1/ D 2
Z Z x
2 6x2 C
2ye3y dy dx y00 D .x C 1/1=2 I y0 . / D 1 , y.24/ D 2572
3
x3 C 4x
Real Estate The rate of change of the value of a house that
Z Z d
24s C 16 cost 350,000 to build can be modeled by D e0:05t , where t is
.ex C 2e!3x ! e5x / dx ds dt
1 C 4s C 3s2
the time in years since the house was built and is the value (in
Z thousands of dollars) of the house. Find .t/.
.6t2 C 4t/.t3 C t2 C 1/6 dt
Life Span If the rate of change of the expected life span, ,
Z at birth of people born in Canada can be modeled by
x.2x2 C 1/!1 dx d 12
D , where t is the number of years after 1 40 and the
Z dt 2t C 50
expected life span was 63 years in 1 40, find the expected life
.45 4
C1 2
C 12/.3 5
C2 3
C 4/!4 d
span for people born in 2000.
642 C nte rat on

Oxygen in Capillary In a discussion of the diffusion of where R is the constant rate at which oxygen diffuses from the
oxygen from capillaries,5 concentric cylinders of radius r are used capillary, and and B1 are constants. Find C. (Write the constant
as a model for a capillary. The concentration C of oxygen in the of integration as B2 .)
% &
capillary is given by Z ! " Find f(2) if f 13 D 2 and f 0 .x/ D e3xC2 ! 3x.
Rr B1
CD C dr
2 r

Objective T I
o sc ss tec n es of an n We turn now to some more di cult integration problems.
ore c a en n nte rat on ro e s
na e a e ra c an at on an When integrating fractions, sometimes a preliminary division is needed to get
tt n t e nte ran to a fa ar familiar integration forms, as the next example shows.
for o nte rate an e onent a
f nct on t a ase erent fro e E AM LE I
an to n t e cons t on f nct on
en t e ar na ro ens t to Z 3
cons e
x Cx
a Find dx.
x2
S A familiar integration form is not apparent. However, we can brea up the
integrand into two fractions by dividing each term in the numerator by the denominator.
We then have
Z 3 Z ! 3 " Z ! "
x Cx x x 1
dx D C dx D x C dx
Here we split up the integrand. x2 x2 x2 x
x2
D C ln jxj C C
2
Z
2x3 C 3x2 C x C 1
b Find dx.
2x C 1
S Here the integrand is a quotient of polynomials in which the degree of the
numerator is greater than or equal to that of the denominator. In such a situation we first
use long division. Recall that if f and g are polynomials, with the degree of f greater than
or equal to the degree of g, then long division allows us to find, uniquely, polynomials
q and r, where either r is the zero polynomial or the degree of r is strictly less than the
degree of g, satisfying
f r
Here we used long division to rewrite the DqC
integrand. g g
Using an obvious, abbreviated notation, we see that
Z Z ! " Z Z
f r r
D qC D qC
g g g
Since integrating a polynomial is easy, we see that integrating rational functions reduces
to the tas of integrating pr per rati na functi ns those for which the degree of the
numerator is strictly less than the degree of the denominator. In the case here we obtain
Z Z ! "
2x3 C 3x2 C x C 1 1
dx D x2 C x C dx
2x C 1 2x C 1
Z
x3 x2 1
D C C dx
3 2 2x C 1
Z
x3 x2 1 1
D C C d.2x C 1/
3 2 2 2x C 1
x3 x2 1
D C C ln j2x C 1j C C
3 2 2
Now ork Problem 1 G
5 W. Simon, athematica echniques f r Physi gy and edicine (New or : Academic Press, Inc., 1 72).
 Section 4.5 ec n es of nte rat on 643

E AM LE I I
Z
1
a Find p p dx.
x. x ! 2/3
Z p
. x ! 2/!3
S We can write this integral as p dx. et us try the power rule for
x
p 1 dx
integration with u D x ! 2. Then du D p dx, so that p D 2 du, and
2 x x
Z p Z ! "
. x ! 2/!3 p !3 dx
p dx D . x ! 2/ p
x x
Here the integral is fit to the form to Z ! !2 "
which the power rule for integration u
applies. D 2 u!3 du D 2 CC
!2
1 1
D! 2 CCD! p CC
u . x ! 2/2
Z
1
b Find dx.
x ln x
1
S If u D ln x, then du D dx, and
x
Z Z ! " Z
1 1 1 1
dx D dx D du
Here the integral fits the familiar form x ln x ln x x u
Z
1 D ln juj C C D ln j ln xj C C
u
du. Z
5
c Find d .
.ln /3=2
1
S If u D ln , then du D d . Applying the power rule for integration, we
have
Z Z ! "
5 !3=2 1
d D5 .ln / d
.ln /3=2
Z
u!1=2
Here the integral is fit to the form to D5 u!3=2 du D 5 " CC
which the power rule for integration 1
applies. !
2
!10 10
D 1=2 C C D ! CC
u .ln /1=2
Now ork Problem 23 G

I bu
In Section 14.4, we integrated an exponential function to the base e:
Z
eu du D eu C C

Now let us consider the integral of an exponential function with an arbitrary base, b.
Z
bu du

To find this integral, we first convert to base e using

bu D e.ln b/u
as we did in many differentiation examples, too. Example 3 will illustrate.
644 C nte rat on

E AM LE A I I bu
Z
Find 23!x dx.

S We want to integrate an exponential function to the base 2. To do this,

we will first convert from base 2 to base e by using Equation (1).

Z Z
3!x
2 dx D e.ln 2/.3!x/ dx

The integrand of the second integral is of the form eu , where u D .ln 2/.3 ! x/. Since
du D ! ln 2dx, we can solve for dx and write
Z Z
1
e.ln 2/.3!x/ dx D ! eu du
ln 2
1 u 1 .ln 2/.3!x/ 1 3!x
D! e CCD! e CCD! 2 CC
ln 2 ln 2 ln 2
Thus,
Z
1 3!x
23!x dx D ! 2 CC
ln 2
Notice that we expressed our answer in terms of an exponential function to the base 2,
the base of the original integrand.
Now ork Problem 27 G
eneralizing the procedure described in Example 3, we can obtain a formula for
integrating bu :
Z Z
b du D e.ln b/u du
u

Z
1
D e.ln b/u d..ln b/u/ ln b is a constant
ln b
1 .ln b/u
D e CC
ln b
1 u
D b CC
ln b
Hence, we have
Z
1 u
bu du D b CC
ln b
Applying this formula to the integral in Example 3 gives
Z
23!x dx b D 2, u D 3 ! x
Z
D ! 23!x d.3 ! x/ !d.3 ! x/ D dx

1 3!x
D! 2 CC
ln 2
which is the same result that we obtained before.

A I
We will now consider an application of integration that relates a consumption function
to the marginal propensity to consume.
 Section 4.5 ec n es of nte rat on 645

E AM LE F C F M
C

For a certain country, the marginal propensity to consume is given by

dC 3 1
D ! p
dI 4 2 3I
where consumption C is a function of national income I. Here, I is expressed in large
denominations of money. etermine the consumption function for the country if it is
nown that consumption is 10 .C D 10/ when I D 12.
S Since the marginal propensity to consume is the derivative of C, we have
Z ! " Z Z
3 1 3 1
C D C.I/ D ! p dI D dI ! .3I/!1=2 dI
4 2 3I 4 2
Z
3 1
D I! .3I/!1=2 dI
4 2
If we let u D 3I, then du D 3dI D d.3I/, and
! " Z
3 1 1
CD I! .3I/!1=2 d.3I/
4 2 3

3 1 .3I/1=2
D I! C
4 6 1
2
p
3 3I
CD I! C
4 3
This is an example of an initial-value
problem. When I D 12, C D 10, so
p
3 3.12/
10 D .12/ ! C
4 3
10 D !2C
Thus, D 3, and the consumption function is
p
3 3I
CD I! C3
4 3

Now ork Problem 61 G

R BLEMS
Z Z
In Pr b ems determine the inde nite integra s ex C 1 6x2 ! 11x C 5
Z Z dx dx
2x6 C x4 ! 4x 4x2 C 3 ex 3x ! 1
dx dx Z
2x2 2x Z
Z .3x C 1/.x C 3/ 5e2x
p dx dx
.3x2 C 2/ 2x3 C 4x C 1 dx xC2 7e2x C 4
Z Z Z Z
x 3 6.e4!3x /2 dx 5e13=x
p dx p dx dx
4 2
x C1 4 ! 5x x2
Z Z Z Z
2
2xex dx 5x
2x4 ! 6x3 C x ! 2 2x3
2 dx dx dx
ex2 ! 2 x!2 x2 C 1
Z Z Z Z p
5 ! 4x2 . x C 2/2
5t dt 2x.7 ! e x2 =4
/ dx dx p dx
3 C 2x 3 x
646 C nte rat on

Z Z
5es 5.x1=3 C 2/4 In Pr b ems and dr=dq is a margina re enue functi n
ds p dx Find the demand functi n
1 C 3es 3 2
x
Z dr 300 dr 00
Z p p ln x D D
aC x dx dq .q C 3/2 dq .2q C 3/3
p dx x
x
Z In Pr b ems and dc=dq is a margina c st functi n Find
p p Z p
r ln.r2 C 1/ the t ta c st functi n if xed c sts in each case are
t.3 ! t t/0:6 dt dr
r2 C 1 dc 20 dc
D D 4e0:005q
Z Z dq qC5 dq
x5 ! 6x4 ! ex3 .11/ln x
dx dx In Pr b ems dC=dI represents the margina pr pensity t
7x2 x
Z Z p c nsume Find the c nsumpti n functi n subject t the gi en
4 c nditi n
dx x2 ex3 C1 dx
x ln.2x2 / dC 1
Z Z D p I C. / D
ax C b dI I
dx c ¤ 0 dx
cx C d .x C 3/ ln.x C 3/ dC 1 1
D ! p C.25=3/ D 2
Z Z dI 3 2 3I
e x3 C x2 ! x ! 3
.ee C xe C ex /dx dx dC 3 1
x2 ! 3 D ! p I C.25/ D 23
dI 4 6 I
Z p Z p Cost Function The marginal-cost function for a
4x ln 1 C x2 12x3 ln.x4 C 1/3
dx dx manufacturer s product is given by
1 C x2 x4 C 1
Z p Z ! 3 " dc 100
x !1 D 10 !
3.x2 C 2/!1=2 xe x2 C2
dx p ! ln 2 dx dq q C 10
x4 ! 4x
Z Z where c is the total cost in dollars when q units are produced.
x ! x!2 2x4 ! x3 ! 6x2 C 4
dx dx When 100 units are produced, the average cost is 50 per unit.
x C 2x!1
2 x3 To the nearest dollar, determine the manufacturer s fixed cost.
Z Z
ex ! e!x x Cost Function Suppose the marginal-cost function for a
dx dx
ex C e!x xC1 manufacturer s product is given by
Z Z
4x3 C 2x xex
2
dc 100q2 ! 3 q C 60
dx p 2 dx D
.x4 C x2 / ln.x4 C x2 / ex C 2 dq q2 ! 40q C 1
Z where c is the total cost in dollars when q units are produced.
5
dx
.3x C 1/Œ1 C ln.3x C 1/$2 a etermine the marginal cost when 40 units are produced.
b If fixed costs are 10,000, find the total cost of producing
Z 40 units.
.e!x C 5/3
dx c Use the results of parts (a) and (b) and differentials to
ex
approximate the total cost of producing 42 units.
Z # $
1 1 Cost Function The marginal-cost function for a
! x dx
x C 1 e . C e!x /2 manufacturer s product is given by
Z q q
p p dc p
2
.x C 2x/ x2 C 2dx D q 0:04q3=4 C 4
dq 10
Z where c is the total cost in dollars when q units are produced.
d
3x ln x .1 C ln x/ dx int
.x ln x/ D 1 C ln x$ Fixed costs are 360.
dx
Z Z a
p q
etermine the marginal cost when 25 units are produced.
3=2
7 b Find the total cost of producing 25 units.
x . x/ C 3 dx dx
x.ln x/! c Use the results of parts (a) and (b) and differentials to
Z p Z approximate the total cost of producing 23 units.
s ln5 x
p ds dx Value of Land It is estimated that t years from now the
e s3 7x
value, (in dollars), of an acre of land near the ghost town of
Z Z Z x
2 ln. ex / onely Falls, .C., will be increasing at the rate of
eln.x C1/ dx dx dx
x t3
p dollars per year. If the land is currently worth
Z 0:2t4 C 000
0
ef .x/Cln. f .x// dx assuming f 0 > 0 600 per acre, how much will it be worth in 15 years ive the
answer to the nearest dollar.
 Section 4.6 e e n te nte ra 647

Revenue Function The marginal-revenue function for a Consumption Function A certain country s marginal
manufacturer s product is of the form propensity to save is given by
dr a
D q dS 2 1:6
dq e Cb D !p
dI 5 3
2I2
for constants a and b, where r is the total revenue received (in
dollars) when q units are produced and sold. Find the demand where S and I represent total national savings and income,
function, and express it in the form p D f .q/. ( int Rewrite respectively, and are measured in billions of dollars.
dr=dq by multiplying both numerator and denominator by e!q .)
a etermine the marginal propensity to consume when total
Savings A certain country s marginal propensity to save is national income is 16 billion.
given by b etermine the consumption function, given that savings are
dS 5 10 billion when total national income is 54 billion.
D c Use the result in part (b) to show that consumption is
dI .I C 2/2
2
where S and I represent total national savings and income, 5
D 16:4 billion when total national income is 16 billion
respectively, and are measured in billions of dollars. If total (a deficit situation).
national consumption is 7.5 billion when total national income d Use differentials and the results in parts (a) and (c) to
is billion, for what value(s) of I is total national savings equal approximate consumption when total national income is
to zero 1 billion.

Objective T I
o ot ate eans of t e conce t Figure 14.2 shows the region, R, bounded by the lines y D f.x/ D 2x, y D 0 (the
of area t e e n te nte ra as a t of
a s ec a s to e a ate s e x-axis), and x D 1. The region is simply a right triangle. If b and h are the lengths of
e n te nte ra s sn a tn the base and the height, respectively, then, from geometry, the area of the triangle is
rocess A D 12 bh D 12 .1/.2/ D 1 square unit. (Henceforth, we will treat areas as pure numbers
and write square unit only if it seems necessary for emphasis.) We will now find this
area by another method, which, as we will see later, applies to more complex regions.
This method involves the summation of areas of rectangles.
et us divide the interval Œ0; 1$ on the x-axis into four subintervals of equal length
by means of the equally spaced points x0 D 0, x1 D 14 , x2 D 24 , x3 D 34 , and x4 D 44 D 1.
1
(See Figure 14.3.) Each subinterval has length !x D . These subintervals determine
4
four subregions of R: R1 , R2 , R3 , and R4 , as indicated.
With each subregion, we can associate a circumscribed rectangle (Figure 14.4)
that is, a rectangle whose base is the corresponding subinterval and whose height is
the maximum value of f.x/ on that subinterval. Since f is an increasing function, the
maximum value of f.x/ on each subinterval occurs when x is the right-hand endpoint.
Thus, the areas of the circumscribed rectangles associated with regions R1 , R2 , R3 , and

y y
y
4
2 f 4
2
f(x) = 2x
f(x) = 2x R4 3
f (x) = 2x f 4

f 2
R3 4

R 1
R2 f 4

x0 R1 x1 x2 x3 x4
x x
1 2 3 4 1 2 3 4
x 4
1 4 4 4 4 4 4 4

FIGURE Region bounded FIGURE Four subregions of R. FIGURE Four circumscribed

by f .x/ D 2x, y D 0, and x D 1. rectangles.
648 C nte rat on

y R4 are 14 f. 14 /, 14 f. 24 /, 14 f. 34 /, and 14 f. 44 /, respectively. The area of each rectangle is an

approximation to the area of its corresponding subregion. Hence, the sum of the areas
of these rectangles, denoted by S4 , and called the fourth upper sum , approximates
the area A of the triangle. We have
f(x) = 2x
f
3
% &
1
% & % & % &
4
S4 D f 14 C 14 f 24 C 14 f 34 C 14 f 44
4
% % & % & % & % &&
D 14 2 14 C 2 24 C 2 34 C 2 44 D 5
4
2
f
P4
4

Using summation notation (see Section 1.5) we can write S4 D iD1 f.xi /!x D 54 .
f
1
4 The fact that S4 is greater than the actual area of the triangle might have been expected,
since S4 includes areas of shaded regions that are not in the triangle. (See Figure 14.4.)
f(0)
n the other hand, with each subregion we can also associate an inscribed rectangle
1 2 3 4
x (Figure 14.5) that is, a rectangle whose base is the corresponding subinterval, but
4 4 4 4 whose height is the minimum value of f.x/ on that subinterval. Since f is an increasing
FIGURE Four inscribed function, the minimum value of f.x/ on each subinterval will occur when x is the left-
rectangles. hand endpoint. Thus, the areas of the four inscribed rectangles associated with R1 , R2 ,
R3 , and R4 are 14 f.0/, 14 f. 14 /, 14 f. 24 /, and 14 f. 34 /, respectively. Their sum, denoted S 4 ,
and called the fourth lower sum , is also an approximation to the area A of the triangle.
We have
y
1
% & % & % &
S4 D 4
f.0/ C 14 f 14 C 14 f 24 C 14 f 34
-
S6 1
% % & % & % &&
D 4
2.0/ C 2 14 C 2 24 C 2 34 D 3
4
6
f 6 P
f(x) = 2x Using summation notation, we can write S 4 D 3iD0 f.xi /!x D 34 . Note that S 4 is less
5
f 6
than the area of the triangle, because the rectangles do not account for the portion of
f
4
6
the triangle that is not shaded in Figure 14.5.
3
Since
f 6
3 5
f 2
D S 4 % A % S4 D
6
4 4
1
f 6
we say that S 4 is an approximation to A from be and S4 is an approximation to A
1 2 3 4 5 6
x from ab e.
6 6 6 6 6 6
If Œ0; 1$ is divided into more subintervals, we expect that better approximations to
FIGURE Six circumscribed A will occur. To test this, let us use six subintervals of equal length !x D 16 . Then S6 ,
rectangles.
the total area of six circumscribed rectangles (see Figure 14.6), and S 6 , the total area
of six inscribed rectangles (see Figure 14.7), are
y
1
% & % & % & % & % & % &
S6 D f 16 C 16 f 26 C 16 f 36 C 16 f 46 C 16 f 56 C 16 f 66
6
S6 % % & % & % & % & % & % &&
D 16 2 16 C 2 26 C 2 36 C 2 46 C 2 56 C 2 66 D 76
-

f(x) = 2x and
f
5
1
% & % & % & % & % &
f.0/ C 16 f 16 C 16 f 26 C 16 f 36 C 16 f 46 C 16 f 56
6
S6 D 6
% % & % & % & % & % &&
4
f 6
D 16 2.0/ C 2 16 C 2 26 C 2 36 C 2 46 C 2 56 D 56
3
f 6

f 2 Note that S 6 % A % S6 , and, with appropriate labeling, both S6 and S 6 will be of

6
the form †f.x/!x. Clearly, using six subintervals gives better approximations to the
f
1
6 area than does four subintervals, as expected.
f(0) ore generally, if we divide Œ0; 1$ into n subintervals of equal length !x, then
x
1 2 3 4 5 6 !x D 1=n, and the endpoints of the subintervals are x D 0; 1=n; 2=n; : : : ; .n!1/=n, and
6 6 6 6 6 6
n=n D 1. (See Figure 14. .) The endpoints of the kth subinterval, for k D 1; : : : n, are
FIGURE Six inscribed rectangles. .k ! 1/=n and k=n and the maximum value of f occurs at the right-hand endpoint k=n. It
 Section 4.6 e e n te nte ra 649

y follows that the area of the kth circumscribed rectangle is 1=n"f.k=n/ D 1=n"2.k=n/ D
2k=n2 , for k D 1; : : : ; n. The total area of a n circumscribed rectangles is
X
n X
n
2k
f
n Sn D f.k=n/!x D
n
kD1 kD1
n2
f(x) = 2x
2 X
n
2
D k by factoring from each term
n2 kD1 n2
2 n.n C 1/
D " from Section 1.5
f 2 n2 2
n
nC1
1 D
f
n n
Pn
0 1 2 n
x (We recall that kD1 k D 1 C 2 C " " " C n is the sum of the first n positive integers and
n n n the formula used above was derived in Section 1.5.)
n-1 For inscribed rectangles, we note that the minimum value of f occurs at the left-
n hand endpoint, .k ! 1/=n, of Œ.k ! 1/=n; k=n$, so that the area of the kth inscribed
FIGURE n circumscribed rectangle is 1=n " f.k ! 1=n/ D 1=n " 2..k ! 1/=n/ D 2.k ! 1/=n2 , for k D 1; : : : n. The
rectangles. total area determined of a n inscribed rectangles (see Figure 14. ) is
X
n X
n
2.k ! 1/
Sn D f..k ! 1/=n/!x D
y
kD1 kD1
n2

2 X
n
2
D k!1 by factoring from each term
n2 kD1 n2

2 X
n-1 f (x) = 2x n!1
f n
D k ad usting the summation
n2 kD0
2 .n ! 1/n
D " adapted from Section 1.5
n2 2
n!1
1 D
f n n
x From Equations (1) and (2), we again see that both Sn and S n are sums of the form
0 1 2 n
X Xn ! " Xn ! "
n n n k k!1
f.x/!x, namely, Sn D f !x and S n D f !x.
n-1
n kD1
n kD1
n

FIGURE n inscribed rectangles. From the nature of S n and Sn , it seems reasonable and it is indeed true that
S n % A % Sn
As n becomes larger, S n and Sn become better approximations to A. In fact, let us ta e
the limits of S n and Sn as n approaches 1 through positive integral values:
! "
n!1 1
lim S n D lim D lim 1 ! D1
n!1 n!1 n n!1 n
! "
nC1 1
lim Sn D lim D lim 1 C D1
n!1 n!1 n n!1 n
Since S n and Sn have the same limit, namely,
lim S n D 1 D lim Sn
n!1 n!1

And since
S n % A % Sn
650 C nte rat on

for all n, where A is the area of the triangle, we conclude that A D 1. This agrees with
our prior finding. It is important to understand that here we developed a de niti n f
the n ti n f area that is applicable to many different regions.
We call the common limit of Sn and S n , namely, 1, the de nite integra of f.x/ D 2x
on the interval from x D 0 to x D 1, and we denote this quantity by writing
Z 1
2xdx D 1
0

The reason for using the term de nite integra and the symbolism in Equation (4) will
becomeR apparent in the next section. The numbers 0 and 1 appearing with the integral
sign in Equation (4) are called the b unds f integrati n 0 is the er b und, and
1 is the upper b und.
In general, for a continuous function f defined on the interval Œa; b$, where a < b,
we can form the sums S n and Sn , which are obtained by considering the minimum and
maximum values, respectively, on each of n subintervals of equal length !x. These
extreme values exist because we have assumed that f is continuous. We can now state
the following:

The common limit of S n and Sn as n ! 1, if it exists, is called the definite integral

of f over Œa; b$ and is written
Z b
f.x/dx
a

The definiteP integral is the limit of sums The numbers a and b are called bounds of integration a is the lower bound, and
of the form f .x/!x. This definition b is the upper bound. The symbol x is called the variable of integration and f.x/
will be useful in later sections. is the integrand.

A L IT I
A company has determined that its In terms of a limiting process, we have
marginal-revenue function is given by X Z b
R 0 .x/ D 600 ! 0:5x, where R is the f.x/ !x ! f.x/ dx
revenue (in dollars) received when x a
units are sold. Find the total revenue
Two points must be made about
P the definite integral. First, the definite integral is
received for selling 10 units by finding
the limit of a sum of the form f.x/!x. In fact, we can thin of the integral sign as
the area in the first quadrant bounded by
y D R 0 .x/ D 600 ! 0:5x and the lines an elongated S , the first letter of Summation . Second, for any continuous function
y D 0; x D 0, and x D 10. f defined on an interval, we may be able to calculate the sums S n and Sn and determine
their common limit. However, some terms in the sums will be negative if f ta es on
In general, over Œa; b$, we have
some negative values in the interval. These terms are not areas of rectangles (an area
is never negative), so the common limit may not represent an area. Thus, the de nite
b!a
!x D integra is n thing m re than a rea number it may r may n t represent an area
n In our discussion of the integral of a function f on an interval Œa; b$ we have limited
ourselves to c ntinu us functions. Integrals can be defined in greater generality than we
y
need but continuity ensures that the sequences S n and Sn have the same limit. Accord-
ingly, we will simplify our calculations in what follows by simply P using the right hand
4
endpoint of each subinterval when computing a sum of the form f.x/!x. f course
such right-hand endpoint values of f may be neither minima nor maxima in general.
f(x) = 4 - x2 The resulting sequence of sums will be denoted simply Sn .

E AM LE C A U R H E

Find the area of the region in the first quadrant bounded by f.x/ D 4 ! x2 and the lines
x D 0 and y D 0.
S A s etch of the region appears in Figure 14.10. The interval over which x
x varies in this region is seen to be Œ0; 2$, which we divide into n subintervals of equal
2
length !x. Since the length of Œ0; 2$ is 2, we ta e !x D 2=n. The endpoints of the
FIGURE Region of Example 1. subintervals are x D 0; 2=n; 2.2=n/; : : : ; .n ! 1/.2=n/,and n.2=n/ D 2, which are
 Section 4.6 e e n te nte ra 651

shown in Figure 14.11. The diagram also shows the corresponding rectangles obtained
y by using the right-hand endpoint of each subinterval. The area of the kth rectangle, for
k D 1; : : : n, is the product of its width, 2=n, and its height, f.k.2=n// D 4 ! .2k=n/2 ,
4 which is the function value at the right-hand endpoint of its base. Summing these areas,
we get
f(x) = 4 - x2
Xn ! ! "" Xn ! "2 !
2 2k 2
Sn D f k" !x D 4!
kD1
n kD1
n n
Xn ! " X n Xn Xn Xn
k2 k2
D ! 3 D ! 3
D 1! 3 k2
kD1
n n kD1
n kD1
n n kD1
n kD1

n.n C 1/.2n C 1/
x D n!
2 2 n n3 6
n n n
! "
4 .n C 1/.2n C 1/
2 2 D !
2 n (n-1) n 3 n2
FIGURE n subintervals and
corresponding rectangles for Example 1. The second line of the preceding computations uses basic summation manipulations
as discussed in Section 1.5. The third line uses two specific summation formulas, also
from Section 1.5: The sum of n copies of 1 is n and the sum of the first n squares is
n.n C 1/.2n C 1/
.
6
Finally, we ta e the limit of the Sn as n ! 1:
!! ""
4 .n C 1/.2n C 1/
lim Sn D lim !
n!1 n!1 3 n2
! 2 "
4 2n C 3n C 1
D ! lim
3 n!1 n2
! "
4 3 1
D ! lim 2 C C 2
3 n!1 n n
16
D ! D
3 3

16
Hence, the area of the region is .
3
Now ork Problem 7 G

E AM LE E I
Z 2
Evaluate .4 ! x2 /dx.
0

S We want to find the definite integral of f.x/ D 4 ! x2 over the interval Œ0; 2$.
16
Thus, we must compute limn!1 Sn . ut this limit is precisely the limit found in
3
Example 1, so we conclude that
Z 2
16
.4 ! x2 /dx D
0 3

Now ork Problem 19 G

652 C nte rat on

E AM LE I F I
0 3 n 3 =3
n n R3
Integrate f.x/ D x ! 5 from x D 0 to x D 3 that is, evaluate 0 .x ! 5/dx.
3
2 n (n -1) 3
n
S We first divide Œ0; 3$ into n subintervals of equal length !x D 3=n. The
FIGURE ividing Œ0; 3$ into n endpoints are 0; 3=n; 2.3=n/; : : : ; .n ! 1/.3=n/; n.3=n/ D 3. (See Figure 14.12.) Using
subintervals in Example 3. right-hand endpoints, we form the sum and simplify

Xn ! "
No units are attached to the answer, since 3 3
Sn D f k
a definite integral is simply a number.
kD1
n n

Xn !! " " X n ! " X 15 X

n n
3 3 15
D k !5 D k! D 2 k! 1
kD1
n n kD1
n2 n n kD1 n kD1
! "
n.n C 1/
15
D .n/ !
y n2 n2
! "
nC1 1
3 (n-1) 3 D ! 15 D 1C ! 15
2 n n 2 n 2 n
3 3 Ta ing the limit, we obtain
! ! " "
n n n
x 1 21
lim Sn D lim 1C ! 15 D ! 15 D !
n!1 n!1 2 n 2 2
-2 Thus,
Z 3
21
f(x) = x - 5 .x ! 5/dx D !
0 2
-5 Note that the definite integral here is a negati e number. The reason is clear from the
graph of f.x/ D x ! 5 over the interval Œ0; 3$. (See Figure 14.13.) Since the value of f.x/
FIGURE f .x/ is negative at each
is negative at each right-hand endpoint, each term in Sn must also be negative. Hence,
right-hand endpoint. limn!1 Sn , which is the definite integral, is negative.
eometrically, each term in Sn is the negative of the area of a rectangle. (Refer
again to Figure 14.13.) Although the definite integral is simply a number, here we can
interpret it as representing the negative of the area of the region bounded by f.x/ D x!5,
y x D 0, x D 3, and the x-axis (y D 0).
Now ork Problem 17 G
y = f(x) In Example 3, it was shown that the de nite integra d es n t ha e t represent
an area. In fact, there the definite integral was negative. However, if f is continuous
and f.x/ & 0 on Œa; b$, then Sn & 0 for all values of n. Therefore, limn!1 Sn & 0, so
Rb
a f.x/dx & 0. Furthermore, this definite integral gives the area of the region bounded
x
a b
by y D f.x/, y D 0, x D a, and x D b. (See Figure 14.14.)
FIGURE If f isRcontinuous and Although the approach that we too to discuss the definite integral is su cient for
b
f .x/ & 0 on Œa; b$, then a f .x/dx repre- our purposes, it is by no means rigorous. he imp rtant thing t remember ab ut the
sents the area under the curve. de nite integra is that it is the imit f a sequence f specia sums

R BLEMS
In Pr b ems sketch the regi n in the rst quadrant that is In Pr b ems and by di iding the indicated inter a int n
b unded by the gi en cur es Appr ximate the area f the regi n subinter a s f equa ength nd Sn f r the gi en functi n se the
by the indicated sum se the right hand endp int f each right hand endp int f each subinter a n t nd limn!1 Sn
subinter a
f .x/ D 4x Œ0; 1$ f .x/ D 2x C 1 Œ0; 2$
f .x/ D x C 1, y D 0, x D 0, x D 1 S4
f .x/ D 3x; y D 0; x D 1 S5 In Pr b ems and (a) simp ify Sn and ( ) nd limn!1 Sn
#! " ! " 'n ($
3
f .x/ D x , y D 0, x D 1 S4 1 1 2
Sn D C1 C C 1 C """ C C1
2
f .x/ D x C 1; y D 0; x D 0; x D 1 S2 n n n n
 Section 4.7 e n a enta eore of Ca c s 653
! "2 ! "2 ! "2 ! Z 3
3 3 3 3
Sn D 1" C 2" C """ C n " Find f .x/ dx without the use of limits, where
n n n n 0
8
In Pr b ems sketch the regi n in the rst quadrant that is < 2 if 0 % x < 1
b unded by the gi en cur es etermine the exact area f the f .x/ D 4 ! 2x if 1 % x < 2
:5x ! 10 if 2 % x % 3
regi n by c nsidering the imit f Sn as n ! 1 se the
right hand endp int f each subinter a Z 3
Find f .x/dx without the use of limits, where
Region as described in Problem 1 !1
Region as described in Problem 2 8
<!x C 2 if x < 1
Region as described in Problem 3 f .x/ D 1 if 1 % x % 2
:
y D x2 , y D 0, x D 1, x D 2 !x C 3 if x > 2
f .x/ D x2 C 1, y D 0, x D 0, x D 3 In each f Pr b ems use a pr grammab e aid such as a
2 ca cu at r r an n ine uti ity t estimate using subdi isi ns
f .x/ D ! x ; y D 0; x D 0
f the re e ant inter a the area f the regi n in the rst quadrant
In Pr b ems e a uate the gi en de nite integra by taking b unded by the gi en cur es R und the ans er t t decima
the imit f Sn se the right hand endp int f each subinter a p aces
Sketch the graph er the gi en inter a f the functi n t be
f .x/ D x3 C 1; y D 0; x D 2; x D 3:7
integrated p
Z 3 Z a f .x/ D 4 ! x; y D 0; x D 1; x D
5x dx b dx
0
f .x/ D ln x, y D 0, x D 1, x D 2
1
Z 3 Z 3 In each f Pr b ems use a pr grammab e aid such as a
!4x dx .!2x C 7/dx ca cu at r r an n ine uti ity t estimate using subdi isi ns
0 1 f the re e ant inter a the a ue f the de nite integra R und
Z 1 Z 2 the ans er t t decima p aces
2
.x C x/ dx .x C 2/ dx Z 5 Z 3! "
xC1 1
0 1 dx 1 C 2 dx
!Z " 2 xC2 1 x
d 1 p Z 2 Z 2
Find 2
1 ! x dx without the use of limits.
dx 0 .4x2 C x ! 13/ dx ln x dx
!1 1

Objective T F T C
o e e o nfor a t e n a enta
eore of Ca c s an to se t to
co te e n te nte ra s T F T
y Thus far, the limiting processes of both the derivative and definite integral have been
considered separately. We will now bring these fundamental ideas together and develop
the important relationship that exists between them. As a result, we will be able to
evaluate definite integrals much more e ciently.
The graph of a function f is given in Figure 14.15. Assume that f is continuous on
y = f(x)
the interval Œa; b$ and that its graph does not fall below the x-axis. That is, f.x/ & 0.
From the preceding section, the area of the region below the graph and above the
Rb
x-axis from x D a to x D b is given by a f.x/dx. We will now consider another
way to determine this area.
x Suppose that there is a function A D A.x/, which we will refer to as an area func-
a b
tion, that gives the area of the region below the graph of f and above the x-axis from
FIGURE n Œa; b$, f is a to x, where a % x % b. This region is shaded in Figure 14.16. o not confuse A.x/,
continuous and f .x/ & 0.
which is an area, with f.x/, which is the height of the graph at x.
From its definition, we can state two properties of A immediately:

A.a/ D 0, since there is no area from a to a

A.b/ is the area from a to b that is,
Z b
A.b/ D f.x/dx
a
654 C nte rat on

y If x is increased by h units, then A.x C h/ is the area of the shaded region in

Figure 14.17. Hence, A.x C h/ ! A.x/ is the difference of the areas in Figures 14.17 and
14.16, namely, the area of the shaded region in Figure 14.1 . For h su ciently close
to zero, the area of this region is the same as the area of a rectangle (Figure 14.1 )
y = f(x) whose base is h and whose height is some value y between f.x/ and f.x C h/. Here y
is a function of h. Thus, on the one hand, the area of the rectangle is A.x C h/ ! A.x/,
and, on the other hand, it is hy, so
A(x)
A.x C h/ ! A.x/ D hy
x
a x b Equivalently,
FIGURE A.x/ is an area A.x C h/ ! A.x/
function. Dy dividing by h
h
Since y is between f.x/ and f.x C h/, it follows that as h ! 0, y approaches f.x/, so
y
A.x C h/ ! A.x/
lim D f.x/
h!0 h
ut the left side is merely the derivative of A. Thus, Equation (1) becomes

A0 .x/ D f.x/

We conclude that the area function A has the additional property that its derivative A0
is f. That is, A is an antiderivative of f. Now, suppose that F is any antiderivative of f.
x Then, since both A and F are antiderivatives of the same function, they differ at most
a x b
by a constant C:
x+h

FIGURE A.x C h/ gives A.x/ D F.x/ C C

the area of the shaded region.
Recall that A.a/ D 0. So, evaluating both sides of Equation (2) when x D a gives
y
0 D F.a/ C C

so that

C D !F.a/

Thus, Equation (2) becomes

A.x/ D F.x/ ! F.a/

x
a x b
If x D b, then, from Equation (3),
x+h
A.b/ D F.b/ ! F.a/
FIGURE Area of shaded
region is A.x C h/ ! A.x/.
ut recall that
y Z b
A.b/ D f.x/dx
a

From Equations (4) and (5), we get

-
y
f(x + h) Z b
f(x) f.x/dx D F.b/ ! F.a/
a

A relationship
R b between a definite integral and antidifferentiation has now become
x clear. To find a f.x/dx, it su ces to find an antiderivative of f, say, F, and subtract the
h value of F at the lower bound a from its value at the upper bound b. We assumed here
FIGURE Area of rectangle is that f was continuous and f.x/ & 0 so that we could appeal to the concept of area. How-
the same as area of shaded region in ever, our result is true for any continuous function and is nown as the Fundamental
Figure 14.1 . heorem of Calculus.
 Section 4.7 e n a enta eore of Ca c s 655

F T C
If f is continuous on the interval Œa; b$ and F is any antiderivative of f on Œa; b$, then
Z b
f.x/dx D F.b/ ! F.a/
a

It is important to understand the difference between a definite integral and an

Rb
The definite integral is a number, and an indefinite integral. The de nite integra a f.x/dx is a number defined to be the limit
R
indefinite integral is a function. of a sum. The Fundamental Theorem states that the inde nite integra f.x/dx (the
most general antiderivative of f ), which is a function of x related to the differentiation
process, can be used to determine this limit. R2
Suppose we apply the Fundamental Theorem to evaluate 0 .4 ! x2 /dx. Here
f.x/ D 4!x2 , a D 0, and b D 2. Since an antiderivative of 4!x2 is F.x/ D 4x!.x3 =3/,
it follows that Z ! "
2
16
.4 ! x2 /dx D F.2/ ! F.0/ D ! ! .0/ D
0 3 3
This confirms our result in Example 2 of Section 14.6. If we had chosen F.x/ to be
4x ! .x3 =3/ C C, then we would have
#! " $
16
F.2/ ! F.0/ D ! C C ! Œ0 C C$ D
3 3
as before. Since the choice of the value of C is immaterial, for convenience we will
always choose it to be 0, as originally done. Usually, F.b/ ! F.a/ is abbreviated by
writing
ˇb
F.b/ ! F.a/ D F.x/ˇa
R
Since F in the Fundamental Theorem of Calculus is any antiderivative of f and f.x/dx
is the most general antiderivative of f, it showcases the notation to write
Z b !Z " ˇb
ˇ
f.x/dx D f.x/dx ˇˇ
a a
ˇb
Using the ˇa notation, we have
Z 2 ! "ˇ ! "
2 x3 ˇˇ2 16
.4 ! x /dx D 4x ! ˇ D ! !0D
0 3 0 3 3

E AM LE A F T
A L IT I
Z 3
The income (in dollars) from a fast-
food chain is increasing at a rate of Find .3x2 ! x C 6/dx.
!1
f .t/ DR 10;000e0:02t , where t is in years.
6
Find 3 10;000e0:02t dt, which gives the S An antiderivative of 3x2 ! x C 6 is
total income for the chain between the x2
third and sixth years. x3 ! C 6x
2
Thus, Z 3
.3x2 ! x C 6/dx
!1
! " ˇ3
x2 ˇ
3
D x ! C 6x ˇˇ
2 !1
# 2 $ # $
3 3 3 .!1/2
D 3 ! C 6.3/ ! .!1/ ! C 6.!1/
2 2
! " ! "
1 15
D ! ! D4
2 2
Now ork Problem 1 G
656 C nte rat on

I
Rb
For a f.x/dx, we have assumed that a < b. We now define the cases in which a > b
or a D b. First,
Z b Z a
If a > b; then f.x/dx D ! f.x/dx:
a b

That is, interchanging the bounds of integration changes the integral s sign. For exam-
ple,
Z 0 Z 2
2
.4 ! x /dx D ! .4 ! x2 /dx
2 0
If the bounds of integration are equal, we have
Z a
f.x/dx D 0
a

Some properties of the definite integral deserve mention. The first of the properties
that follow restates more formally our comment from the preceding section concern-
ing area.

I
Rb
If f is continuous and f.x/ & 0 on Œa; b$, then a f.x/dx can be interpreted as the
area of the region bounded by the curve y D f.x/, the x-axis, and the lines x D a
and x D b.
Rb Rb
a kf.x/dx D k a f.x/dx, where k is a constant
Rb Rb Rb
a . f.x/ ˙ g.x//dx D a f.x/dx ˙ a g.x/dx

Properties 2 and 3 are similar to rules for indefinite integrals because a definite
integral may be evaluated by the Fundamental Theorem in terms of an antiderivative.
Two more properties of definite integrals are as follows.

Rb Rb
a f.x/dx D a f.t/dt
The variable of integration is a dummy variable in the sense that any other variable
produces the same result that is, the same number.

To illustrate Property 4, you can verify, for example, that

Z 2 Z 2
2
x dx D t2 dt
0 0

If f is continuous on an interval I and a, b, and c are in I, then

Z c Z b Z c
f.x/dx D f.x/dx C f.x/dx
a a b

Property 5 means that the definite integral over an interval can be expressed in
terms of definite integrals over subintervals. Thus,
Z 2 Z 1 Z 2
.4 ! x2 /dx D .4 ! x2 /dx C .4 ! x2 /dx
0 0 1
We will loo at some examples of definite integration now and compute some areas
in Section 14. .
 Section 4.7 e n a enta eore of Ca c s 657

E AM LE U F T
Z 1
x3
Find p dx.
0 1 C x4
S To find an antiderivative of the integrand, we will apply the power rule for
integration:
Z 1 Z 1
In Example 2, the value of the x3
1 p dx D x3 .1 C x4 /!1=2 dx
antiderivative .1 C x4 /1=2 at the lower 0 1 C x4 0
2 ˇ1
Z 1 ! " ˇ
1 1=2 4 1=2 ˇ
bound 0 is .1/ . n t assume that 1 1 .1 C x / ˇ
2 D .1 C x4 /!1=2 d.1 C x4 / D ˇ
an evaluation at the bound zero will 4 0 4 1 ˇ
yield 0. ˇ
2 0
ˇ1 ' (
1 ˇ 1
D .1 C x4 /1=2 ˇˇ D .2/1=2 ! .1/1=2
2 2 0
1 p
D . 2 ! 1/
2
Now ork Problem 13 G
E AM LE E I
Z 2
a Find .4t1=3 C t.t2 C 1/3 /dt.
1
S
Z 2 Z 2 Z
1 2 2
.4t1=3 C t.t2 C 1/3 $dt D 4 t1=3 dt C .t C 1/3 d.t2 C 1/
1 1 2 1
ˇ2
ˇ ! " 2 ˇ2
t4=3 ˇˇ 1 .t C 1/4 ˇˇ
D .4/ ˇ C ˇ
4 ˇ 2 4
ˇ 1
3 1
1
D 3.24=3 ! 1/ C .54 ! 24 /
60
D 3 " 24=3 ! 3 C
p
3 5 5
D6 2C
Z 1
b Find e3t dt.
0
S Z Z
1
3t 1 1 3t
e dt D e d.3t/
0 3 0
! " ˇ1
1 3t ˇˇ 1 1
D e ˇ D .e3 ! e0 / D .e3 ! 1/
3 0 3 3
Now ork Problem 15 G
E AM LE F I I
Z 1
Evaluate x3 dx.
!2
S
Z ˇ1
1
x4 ˇˇ 14 .!2/4 1 16 15
x3 dx D ˇ D ! D ! D!
!2 4 !2 4 4 4 4 4
658 C nte rat on

y The reason the result is negative is clear from the graph of y D x3 on the interval Œ!2; 1$.
(See Figure 14.20.) For !2 % x < 0, f.x/ is negative. Since a definite integral is a limit
R0
of a sum of the form †f.x/!x, it follows that !2 x3 dx is not only a negative number
but also the negative of the area of the shaded region in the third quadrant. n the other
-2
x R1
1 hand, 0 x3 dx is the area of the shaded region in the first quadrant, since f.x/ & 0 on
Œ0; 1$. The definite integral over the entire interval Œ!2; 1$ is the a gebraic sum of these
numbers, because, from Property 5,
Z 1 Z 0 Z 1
3 3
y = x3 x dx D x dx C x3 dx
!2 !2 0
R1 3
Thus, !2 x dx does not represent the area between the curve and the x-axis. However,
if area is desired, it can be given by
FIGURE raph of y D x3 on ˇZ 0 ˇ Z 1
the interval Œ!2; 1$. ˇ ˇ
ˇ x 3 ˇ
dx x3 dx
ˇ ˇC
!2 0
Rb
Remember that a f .x/dx is a limit of a
sum. In some cases this limit represents Now ork Problem 25 G
an area. In others it does not. When
f .x/ & 0 on Œa; b$, the integral represents
the area between the graph of f and the
x-axis from x D a to x D b. T I
Since a function f is an antiderivative of f 0 , by the Fundamental Theorem we have
Z b
f 0 .x/dx D f.b/ ! f.a/
a

ut f 0 .x/ is the rate of change of f with respect to x. Hence, if we now the rate of
change of f and want to find the difference in function values f.b/ ! f.a/, it su ces to
Rb
evaluate a f 0 .x/dx.

E AM LE F C F
A L IT I I
A managerial service determines
that the rate of increase in maintenance A manufacturer s marginal-cost function is
costs (in dollars per year) for a par-
ticular apartment complex is given by dc
0 .x/ D 0x2 C5000, where x is the age
D 0:6q C 2
dq
of the apartment complex in years and
.x/ is the total (accumulated) cost of If production is presently set at q D 0 units per wee , how much more would it cost
maintenance for x years. Find the total to increase production to 100 units per wee
cost for the first five years. S The total-cost function is c D c.q/, and we want to find the difference
c.100/ ! c. 0/. The rate of change of c is dc=dq, so, by Equation (6),
Z 100 Z 100
dc
c.100/ ! c. 0/ D dq D .0:6q C 2/dq
0 dq 0
# $ˇ100 ˇ100
0:6q2 ˇ ˇ
D C 2q ˇ D Œ0:3q C 2q$ˇˇ
ˇ 2
2 0 0

D Œ0:3.100/2 C 2.100/$ ! Œ0:3. 0/2 C 2. 0/$

D 3200 ! 20 0 D 1120
If c is in dollars, then the cost of increasing production from 0 units to 100 units
is 1120.
Now ork Problem 59 G
 Section 4.7 e n a enta eore of Ca c s 659

R BLEMS
In Pr b ems e a uate the de nite integra Z p
2
Z 3 Z 5 3.x!2 C x!3 ! x!4 / dx
5 dx .e C 3e/ dx e
0 1 Z 2 ! " Z 2
p 1 2 C6x
Z 2 Z 6
6 x! p dx .x C 1/e3x dx
1 2x 1
5x dx !3xdx
1 !1 Z 5
x
Z 1 Z 1 dx
1 ln ex
.2x ! 3/ dx .4 ! y/ dy
!3 !1 Z 2
x6 C 6x4 C x3 C x2 C x C 5
Z 4 Z 1
dx
0 x3 C 5x C 1
.y2 C 4y C 4/ dy .2t ! 3t2 / dt Z 2
1 4 1 !x
Z Z ultiply the integrand by ee!x .)
dx ( int
!2
1 1 C ex
.3 2
! 2 C 3/d dt Z 2 ) 2
!4 4x if 0 % x < 12
f .x/ dx, where f .x/ D
Z Z 0 2x if 12 % x % 2
3 3
!3 3 !Z 2 "2 Z 2
3t dt dx
1 2 x2 Evaluate xdx ! x2 dx.
Z p Z 1
1
Z
Z 1
1
x
3
x4 dx 4 3 2
.x C x C x C x C 1/dx 1
! 0
Suppose f .x/ D 3 2 dt. Evaluate f .x/ dx.
1 t e
Z 3 Z 36 Z 7 Z 2 p
1 p 2 1
dx . x ! 2/ dx Evaluate ex dx C p dx.
1=2 x2 7 0 3 2
Z 2 Z Z 2 Z 1 Z 3
.z C 1/4 dz .x1=3 ! x!1=3 / dx If f .x/ dx D 5 and f .x/ dx D 2, find f .x/ dx.
!2 1 1 3 2
Z Z Z
Z 1 Z 3
4 4 3
If f .x/ dx D 6; f .x/ dx D 5, and f .x/ dx D 2, find
3x3 .x4 ! 1/4 dx .x C 2/3 dx
0 2 Z 3
1 2 1

Z Z !1
f .x/ dx.
4 2 2
dy dx Z 1 ! Z 1 "
1 y !e! x d x2
Z Evaluate e dx dx ( int It is not necessary to
Z e
1 dx
1
5 dx R1 2 0 0
e dx find 0 ex dx.)
0 2 xC1 Z
Z Z
x
et ! e!t
1 1 Suppose that f .x/ D dt where x > e. Find
5x2 ex dx
3
.3x2 C 4x/.x3 C 2x2 /4 dx e et C e!t
0
0 0 f .x/.
Z 4 Z 20=3 Severity Index In discussing tra c safety, Shonle6
3 p
dx 3x C 5 dx considers how much acceleration a person can tolerate in a crash
3 .x C 3/2 !1=3 so that there is no ma or in ury. The se erity index is defined as
Z 6 p Z 1 p Z
10 ! pdp q q2 C 3 dq S.I. D ˛ 5=2 dt
1 !1 0
Z 1 p Z p ! "
3
2
x where ˛ (a ree letter read alpha ) is considered a constant
x2 7x3 C 1 dx 2x ! dx involved with a weighted average acceleration, and is the
0 0 .x2 C 1/2=3
duration of the crash. Find the severity index.
Z 1 Z b
2x3 C x
dx .my C ny2 /dy
0 x2 C x4 C 1 a
Z Z
1
ex ! e!x 1
dx jxj dx
0 2 !2

6
. I. Shonle, En ir nmenta App icati ns f Genera Physics (Reading, A:
Addison-Wesley Publishing Company, Inc., 1 75).
660 C nte rat on

Statistics In statistics, the mean % (a ree letter read Mineral Consumption If C is the yearly consumption of a
mu ) of the continuous probability density function f defined on mineral at time t D 0, then, under continuous consumption, the
the interval Œa; b$ is given by total amount of the mineral used in the interval Œ0; t$ is
Z b
Z t
%D xf .x/ dx
a Cek" d'
0
and the variance & 2 (& is a ree letter read sigma ) is given by
Z b where k is the rate of consumption. For a rare-earth mineral, it has
&2 D .x ! %/2 f .x/ dx been determined that C D 3000 units and k D 0:05. Evaluate the
a
integral for these data.
Compute % and then & 2 if a D 0, b D 1, and f .x/ D 6.x ! x2 /.
Marginal Cost A manufacturer s marginal-cost function is
Distribution of Incomes The economist Pareto7 has stated
an empirical law of distribution of higher incomes that gives the dc
number, , of persons receiving x or more dollars. If D 0:1q C
dq
d
D !Ax!B If c is in dollars, determine the cost involved to increase
dx
production from 71 to 2 units.
where A and B are constants, set up a definite integral that gives
the total number of persons with incomes between a and b, where Marginal Cost Repeat Problem 5 if
a < b.
dc
Biology In a discussion of gene mutation, the following D 0:004q2 ! 0:5q C 50
integral occurs: dq
Z 10!4
x!1=2 dx and production increases from 0 to 1 0 units.
0 Marginal Revenue A manufacturer s marginal-revenue
Evaluate this integral. function is
Continuous Income Flow The present value (in dollars) of
a continuous ow of income of 2000 a year for five years at 6 dr 2000
D p
compounded continuously is given by dq 300q
Z 5
2000e!0:06t dt If r is in dollars, find the change in the manufacturer s total
0
Evaluate the present value to the nearest dollar. revenue if production is increased from 500 to 00 units.
Biology In biology, problems frequently arise involving Marginal Revenue Repeat Problem 61 if
the transfer of a substance between compartments. An example is
a transfer from the bloodstream to tissue. Evaluate the following dr
D 100 C 50q ! 3q2
integral, which occurs in a two-compartment diffusion problem: dq
Z t
.e!a" ! e!b" /d' and production is increased from 5 to 10 units.
0
Here, ' (read tau ) is a ree letter a and b are constants. Crime Rate A sociologist is studying the crime rate in a
certain city. She estimates that t months after the beginning of
Demography For a certain small population, suppose is a next year, the total number of crimes committed will increase at
function such that .x/ is the number of persons who reach the age the rate of t C 10 crimes per month. etermine the total number
of x in any year of time. This function is called a ife tab e of crimes that can be expected to be committed next year. How
functi n. Under appropriate conditions, the integral many crimes can be expected to be committed during the last six
Z b
months of that year
.t/ dt
a ospital Discharges For a group of hospitalized
gives the expected number of people in the population between individuals, suppose the discharge rate is given by
the exact ages of a and b, inclusive. If
p 1 $ 106
.x/ D 1000 110 ! x for 0 % x % 110 f .t/ D
.300 C t/4
determine the number of people between the exact ages of 10 and
2 , inclusive. ive your answer to the nearest integer, since where f .t/ is the proportion of the group discharged per day at the
fractional answers ma e no sense. What is the size of the end of t days. What proportion has been discharged by the end of
population 500 days

7
. Tintner, eth d gy f athematica Ec n mics and Ec n metrics
(Chicago: University of Chicago Press, 1 67), p. 16.
W. . Ewens, P pu ati n Genetics ( ondon: ethuen Company td., 1 6 ).
W. Simon, athematica echniques f r Physi gy and edicine (New or :
Academic Press, Inc., 1 72).
 Chapter 4 e e 661

Production Imagine a one-dimensional country of length Average Delivered Price In a discussion of a delivered
2R. (See Figure 14.21.10 ) Suppose the production of goods for this price of a good from a mill to a customer, eCanio11 claims that
country is continuously distributed from border to border. If the the average delivered price paid by consumers is given by
amount produced each year per unit of distance is f .x/, then the Z R
country s total yearly production is given by .m C x/Œ1 ! .m C x/$ dx
AD 0 Z R
Z R Œ1 ! .m C x/$ dx
GD f .x/ dx 0
!R where m is mill price, and x is the maximum distance to the point
of sale. eCanio determines that
Evaluate G if f .x/ D i, where i is constant. R R2
m C ! m2 ! mR !
2 3
AD
R
One-dimensional country 1!m!
2
Verify this.
-R 0 R
x In Pr b ems use the Fundamenta he rem f Integra
Ca cu us t determine the a ue f the de nite integra
Z 3:5 Z 1
Border Border 1
.1 C 2x C 3x2 / dx 2
dx
FIGURE 2:5 0 .x C 1/
Z 1
e3t dt Round the answer to two decimal places.
Exports For the one-dimensional country of Problem 65, 0

under certain conditions the amount of the country s exports is In Pr b ems estimate the a ue f the de nite integra by
given by using an appr ximating sum R und the ans er t t decima
p aces
Z 5 2 Z 4
Z x C1 1
R
i !k.R!x/ dx dx
ED Œe C e!k.RCx/ $ dx !1 x2C4
3 x ln x
!R 2 Z 3 p Z 1 p
6 qC1
2 t2 C 3 dt dq
0 !1 q C 3
where i and k are constants .k ¤ 0/. Evaluate E.

Chapter 14 Review
I T S E
S Differentials
differential, dy, dx Ex. 1, p. 620
S he Indefinite Integral R
antiderivative indefinite integral f.x/ dx integral sign Ex. 1, p. 626
integrand variable of integration constant of integration Ex. 2, p. 627
S Integration with Initial Conditions
initial condition Ex. 1, p. 631
S More Integration Formulas
power rule for integration Ex. 1, p. 636
S echniques of Integration
preliminary division Ex. 1, p. 642
S he Definite Integral Rb
definite integral a f.x/ dx bounds of integration Ex. 2, p. 651
S he Fundamental heorem of Calculus
Fundamental Theorem of Integral Calculus F.x/jba D F.b/ ! F.a/ Ex. 1, p. 655

10
R. Taagepera, Why the Trade NP Ratio ecreases with Country Size, S cia Science Research 5 (1 76), 3 5 404.
11 S. . eCanio, elivered Pricing and ultiple asing Point Equationilibria: A Reevaluation, he uarter y urna f Ec n mics CI , no. 2 (1 4), 32 4 .
662 C nte rat on

S
If y D f.x/ is a differentiable function of x, we define the it is important that the integral be written in a form that pre-
differential dy by cisely matches the power rule. ther integration formulas are
dy D f 0 .x/dx Z
where dx D !x is a change in x, which can be any real num- eu du D eu C C
ber. (Thus, dy is a function of two variables, namely x and Z
1
dx.) If dx is close to zero, then dy is an approximation to and du D ln juj C C u¤0
!y D f.x C dx/ ! f.x/. u

!y # dy If the rate of change of a function f is nown that is, if f 0

oreover, dy can be used to approximate a function value is nown then f is an antiderivative of f 0 . In addition, if we
using now that f satisfies an initial condition, then we can find
the particular antiderivative. For example, if a marginal-cost
f.x C dx/ # f.x/ C dy function dc=dq is given to us, then by integration we can find
the most general form of c. That form involves a constant of
An antiderivative of a function f is a function F such that integration. However, if we are also given fixed costs (that is,
F0 .x/ D f.x/. Any two antiderivatives of f differ at most by costs involved when q D 0), then we can determine the value
a constant. The most general antiderivative
R of f is called the of the constant of integration and, thus, find the particular
indefinite integral of f and is denoted f.x/dx. Thus, cost function, c. Similarly, if we are given a marginal-revenue
Z function dr=dq, then by integration and by using the fact that
f.x/dx D F.x/ C C r D 0 when q D 0, we can determine the particular rev-
enue function, r. nce r is nown, the corresponding demand
equation can be found by using the equation p D r=q.
where C is called the constant of integration, if and only if It is helpful at this point to review summation notation
F0 D f. Z from Section 1.5. This notation is especially useful in deter-
mining areas. For continuous f & 0, to find the area of the
It is important to remember that . /dx is an operation,
region bounded by y D f.x/, y D 0, x D a, and x D b, we
d divide the interval Œa; b$ into n subintervals of equal length
li e . /, that applies to functions to produce new func-
dx dx D .b!a/=n. If xi is the right-hand endpoint of an arbitrary
tions. The aptness of these strange notations becomes appar- subinterval, then the product f.xi / dx is the area of a rectan-
ent only after considerable study. gle. enoting the sum of all such areas of rectangles for the
Some elementary integration formulas are as follows: n subintervals by Sn , we define the limit of Sn as n ! 1 as
the area of the entire region:
Z
k dx D kx C C k a constant
X
n
Z lim Sn D lim f.xi / dx D area
xaC1 n!1 n!1
xa dx D CC a ¤ !1 iD1
aC1
Z
1
dx D ln x C C for x > 0 If the restriction that f.x/ & 0 is omitted, this limit is defined
x as the definite integral of f over Œa; b$:
Z
ex dx D ex C C
X
n Z b
Z Z
lim f.xi / dx D f.x/ dx
kf.x/ dx D k f.x/ dx k a constant n!1
iD1 a
Z Z Z
Œf.x/ ˙ g.x/$ dx D f.x/ dx ˙ g.x/ dx Instead of evaluating definite integrals by using limits,
we may be able to employ the Fundamental Theorem of
Calculus. In symbols,
Another formula is the power rule for integration:
Z Z ˇb
uaC1 b ˇ
ua du D C C; if a ¤ !1 f.x/ dx D F.x/ˇˇ D F.b/ ! F.a/
aC1 a a

Here u represents a differentiable function of x, and du is its

differential. In applying the power rule to a given integral, where F is any antiderivative of f.
 Chapter 4 e e 663
Z
Some properties of the definite integral are
It must be stressed that f.x/dx is a number, which if
Z b Z b f.x/ & 0 on Œa; b$ gives the area of the region bounded by
kf.x/ dx D k f.x/ dx k a constant y D f.x/, y D 0 and the vertical lines x D a and x D b.
a a
Z b Z b Z b
Œf.x/ ˙ g.x/$ dx D f.x/ dx ˙ g.x/ dx
a a a

and
Z c Z b Z c
f.x/ dx D f.x/ dx C f.x/ dx
a a b

R
In Pr b ems determine the integra s
Z Z Z Z
0
.x3 C 2x ! 7/ dx dx x2 C 4x ! 1 .x2 C 4/2
dx dx
!1 xC2 x2
Z Z Z
12 p 4 p
x Z p
. 3x C 3x2 / dx dx e Cx e 5x
0 5 ! 3x p dx p dx
2 x 3x
Z Z
3 Z 2 Z
dx .y ! 6/301 dy eln x 6x2 C 4
.x C 2/4 dx dx
3
1 x3 ex3 C2x
Z 2 Z 3 Z Z
6x ! 12 2
.1 C e2x /3 c
dx 2xe5!x dx dx dx
x3 ! 6x C 1 0 e!2x ebx .a C e!bx /n
Z 1 Z for n ¤ 1 and b ¤ 0
p 3 ! 4x
3
3t C dt dx Z p
0 2
3 103x dx
Z Z
y.y C 1/2 dy
1
Z
10! dx 3x3 C 6x2 C 17x C 2
0 dx
x2 C 2x C 5
Z p p Z
7
t! t .0:5x ! 0:1/4
p dt dx
3
t 0:4 In Pr b ems and nd y subject t the gi en c nditi n
Z 2 Z xC5
6t2 4x2 ! x y0 D e2x C 3; y.0/ D ! 12 y0 D ; y.1/ D 3
3
dt dx x
0 5 C 2t x
Z p Z In Pr b ems determine the area f the regi n b unded by
x2 3x3 C 2 dx .6x2 C 4x/.x3 C x2 /3=2 dx the gi en cur e the x axis and the gi en ines
Z Z y D x3 , x D 0, x D 2 y D 4ex ; x D 0; x D 3
4x
.e2y ! e!2y / dy p dx p
4
5 7 ! x2 yD x C 1, xD0
Z ! " Z 2 y D x2 ! x ! 6; x D !4; x D 3
1 2 3e3x
C 2 dx dx
x x 0 1 C e3x y D 5x ! x 2
Z 2 Z 70 p
.y4 C y3 C y2 C y/ dy dx y D 3 x, x D , x D 16
!2 7 1
Z Z y D C 2; x D 1; x D 4 y D x3 ! , x D 0
1 p 1 x
4x 5 ! x2 dx .2x C 1/.x2 C x/4 dx
0 0 Marginal Revenue If marginal revenue is given by
Z 1# $ Z 1 p
1
2x !
.x C 1/2=3
dx .2x ! 3 2x C 1/ dx dr 3p
0 0 D 100 ! 2q
Z p Z dq 2
t!3 z2
dt dz
t2 z!1 determine the corresponding demand equation.
664 C nte rat on

Marginal Cost If marginal cost is given by Biology In a discussion of gene mutation,12 the equation
Z qn Z n
dq
D !.u C / dt
dc q0 q ! bq
D q2 C 7q C 6 0
dq occurs, where u and are gene mutation rates, the q s are
gene frequencies, and n is the number of generations. Assume
and fixed costs are 2500, determine the total cost of that all letters represent constants, except q and t. Integrate
producing six units. Assume that costs are in dollars. both sides and then use your result to show that
ˇ ˇ
1 ˇ q0 ! b
q ˇˇ
Marginal Revenue A manufacturer s marginal-revenue nD ˇ
ln ˇ
function is uC q !b
n qˇ
Fluid Flow In studying the ow of a uid in a tube of
dr constant radius R, such as blood ow in portions of the body,
D 250 ! q ! 0:2q2
dq we can thin of the tube as consisting of concentric tubes of
radius r, where 0 % r % R. The velocity of the uid is a
If r is in dollars, find the increase in the manufacturer s total function of r and is given by13
revenue if production is increased from 15 to 25 units. .P1 ! P2 /.R2 ! r2 /
D
Marginal Cost A manufacturer s marginal-cost function is 4#
where P1 and P2 are pressures at the ends of the tube, # (a
dc 1000 ree letter read eta ) is the uid viscosity, and is the
D p length of the tube. The volume rate of ow through the tube,
dq 3q C 70
, is given by
Z R
If c is in dollars, determine the cost involved to increase D 2"r dr
production from 10 to 33 units. 0
ospital Discharges For a group of hospitalized "R 4 .P1 ! P2 /
Show that D . Note that R occurs as a factor
individuals, suppose the discharge rate is given by #
to the fourth power. Thus, doubling the radius of the tube has
f .t/ D 0:00 e!0:00 t the effect of increasing the ow by a factor of 16. The
formula that you derived for the volume rate of ow is called
P iseui e s a after the French physiologist ean Poiseuille.
where f .t/ is the proportion discharged per day at the end of t
days of hospitalization. What proportion of the group is Inventory In a discussion of inventory, arbosa and
discharged at the end of 100 days Friedman14 refer to the function
Z
1 1=x r
Business Expenses The total expenditures (in dollars) of a g.x/ D ku du
business over the next five years are given by k 1
where k and r are constants, k > 0 and r > !2, and x > 0.
Z 5 Verify the claim that
4000e0:05t dt 1
0 g0 .x/ D !
xrC2
Evaluate the expenditures. ( int Consider two cases: when r ¤ !1 and when r D !1.)

12
W. . ather, Princip es f uantitati e Genetics ( inneapolis: urgess
Publishing Company, 1 64).
13 R. W. Stacy et al., Essentia s f Bi gica and edica Physics (New or :
c raw-Hill, 1 55).
14 . C. arbosa and . Friedman, eterministic Inventory ot Size
odels a eneral Root aw, anagement Science 24, no. (1 7 ),
1 26.
 cat ons
of nte rat on

A
ny function, f, constructed from polynomials, exponentials, and logarithms
15.1 nte rat on a es using algebraic operations and composition can be differentiated and the
15.2 ro ate nte rat on resulting function, f 0 , is again of the same ind one that can be con-
structed from polynomials, exponentials, and logarithms using algebraic
15.3 rea et een C r es operations and composition. et us call such functions e ementary (although the term
15.4 Cons ers an usually has a slightly different meaning). In this terminology, the derivative of an ele-
Pro cers r s mentary function is also elementary. Integration is more complicated. There are fairly
2
simple elementary functions, for example f.x/ D ex , that do not have an elementary
15.5 era e a e of antiderivative. Said otherwise, there are integrals of elementary functions that we can-
a nct on
not d .
15.6 erent a at ons However, there are many integrals (of elementary functions) that can be d ne using
techniques that are beyond the scope of this text. A great many nown integrals have
15.7 ore cat ons of been collected together in tab es and a very short such table appears in our Appendix .
erent a at ons
The first section of this chapter is devoted to explaining the use of such tables. For prac-
15.8 ro er nte ra s tical problems their use saves a lot of time. Even those who now advanced integration
techniques may wrestle for some time with an integral that loo s deceptively simple.
C er 15 e e f course one of the main practical reasons for d ing an integral is to compute a
definite integral (a number) using the Fundamental Theorem of Calculus. However, it
turns out that the numbers given by definite integrals can be calculated approximately,
but to any desired degree of accuracy, by techniques that are not too far removed from
the definition of the definite integral as a limit of sums. We explore the two most widely
used such techniques for approximate integration.
With the possibility of approximating definite integrals and d ing indefinite inte-
grals using tables available from now on, we pursue further applications of integration.
The area warrants further study but perhaps the most important use of integration is
its role in studying and solving di erentia equati ns. These are equations where the
variable is a function, say y, that involve the first derivative y0 or perhaps higher order
derivatives of y. To solve such an equation is to find all functions, y, that satisfy the
given equation. A very common and simple example is the equation y0 D ky, where y
is understood to be a function of x and k is a constant.

665
666 C cat ons of nte rat on

Objective I T
o strate t e se of t e ta e of Certain forms of integrals that occur frequently can be found in standard tables of inte-
nte ra s n en
gration formulas. See, for example, W. H. eyer (ed.), CRC Standard athematica
ab es and F rmu ae 30th ed. ( oca Raton, F : CRC Press, 1 6). A very short table
appears in Appendix , and its use will be illustrated in this section.
A given integral may have to be replaced by an equivalent form before it will fit
a formula in the table. The equivalent form must match the formula exactly. Conse-
quently, the steps performed to get the equivalent form should be written carefully
rather than performed mentally. efore proceeding with the exercises that use tables,
we recommend studying the examples of this section carefully.
In the following examples, the formula numbers refer to the Table of Selected Inte-
grals given in Appendix . efore passing to such examples, though, we want to write
out here another basic integration formula.
ur rules for differentiation were of two types: asic Rules that dealt with specific
function types or functions (namely, constants, c powers of x, xa and the logarithm
function, ln x) and Combining Rules that dealt with arithmetic operations, composition
of functions, and the Inverse Function Rule. For integration there are fewer combining
rules of universal applicability. However, we can state

I I R Z
If f is invertible and differentiable and f.x/dx D F.x/ C C then
Z
f !1 .x/dx D x f !1 .x/ ! F. f !1 .x// C C

This rule can be deduced using one of the general techniques for integration that we
will not cover, but notice that, li e any putative antidifferention formula, it can easily
be eri ed by differentiation. Consider
d ! !1 " x f. f !1 .x//
x f .x/ C F. f !1 .x// D f !1 .x/ C 0 !1 ! 0 !1
dx f . f .x// f . f .x//
x x
D f !1 .x/ C 0 !1 ! 0 !1
f . f .x// f . f .x//
D f !1 .x/
Here, the derivative of the first term on the left is given by the Product Rule, ma ing
use of the Inverse Function Rule for differentiation, and the derivative of the second
term is given by the Chain Rule, ma ing use of F0 D f and again the Inverse Function
Rule. Finally, the definition of inverse functions is used to replace f. f !1 .x// by x.
If we apply this new rule to the case f.x/ D ex , where f !1 .x/ D ln x, we obtain
Z
ln xdx D x ln x ! eln x C C D x ln x ! x C C

It is worth generalizing this result to a logarithm of a function:

I L
Z
ln udu D u ln u ! u C C

This Integral of a ogarithm formula appears as Formula (41) in Appendix .

 Section 5. nte rat on a es 667

E AM LE I T
Z
xdx
Find
.2 C 3x/2
S Scanning the table, we identify the integrand with Formula (7):
Z # $
udu 1 a
D ln ja C buj C CC
.a C bu/2 b2 a C bu
Now we see if we can exactly match the given integrand with that in the formula. If we
replace x by u, 2 by a, and 3 by b, then du D dx, and by substitution we have
Z Z # $
xdx udu 1 a
D D 2 ln ja C buj C CC
.2 C 3x/2 .a C bu/2 b a C bu
Returning to the variable x and replacing a by 2 and b by 3, we obtain
Z # $
xdx 1 2
D ln j2 C 3xj C CC
.2 C 3x/2 2 C 3x
Note that the answer must be given in terms of x, the rigina variable of integration.
Now ork Problem 5 G

E AM LE I T
Z p
Find x2 x2 ! 1 dx.

S This integral is identified with Formula (24):

Z p p p
u a4
u2 u2 ˙ a2 du D .2u2 ˙ a2 / u2 ˙ a2 ! ln ju C u2 ˙ a2 j C C

In the preceding formula, if the bottommost sign in the dual symbol ˙ on the left
side is used, then the bottommost sign in the dual symbols on the right side must also
be used. In the original integral, we let u D x and a D 1. Then du D dx, and by
substitution the integral becomes
Z p Z p
x2 x2 ! 1 dx D u2 u2 ! a2 du

u p a4 p
D .2u2 ! a2 / u2 ! a2 ! ln ju C u2 ! a2 j C C

Since u D x and a D 1,
Z p p p
x 1
x2 x2 ! 1 dx D .2x2 ! 1/ x2 ! 1 ! ln jx C x2 ! 1j C C

Now ork Problem 17 G

This example, as well as Examples 4, 5, E AM LE I T

and 7, shows how to ad ust an integral so Z
that it conforms to one in the table. dx
Find p .
x 16x2 C 3
S The integrand can be identified with Formula (2 ):
Z ˇp ˇ
du 1 ˇˇ u2 C a2 ! a ˇˇ
p D ln ˇ ˇCC
u u2 C a2 a ˇ u ˇ
p
If we let u D 4x and a D 3, then du D 4 dx. Watch closely how, by inserting 4 s
in the numerator and denominator, we transform the given integral into an equivalent
668 C cat ons of nte rat on

form that matches Formula (2 ):

Z Z Z
dx .4 dx/ du
p D q p D p
2
x 16x C 3 u u2 C a2
.4x/ .4x/2 C . 3 /2
ˇp ˇ
1 ˇˇ u2 C a2 ! a ˇˇ
D ln ˇ ˇCC
a ˇ u ˇ
ˇp ˇ
ˇ 16x2 C 3 ! p3 ˇ
1 ˇ ˇ
D p ln ˇ ˇCC
3 ˇ 4x ˇ
Now ork Problem 7 G
E AM LE I T
Z
dx
Find .
x2 .2 ! 3x2 /1=2
S The integrand is identified with Formula (21):
Z p
du a2 ! u2
p D! CC
u2 a2 ! u2 a2 u
p p
etting u D 3x and a2 D 2, we have du D 3 dx. Hence, by inserting two factors
p
of 3 in both the numerator and denominator of the original integral, we have
Z p Z
p
p Z
dx . 3 dx/ du
2 2 1=2
D 3 p p D 3
x .2 ! 3x / . 3x/2 Œ2 ! . 3x/2 !1=2 u .a ! u2 /1=2
2 2

" p # " p #
p a2 ! u2 p 2 ! 3x2
D 3 ! CCD 3 ! p CC
a2 u 2. 3x/
p
2 ! 3x2
D! CC
2x
Now ork Problem 35 G
E AM LE I T
Z
Find 7x2 ln.4x/ dx.

S This is similar to Formula (42) with n D 2:

Z
unC1 ln u unC1
un ln u du D ! CC
nC1 .n C 1/2
If we let u D 4x, then du D 4 dx. Hence,
Z Z
7
7x ln.4x/ dx D 3 .4x/2 ln.4x/.4 dx/
2
4
Z # $
7 2 7 u3 ln u u3
D u ln u du D ! CC
64 64 3
# $
7 .4x/3 ln.4x/ .4x/3
D ! CC
64 3
# $
ln.4x/ 1
D 7x3 ! CC
3
7x3
D .3 ln.4x/ ! 1/ C C
Now ork Problem 45 G
 Section 5. nte rat on a es 669

E AM LE I T N N
Z
e2x dx
Find .
7 C e2x
S At first glance, we do not identify the integrand with any form in the table.
Perhaps rewriting the integral will help. et u D 7 C e2x , then du D 2e2x dx. So
Z Z Z
e2x dx 1 .2e2x dx/ 1 du 1
2x
D 2x
D D ln juj C C
7Ce 2 7Ce 2 u 2
1 1
D ln j7 C e2x j C C D ln.7 C e2x / C C
2 2
Thus, we had only to use our nowledge of basic integration forms. (Actually, this form
appears as Formula (2) in the table, with a D 0 and b D 1.)
Now ork Problem 39 G

E AM LE F I U T
Z 4
dx
Evaluate .
1 .4x2C 2/3=2
S We will use Formula (32) to get the indefinite integral first:
Z
du ˙u
D p CC
.u2 ˙ a2 /3=2 a u2 ˙ a2
2

etting u D 2x and a2 D 2, we have du D 2 dx. Thus,

Z Z Z
dx 1 .2 dx/ 1 du
D D
2
.4x C 2/ 3=2 2 2
..2x/ C 2/ 3=2 2 .u C 2/3=2
2

# $
1 u
D p CC
2 2 u2 C 2
Instead of substituting bac to x and evaluating from x D 1 to x D 4, we can deter-
Here we determine the bounds of mine the corresponding limits of integration with respect to u and then evaluate the last
integration with respect to u. expression between those limits. Since u D 2x, when x D 1, we have u D 2 when
x D 4, we have u D . Hence,
Z 4 Z
When changing the variable of dx 1 du
integration x to the variable of integration
D
1 .4x 2 C 2/3=2 2 2 .u2 C 2/3=2
u, be sure to change the bounds of
# $ˇ
integration so that they agree with u. 1 u ˇ
D p ˇ D p2 ! p 1
2 2 u2 C 2 ˇ2 66 2 6
Now ork Problem 15 G

I A A
Tables of integrals are useful when we deal with integrals associated with annuities.
Suppose that you must pay out 100 at the end of each year for the next two years.
Recall from Chapter 5 that a series of payments over a period of time, such as this,
is called an annuity. If you were to pay off the debt now instead, you would pay the
present value of the 100 that is due at the end of the first year, plus the present value
of the 100 that is due at the end of the second year. The sum of these present values
is the present value of the annuity. (The present value of an annuity is discussed in
Section 5.4.) We will now consider the present value of payments made continuously
over the time interval from t D 0 to t D , with t in years, when interest is compounded
continuously at an annual rate of r.
670 C cat ons of nte rat on

Suppose a payment is made at time t such that on an annual basis this payment is
f.t/. If we divide the interval Œ0; ! into subintervals Œti!1 ; ti ! of length dt (where dt is
small), then the total amount of all payments over such a subinterval is approximately
f.ti / dt. (For example, if f.t/ D 2000 and dt were one day, the total amount of the
1
payments would be 2000. 365 /.) The present value of these payments is approximately
!rti
e f.ti / dt. (See Section 5.3.) ver the interval Œ0; !, the total of all such present val-
ues is
X
e!rti f.ti /dt
This sum approximates the present value, A, of the annuity. The smaller dt is, the bet-
ter the approximation. That is, as dt ! 0, the limit of the sum is the present value.
However, this limit is also a definite integral. That is,

Z
AD f.t/e!rt dt
0
where A is the present value of a continuous annuity at an annual rate r (com-
pounded continuously) for years if a payment at time t is at the rate of f.t/ per year.

We say that Equation (1) gives the present value of a continuous income stream.
Equation (1) can also be used to find the present value of future profits of a business.
In this situation, f.t/ is the annual rate of profit at time t.
We can also consider the future value of an annuity rather than its present value. If
a payment is made at time t, then it has a certain value at the end of the period of the
annuity that is, ! t years later. This value is
# $ # $
amount of interest on this
C
payment payment for ! t years
If S is the total of such values for all payments, then S is called the accumu ated am unt
f a c ntinu us annuity and is given by the formula
Z
!t/
SD f.t/er. dt
0
where S is the accumulated amount of a continuous annuity at the end of years
at an annual rate r (compounded continuously) when a payment at time t is at the
rate of f.t/ per year.

E AM LE C A

Find the present value (to the nearest dollar) of a continuous annuity at an annual rate
of for 10 years if the payment at time t is at the rate of t2 dollars per year.
S The present value is given by
Z Z 10
!rt
AD f.t/e dt D t2 e!0:0 t dt
0 0
We will use Formula (3 ),
Z Z
un eau n
un eau du D ! un!1 eau du
a a
This is called a reducti n f rmu a since it reduces one integral to an expression that
involves another integral that is easier to determine. If u D t; n D 2, and a D !0:0 ,
then du D dt, and we have
ˇ10 Z 10
t2 e!0:0 t ˇˇ 2
AD ! te!0:0 t dt
!0:0 ˇ0 !0:0 0
 Section 5. nte rat on a es 671

In the new integral, the exponent of t has been reduced to 1. We can match this integral
with Formula (3 ),
Z
eau
ueau du D 2 .au ! 1/ C C
a
by letting u D t and a D !0:0 . Then du D dt, and
Z 10 ˇ10 # !0:0 t $ˇ10
t2 e!0:0 t ˇˇ 2 e ˇ
AD 2 !0:0 t
te dt D ˇ ! 2
.!0:0 t ! 1/ ˇˇ
0 !0:0 0 !0:0 .!0:0 / 0

!0: # !0: $
100e 2 e 1
D ! 2
.!0: ! 1/ ! .!1/
!0:0 !0:0 .!0:0 / .!0:0 /2
"1 5
The present value is 1 5.
Now ork Problem 59 G

R BLEMS
Z Z
In Pr b ems and use F rmu a in Appendix B t x dx 2dx
p
determine the integra s .1 C 3x/2 .2 C 2x/.5 C 2x/
Z Z
dx dx Z Z
dx p
. ! x2 /3=2 .25 ! 4x2 /3=2 7x2 3x2 ! 6 dx
7 ! 5x2
In Pr b ems and use F rmu a in Appendix B t Z Z
5 dx
determine the integra s 36x5 ln.3x/ dx
Z Z x2 .3 C 2x/2
dx 3 dx Z Z
p p p
x 16x2 C 3
2 x3 x4 ! 2x 1 C 3xdx x2 ln x dx
In Pr b ems nd the integra s by using the tab e in Z Z
dx dx
Appendix B p
Z Z 4x2 ! 13 x ln.2x/
dx 3x2 dx Z Z p
x.6 C 7x/ .1 C 2x/2 2 dx 5 ! x2
p dx
Z Z x2 16 ! x2 x
dx dx
p Z Z 1
x x2 C .x2 C 7/3=2 dx 3x2 dx
Z Z p p
x dx x." C 7e4 x / 0 1 C 2x
3
25x dx
.2 C 3x/.4 C 5x/
In Pr b ems nd the integra s by any meth d
Z Z Z Z
dx p x dx p 52
2x
x2 1 C x dx 3x xex dx
3Ce x2 C 1
Z Z Z Z p
7 dx dx .ln x/2 5x3 ! x
p dx dx
x.5 C 2x/2 x 5 ! 11x2 x 2x
Z 1 Z Z Z
x dx !2x2 dx dx e2x
p dx
0 2Cx 3x ! 2 x2 ! 5x C 6 e2x C 3
Z p Z Z Z
dx
x2 ! 3 dx 3
x ln x dx .4x C 2/e6xC3 dx
.1 C 5x/.2x C 3/
Z 1=12 Z r Z Z 2
2 C 3x 2 p
xe12x dx dx 4x3 e3x dx 35x2 3 C 2x dx
0 5 C 3x 1
Z Z 2 Z Z e
4 dx
x2 e2x dx ln2 x dx 3x ln x2 dx
2
1 x .1 C x/ 1
Z p 2 Z Z 1 Z 3
5x C 1 dx 2xdx p
dx p p x 2 C 3x dx
2x 2 x 2!x !1 5 C 2x 2
672 C cat ons of nte rat on

Z Z
1
2x dx ln 2 Continuous Annuity Find the present value, to the nearest
p x2 e3x dx dollar, of a continuous annuity at an annual rate of r for years if
0 ! x2 0
the payment at time t is at the annual rate of f .t/ dollars, given that
Z 2 Z 1
x ln.2x/ dx dX a r D 0:04 D f .t/ D 1000
1 !1 b r D 0:06 D 10 f .t/ D 500t
Biology In a discussion about gene frequency,1 the integral If f .t/ D k, where k is a positive constant, show that the value
of the integral in Equation (1) of this section is
Z # $
qn
dq 1 ! e!r
k
q0 q.1 ! q/ r
Continuous Annuity Find the accumulated amount, to the
occurs, where the q s represent gene frequencies. Evaluate this nearest dollar, of a continuous annuity at an annual rate of r for
integral. years if the payment at time t is at an annual rate of f .t/ dollars,
given that
Biology Under certain conditions, the number, n, of
generations required to change the frequency of a gene from 0.3 a r D 0:05 D 20 f .t/ D 100
to 0.1 is given by2 b r D 0:06 D 25 f .t/ D 100
Value of Business ver the next five years, the profits of
Z 0:1 a business at time t are estimated to be 50,000t dollars per year.
1 dq
nD! The business is to be sold at a price equal to the present value of
0:4 0:3 q2 .1 ! q/ these future profits. To the nearest 10 dollars, at what price should
the business be sold if interest is compounded continuously at the
Find n (to the nearest integer). annual rate of 7

Objective A I
o est ate t e a e of a e n te
nte ra s n ot t e tra e o a
r e an son s r e T R
We mentioned in the opening paragraphs for this chapter that there are seemingly sim-
ple functions that cannot be integrated in terms of elementary functions. No table con-
tains, for example, a formula for
Z
2
e!x dx

even though, as we will see, de nite integrals with integrands very similar to that above
are extremely important in practical applications.
n the other hand, consider a function f that is continuous on a closed interval Œa; b!
Rb
with f.x/ # 0 for all x in Œa; b!. The de nite integra a f.x/dx is simply the number
that gives the area of the region bounded by the curves y D f.x/, y D 0, x D a,
and x D b. It is unsatisfying, and perhaps impractical, not to say anything about the
Rb R
number a f.x/dx because of an inability to do the integral f.x/dx. This also applies
R
when the integral f.x/dx is merely too di cult for the person who needs to find the
Rb
number a f.x/dx, and these remar s apply to any continuous function, not ust those
with f.x/ # 0.
Rb P
Since a f.x/dx is defined as a limit of sums of the form f.x/#x, any partic-
P
ular well-formed sum of the form f.x/#x can be regarded as an approximation of
Rb
a f.x/dx. At least for nonnegative f, such sums can be regarded as sums of areas of
thin rectangles. Consider, for example, Figure 14.11 in Section 14.6, in which two rect-
angles are explicitly shown. It is clear that the error that arises from such rectangles is
associated with the small side at the top. The error would be reduced if we replaced

1 W. . ather, Princip es f uantitati e Genetics ( inneapolis: urgess Publishing Company, 1 64).

2 E. . Wilson and W. H. ossert, A Primer f P pu ati n Bi gy (Stamford, CT: Sinauer Associates, Inc., 1 71).
 Section 5.2 ro ate nte rat on 673

the rectangles by shapes that have a top side that is closer to the shape of the curve.
We will consider two possibilities: using thin trapezoids rather than thin rectangles,
the trape oidal rule and using thin regions surmounted by parabolic arcs, Simpson’s
rule. In each case only a finite number of numerical values of f.x/ need to be nown,
and the calculations involved are especially suitable for computers or calculators. In
both cases, we assume that f is continuous on Œa; b!.
In developing the trapezoidal rule, for convenience we will also assume that
f.x/ # 0 on Œa; b!, so that we can thin in terms of area. This rule involves approx-
imating the graph of f by straight-line segments.

y = f(x)

x
x0 x1 x2 x3 xn - 1 xn
=a =b

FIGURE Approximating an area by using trapezoids.

In Figure 15.1, the interval Œa; b! is divided into n subintervals of equal length by
the points a D x0 , x1 , x2 , : : :, and xn D b. Since the length of Œa; b! is b ! a, the length
of each subinterval is .b ! a/=n, which we will call h.
Clearly,

x1 D a C h; x2 D a C 2h; : : : ; xn D a C nh D b

With each subinterval, we can associate a trapezoid (a four-sided figure with two par-
allel sides). The area A of the region bounded by the curve, the x-axis, and the lines
f(a + h) Rb
f(a) x D a and x D b is a f.x/dx and can be approximated by the sum of the areas of the
trapezoids determined by the subintervals.
h
Consider the first trapezoid, which is redrawn in Figure 15.2. Since the area of a
a a+h
trapezoid is equal to one-half the base times the sum of the lengths of the parallel sides,
this trapezoid has area
FIGURE First trapezoid.
1
2
h. f.a/ C f.a C h//

Similarly, the second trapezoid has area

1
2
h. f.a/ C f.a C h//

Similarly, the second trapezoid has area

1
2
h. f.a C h/ C f.a C 2h//

The area, A, under the curve is approximated by the sum of the areas of n trapezoids:

A " 12 h. f.a/ C f.a C h// C 12 h. f.a C h/ C f.a C 2h//

C 12 h. f.a C 2h/ C f.a C 3h// C $ $ $ C 12 h. f.a C .n ! 1/h/ C f.b//
Rb
Since A D a f.x/dx, by simplifying the preceding formula we have the trapezoidal
rule:
674 C cat ons of nte rat on

T T R
Z b
h
f.x/dx "
. f.a/C2f.a C h/C2f.aC2h/C$ $ $C2f.aC.n ! 1/h/Cf.b//
a 2
where h D .b ! a/=n

The pattern of the coe cients inside the braces is 1, 2, 2, : : : , 2, 1. Usually, the
more subintervals, the better is the approximation. In our development, we assumed
for convenience that f.x/ # 0 on Œa; b!. However, the trapezoidal rule is valid without
this restriction.

E AM LE T R

Use the trapezoidal rule to estimate the value of

Z 1
1
dx
0 1 C x2

A L IT I
for n D 5. Compute each term to four decimal places, and round the answer to three
decimal places.
An oil tan er is losing oil at a rate
60 S Here, f.x/ D 1=.1 C x2 /; n D 5; a D 0; and b D 1. Thus,
of R0 .t/ D p , where t is the
t2 C
time in minutes and R.t/ is the radius b!a 1!0 1
hD D D D 0:2
of the oil slic in feet. Use the trape- n 5 5
zoidal rule with n D 5 to approximate
Z 5 The terms to be added are
60
p dt, the size of the radius
0 2
t C f.a/ D f.0/ D 1:0000
after five seconds.
2f.a C h/ D 2f.0:2/ D 1: 231
2f.a C 2h/ D 2f.0:4/ D 1:7241
2f.a C 3h/ D 2f.0:6/ D 1:4706
2f.a C 4h/ D 2f.0: / D 1:21 5
f.b/ D f.1/ D 0:5000 a C nh D b
7: 373 D sum

Hence, our estimate for the integral is

Z 1
1 0:2
2
dx " .7: 373/ " 0:7 4
0 1Cx 2

The actual value of the integral is approximately 0.7 5.

Now ork Problem 1 G
S R
Rb
Another method for estimating a f.x/ dx is given by Simpson s rule, which involves
approximating the graph of f by parabolic segments. We will omit the derivation.

S R
Z b
h
f.x/dx "
. f.a/C4f.aCh/C2f.aC2h/C$ $ $C4f.aC.n ! 1/h/Cf.b//
a 3
where h D .b ! a/=n and n is even.
 Section 5.2 ro ate nte rat on 675

The pattern of coe cients inside the braces is 1; 4; 2; 4; 2; : : : ; 2; 4; 1, which requires

that n be even. et us use this rule for the integral in Example 1.

E AM LE S R
A L IT I Z 1
1
A yeast culture is growing at the Use Simpson s rule to estimate the value of 2
dx for n D 4. Compute each
2 0 1Cx
rate of A0 .t/ D 0:3e0:2t , where t is the
term to four decimal places, and round the answer to three decimal places.
time in hours and A.t/ is the amount in
grams. Use SimpsonR s rule with n D S Here f.x/ D 1=.1 C x2 /, n D 4, a D 0, and b D 1. Thus, h D .b ! a/=n D
4 2
to approximate 0 0:3e0:2t dt, the 1=4 D 0:25. The terms to be added are
amount the culture grew over the first
four hours. f.a/ D f.0/ D 1:0000
4f.a C h/ D 4f.0:25/ D 3:7647
2f.a C 2h/ D 2f.0:5/ D 1:6000
4f.a C 3h/ D 4f.0:75/ D 2:5600
f.b/ D f.1/ D 0:5000
:4247 D sum

Therefore, by Simpson s rule,

Z 1
1 0:25
dx " . :4247/ " 0:7 5
0 1 C x2 3

This is a better approximation than that which we obtained in Example 1 by using the
trapezoidal rule.
Now ork Problem 5 G
oth Simpson s rule and the trapezoidal rule can be used if we now only f.a/,
f.a C h/, and so on we do not need to now f.x/ for all x in Œa; b!. Example 3 will
illustrate.

In Example 3, a definite integral is

E AM LE
estimated from data points the function
itself is not nown. A function often used in demography (the study of births, marriages, mortality, etc.,
in a population) is the life table function, denoted . In a population having 100,000
births in any year of time, .x/ represents the number of persons who reach the age of
x in any year of time. For example, if .20/ D ; 57, then the number of persons
who attain age 20 in any year of time is , 57. Suppose that the function applies to
all people born over an extended period of time. It can be shown that, at any time, the
expected number of persons in the population between the exact ages of x and x C m,
inclusive, is given by
Z xCm
.t/dt
x

The following table gives values of .x/ for males and females in the United States.3
Approximate the number of women in the 20 35 age group by using the trapezoidal
rule with n D 3.

3 ati na ita Statistics Rep rt, vol. 4 , no. 1 , February 7, 2001.

676 C cat ons of nte rat on

L T
x x
Age D x ales Females Age D x ales Females

0 100,000 100,000 45 3,717 6,5 2

5 ,066 ,220 50 1,616 5,3 2
10 , 67 ,144 55 ,646 3,562
15 , 34 ,05 60 4,1 0,700
20 ,346 , 57 65 77,547 6,2
25 7,64 ,627 70 6 ,375 7 , 26
30 6, 70 ,350 75 56,2 70,761
35 6,1 4 7, 64 0 42,127 5 ,573
40 5,163 7,3

S We want to estimate
Z 35
.t/dt
20
b!a 35 ! 20
We have h D D D 5. The terms to be added under the trapezoidal
n 3
rule are
.20/ D ; 57
2 .25/ D 2. ; 627/ D 1 7; 254
2 .30/ D 2. ; 350/ D 1 6; 700
.35/ D 7; 64
5 0; 775 D sum
y the trapezoidal rule,
Z 35
5
.t/dt " .5 0; 775/ D 1; 476; 37:5
2
20
Now ork Problem 17 G
Formulas used to determine the accuracy of answers obtained with the trapezoidal
rule or Simpson s rule can be found in standard texts on numerical analysis. For exam-
ple, one such formula tells us that the error committed by using the trapezoidal rule to
Rb
estimate a f.x/dx is given by
.b ! a/3 00
! f .Nx/ for some xN in .a; b/
12n2
The point of such formulas is that, given a required degree of accuracy, we can deter-
mine how big n needs to be to ensure that the trapezoidal approximation is adequate.

R BLEMS
In Pr b ems and use the trapez ida ru e r Simps n s ru e t three decima p aces In Pr b ems a s e a uate the
as indicated and the gi en a ue f n t estimate the integra integra by antidi erentiati n the Fundamenta he rem f
Z 4 Ca cu us
170 Z 1
2
dx trapezoidal rule, n D 6
!2 1 C x x3 dx trapezoidal rule, n D 5
Z 4 0
170 Z
dx Simpson s rule, n D 4 1
!2 1 C x2 x2 dx Simpson s rule, n D 4
0
In Pr b ems use the trapez ida ru e r Simps n s ru e as Z 4
indicated and the gi en a ue f n t estimate the integra dx
Simpson s rule, n D 4
C mpute each term t f ur decima p aces and r und the ans er 1 x2
 Section 5.2 ro ate nte rat on 677
Z 4
dx Area of Pool exter ri th, who is een on mathematics,
trapezoidal rule, n D 6 would li e to determine the surface area of his family s curved,
1 x
Z irregularly shaped, swimming pool. (All the tiles on the bottom of
2
xdx the pool need to be replaced and nobody has been able to
trapezoidal rule, n D
0 xC1 determine how many boxes of tiles to buy.) There is a straight
Z 4 fence that runs along the side of the pool dec . exter mar s off
dx
Simpson s rule, n D 6 points a and b on the fence as shown in Figure 15.4. He notes that
1 x the distance from a to b is m and subdivides the interval into
eight equal subintervals, naming the resulting points on the fence
In Pr b ems and use the ife tab e in Examp e t estimate
x1 , x2 , x3 , x4 , x5 , x6 , and x7 . exter ( ) stands at point x1 , holds a
the gi en integra s by the trapez ida ru e
Z 70 tape measure, and has his little brother Remy (R) ta e the free end
.t/ dt, males, n D 5 of the tape measure to the point P1 on the far side of the pool. He
45 as s his um, esley ( ) to stand at point 1 on the near side of
Z 55 the pool and note the distance on the tape measure. See
.t/ dt, females, n D 4 Figure 15.4.
35
In Pr b ems and supp se the graph f a c ntinu us
functi n f here f .x/ # 0 c ntains the gi en p ints se
Simps n s ru e and a f the p ints t appr ximate the area
P2
bet een the graph and the x axis n the gi en inter a R und P1
the ans er t ne decima p ace R B
.1; 0:4/; .2; 0:6/; .3; 1:2/; .4; 0: /; .5; 0:5/ 1,5
.2; 1/, .2:5; 3/, .3; 6/, .3:5; 10/, .4; 6/, .4:5; 3/, .5; 1/ Œ2; 5! A

R 3 Using all the information given in Figure 15.3, estimate L

1 f .x/dx by using Simpson s rule. ive the answer in Q1 Q2
fractional form.

D
y = f(x) a x1 x2 x3 x4 x5 x6 x7 b

FIGURE

2 ( , 2)
3
2 (2, 2)

exter then moves to point x2 and as s his brother to move to P2 ,

and his um to move to 2 and repeat the procedure. They do
1 (1, 1) (3, 1) this for each of the remaining points x3 to x7 . exter tabulates
their measurements in the following table:
( 5
2 ,
1
2 )
3 5
x
1 2 3
2 2 istance 0 1 2 3 4 5 6 7
along fence (m)
FIGURE
istance 0 3 4 3 3 2 2 2 0
In Pr b ems and use Simps n s ru e and the gi en a ue f across pool (m)
n t estimate the integra C mpute each term t f ur decima
p aces and r und the ans er t three decima p aces
Z 3
2
p dx n D 4 Also, evaluate the integral by the exter says that Simpson s rule now allows them to approximate
1 1Cx the area of the pool as
Fundamental Theorem of Calculus.
Z 1p
1 ! x2 dxI n D 4 1
0 .4.3/ C 2.4/ C 4.3/ C 2.3/ C 4.2/ C 2.2/ C 4.2//
Revenue Use Simpson s rule to approximate the total 3
revenue received from the production and sale of 0 units of a
product if the values of the marginal-revenue function dr=dq are
as follows: square meters. Remy says that this is not how he remembers
Simpson s rule. esley thin s that some terms are missing, but
q (units) 0 10 20 30 40 50 60 70 0 Remy gets bored and goes for a swim. Is exter s calculation
correct Explain, calculate the area, and then determine how many
dr
( per unit) 10 .5 .5 7.5 7 6.5 7 tiles, each with area 6:25cm2 , are needed to tile the bottom of the
dq pool.
678 C cat ons of nte rat on

Manufacturing A manufacturer estimated both marginal a Using the trapezoidal rule, estimate the total variable costs of
cost ( C) and marginal revenue ( R) at various levels of output production for 100 units.
(q). These estimates are given in the following table: b Using the trapezoidal rule, estimate the total revenue from the
sale of 100 units.
q(units) 0 20 40 60 0 100 c If we assume that maximum profit occurs when R D C
C ( per unit) 260 250 240 200 240 250 (that is, when q D 100), estimate the maximum profit if fixed
costs are 2000.
R ( per unit) 410 350 300 250 270 250

Objective A B C
o n t e area of a re on o n e In Sections 14.6 and 14.7 we saw that the area of a region bounded by the lines x D a,
c r es s n nte rat on o er ot
ert ca an or onta str s x D b, y D 0, and a curve y D f.x/ with f.x/ # 0 for a % x % b can be found
Z b
by evaluating the definite integral f.x/dx. Similarly, for a function f.x/ % 0 on an
a
interval Œa; b!, the area of the region bounded by x D a, x D b, y D 0, and y D f.x/ is
Z b Z b
given by ! f.x/dx D !f.x/dx. ost of the functions, f, we have encountered,
a a
and will encounter, are continuous and have a finite number of roots of f.x/ D 0. For
such functions, the roots of f.x/ D 0 partition the domain of f into a finite number of
intervals on each of which we have either f.x/ # 0 or f.x/ % 0. For such a function
we can determine the area bounded by y D f.x/, y D 0 and any pair of vertical lines
x D a and x D b, with a and b in the domain of f. We have only to find all the roots
Z c1
c1 < c2 < $ $ $ < ck with a < c1 and ck < b calculate the integrals f.x/ dx,
Z c2 Z b a

y f.x/ dx, $ $ $, f.x/ dx attach to each integral the correct sign to correspond to an
c1 ck
area and add the results. Example 1 will provide a modest example of this idea.
For such an area determination, a rough s etch of the region involved is extremely
valuable. To set up the integrals needed, a sample rectangle should be included in the
-1 ¢x = dx
-2
x s etch for each individual integral as in Figure 15.5. The area of the region is a limit
2
¢x =
dx
of sums of areas of rectangles. A s etch helps to understand the integration process
and it is indispensable when setting up integrals to find areas of complicated regions.
y = x2 - x - 2 Such a rectangle (see Figure 15.5) is called a vertical strip. In the diagram, the width
of the vertical strip is #x. We now from our wor on differentials in Section 14.1 that
FIGURE iagram for we can consistently write #x D dx, for x the independent variable. The height of the
Example 1.
vertical strip is the y-value of the curve. Hence, the rectangle has area y#x D f.x/dx.
The area of the entire region is found by summing the areas of all such vertical strips
between x D a and x D b and finding the limit of this sum, which is the definite
integral. Symbolically, we have
Z b
†y#x ! f.x/dx
a

For f.x/ # 0 it is helpful to thin of dx as a length differential and f.x/dx as an area

dA
differential dA. Then, as we saw in Section 14.7, we have D f.x/ for some area
dx
function A and
Z b Z b
f.x/ dx D dA D A.b/ ! A.a/
a a

(If our area function A measures area starting at the line x D a, as it did in Section 14.7,
then A.a/ D 0 and the area under f (and over 0) from a to b is ust A.b/.) It is important
to understand here that we need f.x/ # 0 in order to thin of f.x/ as a length and, hence,
f.x/dx as a differential area. ut if f.x/ % 0 then !f.x/ # 0 so that !f.x/ becomes a
length and !f.x/dx becomes a differential area.
 Section 5.3 rea et een C r es 679

E AM LE A A R T I
It is wrong to write hastily that the area is
R2 Find the area of the region bounded by the curve
!2 ydx, for the following reason: For the
left rectangle, the height is y. However, y D x2 ! x ! 2
for the rectangle on the right, y is
negative, so its height is the positive and the line y D 0 (the x-axis) from x D !2 to x D 2.
number !y. This points out the
S A s etch of the region is given in Figure 15.5. Notice that the x-intercepts
importance of s etching the region.
are .!1; 0/ and .2; 0/.
n the interval Œ!2; !1!, the area of the vertical strip is
ydx D .x2 ! x ! 2/dx
n the interval Œ!1; 2!, the area of the vertical strip is
.!y/dx D !.x2 ! x ! 2/dx
Thus,
Z !1 Z 2
area D .x2 ! x ! 2/dx C !.x2 ! x ! 2/dx
!2 !1
# $ ˇ!1 # 3 $ ˇ2
x3 x2 ˇ x x2 ˇ
D ! ! 2x ˇˇ ! ! ! 2x ˇˇ
3 2 !2 3 2 !1
## $ # $$
1 1 4
D ! ! C2 ! ! ! C4
3 2 3 2
## $ # $$
4 1 1
! ! !4 ! ! ! C2
3 2 3 2
1
D
3
Now ork Problem 22 G
efore embar ing on more complicated area problems, we motivate the further
study of area by seeing the use of area as a probability in statistics.

E AM LE S A

In statistics, a (probability) density function, f, of a variable, x, where x assumes all

values in the interval Œa; b!, has the following properties:
i f.x/ # 0
Rb
ii a f.x/dx D 1
The probability that x assumes a value between c and d, which is written P.c % x % d/,
where a % c % d % b, is represented by the area of the region bounded by the graph
of f and the x-axis between x D c and x D d. Hence (see Figure 15.6),
Z d
P.c % x % d/ D f.x/dx
c

y
y = f (x)

d
P (c … x … d) = f (x) dx
c

x
a c d b

FIGURE Probability as an area.

680 C cat ons of nte rat on

(In the terminology of Chapters and , the condition c ! x ! d defines an e ent,

and P.c ! x ! d/ is consistent with the notation of the earlier chapters. Note, too, that
the hypothesis (ii) above ensures that a ! x ! b is the certain e ent.)
For the density function f.x/ D 6.x " x2 /, where 0 ! x ! 1, find each of the
following probabilities.
a P.0 ! x ! 14 /
S Here Œa; b! is Œ0; 1!, c is 0, and d is 14 . We have
Z 1=4 Z
! " 1=4
P 0 ! x ! 14 D 6.x " x2 /dx D 6 .x " x2 /dx
0 0
# ˇ
3 $ ˇ1=4
ˇ1=4
x x2 ˇ
D6 " 2 ˇ 3 ˇ
D .3x " 2x /ˇ
2 3 0 ˇ
0
# $2 # $3 !
1 1 5
D 3 "2 "0D
4 4 32
! "
b P x # 12
1 1
S Since the domain of f is 0 ! x ! 1, to say that x # 2
means that 2
! x ! 1.
Thus,
# $ Z 1 Z 1
1
P x# D 6.x " x2 /dx D 6 .x " x2 /dx
2 1=2 1=2
# 2 $ˇ ˇ1
x x3 ˇˇ1 2
ˇ
3 ˇ 1
D6 " ˇ D .3x " 2x /ˇ D
2 3 1=2 1=2 2

Now ork Problem 27 G

S
We will now find the area of a region enclosed by several curves. As before, our proce-
dure will be to draw a sample strip of area and use the definite integral to add together
the areas of all such strips.
For example, consider the area of the region in Figure 15.7 that is bounded on the
top and bottom by the curves y D f.x/ and y D g.x/ and on the sides by the lines
x D a and x D b. The width of the indicated vertical strip is dx, and the height is the
y-value of the upper curve minus the y-value of the lower curve, which we will write
as yupper " ylower . Thus, the area of the strip is

.yupper " ylower /dx

(x, yupper)

y = f(x)

y = g(x)

(x, ylower)
x
a b
dx

FIGURE Region between curves.

 Section 5.3 rea et een C r es 681

which is

. f.x/ " g.x//dx

Summing the areas of all such strips from x D a to x D b by the definite integral gives
the area of the region:
X Z b
. f.x/ " g.x//dx ! . f.x/ " g.x//dx D area
a

We remar that there is another way to view this area problem. In Figure 15.7 both
f and g are above y D 0 and it is clear that the area we see is also the area under f
minus the area under g. That approach tells us that the required area is
Z b Z b Z b
f.x/dx " g.x/dx D . f.x/ " g.x//dx
a a a

However, our first approach does not require that either f or g lie above 0. ur usage of
yupper and ylower is really ust a way of saying that f # g on Œa; b!. This is equivalent to
saying that f " g # 0 on Œa; b! so that each differential . f.x/ " g.x// dx is meaningful
as an area.

y E AM LE F A C T
(x, yupper) p
Find the area of the region bounded by the curves y D x and y D x.
(1, 1)
S A s etch of the region appears in Figure 15. . To determine
p where the curves
y= x
y=x intersect, we solve the system formed by the equations y D x and y D x. Eliminating
(x, ylower)
y by substitution, we obtain
x p
(0, 0) dx 1 xDx
FIGURE iagram for Example 3. x D x2 squaring both sides
0 D x2 " x D x.x " 1/
x D 0 or xD1

Since we squared both sides, we must chec the solutions found with respect to the
It should be obvious that nowing the rigina equation. It is easily determined that both x D 0 and x D 1 are solutions of
points of intersection is important in p
determining the bounds of integration. x D x. If x D 0, then y D 0 if x D 1, then y D 1. Thus, the curves intersect at .0; 0/
and .1; 1/. The width of the indicated strip of area is dx. The height is the y-value on
the upper curve minus the y-value on the lower curve:
p
yupper " ylower D x " x
p
Hence, the area of the strip is . x " x/dx. Summing the areas of all such strips from
x D 0 to x D 1 by the definite integral, we get the area of the entire region:
Z 1
p
area D . x " x/dx
0
0 1ˇ1
Z ˇ
1 ˇ 3=2 2
B x x Cˇ
D .x1=2 " x/dx D @ " Aˇ
3 2 ˇ
0
ˇ
2 0
# $
2 1 1
D " " .0 " 0/ D
3 2 6
Now ork Problem 47 G
682 C cat ons of nte rat on

y E AM LE F A T C
y = 4x - x2 + 8
Find the area of the region bounded by the curves y D 4x " x2 C and y D x2 " 2x.
(4, 8) S A s etch of the region appears in Figure 15. . To find where the curves
2
y = x - 2x
intersect, we solve the system of equations y D 4x " x2 C and y D x2 " 2x:
(-1, 3) 4x " x2 C D x2 " 2x
dx
x "2x2 C 6x C D0
2
x " 3x " 4 D 0
FIGURE iagram for Example 4.
.x C 1/.x " 4/ D 0 factoring
x D "1 or xD4
When x D "1, then y D 3 when x D 4, then y D . Thus, the curves intersect at
."1; 3/ and .4; /. The width of the indicated strip is dx. The height is the y-value on
the upper curve minus the y-value on the lower curve:
yupper " ylower D .4x " x2 C / " .x2 " 2x/
Therefore, the area of the strip is
..4x " x2 C / " .x2 " 2x//dx D ."2x2 C 6x C /dx
Summing all such areas from x D "1 to x D 4, we have
Z 4
area D ."2x2 C 6x C /dx D 41 23
!1

Now ork Problem 51 G

y E AM LE A R H T U C

dx dx Find the area of the region between the curves y D " x2 and y D x2 C 1 from x D 0
to x D 3.
9 S The region is s etched in Figure 15.10. The curves intersect when
y = 9 - x2 y = x2 + 1
" x2 D x2 C 1
D 2x2
4 D x2
x D ˙2 two solutions
(2, 5) When x D ˙2, then y D 5, so the points of intersection are .˙2; 5/. ecause we are
x interested in the region from x D 0 to x D 3, the intersection point that is of concern to
2 3
us is .2; 5/. Notice in Figure 15.10 that in the region to the eft of the intersection point
FIGURE yupper is " x2 on .2; 5/, a strip has
Œ0; 2! and is x2 C 1 on Œ2; 3!.
yupper D " x2 and ylower D x2 C 1
but for a strip to the right of .2; 5/ the reverse is true, namely,
yupper D x2 C 1 and ylower D " x2
Thus, from x D 0 to x D 2, the area of a strip is
.yupper " ylower /dx D .. " x2 / " .x2 C 1/dx
D . " 2x2 /dx
but from x D 2 to x D 3, it is
.yupper " ylower /dx D ..x2 C 1/ " . " x2 //dx
D .2x2 " /dx
 Section 5.3 rea et een C r es 683

Therefore, to find the area of the entire region, we need t integrals:

Z 2 Z 3
2
area D . " 2x /dx C .2x2 " /dx
0 2
# $ˇ2 # 3 $ˇ3
2x3 ˇˇ 2x ˇ
ˇ
D x" ˇ C " x ˇ
3 0 3 2
## $ $ # # $$
16 16
D 16 " " 0 C .1 " 24/ " " 16
3 3
46
D
3
Now ork Problem 42 G

H S
Sometimes, area can more easily be determined by summing areas of horizontal strips
rather than vertical strips. In the following example, an area will be found by both
methods. In each case, the strip of area determines the form of the integral.

y
E AM LE S H S
y=3 9, 3
3 4 Find the area of the region bounded by the curve y2 D 4x and the lines y D 3 and x D 0
y2 = 4x
(the y-axis).
S The region is s etched in Figure 15.11. When the curves y D 3 and y2 D 4x
intersect, D 4x, so x D 4 . Thus, the intersection point is . 4 ; 3/. Since the width of
dx
the vertical strip is dx, we integrate with respect to the variable x. Accordingly, yupper
x
9 and ylowerpmust be expressed as functions of x. For the lower curve, y2 D 4x, we have
4 y D ˙2 x. ut y # 0 for the portion of this curve that bounds the region, so we use
p
FIGURE Vertical strip y D 2 x. The upper curve is y D 3. Hence, the height of the strip is
of area. p
yupper " ylower D 3 " 2 x
p
Therefore, the strip has an area of .3 " 2 x/"x, and we wish to sum all such areas
from x D 0 to x D 4 . We have
Z =4 !ˇ =4
p 4x3=2 ˇˇ
area D .3 " 2 x/dx D 3x " ˇ
With horizontal strips, the width is dy. 0 3 ˇ
0

# $ # $3=2 !
4
D 3 " " .0/
4 3 4

# $1=2 !3 # $3
y 27 4 27 4 3
D " D " D
y=3 9, 3 4 3 4 4 3 2 4
3 4
et us now approach this problem from the point of view of a hori ontal strip as
dy y2 = 4x shown in Figure 15.12. The width of the strip is dy. The length of the strip is the x a ue
n the rightm st cur e minus the x a ue n the eftm st cur e. Thus, the area of the
x=0 strip is

x .xright " xleft /dy

9
4 We wish to sum all such areas from y D 0 to y D 3:
FIGURE Horizontal X Z 3
strip of area. .xright " xleft /dy ! .xright " xleft /dy
0
684 C cat ons of nte rat on

Since the variable of integration is y, we must express xright and xleft as functions of y.
The rightmost curve is y2 D 4x so that x D y2 =4. The left curve is x D 0. Thus,
Z 3
area D .xright " xleft /dy
0
Z # $ ˇ3
3
y2 y3 ˇˇ
D " 0 dy D D
0 4 12 ˇ0 4

Note that for this region, horizontal strips ma e the definite integral easier to evaluate
(and set up) than an integral with vertical strips. In any case, remember that the bounds
of integration are bounds for the variable of integration
Now ork Problem 56 G

E AM LE A H E

Find the area of the region bounded by the graphs of y2 D x and x " y D 2.
S The region is s etched in Figure 15.13. The curves intersect when
y2 " y D 2. Thus, y2 " y " 2 D 0 equivalently, .y C 1/.y " 2/ D 0, from which
it follows that y D "1 or y D 2. This gives the intersection points .1; "1/ and .4; 2/.
try vertical strips of area. See Figure 15.13(a). Solving y2 D x for y gives
et us p
y D ˙ x. As p seen in Figure 15.13(a), to the eftpof x D 1, the upper end of the strip
lies on y D px and the lower end lies on y D " x. To the right of x D 1, the upper
curve is y D x and the lower curve is x " y D 2 (equivalently y D x " 2). Thus, with
vertical strips, t integrals are needed to evaluate the area:
Z 1 Z 4
p p p
area D . x " ." x//dx C . x " .x " 2//dx
0 1

y y

y2 = x
y2 = x (4, 2)
(4, 2)
dy
x-y=2
x x
dx x-y=2
(1, -1) (1, -1)
(a) (b)

FIGURE Region of Example 7 with vertical and horizontal strips.

Perhaps the use of horizontal strips can simplify our wor . In Figure 15.13(b), the width
of the strip is "y. The rightmost curve is a ays x " y D 2 (equivalently x D y C 2),
and the leftmost curve is always x D y2 . Therefore, the area of the horizontal strip is
Œ.y C 2/ " y2 !"y, so the total area is
Z 2
area D .y C 2 " y2 /dy D
!1 2

Clearly, the use of horizontal strips is the most desirable approach to solving the prob-
lem. nly a single integral is needed, and it is much simpler to compute.
Now ork Problem 57 G
 Section 5.3 rea et een C r es 685

R BLEMS
In Pr b ems use a de nite integra t nd the area f the a et r be a real number, where r > 1. Evaluate
regi n b unded by the gi en cur e the x axis and the gi en ines
In each case rst sketch the regi n atch ut f r areas f regi ns Z r
1
that are be the x axis dx
1 x2
y D 5x C 2; x D 1; xD4 y D x C 5; x D 2; xD4
y D 5x2 , x D 2, x D 6 y D x2 ; x D 2; xD3 b our answer to part (a) can be interpreted as the area of a
2 3 certain region of the#Z
plane. S etch
yDxCx Cx ; xD1
r $ this region.
y D x2 " 2x; x D "3; x D "1 1
c Evaluate lim 2
dx .
r!1 1 x
y D 3x2 " 4x; x D "2; x D "1
4 d our answer to part (c) can be interpreted as the area of a
2
yD2"x"x yD ; x D 1; xD2 certain region of the plane. S etch this region.
x
y D 2 " x " x3 ; x D "3; xD0 In Pr b ems use de nite integrati n t estimate the area f
the regi n b unded by the gi en cur e the x axis and the gi en
y D ex ; x D 1; x D 3
ines R und the ans ers t t decima p aces
1
yD ; x D 2; x D 3 1
.x " 1/2 yD 2 ; x D "2; x D 1
x C1
1 x
y D " , x D "e, x D "1 yD p ; x D 2; x D 7
x xC5
p
yD xC ; xD" ; xD0 y D x4 " 2x3 " 2, x D 1, x D 4
y D x2 " 4x; x D 2; x D 6 y D 1 C 3x " x4
p
y D 2x " 1; x D 1; x D 5 In Pr b ems express the area f the shaded regi n in terms
y D x3 C 3x2 ; x D "2; xD2 f an integra r integra s n t e a uate y ur expressi n
p
y D 3 x, x D See Figure 15.14.
x
y D e C 1; x D 0; xD1
y
y D jxj; x D "2; xD2
2 x=4
y D x C ; x D 1; x D 2
x y = 2x
y D x3 ; x D "2; x D 4
p
y D x " 3, x D 3, x D 2
y = x2 - x
y D x2 C 1; x D 0; xD4
x
iven that 0 4
%
3x2 if 0 ! x < 2 FIGURE
f .x/ D
16 " 2x if x # 2
See Figure 15.15.
determine the area of the region bounded by the graph of y D f .x/,
the x-axis, and the line x D 3. Include a s etch of the region. y
Under conditions of a continuous uniform distribution (a topic
in statistics), the proportion of persons with incomes between a
and t, where a ! t ! b, is the area of the region between the curve
y D 1=.b " a/ and the x-axis from x D a to x D t. S etch the y = 2x
graph of the curve and determine the area of the given region.
Suppose f .x/ D x= , where 0 ! x ! 4. If f is a density
function (refer to Example 2), find each of the following.
a P.0 ! x ! 1/ b P.2 ! x ! 4/ c P.x # 3/ y = x(x - 3)2
1
Suppose f .x/ D .1 " x/2 , where 0 ! x ! 3. If f is a density
3
function (refer to Example 2), !find each of" the following.
a P .1 ! x ! 3/ b P 1 ! x ! 32 c P .x ! 2/

d P .x # 2/ using the result from part (c)

Suppose f .x/ D 1=x, where e ! x ! e2 . If f is a density
function (refer to Example 2), find each of the following. x

a P.3 ! x ! 7/ b P.x ! 5/ c P.x # 4/

d Verify that P.e ! x ! e2 / D 1. FIGURE
686 C cat ons of nte rat on

See Figure 15.16. Find the area of the region that is between the curves

y y D x2 " 4x C 4 and y D 10 " x2

y=1 from x D 1 to x D 5.
Loren Curve A renz cur e is used in studying income
distributions. If x is the cumulative percentage of income
y = 1 - x2 y=x-1
recipients, ran ed from poorest to richest, and y is the cumulative
percentage of income, then equality of income distribution is
x
given by the line y D x in Figure 15.1 , where x and y are
expressed as decimals. For example, 10 of the people receive
FIGURE 10 of total income, 20 of the people receive 20 of the
See Figure 15.17. income, and so on. Suppose the actual distribution is given by the
y
orenz curve defined by
14 2 1
y=4 yD x C x
15 15
y
y = 2x
y = -2x - 8 1

Cumulative percentage of income

x

y=x

14 2 1 x
y= x +
FIGURE 15 15
Express, in terms of a single integral, the total area of the Lorentz curve
region to the right of the line x D 1 that is between the curves
y D x2 " 5 and y D 7 " 2x2 . o n t evaluate the integral. 0.104
Express, in terms of a single integral, the total area of the x
0.10 0.30 1
region in the first quadrant bounded by the x-axis and the graphs
Cumulative percentage of income recipients
of y2 D x and 2y D 3 " x. o n t evaluate the integral.
In Pr b ems nd the area f the regi n b unded by the FIGURE
graphs f the gi en equati ns Be sure t nd any needed p ints f Note, for example, that 30 of the people receive only 10.4 of
intersecti n C nsider hether the use f h riz nta strips makes total income. The degree of deviation from equality is measured
the integra simp er than hen ertica strips are used by the c e cient f inequa ity4 for a orenz curve. This
y D x2 ; y D 2x y D x; y D "x C 3; y D 0 coe cient is defined to be the area between the curve and
y D 12 " x2 , y D 3 y2 D x C 1; xD1 the diagonal, divided by the area under the diagonal:

x D C 2y; x D 0; y D "1; y D 3 area between curve and diagonal

y D x " 6; y2 D x y2 D 4x; y D 2x " 4 area under diagonal
y D x3 , y D 6x C , xD0 For example, when all incomes are equal, the coe cient of
2
2y D 4x " x ; 2y D x " 4 inequality is zero. Find the coe cient of inequality for the orenz
p curve ust defined.
y D x; y D x2
yD " x2 ; y D x2 ; x D "1; xD1 Loren curve Find the coe cient of inequality as in
Problem 5 for the orenz curve defined by y D 11 1
x2 C 12 x.
y D x3 C x, y D 0, x D "1, x D 2 12
p Find the area of the region bounded by the graphs of the
y D x3 , y D x y D x3 ; y D x
equations y2 D 3x and y D mx, where m is a positive constant.
1
4x C 4y C 17 D 0; y D a Find the area of the region bounded by the graphs of
x
y2 D "x " 2; x " y D 5; y D "1; y D 1 y D x2 " 1 and y D 2x C 2.
b What percentage of the area in part (a) lies above the x-axis
Find the area of the region that is between the curves
The region bounded by the curve y D x2 and the line y D 1 is
y D x " 1 and y D 5 " 2x divided into two parts of equal area by the line y D k, where k is a
from x D 0 to x D 4. constant. Find the value of k.

4
. Stigler, he he ry f Price 3rd ed. (New or : The acmillan
Company, 1 66), pp. 2 3 4.
 Section 5.4 Cons ers an Pro cers r s 687
p
In Pr b ems estimate the area f the regi n b unded by the yD 25 " x2 ; y D 7 " 2x " x4
graphs f the gi en equati ns R und y ur ans er t t decima
p aces y D x3 " x C 1; y D x2 " 5
6 y D x5 " 3x3 C 2x; y D 3x2 " 4
y D x2 " 4x C 1; y D "
x
y D x4 " 3x3 " 15x2 C 1 x C 30; y D x3 C x2 " 20x

Objective C S
o e e o t e econo c conce ts of etermining the area of a region has applications in economics. Figure 15.1 shows
cons ers s r s an ro cers
s r s c are re resente a supply curve for a product. The curve indicates the price, p, per unit at which the
areas manufacturer will sell (supply) q units. The diagram also shows a demand curve for the
product. This curve indicates the price, p, per unit at which consumers will purchase
p (demand) q units. The point .q0 ; p0 / where the two curves intersect is called the p int
f equi ibrium. Here p0 is the price per unit at which consumers will purchase the same
quantity, q0 , of a product that producers wish to sell at that price. In short, p0 is the
Demand price at which stability in the producer consumer relationship occurs.
curve
p1 et us assume that the mar et is at equilibrium and the price per unit of the product
p Supply
curve is p0 . According to the demand curve, there are consumers who would be willing to pay
p0 m re than p0 . For example, at the price per unit of p1 , consumers would buy q1 units.
These consumers are benefiting from the lower equilibrium price p0 .
The vertical strip in Figure 15.1 has area pdq. This expression can also be thought
of as the total amount of money that consumers would spend by buying dq units of the
q
q1
dq
q0 product if the price per unit were p. Since the price is actually p0 , these consumers
spend only p0 dq for the dq units and, thus, benefit by the amount pdq " p0 dq. This
FIGURE Supply and expression can be written .p " p0 /dq, which is the area of a rectangle of width dq and
demand curves. height p " p0 . (See Figure 15.20.) Summing the areas of all such rectangles from q D 0
to q D q0 by definite integration, we have
Z q0
p
.p " p0 /dq
0
Demand This integral, under certain conditions, represents the total gain to consumers who are
curve willing to pay more than the equilibrium price. This total gain is called consumers’
p Supply
curve
surplus, abbreviated CS. If the demand function is given by p D f.q/, then
p0 Z q0
CS D . f.q/ " p0 /dq
0
q
q0 eometrically (see Figure 15.21), consumers surplus is represented by the area between
dq
the line p D p0 and the demand curve p D f.q/ from q D 0 to q D q0 .
FIGURE enefit to Some of the producers also benefit from the equilibrium price, since they are will-
consumers for dq units. ing to supply the product at prices ess than p0 . Under certain conditions, the total gain
to the producers is represented geometrically in Figure 15.22 by the area between the

p p

Supply
curve
p = f(q )
p = g (q)
CS
p0 p0
PS
Demand
curve
q q
q0 q0

FIGURE Consumers surplus. FIGURE Producers surplus.

688 C cat ons of nte rat on

line p D p0 and the supply curve p D g.q/ from q D 0 to q D q0 . This gain, called
producers’ surplus and abbreviated PS, is given by
Z q0
PS D .p0 " g.q//dq
0

E AM LE F C S S

The demand function for a product is

p D f.q/ D 100 " 0:05q
where p is the price per unit (in dollars) for q units. The supply function is
p D g.q/ D 10 C 0:1q
etermine consumers surplus and producers surplus under mar et equilibrium.
S First we must find the equilibrium point .p0 ; q0 / by solving the system
formed by the functions p D 100 " 0:05q and p D 10 C 0:1q. We thus equate the
two expressions for p and solve:
10 C 0:1q D 100 " 0:05q
0:15q D 0
q D 600
When q D 600 then p D 10 C 0:1.600/ D 70. Hence, q0 D 600 and p0 D 70.
Consumers surplus is
Z q0 Z 600
CS D . f.q/ " p0 /dq D .100 " 0:05q " 70/dq
0 0

# $ˇ600
q2 ˇˇ
D 30q " 0:05 D 000
2 ˇ0
Producers surplus is
Z q0 Z 600
PS D .p0 " g.q//dq D .70 " .10 C 0:1q//dq
0 0

# $ˇ600
q2 ˇˇ
D 60q " 0:1 D 1 ; 000
2 ˇ0
Therefore, consumers surplus is 000 and producers surplus is 1 ,000.
Now ork Problem 1 G
E AM LE U H S F C S
S

The demand equation for a product is

0
q D f.p/ D "2
p
and the supply equation is q D g.p/ D p " 1. etermine consumers surplus and
producers surplus when mar et equilibrium has been established.
S etermining the equilibrium point, we have
0
p"1D "2
p
p2 C p " 0 D 0
.p C 10/.p " / D 0
 Section 5.4 Cons ers an Pro cers r s 689

45
q=p-1

dp
CS
9 PS q = 90
p -2
1 q
8

FIGURE iagram for Example 2.

Thus, p0 D , so q0 D " 1 D . (See Figure 15.23.) Note that the demand equation
expresses q as a function of p and that when q D 0, p D 45. Since consumers surplus
can be considered an area, this area can be determined by means of horizontal strips
of width dp and length q D f.p/. The areas of these strips are summed from p D to
p D 45 by integrating with respect to p:
Z 45 # $ ˇ45
0 ˇ
CS D " 2 dp D . 0 ln jpj " 2p/ˇˇ
p
D 0 ln 5 " 72 $ 72: 5
Using horizontal strips for producers surplus, we have
Z ˇ
.p " 1/2 ˇˇ
PS D .p " 1/ dp D ˇ D 32
1 2 1

Now ork Problem 5 G

R BLEMS
In Pr b ems the rst equati n is a demand equati n and the and the supply equation is
sec nd is a supp y equati n f a pr duct In each case determine
c nsumers surp us and pr ducers surp us under market q
pD C5
equi ibrium 60
p D 22 " 0: q p D 2200 " q2 Find producers surplus and consumers surplus under mar et
p D 6 C 1:2q p D 400 C q2 equilibrium.
50 The demand equation for a product is p D 2 !q , and the
pD p D 1000 " q2
qC5 p D 10q C 400 supply equation is p D 2qC3 , where p is the price per unit (in
q hundreds of dollars) when q units are demanded or supplied.
pD C 4:5
10 etermine, to the nearest thousand dollars, consumers surplus
p under mar et equilibrium.
q D 100.10 " 2p/ qD 100 " p
q D 50.2p " 1/ p The demand equation for a product is
q D " 10
2
The demand equation for a product is .p C 10/.q C 20/ D 1000
p
q D 10 100 " p and the supply equation is

Calculate consumers surplus under mar et equilibrium, which q " 4p C 10 D 0

occurs at a price of 4.
The demand equation for a product is a Verify, by substitution, that mar et equilibrium occurs when
p D 10 and q D 30.
q D 400 " p2 b etermine consumers surplus under mar et equilibrium.
690 C cat ons of nte rat on

The demand equation for a product is Producers’ Surplus The supply function for a product
50q is given by the following table, where p is the price per unit
p D 60 " p (in dollars) at which q units are supplied to the mar et:
q2 C 3600
and the supply equation is q 0 10 20 30 40 50
p 25 4 5 71 0 4
p D 10 ln.q C 20/ " 26
etermine consumers surplus and producers surplus under Use the trapezoidal rule to estimate the producers surplus if the
mar et equilibrium. Round the answers to the nearest integer. selling price is 0.

Objective A F
o e e o t e conce t of t e a era e If we are given the three numbers 1, 2, and , then their average value, also nown as
a e of a f nct on their mean, is their sum divided by 3. enoting this average by y, we have
1C2C
yD D4
3
Similarly, suppose we are given a function f defined on the interval Œa; b!, and the
points x1 ; x2 ; : : : ; xn are in the interval. Then the average value of the n corresponding
function values f.x1 /; f.x2 /; : : : ; f.xn / is
Pn
f.x1 / C f.x2 / C % % % C f.xn / f.xi /
yD D iD1
n n
We can go a step further. et us divide the interval Œa; b! into n subintervals of equal
length. We will choose xi to be the right-hand endpoint of the ith subinterval. ecause
b"a
Œa; b! has length b " a, each subinterval has length , which we will call dx. Thus,
n
Equation (1) can be written
X # $
1 X
n n
dx
f.xi / f.xi /dx
dx dx iD1 1 X
n
iD1
yD D D f.xi /dx
n n n dx iD1
b"a 1
Since dx D , it follows that ndx D b " a. So the expression in Equation (2)
n ndx
1
can be replaced by . oreover, as n ! 1, the number of function values used in
b"a
computing y increases, and we get the so-called a erage a ue f the functi n f, denoted
by f:
!
1 X X
n n
1
f D lim f.xi /dx D lim f.xi /dx
n!1 b " a b " a n!1 iD1
iD1
Rb
ut the limit on the right is ust the definite integral a f.x/dx. This motivates the fol-
lowing definition:

The average value of a function f.x/ over the interval Œa; b! is denoted f and is
given by
Z b
1
fD f.x/dx
b"a a
 Section 5.5 era e a e of a nct on 691

E AM LE A F

Find the average value of the function f.x/ D x2 over the interval Œ1; 2!.
S
y
Z b
1
fD f.x/dx
b"a a
4
Z ˇ2
1 2
2 x3 ˇˇ 7
D x dx D ˇ D
2"1 1 3 1 3
f(x) = x2
3
Now ork Problem 1 G
7
f = 3
In Example 1, we found that the average value of y D f.x/ D x2 over the interval
2
Œ1; 2! is 73 . We can interpret this value geometrically. Since
Z 2
7 1 7
1
3
x2 dx D
2"1 1 3
by solving for the integral we have
x Z 2
1 2 7
x2 dx D .2 " 1/
1 3
FIGURE eometric
interpretation of the average value However, this integral gives the area of the region bounded by f.x/ D x2 and the x-axis
of a function. from x D 1 to x D 2. (See Figure 15.24.) From the preceding equation, this area is
!7"
3
.2 " 1/, which is the area of a rectangle whose height is the average value f D 73
and whose width is b " a D 2 " 1 D 1.

E AM LE A F B

Suppose the ow of blood at time t in a system is given by

F1
F.t/ D 0!t!
.1 C ˛t/2
where F1 and ˛ (a ree letter read alpha ) are constants.5 Find the average ow F
on the interval Œ0; !.
S
Z
1
FD F.t/dt
"0 0
Z Z
1 F1 F1
D 2
dt D .1 C ˛t/!2 .˛dt/
0 .1 C ˛t/ ˛ 0
# $ ˇ # $
F1 .1 C ˛t/!1 ˇˇ F1 1
D ˇ D˛ " C1
˛ "1 0 1C˛
# $ # $
F1 "1 C 1 C ˛ F1 ˛ F1
D D D
˛ 1C˛ ˛ 1C˛ 1C˛

Now ork Problem 11 G

5 W. Simon, athematica echniques f r Physi gy and edicine (New or : Academic Press, Inc., 1 72).
692 C cat ons of nte rat on

R BLEMS
In Pr b ems nd the a erage a ue f the functi n er the value, S (in dollars), is given by S D 3000e0:05t . Find the average
gi en inter a value of a two-year investment.
f .x/ D x2 Œ"1; 3! f .x/ D 2x C 1 Œ0; 1! Medicine Suppose that colored dye is in ected into the
2
f .x/ D 2 " 3x I Œ"1; 2! 2
f .x/ D x C x C 1 1, 3 bloodstream at a constant rate, R. At time t, let
4
p R
f .t/ D 3t Œ"1; 2! 2
f .t/ D t t C 0, 4 C.t/ D
p F.t/
f .x/ D x Œ0; 1! f .x/ D 5=x2 Œ1; 3!
be the concentration of dye at a location distant (distal) from the
Profit The profit (in dollars) of a business is given by point of in ection, where F.t/ is as given in Example 2. Show that
the average concentration on 0, is
P D P.q/ D 36 q " 2:1q2 " 400 ! "
R 1 C ˛ C 13 ˛ 2 2
where q is the number of units of the product sold. Find the CD
F1
average profit on the interval from q D 0 to q D 100.
Revenue Suppose a manufacturer receives revenue, r, from
Cost Suppose the cost (in dollars) of producing q units the sale of q units of a product. Show that the average value of the
of a product is given by marginal-revenue function over the interval 0, q0 is the price per
c D 5000 C 12q C 0:3q2 unit when q0 units are sold.
1
Find the average cost on the interval from q D 200 to q D 500. Find the average value of f .x/ D 2 over the
x " 4x C 5
Investment An investment of 3000 earns interest at an interval Œ0; 1! using an approximate integration technique. Round
annual rate of 5 compounded continuously. After t years, its your answer to two decimal places.

Objective E
o so e a erent a e at on sn Frequently, we have to solve an equation that involves the derivative of an un nown
t e et o of se arat on of ar a es function. Such an equation is called a differential equation. An example is
o sc ss art c ar so t ons an
enera so t ons o e e o nterest y0 D xy2
co o n e cont n o s n ter s of
a erent a e at on o sc ss ore precisely, Equation (1) is called a first order differential equation, because
e onent a ro t an eca
it involves a derivative of the first order and none of any higher order. A solution of
Equation (1) is any function y D f.x/ that is defined on an interval and satisfies the
equation for all x in the interval.
To solve y0 D xy2 , equivalently,
dy
D xy2
dx
we thin of dy=dx as a quotient of differentials, and algebraically we separate the
variables by rewriting the equation so that each side contains only one variable and
all differentials appear as numerators:
dy
D xdx
y2
Integrating both sides and combining the constants of integration, we obtain
Z Z
1
dy D xdx
y2
1 x2
" D C C1
y 2
1 x2 C 2C1
" D
y 2
Since 2C1 is an arbitrary constant, we can replace it by C.
1 x2 C C
" D
y 2
 Section 5.6 erent a at ons 693

Solving Equation (3) for y, we have

2
yD"
x2 CC
We can verify that y is a solution to the differential Equation (2):
For if y is given by Equation (4), then
dy 4x
D 2
dx .x C C/2
while also
# $2
2 2 4x
xy D x " 2
D
x CC .x2 C C/2
showing that our y satisfies (2). Note in Equation (4) that, for each value of C, a differ-
ent solution is obtained. We call Equation (4) the general solution of the differential
equation. The method that we used to find it is called separation of variables.
In the foregoing example, suppose we are given the condition that y D " 23 when
x D 1 that is, y.1/ D " 23 . Then the particu ar function that satisfies both Equation (2)
and this condition can be found by substituting the values x D 1 and y D " 23 into
Equation (4) and solving for C:
2 2
" D" 2
3 1 CC
CD2
Therefore, the solution of dy=dx D xy2 such that y.1/ D " 23 is
2
yD"
x2 C2
We call Equation (5) a particular solution of the differential equation.

E AM LE S
A L IT I
y
For a clear liquid, light intensity Solve y0 D " if x; y > 0.
dI x
diminishes at a rate of D "kI, where
dx S Writing y0 as dy=dx, separating variables, and integrating, we have
I is the intensity of the light and x is the
number of feet below the surface of the dy y
liquid. If k D 0:00 5 and I D I0 when
D"
dx x
x D 0, find I as a function of x. dy dx
D"
y x
Z Z
1 1
dy D " dx
y x
ln jyj D C1 " ln jxj
Since x; y > 0, we can omit the absolute-value bars:
ln y D C1 " ln x
To solve for y, we convert Equation (6) to exponential form:
y D eC1 !ln x
So
eC1
y D eC1 e! ln x D
eln x
694 C cat ons of nte rat on

Replacing eC1 by C, where C > 0, and rewriting eln x as x gives

C
yD C; x > 0
x

Now ork Problem 1 G

In Example 1, note that Equation (6) expresses the solution implicitly, whereas
the final equation, y D C=x, states the solution y explicitly in terms of x. Solutions of
certain differential equations are often expressed in implicit form for convenience (or
necessity if there is an insurmountable di culty in obtaining an explicit form).

E G
In Section 5.3, the notion of interest compounded continuously was developed. et us
now ta e a different approach to this topic that involves a differential equation. Sup-
pose P dollars are invested at an annual rate, r, compounded n times a year. et the
function S D S.t/ give the compound amount, S, that is the total amount present, after
t years from the date of the initial investment. Then the initial principal is S.0/ D P.
Furthermore, since there are n interest periods per year, each period has length 1=n
years, which we will denote by dt. At the end of the first period, the accrued interest for
that period is added to the principal, and the sum acts as the principal for the second
period, and so on. Hence, if the beginning of an interest period occurs at time t, then the
increase in the amount present at the end of a period, dt, is S.t C dt/ " S.t/, which we
write as "S. This increase, "S, is also the interest earned for the period. Equivalently,
the interest earned is principal times rate times time:
"S D S % r % dt
ividing both sides by dt, we obtain
"S
D rS
dt
1
As dt ! 0, then n D ! 1, and consequently interest is being c mp unded
dt
c ntinu us y that is, the principal is sub ect to continuous growth at every instant.
However, as dt ! 0, then "S=dt ! dS=dt, and Equation (7) ta es the form
dS
D rS
dt
This differential equation means that hen interest is c mp unded c ntinu us y the
rate f change f the am unt f m ney present at time t is pr p rti na t the am unt
present at time t. The constant of proportionality is r.
To determine the actual function S, we solve the differential Equation ( ) by the
method of separation of variables:
dS
D rS
dt
dS
D rdt
S
Z Z
1
dS D rdt
S
ln jSj D rt C C1
We assume that S > 0, so ln jSj D ln S. Thus,
ln S D rt C C1
To get an explicit form, we can solve for S by converting to exponential form.
S D ertCC1 D eC1 ert
 Section 5.6 erent a at ons 695

S For simplicity, eC1 can be replaced by C (and then necessarily C > 0) to obtain the
general solution
S D Cert
3P
The condition S.0/ D P allows us to find the value of C:
2P S = Pert
P D Cer.0/ D C % 1
P
Hence, C D P, so
t S D Pert
Equation ( ) gives the total value after t years of an initial investment of P dollars
compounded continuously at an annual rate, r. (See Figure 15.25.)
FIGURE Compounding
continuously.
In our discussion of compound interest, we saw from Equation ( ) that the rate of
change in the amount present was proportional to the amount present. There are many
natural quantities, such as population, whose rate of growth or decay at any time is
considered proportional to the amount of that quantity present. If denotes the amount
of such a quantity at time t, then this rate of growth means that
d
Dk
dt
where k is a constant. If we separate variables and solve for as we did for Equation ( ),
we get
kt
D 0e

where 0 is a constant. In particular, if t D 0, then D 0 e0 D 0 % 1 D 0 . Thus, the

constant 0 is simply .0/. ue to the form of Equation (10), we say that the quantity
follows an exponential law of growth if k is positive and an exponential law of decay
if k is negative.

E AM LE G

In a certain city, the rate at which the population grows at any time is proportional to
the size of the population. If the population was 125,000 in 1 70 and 140,000 in 1 0,
what was the expected population in 2010
S et be the size of the population at time t. Since the exponential law of
growth applies,
kt
D 0e

To find the population in 2010, we must first find the particular law of growth involved
by determining the values of 0 and k. et the year 1 70 correspond to t D 0. Then
t D 20 in 1 0 and t D 40 in 2010. We have
0 D .0/ D 125; 000
Thus,
D 125; 000ekt
To find k, we use the fact that D 140; 000 when t D 20:
140; 000 D 125; 000e20k
Hence,
140; 000
e20k D D 1:12
125; 000
Therefore, the law of growth is
D 125; 000ekt
D 125; 000.e20k /t=20
D 125; 000.1:12/t=20
696 C cat ons of nte rat on

Setting t D 40 gives the expected population in 2010:

D .40/ D 125; 000.1:12/2 D 156; 00
ln.1:12/
We remar that from e20k D 1:12 we have 20k D ln.1:12/ and hence k D $
20
0:0057, which can be placed in D 125; 000ekt to give
$ 125; 000e0:0057t

Now ork Problem 23 G

In Chapter 4, radioactive decay was discussed. Here we will consider this topic
from the perspective of a differential equation. The rate at which a radioactive element
decays at any time is found to be proportional to the amount of that element present. If
is the amount of a radioactive substance at time t, then the rate of decay is given by
d
D "# :
dt
The positive constant # (a ree letter read lambda ) is called the decay constant, and
the minus sign indicates that is decreasing as t increases. Thus, we have exponential
decay. From Equation (10), the solution of this differential equation is
!!t
D 0e

If t D 0, then D 0 %1 D 0 , so 0 represents the amount of the radioactive substance

present when t D 0.
The time for one-half of the substance to decay is called the half life of the sub-
stance. In Section 4.2, it was shown that the half-life is given by
ln 2 0:6 315
half-life D
$
# #
Note that the half-life depends on #. In Chapter 4, Figure 4.13 shows the graph of
radioactive decay.

E AM LE F C H L

If 60 of a radioactive substance remains after 50 days, find the decay constant and
the half-life of the element.
S From Equation (14),
!!t
D 0e

where 0 is the amount of the element present at t D 0. When t D 50, then D 0:6 0,
and we have
!50!
0:6 0 D 0e

0:6 D e!50!
"50# D ln.0:6/ logarithmic form
ln.0:6/
#D" $ 0:01022
50
!0:01022t
Thus, $ 0e . The half-life, from Equation (15), is
ln 2
$ 67: 2days
#

Now ork Problem 27 G

 Section 5.6 erent a at ons 697

Radioactivity is useful in dating such things as fossil plant remains and archaeo-
logical remains made from organic material. Plants and other living organisms con-
tain a small amount of radioactive carbon 14 (14 C/ in addition to ordinary carbon
.12 C/. The 12 C atoms are stable, but the 14 C atoms are decaying exponentially. How-
ever, 14 C is formed in the atmosphere due to the effect of cosmic rays. This 14 C is
ta en up by plants during photosynthesis and replaces what has decayed. As a result,
the ratio of 14 C atoms to 12 C atoms is considered constant in living tissues over a
long period of time. When a plant dies, it stops absorbing 14 C, and the remaining
14
C atoms decay. y comparing the proportion of 14 C to 12 C in a fossil plant to that
of plants found today, we can estimate the age of the fossil. The half-life of 14 C is
approximately 5730 years. Thus, if a fossil is found to have a 14 C-to-12 C ratio that
is half that of a similar substance found today, we would estimate the fossil to be
5730 years old.

E AM LE E A A T

A wood tool found in a iddle East excavation site is found to have a 14 C-to-12 C ratio
that is 0.6 of the corresponding ratio in a present-day tree. Estimate the age of the tool
to the nearest hundred years.
S et be the amount of 14 C present in the wood t years after the tool was
made. Then D 0 e!!t , where 0 is the amount of 14 C when t D 0. Since the ratio
of 14 C to 12 C is 0.6 of the corresponding ratio in a present-day tree, this means that we
want to find the value of t for which D 0:6 0 . Thus, we have
!!t
0:6 0 D 0e

0:6 D e!!t
"#t D ln.0:6/ logarithmic form
1
t D " ln.0:6/
#
From Equation (15), the half-life is .ln 2/=#, which is approximately 5730, so
# $ .ln 2/=5730. Consequently,
1
t$" ln.0:6/
.ln 2/=5730
5730 ln.0:6/
$"
ln 2
$ 4200 years
Now ork Problem 29 G
R BLEMS
In Pr b ems s e the di erentia equati ns y0 D ex!y y.0/ D 0 ( int ex!y D ex =ey .)
0 2 0 2 2
y D 3xy y Dx y ey y0 " x3 D 0 y D 1 when x D 0
dy dy x 1
" 2x ln .x2 C 1/ D 0 D x2 y0 C D 0I y.1/ D 2
dx dx y y2
dy
D y, y > 0 y0 D ex y2 .3x2 C 2/3 y0 " xy2 D 0I y.0/ D 2
dx
y dy y0 C x3 y D 0I y D e when x D 0
y0 D ; x; y > 0 " x ln x D 0 p
x dx p
dy 3x 1 C y2
D I y > 0; y.1/ D
In Pr b ems s e each f the di erentia equati ns subject dx y
t the gi en c nditi ns dy 3x2 C 1
1 2y.x3 C x C 1/ D p y.0/ D 0
y0 D 2 y.1/ D 1 dx y2 C 4
y
698 C cat ons of nte rat on

dy xe!y Radioactivity If 20 of the initial amount of a radioactive

2 D p I y.1/ D 0
dx x2 C 3 sample has decayed after 100 seconds, find the decay constant and
2 the half-life of the element.
dy D 2xyex dx; y > 0I y.0/ D e
Carbon Dating An Egyptian scroll was found to have
Cost Find the manufacturer s cost function c D f .q/ a 14 C-to-12 C ratio 0.7 of the corresponding ratio in similar
given that present-day material. Estimate the age of the scroll, to the nearest
dc hundred years.
.q C 1/2 D cq
dq
and fixed cost is e. Carbon Dating A recently discovered archaeological
specimen has a 14 C-to-12 C ratio 0.1 of the corresponding ratio
Find f .2/, given that f .1/ D 0 and that y D f .x/ satisfies the found in present-day organic material. Estimate the age of the
differential equation specimen, to the nearest hundred years.
dy
D xex!y Population Growth Suppose that a population follows
dx exponential growth given by d =dt D k for t # t0 . Suppose
Circulation of Money A country has 1.00 billion dollars also that D 0 when t D xt0 . Find , the population size at
of paper money in circulation. Each wee 25 million dollars is time t.
brought into the ban s for deposit, and the same amount is paid
Radioactivity Polonium-210 has a half-life of about
out. The government decides to issue new paper money whenever
140 days. a Find the decay constant in terms of ln 2. b What
the old money comes into the ban s, it is destroyed and replaced
fraction of the original amount of a sample of polonium-210
by new money. et y be the amount of old money (in millions of
remains after one year
dollars) in circulation at time t (in wee s). Then y satisfies the
differential equation Radioactivity Radioactive isotopes are used in medical
diagnoses as tracers to determine abnormalities that may exist in
dy
D "0:025y an organ. For example, if radioactive iodine is swallowed, after
dt some time it is ta en up by the thyroid gland. With the use of a
How long will it ta e for 5 of the paper money in circulation to detector, the rate at which it is ta en up can be measured, and a
be new Round your answer to the nearest wee . ( int If 5 of determination can be made as to whether the upta e is normal.
money is new, then y is 5 of 1000.) Suppose radioactive technetium- m, which has a half-life of six
Marginal Revenue and Demand Suppose that a hours, is to be used in a brain scan two hours from now. What
monopolist s marginal-revenue function is given by the should be its activity now if the activity when it is used is to be
differential equation 12 units ive your answer to one decimal place. int In
Equation (14), let D activity t hours from now and
dr
D .50 " 4q/e!r=5 0 D activity now.
dq
Radioactivity A radioactive substance that has a half-life
Find the demand equation for the monopolist s product. of eight days is to be temporarily implanted in a hospital patient
Population Growth In a certain town, the population until three-fifths of the amount originally present remains. How
at any time changes at a rate proportional to the population. long should the implant remain in the patient
If the population in 1 0 was 60,000 and in 2000 was 64,000, Ecology In a forest, natural litter occurs, such as fallen
find an equation for the population at time t, where t is the leaves and branches, dead animals, and so on.6 et A D A.t/
number of years past 1 0. What is the expected population denote the amount of litter present at time t, where A.t/ is
in 2010 expressed in grams per square meter and t is in years. Suppose
Population Growth The population of a town increases by that there is no litter at t D 0. Thus, A.0/ D 0. Assume that
natural growth at a rate proportional to the number, , of persons (i) itter falls to the ground continuously at a constant rate of
present. If the population at time t D 0 is 50,000, find two 200 grams per square meter per year.
expressions for the population , t years later, if the population (ii) The accumulated litter decomposes continuously at the rate of
doubles in 50 years. Assume that ln 2 D 0:6 . Also, find for 50 of the amount present per year (which is 0:50A).
t D 100. The difference of the two rates is the rate of change of the amount
of litter present with respect to time:
Population Growth Suppose that the population of
the world in 1 30 was 2 billion and in 1 60 was 3 billion. If the # $ # $ # $
rate of change rate of falling rate of
exponential law of growth is assumed, what is the expected D "
of litter present to ground decomposition
population in 2015 ive your answer in terms of e.
Population Growth If exponential growth is assumed, Therefore,
in approximately how many years will a population double if
it triples in 1 years dA
D 200 " 0:50A
Radioactivity If 30 of the initial amount of a radioactive dt
sample remains after 100 seconds, find the decay constant and the Solve for A. To the nearest gram, determine the amount of litter
half-life of the element. per square meter after one year.

6
R. W. Poole, An Intr ducti n t uantitati e Ec gy (New or :
c raw-Hill oo Company, 1 74).
 Section 5.7 ore cat ons of erent a at ons 699

Profit and Advertising A company determines that the Value of a Machine The value of a certain machine
rate of change of monthly net profit P, as a function of monthly depreciates 25 in the first year after the machine is purchased.
advertising expenditure x, is proportional to the difference The rate at which the machine subsequently depreciates is
between a fixed amount, 250,000, and 2P that is, dP=dx is proportional to its value. Suppose that such a machine was
proportional to 250; 000 " 2P. Furthermore, if no money is spent purchased new on uly 1, 1 5, for 0,000 and was valued at
on monthly advertising, the monthly net profit is 10,000 if 3 , 00 on anuary 1, 2006.
1000 is spent on monthly advertising, the monthly net profit is a etermine a formula that expresses the value of the machine
50,000. What would the monthly net profit be if 5000 were in terms of t, the number of years after uly 1, 1 6.
spent on advertising each month b Use the formula in part (a) to determine the year and month in
which the machine has a value of exactly 14,000.

Objective M A E
o e e o t e o st c f nct on as a
so t on of a erent a e at on o
o e t e s rea of a r or o L G
sc ss an a e ton s a of
coo n In the previous section, we found that if the number of individuals in a population at
time t follows an exponential law of growth, then D 0 ekt , where k > 0 and 0 is
the population when t D 0. This law assumes that at time t the rate of growth, d =dt,
of the population is proportional to the number of individuals in the population. That
is, d =dt D k .
Under exponential growth, a population would get infinitely large as time goes
on, meaning that limt!1 0 ekt D 1. In reality, however, when the population gets
large enough, environmental factors slow down the rate of growth. Examples are food
supply, predators, overcrowding, and so on. These factors cause d =dt to decrease
eventually. It is reasonable to assume that the size of a population is limited to some
maximum number , where 0 < < , and as ! , d =dt ! 0, and the popula-
tion size tends to be stable.
In summary, we want a population model that has exponential growth initially but
also includes the effects of environmental resistance to large population growth. Such a
model is obtained by multiplying the right side of d =dt D k by the factor . " /= :
# $
d "
Dk
dt
Notice that if is small, then . " /= is close to 1, and we have growth that is
approximately exponential. As ! , then " ! 0 and d =dt ! 0, as we
wanted in our model. ecause k= is a constant, we can replace it by . Thus,
d
D . " /
dt
This states that the rate of growth is proportional to the product of the size of the pop-
ulation and the difference between the maximum size and the actual size of the popu-
lation. We can solve for in the differential Equation (1) by the method of separation
of variables:
d
D dt
. " /
Z Z
1
d D dt
. " /
The integral on the left side can be found by using Formula (5) in the table of integrals
in Appendix . Thus, Equation (2) becomes
ˇ ˇ
1 ˇˇ ˇ
ˇD tCC
ln ˇ
" ˇ
so
ˇ ˇ
ˇ ˇ
ln ˇˇ ˇD
ˇ tC C
"
700 C cat ons of nte rat on

Since > 0 and " > 0, we can write

ln D tC C
"
In exponential form, we have

tC C t C
De De e
"
C
Replacing the positive constant e by A and solving for gives

t
D Ae
"
t
D. " /Ae
t t
D Ae " Ae
t t
Ae C D Ae
t t
.Ae C 1/ D Ae
t
Ae
D t
Ae C1
t
ividing numerator and denominator by Ae , we have

D D
1 1 ! t
1C t
1C e
Ae A
Replacing 1=A by b and by c yields the so-called gistic functi n
N
L F
M The function defined by

D
M
1 C be!ct
2 N= M is called the logistic function or the Verhulst Pearl logistic function
1 + be-ct

t
The graph of Equation (3), called a gistic cur e is S-shaped and appears in
FIGURE ogistic
Figure 15.26. Notice that the line D is a horizontal asymptote that is,
curve.

lim D D
t!1 1 C be!ct 1 C b.0/
oreover, from Equation (1), the rate of growth is

. " /

which can be considered a function of . To find when the maximum rate of growth
d
occurs, we solve . . " // D 0 for :
d
d d
. . " // D . . " 2 //
d d
D . "2 /D0

Thus, D =2. In other words, the rate of growth increases until the population size
is =2 and decreases thereafter. The maximum rate of growth occurs when D =2
 Section 5.7 ore cat ons of erent a at ons 701

and corresponds to a point of in ection in the graph of . To find the value of t for
which this occurs, we substitute =2 for in Equation (3) and solve for t:

D
2 1 C be!ct
1 C be!ct D 2
1
e!ct D
b
ct
e Db
ct D ln b logarithmic form
ln b
tD
c
Therefore, the maximum rate of growth occurs at the point ..ln b/=c; =2/.
We remar that in Equation (3) we can replace e!c by C, and then the logistic
function has the following form:

A F L F

D
1 C bCt

E AM LE L G C M

Suppose the membership in a new country club is to be a maximum of 00 persons,

due to limitations of the physical plant. ne year ago the initial membership was 50
persons, and now there are 200. Provided that enrollment follows a logistic function,
how many members will there be three years from now
S et be the number of members enrolled t years after the formation of the
club. Then, from Equation (3),

D
1 C be!ct
Here D 00, and when t D 0, we have D 50. So
00
50 D
1Cb
00
1CbD D 16
50
b D 15

Thus,
00
D
1 C 15e!ct
When t D 1, then D 200, so we have
00
200 D
1 C 15e!c
00
1 C 15e!c D D4
200
3 1
e!c D D
15 5
702 C cat ons of nte rat on

Hence, c D " ln 15 D ln 5. Rather than substituting this value of c into Equation (4),
it is more convenient to substitute the value of e!c there:
00
D ! "t
1 C 15 15
Three years from now, t will be 4. Therefore,
00
D ! "4 $ 7 1
1 C 15 15
Now ork Problem 5 G
M S R
et us now consider a simple model of how a rumor spreads in a population of size .
A similar situation would be the spread of an epidemic or a new fad.
et D .t/ be the number of persons who now the rumor at time t. We will
assume that those who now the rumor spread it randomly in the population and that
those who are told the rumor become spreaders of the rumor. Furthermore, we will
assume that each nower tells the rumor to k individuals per unit of time. (Some of
these k individuals may already now the rumor.) We want an expression for the rate
of increase of the nowers of the rumor. ver a unit of time, each of approximately
persons will tell the rumor to k persons. Thus, the total number of persons who are
told the rumor over the unit of time is (approximately) k. However, we are interested
only in ne nowers. The proportion of the population that does not now the rumor
is . " /= . Hence, the total number of new nowers of the rumor is
# $
"
k

which can be written .k= / . " /. Therefore,

d k
D . " /
dt
k
D . " /; where D
This differential equation has the form of Equation (1), so its solution, from
Equation (3), is a gistic functi n

D
1 C be!ct
E AM LE C R

In a large university of 45,000 students, a sociology ma or is researching the spread of

a new campus rumor. When she begins her research, she determines that 300 students
now the rumor. After one wee , she finds that 00 now it. Estimate the number of
students who now it four wee s after the research begins by assuming logistic growth.
ive the answer to the nearest thousand.
S et be the number of students who now the rumor t wee s after the
research begins. Then

D
1 C be!ct
Here , the size of the population, is 45,000, and when t D 0; D 300. So we have
45; 000
300 D
1Cb
45; 000
1CbD D 150
300
b D 14
 Section 5.7 ore cat ons of erent a at ons 703

Thus,
45; 000
D
1 C 14 e!ct
When t D 1, then D 00. Hence,
45; 000
00 D
1 C 14 e!c
45; 000
1 C 14 e!c D D 50
00
4
Therefore, e!c D 14
, so
45; 000
D ! 4 "t
1 C 14 14
When t D 4,
45; 000
D ! 4 "4 $ 16; 000
1 C 14 14
After four wee s, approximately 16,000 students now the rumor.
Now ork Problem 3 G
N L C
We conclude this section with an interesting application of a differential equation. If a
homicide is committed, the temperature of the victim s body will gradually decrease
from 37ı C (normal body temperature) to the temperature of the surroundings (ambi-
ent temperature). In general, the temperature of the cooling body changes at a rate
proportional to the difference between the temperature of the body and the ambient
temperature. This statement is nown as ewton’s law of cooling. Thus, if .t/ is
the temperature of the body at time t and the ambient temperature is a, then
d
D k. " a/
dt
where k is the constant of proportionality. Therefore, Newton s law of cooling is a
differential equation. It can be applied to determine the time at which a homicide was
committed, as the next example illustrates.

E AM LE T M

A wealthy industrialist was found murdered in his home. Police arrived on the scene at
11:00 . . The temperature of the body at that time was 31ı C, and one hour later it was
30ı C. The temperature of the room in which the body was found was 22ı C. Estimate
the time at which the murder occurred.
S et t be the number of hours after the body was discovered and .t/ be the
temperature (in degrees Celsius) of the body at time t. We want to find the value of t for
which D 37 (normal body temperature). This value of t will, of course, be negative.
y Newton s law of cooling,
d
D k. " a/
dt
where k is a constant and a (the ambient temperature) is 22. Thus,
d
D k. " 22/
dt
704 C cat ons of nte rat on

Separating variables, we have

d
D kdt
" 22
Z Z
d
D kdt
" 22
ln j " 22j D kt C C
ecause " 22 > 0,
ln. " 22/ D kt C C
When t D 0, then D 31. Therefore,
ln.31 " 22/ D k % 0 C C
C D ln
Hence,
ln. " 22/ D kt C ln
ln. " 22/ " ln D kt
" 22 a
ln D kt ln a " ln b D ln
b
When t D 1, then D 30, so
30 " 22
ln Dk%1

k D ln
Thus,
" 22
ln D t ln
Now we find t when D 37:
37 " 22
ln D t ln

ln.15= /
tD $ "4:34
ln. = /
Accordingly, the murder occurred about 4.34 hours bef re the time of discovery of the
body (11:00 . .). Since 4.34 hours is (approximately) 4 hours and 20 minutes, the
industrialist was murdered about 6:40 . .
Now ork Problem 9 G
R BLEMS
Population The population of a city follows logistic growth administrative council spread information about the event as a
and is limited to 100,000. If the population in 1 5 was 50,000 rumor. At the end of one wee , 100 people now the rumor.
and in 2000 was 60,000, what will the population be in 2005 Assuming logistic growth, how many people now the rumor
ive your answer to the nearest hundred. after two wee s ive your answer to the nearest hundred.
Production A company believes that the production of its Spread of a Fad At a university with 50,000 students,
product in present facilities will follow logistic growth. Presently, it is believed that the number of students with a particular ring
300 units per day are produced, and production will increase tone on their mobile phones is following a logistic growth pattern.
to 500 units per day in one year. If production is limited to 00 The student newspaper investigates when a survey reveals that
units per day, what is the anticipated daily production in two 500 students have the ring tone. ne wee later, a similar survey
years ive the answer to the nearest unit. reveals that 1500 students have it. The newspaper writes a story
Spread of Rumor In a university of 40,000 students, the about it and includes a formula predicting the number
administration holds meetings to discuss the idea of bringing D .t/ of students who will have the ring tone t wee s
in a ma or roc band for homecoming wee end. efore the plans after the first survey. What is the formula that the newspaper
are o cially announced, student representatives on the publishes
 Section 5.7 ore cat ons of erent a at ons 705

Flu Outbrea In a city whose population is 100,000, an hour later, the body temperature was 1 ı C and the warehouse
outbrea of u occurs. When the city health department begins its temperature was unchanged. The police forensic mathematician
record eeping, there are 500 infected persons. ne wee later, calculates using Newton s law of cooling. What is the time she
there are 1000 infected persons. Assuming logistic growth, reports as the time of the murder
estimate the number of infected persons two wee s after En yme Formation An enzyme is a protein that acts as a
record eeping begins. catalyst for increasing the rate of a chemical reaction that occurs
Sigmoid Function A very special case of the logistic in cells. In a certain reaction, an enzyme A is converted to another
function defined by Equation (3) is the sigm id functi n obtained enzyme, . Enzyme acts as a catalyst for its own formation. et
by ta ing D b D c D 1 so that we have p be the amount of enzyme at time t and I be the total amount of
1 both enzymes when t D 0. Suppose the rate of formation of is
.t/ D proportional to p.I " p/. Without directly using calculus, find the
1 C e!t
a Show directly that the sigmoid function is the solution of the value of p for which the rate of formation will be a maximum.
differential equation Fund Raising A small town decides to conduct a
d fund-raising drive for a fire engine, the cost of which is 200,000.
D .1 " / The initial amount in the fund is 50,000. n the basis of past
dt
and the initial condition .0/ D 1=2. drives, it is determined that t months after the beginning of the
b Show that .0; 1=2/ is an in ection point on the graph of the drive, the rate, dx=dt, at which money is contributed to such a
sigmoid function. fund is proportional to the difference between the desired goal of
c Show that the function 200,000 and the total amount, x, in the fund at that time. After
one month, a total of 100,000 is in the fund. How much will be
1 1 in the fund after three months
f .t/ D !t
"
1Ce 2
is symmetric about the origin.
d Explain how (c) above shows that the sigmoid function is
symmetric ab ut the p int .0; 1=2/, explaining at the same time
what this means.
e S etch the graph of the sigmoid function.
Biology In an experiment,7 five Paramecia were placed
in a test tube containing a nutritive medium. The number
Birthrate In a discussion of unexpected properties of
of Paramecia in the tube at the end of t days is given
mathematical models of population, ailey considers the case in
approximately by
which the birthrate per indi idua is proportional to the population
375 1d
D size at time t. Since the growth rate per individual is , this
1 C e5:2!2:3t dt
a Show that this equation can be written as means that
1d
375 Dk
D dt
1 C 1 1:27e!2:3t
so that
and, hence, is a logistic function.
d
b Find limt!1 . Dk 2 sub ect to D 0 at t D 0
c How many days will it ta e for the number of Paramecia to dt
exceed 370 where k > 0. Show that
Biology In a study of the growth of a colony of unicellular D
0
organisms, the equation 1"k 0t

0:2524 Use this result to show that

D 0!x!5 # $!
e!2:12 x C 0:005125 1
lim D1 as t!
was obtained, where is the estimated area of the growth in k 0
square centimeters and x is the age of the colony in days after This means that over a finite interval of time, there is an infinite
being first observed. amount of growth. Such a model might be useful only for rapid
a Put this equation in the form of a logistic function. growth over a short interval of time.
b Find the area when the age of the colony is 0. Population Suppose that the rate of growth of a population
ime of a Murder A murder was committed in an is proportional to the difference between some maximum size
abandoned warehouse, and the victim s body was discovered at and the number of individuals in the population at time t.
3:17 . . by the police. At that time, the temperature of the body Suppose that when t D 0, the size of the population is 0 .
was 27ı C and the temperature in the warehouse was "5ı C. ne Find a formula for .

7
. F. ause, he Strugg e f r Existence (New or : Hafner Publishing Co.,
1 64).
A. . ot a, E ements f athematica Bi gy (New or : over Publications, N. T. . ailey, he athematica Appr ach t Bi gy and edicine (New
Inc., 1 56). or : ohn Wiley Sons, Inc., 1 67).
706 C cat ons of nte rat on

Objective I I
o e ne an e a ate ro er Suppose f.x/ is continuous
R b and nonnegative for a ! x < 1. (See Figure 15.27.) We
nte ra s now that the integral a f.x/dx is the area of the region between the curve y D f.x/
and the x-axis from x D a to x D b. As b ! 1, we can thin of
Z b
lim f.x/dx
b!1 a

y y

y = f (x) y = f(x)

x x
a b a b q

FIGURE Area from a to b. FIGURE Area from a to b as b ! 1.

as the area of the unbounded region that is shaded in Figure 15.2 . This limit is abbre-
viated by
Z 1
f.x/dx
a
R1
and called an improper integral. If the limit exists, a f.x/dx is said to be convergent
and the improper integral c n erges to that limit. In this case the unbounded region is
R1
considered to have a finite area, and this area is represented by a f.x/dx. If the limit
does not exist, the improper integral is said to be divergent, and the region does not
have a finite area.
R 1 We can remove the restriction that f.x/ # 0. In general, the improper integral
a f.x/dx is defined by

Z 1 Z b
f.x/dx D lim f.x/dx
a b!1 a

ther types of improper integrals are

Z b
f.x/ dx
!1

and
Z 1
f.x/dx
!1
A L IT I
The rate at which the human In each of the three types of improper integrals (5), (2), and (3) , the interval over which
body eliminates a certain drug from the integral is evaluated has infinite length. The improper integral in (2) is defined by
its system may be approximated by
Z b Z b
R.t/ D 3e!0:1t " 3e!0:3t , where R.t/ is
in milliliters per minute and t is the time
f.x/dx D lim f.x/dx
!1 a!!1 a
in minutes since the drug was ta en.
R1 Rb
Find 0 .3e!0:1t " 3e!0:3t / dt, the total
If this limit exists, !1 f.x/dx is said to be convergent. therwise, it is divergent. We
amount of the drug that is eliminated.
will define the improper integral in (3) after the following example.
 Section 5.8 ro er nte ra s 707
R1 Rb
E AM LE I I F a f.x/dx !1 f.x/dx

etermine whether the following improper integrals are convergent or divergent. For
any convergent integral, determine its value.
Z 1
1
a dx
1 x3
Z 1 Z b ˇb
1 !3 x!2 ˇˇ
S dx D lim x dx D lim "
x3 b!1
1 b!1 2 ˇ 1 1
# $
1 1 1 1
D lim " 2 C D "0 C D
b!1 2b 2 2 2
Z 1
1 1
Therefore, 3
dx converges to .
1 x 2
Z 0
b ex dx
!1
Z 0 Z 0 ˇ0
ˇ
S ex dx D lim ex dx D lim ex ˇ
!1 a!!1 a a!1 a

D lim .1 " ea / D 1 " 0 D 1 e0 D 1

a!!1
(Here we used the fact that as xR! "1, the graph of y D ex approaches the x-axis, so
0
lima!!infty ea D 0.) Therefore, !1 ex dx converges to 1.
Z 1
1
c p dx
1 x
Z 1 Z b ˇb
1 !1=2
ˇ
1=2 ˇ
S p dx D lim x dx D lim 2x ˇ
x b!11 b!1 1 1
p
D lim 2. b " 1/ D 1
b!1
Therefore, the improper integral diverges.
Now ork Problem 3 G
R1
The improper integral !1 f.x/dx is defined in terms of improper integrals of the forms
(5) and (2):
Z 1 Z 0 Z 1
f.x/dx D f.x/dx C f.x/dx
!1 !1 0
R1
If b th integrals on the right side of Equation (4) are convergent, then !1 f.x/dx is
said to be convergent otherwise, it is divergent.
R1
E AM LE A I I F !1 f.x/dx
Z 1
etermine whether ex dx is convergent or divergent.
!1
Z 1 Z 0 Z 1
x x
S e dx D e dx C ex dx
!1 !1 0
Z 0
y Example 1(b), ex dx D 1. n the other hand,
!1
Z Z ˇb
1
x
b ˇ xˇ
e dx D lim e dx D lim e ˇ D lim .eb " 1/ D 1
x
0 b!1 0 b!1 b!1 0
R1 x
R1
Since 0 e dx is divergent, !1 ex dx is also divergent.
Now ork Problem 11 G
708 C cat ons of nte rat on

E AM LE F

In statistics, a function, f, is called a density function if f.x/ # 0 and

Z 1
f.x/dx D 1
!1
Suppose
% !x
ke for x # 0
f.x/ D
0 elsewhere
is a density function. Find k.
R1
S We write the equation !1 f.x/dx D 1 as
Z 0 Z 1
f.x/dx C f.x/dx D 1
!1 0
R0
Since f.x/ D 0 for x < 0, !1 f.x/dx D 0. Thus,
Z 1
ke!x dx D 1
0
Z b
lim ke!x dx D 1
b!1 0
ˇb
ˇ !x ˇ
lim "ke ˇ D 1
b!1 0
!b
lim ."ke C k/ D 1
b!1

0CkD1 lim e!b D 0

b!1
kD1

Now ork Problem 13 G

R BLEMS
In Pr b ems determine the integra s if they exist Indicate Density Function The density function for the life x, in
th se that are di ergent hours, of an electronic component in a radiation meter is given by
Z 1 Z 1 (
1 1 k
dx dx
3 x3 1 .3x " 1/2 f .x/ D x2 for x # 500
0 for x < 500
Z 1 Z 1 R1
1 1 a If k satisfies the condition that 500 f .x/dx D 1, find k.
dx p dx
e1000 x 2
3
.x C 2/2 b The probability that the component will last at least 1000
R1
Z Z hours is given by 1000 f .x/dx. Evaluate this integral.
1 1
e!x dx .5 C e!x / dx Density Function iven the density function
37 0 % !2x
ke for x # 1
f .x/ D
Z 1 Z 1 0 elsewhere
1 xdx
p dx p find k. ( int See Example 3.)
1 x 5 .x2 " /3
Future Profits For a business, the present value of
Z !3
Z 1 all future profits at an annual interest rate, r, compounded
1 1
dx p dx continuously is given by
!1 .x C 1/2 1
3
x"1 Z 1
p.t/e!rt dt
Z 1 Z 1 0
2
2xe!x dx .5 " 3x/ dx where p.t/ is the profit per year in dollars at time t. If
!1 !1
p.t/ D 500; 000 and r D 0:02, evaluate this integral.
 Chapter 5 e e 709

Psychology In a psychological model for signal Economics In discussing entrance of a firm into an
detection,10 the probability ˛ (a ree letter read alpha ) industry, Stigler11 uses the equation
of reporting a signal when no signal is present is given by Z 1
Z 1
D $0 e! t e!"t dt
˛D e!x dx x # 0 0
xc
where $0 ; % (a ree letter read theta ), and & (a ree letter
The probability ˇ (a ree letter read beta ) of detecting a signal read rho ) are constants. Show that D $0 =.& " %/ if % < &.
when it is present is
Z 1 Population The predicted rate of growth per year of the
ˇD ke!kx dx x # 0 population of a certain small city is given by
xc 40; 000
In both integrals, xc is a constant (called a criterion value in this .t C 2/2
model). Find ˛ and ˇ if k D 1 . where t is the number of years from now. In the long run (that is,
Find the area of the region in the third quadrant bounded by as t ! 1), what is the expected change in population from
the curve y D e3x and the x-axis. today s level

Chapter 15 Review
I T S E
S Integration by ables
present value and accumulated amount of a continuous annuity Ex. , p. 670
S Approximate Integration
trapezoidal rule Simpson s rule Ex. 2, p. 675
S Area Between Curves
vertical strip of area Ex. 1, p. 67
horizontal strip of area Ex. 6, p. 6 3
S Consumers’ and Producers’ Surplus
consumers surplus producers surplus Ex. 1, p. 6
S Average Value of a Function
average value of a function Ex. 1, p. 6 1
S Differential Equations
first-order differential equation separation of variables Ex. 1, p. 6 3
exponential growth and decay decay constant half-life Ex. 3, p. 6 6
S More Applications of Differential Equations
logistic function Ex. 1, p. 701
Newton s law of cooling Ex. 3, p. 703
S Improper Integrals
improper integral convergent divergent Ex. 1, p. 707
R1 Rb R1
a f.x/dx, !1 f.x/dx, !1 f.x/dx Ex. 2, p. 707

S
An integral that does not have a familiar form may have been that a payment at time t is at the rate of f.t/ per year. If the
done by others and recorded in a table of integrals. However, annual rate of interest is r compounded continuously, then
it may be necessary to transform the given integral into an the present value of the continuous annuity is given by
equivalent form before the matching can occur.
An annuity is a series of payments over a period of time. Z
Suppose payments are made continuously for years such AD f.t/e!rt dt
0

10 11
. aming, athematica Psych gy (New or : Academic Press, Inc., . Stigler, he he ry f Price 3rd ed. (New or : acmillan Publishing
1 73). Company, 1 66), p. 344.
710 C cat ons of nte rat on

and the accumulated amount is given by ables. In that method, by considering the derivative to be a
Z quotient of differentials, we rewrite the equation so that each
!t/ side contains only one variable and a single differential in the
SD f.t/er. dt
0
numerator. Integrating both sides of the resulting equation
gives the solution. This solution involves a constant of inte-
Rb
If the integrand of a definite integral, a f.x/dx, does not gration and is called the general solution of the differential
have an elementary antiderivative, or even if the antideriva- equation. If the un nown function must satisfy the condition
tive is merely daunting, the required number can be found, that it has a specific function value for a given value of the
approximately, with either the Trapezoidal Rule: independent variable, then a particular solution can be found.
ifferential equations arise when we now a relation
h involving the rate of change of a function. For example, if
. f.a/ C 2f.a C h/ C % % % C 2f.a C .n " 1/h/ C f.b// a quantity, , at time t is such that it changes at a rate pro-
2
portional to the amount present, then
or Simpson s Rule: d
Dk ; where k is a constant
dt
h
. f.a/C4f.aCh/C2f.aC2h/C% % %C4f.aC.n"1/h/Cf.b// The solution of this differential equation is
3
kt
where for both rules we have D .b " a/=n, but in the case of D 0e
Simpson s Rule n must be even.
If f.x/ # 0 is continuous on Œa; b!, then the definite inte- where 0 is the quantity present at t D 0. The value of k
gral can be used to find the area of the region bounded by may be determined when the value of is nown for a given
y D f.x/, the x-axis, x D a, and x D b. The definite integral value of t other than t D 0. If k is positive, then follows
can also be used to find areas of more complicated regions. In an exponential law of growth if k is negative, follows an
these situations, a strip of area should be drawn in the region. exponential law of decay. If represents a quantity of a
This allows us to set up the proper definite integral. In this radioactive element, then
regard, both vertical strips and horizontal strips have their d
uses. D "# ; where # is a positive constant
dt
ne application of finding areas involves consumers
surplus and producers surplus. Suppose the mar et for a Thus, follows an exponential law of decay, and hence,
product is at equilibrium and .q0 ; p0 / is the equilibrium point !!t
(the point of intersection of the supply curve and the demand D 0e
curve for the product). Then consumers surplus, CS, corre-
sponds to the area from q D 0 to q D q0 , bounded above by The constant # is called the decay constant. The time for one-
the demand curve and below by the line p D p0 . Thus, half of the element to decay is the half-life of the element:
Z q0 ln 2 0:6 315
half-life D $
CS D . f.q/ " p0 /dq # #
0
A quantity, , may follow a rate of growth given by
where f is the demand function. Producers surplus, PS, cor-
responds to the area from q D 0 to q D q0 , bounded above d
by the line p D p0 and below by the supply curve. Therefore, D . " /; where ; are constants
dt
Z q0
Solving this differential equation gives a function of the form
PS D .p0 " g.q//dq
0
D ; where b; c are constants
where g is the supply function. 1 C be!ct
The average value, f, of a function, f, over the interval
Œa; b! is given by which is called a logistic function. any population sizes can
be described by a logistic function. In this case, represents
Z
1 b the limit of the size of the population. A logistic function is
fD f.x/dx also used in analyzing the spread of a rumor.
b"a a
Newton s law of cooling states that the temperature, , of
An equation that involves the derivative of an un nown a cooling body at time t changes at a rate proportional to the
function is called a differential equation. If the highest-order difference " a, where a is the ambient temperature. Thus,
derivative that occurs is the first, the equation is called a
first-order differential equation. Some first-order differential d
D k. " a/; where k is a constant
equations can be solved by the method of separation of vari- dt
 Chapter 5 e e 711

The solution of this differential equation can be used to and

determine, for example, the time at which a homicide was Z b Z b
committed. f.x/dx D lim f.x/dx
An integral of the form !1 a!!1 a

Z 1 Z b Z 1
R1 Rb
If a f.x/dx !1 f.x/ dx is a finite number, we say that
f.x/ dx f.x/ dx or f.x/ dx the integral is convergent otherwise, it is divergent. The
a !1 !1 R1
improper integral !1 f.x/dx is defined by
is called an improper integral. The first two integrals are Z 1 Z 0 Z 1
defined as follows: f.x/dx D f.x/dx C f.x/dx
!1 !1 0
Z 1 Z b
f.x/dx D lim f.x/dx R1
a b!1 a If both integrals on the right side are convergent, !1 f.x/dx
is said to be convergent otherwise, it is divergent.

R
In Pr b ems determine the integra s Consumers’ and Producers’ Surplus The demand
Z Z
1 equation for a product is
x2 ln x dx p dx
4x2 C 1
p D 0:01q2 " 1:1q C 30
Z 2p Z
5x " 13
2
x C 16 dx dx and the supply equation is
0
x"3
Z Z eb p D 0:01q2 C
15x " 2 1
dx dx
.3x C 1/.x " 2/ ea x ln x
etermine consumers surplus and producers surplus when
Z Z mar et equilibrium has been established.
dx dx
x.x C 2/2 x2 " 1 Consumers’ Surplus The demand equation for a
Z Z product is
dx
p x3 ln x2 dx
x2 4 " x2 p D .q " 4/2
Z Z
dx x and the supply equation is
p dx
x2 " a2 2 C 5x
Z Z p D q2 C q C 7
7x dx
4 xe dx
5 C 2e3x where p (in thousands of dollars) is the price per 100 units when
Z Z q hundred units are demanded or supplied. etermine consumers
dx dx
surplus under mar et equilibrium.
2x ln x2 x.x C a/
Z Z Find the average value of f .x/ D x3 " 3x2 C 2x C 1 over the
2x dx interval Œ0; 5!.
dx p
3 C 2x x2 4x2 " Find the average value of f .t/ D t2 et over the interval 0, 1 .

In Pr b ems nd the area f the regi n b unded by the In Pr b ems and s e the di erentia equati ns
gi en cur es 0
y D 3x y C 2xy 2
y>0
y D "x.x " a/, y D 0 for 0 < a y0 " f 0 .x/ef .x/!y D 0 y.0/ D f .0/
y D 2x2 ; y D x2 C y D x2 " x; y D 10 " x2
p In Pr b ems determine the impr per integra s if they exist
y D x; x D 0; y D 3 Z 1 Z 0
1
y D ln x; x D 0; y D 0; yD1 2:5
dx e3x dx
1 x !1
y D 3 " x, y D x " 4, y D 0, y D 3
Z b Z 1 Z 1
dx 1 2
Show that ln b D . Use the trapezoidal rule, with n D dx xe1!x dx
1 x 1 2x !1
to approximate ln 2. Express ust those digits which agree with the
Population The population of a fast-growing city was
true value of ln 2.
500,000 in 1 0 and 1,000,00 in 2000. Assuming exponential
Repeat Problem 25 using Simpson s rule with n D . growth, pro ect the population in 2020.
712 C cat ons of nte rat on

Population The population of a city doubles every Product Consumption Suppose that A.t/ is the amount of
10 years due to exponential growth. At a certain time, the a product that is consumed at time t and that A follows an
population is 40,000. Find an expression for the number of exponential law of growth. If t1 < t2 and at time t2 the amount
people, , at time t years later. ive your answer in terms of ln 2. consumed, A.t2 /, is double the amount consumed at time t1 ; A.t1 /,
Radioactive If of a radioactive substance remains then t2 " t1 is called a doubling period. In a discussion of
after 1000 years, find the decay constant, and, to the nearest exponential growth, Shonle12 states that under exponential
percent, give the percentage of the original amount present after growth, the amount of a product consumed during one doubling
5000 years. period is equal to the total used for all time up to the beginning of
the doubling period in question. To ustify this statement,
Medicine Suppose q is the amount of penicillin in the body reproduce his argument as follows. The amount of the product
at time t, and let q0 be the amount at t D 0. Assume that the rate used up to time t1 is given by
of change of q with respect to t is proportional to q and that q Z t1
decreases as t increases. Then we have dq=dt D "kq, where
A0 ekt dt k > 0
k > 0. Solve for q. What percentage of the original amount !1
present is there when t D 7=k
where A0 is the amount when t D 0. Show that this is equal to
Biology Two organisms are initially placed in a medium .A0 =k/ekt1 . Next, the amount used during the time interval from t1
and begin to multiply. The number, , of organisms that are to t2 is
present after t days is recorded on a graph with the horizontal axis Z t2
labeled t and the vertical axis labeled . It is observed that the
points lie on a logistic curve. The number of organisms present A0 ekt dt
t1
after 6 days is 300, and beyond 10 days the number approaches a
limit of 450. Find the logistic equation. Show that this is equal to
College Enrollment A university believes that its A0 kt1 k.t2 !t1 /
enrollment follows logistic growth. ast year enrollment was e Œe " 1!
k
10,000, and this year it is 11,000. If the university can
accommodate a maximum of 20,000 students, what is the If the interval t1 ; t2 is a doubling period, then
anticipated enrollment next year
A0 ekt2 D 2A0 ekt1
ime of Murder A coroner is called in on a murder
case. He arrives at 6:00 . . and finds that the victim s temperature Show that this relationship implies that ek.t2 !t1 / D 2. Substitute
is 35ı C. ne hour later the body temperature is 34ı C. The this value into Equation (5) your result should be the same as the
temperature of the room is 25ı C. About what time was the murder total used during all time up to t1 , namely, .A0 =k/ekt1 .
committed (Assume that normal body temperature is 37ı C.) Revenue Cost and Profit The following table gives
Annuity Find the present value, to the nearest dollar, of values of a company s marginal-revenue ( R) and marginal-cost
a continuous annuity at an annual rate of 5 for 10 years if the ( C) functions:
payment at time t is at the annual rate of f .t/ D 100t dollars.
q 0 3 6 12 15 1
ospital Discharges For a group of hospitalized
individuals, suppose the proportion that has been discharged R 25 22 1 13 7 3 0
at the end of t days is given by C 15 14 12 10 7 4 2
Z t
f .x/ dx
0 The company s fixed cost is 25. Assume that profit is a maximum
where f .x/ D 0:007e!0:01x C 0:00005e!0:0002x . Evaluate when R D C and that this occurs when q D 12. oreover,
Z 1 assume that the output of the company is chosen to maximize the
f .x/ dx profit. Use the trapezoidal rule and Simpson s rule for each of the
0 following parts.
Integration by Parts et f and g be differentiable a Estimate the total revenue by using as many data values as
functions. Show that if either f 0 g or fg0 has an antiderivative then possible.
the other one does. It su ces to show it in one case, so for b Estimate the total cost by using as few data values as possible.
definiteness, assume that 0 .x/ D f 0 .x/g.x/ (equivalently c etermine how the maximum profit is related to the area
R 0
f .x/g.x/dx D .x/ C C) and show that enclosed by the line q D 0 and the R and C curves, and use
R this relation to estimate the maximum profit as accurately as
f .x/g0 .x/ D f .x/g.x/ " .c/ C C. This is often written as
Z Z possible.
f .x/g0 .x/dx D f .x/g.x/ " f 0 .x/g.x/dx

Writing u D f .x/ and D g.x/ we have, equivalently,

Z Z
ud D u " du

12
. I. Shonle, En ir nmenta App icati ns f Genera Physics (Reading, A:
Addison-Wesley Publishing Company, Inc., 1 75).
 Cont n o s an o
ar a es

C
onsider the problem of designing a cellular telephone networ for a large
16.1 Cont n o s an o urban area. Ideally, the system would have enough capacity to meet all pos-
ar a es
sible demands. However, demand uctuates, sometimes wildly. Some uc-
16.2 e or a str t on tuation is predictable. For example, at times when most people are asleep
there is less demand, and on wee ends and holidays, when many people call their fam-
16.3 e or a ilies and friends, there is more demand. However, some increases in demand are not
ro at on to t e
no a str t on predictable. For example, after an earthqua e or some other natural disaster, even a
severe storm, many people call emergency services and many call their family and
C er 16 e e friends to chec that they are all right. It is usually prohibitively expensive to build
and operate a system that will handle any sudden increase in demand. Stri ing a bal-
ance between the goal of serving customers and the need to limit costs to maintain
profitability is a serious problem.
A sensible approach is to design and build a system capable of handling the load
of telephone tra c under normally busy conditions, and to accept the fact that on rare
occasions, heavy tra c will lead to overloads. We cannot always predict when over-
loads will occur since disasters, such as earthqua es, are unforeseen occurrences. ut
some good pr babi istic predictions of future tra c volume will su ce. ne could
build a system that would meet demand .4 of the time, for example. The remain-
ing 0.6 of the time, customers would simply have to put up with intermittent delays
in service.
A probabilistic description of tra c on a phone networ is an example of a proba-
bility density function. Such functions are the focus of this chapter. Probability density
functions have a wide range of applications not only calculating how often a system
will be overloaded, for example, but also calculating the system s average load. Aver-
age load allows prediction of such things as average power consumption and average
volume of system maintenance activity. Such considerations are vital to the profitabil-
ity of a business.

713
714 C Cont n o s an o ar a es

Objective C R
o ntro ce cont n o s ran o
ar a es to sc ss ens t f nct ons F
nc n n for an e onent a
str t ons to sc ss c at e In Chapter , the random variables that we considered were discrete. Now we will con-
str t on f nct ons an to co te cern ourselves with ontinuous random variables. A random variable is continuous
t e ean ar ance an stan ar
e at on for a cont n o s ran o if it can assume any value in some interval or intervals. A continuous random variable
ar a e usually represents data that are measured such as heights, weights, distances, and peri-
ods of time. y contrast, the discrete random variables of Chapter usually represent
data that are c unted.
For example, the number of hours of life of a calculator battery is a continuous
random variable, X. If the maximum possible life is 1000 hours, then X can assume any
value in the interval Œ0; 1000!. In a practical sense, the li elihood that X will assume a
single specified value, such as 764.123 , is extremely remote. It is more meaningful
to consider the li elihood of X lying within an inter a such as that between 764 and
765. Thus, 764 < X < 765. (For that matter, the nature of measurement of physical
quantities, li e time, tells us that a statement such as X D 764:123 is really one of the
form 764:123750 < X < 764:123 4 .) In general, ith a c ntinu us rand m ariab e
ur c ncern is the ike ih d that it fa s ithin an inter a and n t that it assumes a
particu ar a ue
As another example, consider an experiment in which a number X is randomly
selected from the interval Œ0; 2!. Then X is a continuous random variable. What is the
probability that X lies in the interval Œ0; 1! ecause we can thin of Œ0; 1! as being
half the interval Œ0; 2!, a reasonable (and correct) answer is 12 . Similarly, if we thin
of the interval Œ0; 12 ! as being one-fourth of Œ0; 2!, then P.0 ! X ! 12 / D 14 . Actually,
each one of these probabilities is simply the length of the given interval divided by the
length of Œ0; 2!. For example,
! "
1 length of Œ0; 12 ! 1
1
P 0!X! D D 2 D
2 length of Œ0; 2! 2 4
et us now consider a similar experiment in which X denotes a number chosen
at random from the interval Œ0; 1!. As might be expected, the probability that X will
assume a value in any given interval within Œ0; 1! is equal to the length of the given
interval divided by the length of Œ0; 1!. ecause Œ0; 1! has length 1, we can simply say
that the probability of X falling in an interval is the length of the interval. For example,
P.0:2 ! X ! 0:5/ D 0:5 " 0:2 D 0:3
and P.0:2 ! X ! 0:2001/ D 0:0001. Clearly, as the length of an interval approaches
0, the probability that X assumes a value in that interval approaches 0. eeping this in
mind, we can thin of a single number such as 0:2 as the limiting case of an interval
as the length of the interval approaches 0. (Thin of Œ0:2; 0:2 C x! as x ! 0.) Thus,
P.X D 0:2/ D 0. In general, the pr babi ity that a c ntinu us rand m ariab e X
assumes a particu ar a ue is 0. As a result, the pr babi ity that X ies in s me inter a
is n t a ected by hether r n t either f the endp ints f the inter a is inc uded r
exc uded. For example,
P.X ! 0:4/ D P.X < 0:4/ C P.X D 0:4/
D P.X < 0:4/ C 0
D P.X < 0:4/
Similarly, P.0:2 ! X ! 0:5/ D P.0:2 < X < 0:5/.
We can geometrically represent the probabilities associated with a continuous ran-
dom variable X. This is done by means of the graph of a function y D f.x/ # 0 such
that the area under this graph and above the x-axis, between the lines x D a and x D b,
represents the probability that X assumes a value between a and b. (See Figure 16.1.)
 Section 6. Cont n o s an o ar a es 715

P(a … X … b) = area of shaded region

y = f(x)

x
a b

FIGURE Probability density function.

Rb
Since this area is given by the definite integral a f.x/dx, we have
Z b
P.a ! X ! b/ D f.x/dx
a

We call the function f the pr babi ity density functi n for X (more simply the
density functi n for X) and say that it defines the distributi n f X. ecause
probabilities are always nonnegative, we must have f.x/ # 0. Also, because the event
"1 < X < 1 must occur, the total area under the density function curve must be 1.
R1
That is, !1 f.x/dx D 1. In summary, we have the following definition.

A continuous function y D f.x/ is called a probability density function, for a

continuous random variable, if and only if it has the following properties:
f.x/ # 0
R1
!1 f.x/dx D 1
If X is a continuous random variable, such an f is a density function for X, if
Rb
P.a ! X ! b/ D a f.x/dx

To illustrate a density function, we return to the previous experiment in which a

number X is chosen at random from the interval Œ0; 1!. Recall that
P.a ! X ! b/ D length of Œa; b! D b " a
where a and b are in Œ0; 1!. We will show that the function
#
1 if 0 ! x ! 1
f.x/ D
0 otherwise
whose graph appears in Figure 16.2(a), is a density function for X. To do this, we must
verify that f.x/ satisfies the conditions for a density function. First, f.x/ is either 0 or 1,

f(x) f(x)
1 if 0 … x … 1
f(x) =
0 otherwise
1 1

x x
1 a b 1

(a) (b)

FIGURE Probability density function.

716 C Cont n o s an o ar a es

so f.x/ # 0. Next, since f.x/ D 0 for x outside Œ0; 1!,

Z 1 Z 1 ˇ1
ˇ
f.x/dx D 1dx D xˇˇ D 1
!1 0 0
Finally, for the X under consideration and a and b in Œ0; 1! with a < b, we must verify
Rb
that P.a ! X ! b/ D a f.x/dx. We compute the area under the graph between x D a
and x D b (Figure 16.2(b)). We have
Z b Z b ˇb
ˇ
f.x/dx D 1dx D xˇˇ D b " a
a a a
which, as stated in Equation (1), is P.a ! X ! b/.
The function in Equation (2) is called the uniform density function over Œ0; 1!,
and X is said to have a uniform distribution. The word unif rm is meaningful in the
sense that the graph of the density function is horizontal, at , over Œ0; 1!. As a result,
X is ust as li ely to assume a value in one interval within Œ0; 1! as in another of equal
length. A more general uniform distribution is given in Example 1.

f (x) E AM LE U F

The uniform density function over Œa; b! for the random variable X is given by
1 8
b-a < 1
if a ! x ! b
f.x/ D b " a
x : 0 otherwise
a b

FIGURE Uniform density See Figure 16.3. Note that over Œa; b!, the region under the graph is a rectangle with
function over Œa; b!. height 1=.b " a/ and width b " a. Thus, its area is given by .1=.b " a//.b " a/ D 1
R1
so !1 f.x/dx D 1, as must be the case for a density function. If Œc; d! is any interval
within Œa; b!, then
Z d Z d
1
P.c ! X ! d/ D f.x/dx D dx
c c b"a
x ˇˇd d"c
D ˇ D
b"a c b"a
For example, suppose X is uniformly distributed over the interval Œ1; 4! and we need to
find P.2 < X < 3/. Then a D 1; b D 4; c D 2, and d D 3. Therefore,
3"2 1
P.2 < X < 3/ D D
4"1 3
Now ork Problem 3(a)--(g) G
E AM LE F
A L IT I
Suppose the time (in minutes) pas- The density function for a random variable X is given by
sengers must wait for an airplane is #
kx if 0 ! x ! 2
uniformly distributed with density func- f.x/ D
tion f .x/ D 1
where 0 ! x ! 60;
0 otherwise
60 ;
and f .x/ D 0 elsewhere. What is the where k is a constant.
probability that a passenger must wait a Find k.
between 25 and 45 minutes R1
S Since !1 f.x/dx must be 1 and f.x/ D 0 outside Œ0; 2!, we have
Z 1 Z 2 ˇ2
kx2 ˇˇ
f.x/dx D kxdx D D 2k D 1
!1 0 2 ˇ0
Thus, k D 12 , so f.x/ D 12 x on Œ0; 2!.
 Section 6. Cont n o s an o ar a es 717

b Find P. 12 < X < 1/.

S
! " Z 1 ˇ1
1 1 x2 ˇˇ 1 1 3
P <X<1 D xdx D ˇ D " D
2 1=2 2 4 1=2 4 16 16
c Find P.X < 1/.
S Since f.x/ D 0 for x < 0, we need only compute the area under the density
function between 0 and 1. Thus,
Z 1 ˇ1
1 x2 ˇˇ 1
P.x < 1/ D xdx D ˇ D
0 2 4 0 4

Now ork Problem 9(a)--(d), (g), (h) G

E AM LE E F
A L IT I
The life expectancy (in years) of an The exponential density function is defined by
automobile s bra e pads is distributed # !kx
exponentially with k D 10 1
. If the bra e ke if x # 0
f.x/ D
pads warranty lasts five years, what is 0 if x < 0
the probability that the bra e pads will
brea down after the warranty period where k is a positive constant, called a parameter, whose value depends on the exper-
iment under consideration. If X is a random variable with this density function, then X
is said to have an exponential distribution. The case k D 1 is shown in Figure 16.4.
a Verify that f is a density function.
f(x)
S
y definition, f # 0 on ."1; 1/, and because f D 0 on ."1; 0/ we have, for any
e-xif x Ú 0
f (x) = 0 if x 6 0 positive k,
Z 1 Z 1
f.x/ dx D ke!kx dx
!1 0
x Z b
D lim ke!kx dx
FIGURE Exponential density b!1 0
function. ˇb
D lim "e!kx ˇ0
b!1

D lim .."e!kb / " ."e0 //

b!1

D .0/ " ."1/

D1
b For k D 1, find P.2 < X < 3/.
S
Z 3
P.2 < X < 3/ D e!x dx D "e!x j32
2

D "e!3 " ."e!2 / D e!2 " e!3 $ 0:0 6

c For k D 1, find P.X > 4/. Z Z
1 b
!x
S P.X > 4/ D e dx D lim e!x dx
4 b!1 4
ˇb
ˇ !x ˇ
D lim "e ˇ D lim ."e!b C e!4 /
b!1 b!1 4

D 0 C e!4
$ 0:01
718 C Cont n o s an o ar a es

Alternatively, we can avoid an improper integral because

Z 4
P.X > 4/ D 1 " P.X ! 4/ D 1 " e!x dx
0

Now ork Problem 7(a)--(c), (e) G

The cumulative distribution function F for a continuous random variable X with
density function f is defined by
Z x
F.x/ D P.X ! x/ D f.t/dt
!1

For example, F(2) represents the entire area under the density curve that is to the left
of the line x D 2 (Figure 16.5). Where f.x/ is continuous, it can be shown that

F0 .x/ D f.x/
That is, the derivative of the cumulative distribution function is the density function.
Thus, F is an antiderivative of f, and by the Fundamental Theorem of Calculus,
Z b
P.a < X < b/ D f.x/dx D F.b/ " F.a/
a

This means that the area under the density curve between a and b (Figure 16.6) is simply
the area to the left of b minus the area to the left of a.

f(x) f(x)

x x
2 a b

FIGURE F.2/ D P.X ! 2/ D area of FIGURE P.a < X < b/.

shaded region.

1
f(x) x if 0 … x … 2
f(x) = 2
0 otherwise E AM LE F A C
F
1

Suppose X is a random variable with density function given by

2
x
(
1
x if 0 ! x ! 2
f.x/ D 2
0 otherwise
FIGURE ensity function for
Example 4. as shown in Figure 16.7.

a Find and s etch the cumulative distribution function.

S ecause f.x/ D 0 if x < 0, the area under the density curve to the left of
x D 0 is 0. Hence, F.x/ D 0 if x < 0. If 0 ! x ! 2, then
Z x Z x ˇx
1 t2 ˇˇ x2
F.x/ D f.t/dt D tdt D ˇ D
!1 0 2 4 0 4
 Section 6. Cont n o s an o ar a es 719

Since f is a density function and f.x/ D 0 for x < 0 and also for x > 2, the area under
the density curve from x D 0 to x D 2 is 1. Thus, if x > 2, the area to the left of x is 1,
F (x) so F.x/ D 1. Hence, the cumulative distribution function is
8̂
0 if x < 0
1 ˆ
< 2
x
F.x/ D if 0 ! x ! 2
ˆ4
x :̂
2 1 if x > 2
which is shown in Figure 16. .
FIGURE Cumulative b Find P.X < 1/ and P.1 < X < 1:1/.
distribution function for Example 4.
S Using the results of part (a), we have
12 1
P.X < 1/ D F.1/ D D
4 4
From Equation (3),
1:12 1
P.1 < X < 1:1/ D F.1:1/ " F.1/ D " D 0:0525
4 4
Now ork Problem 1 G
M S
For a random variable X with density function f, the mean " (also called the expecta
tion of X), E.X/ is given by
Z 1
" D E.X/ D xf.x/dx
!1
if the integral is convergent, and can be thought of as the average value of X in the long
run. The variance # 2 (also written Var.X/) is given by
Z 1
# 2 D Var.X/ D .x " "/2 f.x/dx
!1
if the integral is convergent. Noticed that these formulas are similar to the corresponding
ones in Chapter for a discrete random variable. It is easy to show that an alternative
formula for the variance is
Z 1
# 2 D Var.X/ D x2 f.x/dx " "2
!1
The standard deviation is
p
#D Var.X/
For example, it can be shown that if X is exponentially distributed (see Example 3),
then " D 1=k and # D 1=k. As with a discrete random variable, the standard deviation
of a continuous random variable X is small if X is li ely to assume values close to the
mean but unli ely to assume values far from the mean. The standard deviation is large
if the opposite is true.

E AM LE F M S
A L IT I
The life expectancy (in years) of If X is a random variable with density function given by
(
patients after they have contracted a cer- 1
x if 0 ! x ! 2
tain disease is exponentially distributed f.x/ D 2
with k D 0:2. Use the information in 0 otherwise
the paragraph that precedes Example 5
find its mean and standard deviation.
to find the mean life expectancy and the
standard deviation. S The mean is given by
Z 1 Z ˇ2
2
1 x3 ˇ 4
"D xf.x/dx D x % xdx D ˇˇ D
!1 0 2 6 0 3
720 C Cont n o s an o ar a es

y the alternative formula for variance, we have

Z 1 Z 2 ! "2
2 2 2 2 1 4
# D x f.x/dx " " D x % xdx "
!1 0 2 3
ˇ
4 ˇ2
x 16 16 2
D ˇˇ " D2" D
0
Thus, the standard deviation is
r p
2 2
#D D
3
Now ork Problem 5 G
We conclude this section by emphasizing that a density function for a continuous ran-
dom variable must not be confused with a probability distribution function for a discrete
random variable. Evaluating such a probability distribution function at a p int gives a
probability. ut evaluating a density function at a point does not. Instead, the area
under the density function curve over an inter a is interpreted as a probability. That
is, probabilities associated with a continuous random variable are given by integrals.

R BLEMS
Suppose X is a continuous random variable with density Find the cumulative distribution function F and s etch its
function given by graph. Use F to find P.1 < X < 3:5/.
#1
.x C 1/ if 1 < x < 3 If X is a random variable with density function f, then the
f .x/ D 6 R1
0 otherwise expectation of X is given by " D E.X/ D !1 xf .x/dx. Now, we
R 1
a Find P.1 < X < 2/. b Find P.X < 2:5/. will also write E.X2 / D !1 x2 f .x/dx and
R 1
c Find P.X # 32 /. E..X " "/2 / D !1 .x " "/2 f .x/dx. In the text it was claimed
R1
d Find c such that P.X < c/ D 12 . ive your answer in that the variance of X, Var.X/ D !1 .x " "/2 f .x/dx is also given
R1 2
radical form. by Var.X/ D !1 x f .x/dx " "2 , so that
Suppose X is a continuous random variable with density
function given by E..X " E.X//2 / D E.X2 / " E.X/2
( 1000
if x > 1000 Prove this.
f .x/ D x2
0 otherwise
Suppose X is a continuous random variable with density
a Find P.1000 < X < 2000/. b Find P.X > 5000/. function given by
Suppose X is a continuous random variable that is uniformly #
distributed on 1, 4 . k if a ! x ! b
f .x/ D
a What is the formula of the density function for X S etch its 0 otherwise
graph. $ %
b Find P 32 < X < 72 . c Find P.0 < X < 1/. 1
a Show that k D and thus X is uniformly distributed.
d Find P.X ! 3:5/. e Find P.X > 3/. b"a
f Find P.X D 2/. g Find P.X < 5/. b Find the cumulative distribution function F.
h Find ". i Find #. Suppose the random variable X is exponentially distributed
Find the cumulative distribution function F and s etch its with k D 2.
graph. Use F to find P.X < 2/ and P.1 < X < 3/. a Find P.1 < X < 2/. b Find P.X < 3/.
Suppose X is a continuous random variable that is uniformly c Find P.X > 5/.
distributed on 0, 5 . d Find P." " 2# < X < " C 2#/.
e Find the cumulative distribution function F.
a What is the formula of the density function for X S etch its
graph. Suppose the random variable X is exponentially distributed
b Find P.1 < X < 3/. c Find P.4:5 ! X < 5/. with k D 0:5.
d Find P.X D 4/. e Find P.X > 2/. a Find P.X > 4/. b Find P.0:5 < X < 2:6/.
f Find P.X < 5/. g Find P.X > 5/. c Find P.X < 5/. d Find P.X D 4/.
h Find ". i Find #. e Find c such that P.0 < X < c/ D 12 .
 Section 6.2 e or a str t on 721

The density function for a random variable X is given by and f .x/ D 0 otherwise. What is the probability that a person must
# wait at most seven minutes What is the average time that a
kx if 0 ! x ! 4
f .x/ D person must wait
0 otherwise
Soft Drin Dispensing An automatic soft-drin dispenser
a Find k. b Find P.2 < X < 3/. at a fast-food restaurant dispenses X ounces of cola in a 12-ounce
c Find P.X > 2:5/. d Find P.X > 0/. drin . If X is uniformly distributed over Œ11: 2; 12:0 !, what is the
e Find ". f Find #. probability that less than 12 ounces will be dispensed What is
the probability that exactly 12 ounces will be dispensed What
g Find c such that P.X < c/ D 12 . is the average amount dispensed
h Find P.3 < X < 5/.
Emergency Room Arrivals At a particular hospital, the
The density function for a random variable X is given by length of time X (in hours) between successive arrivals at the
8 x emergency room is exponentially distributed with k D 3. What is
< C k if 1 ! x ! 5
the probability that more than one hour passes without an
f .x/ D 16
: arrival
0 otherwise
Electronic Component Life The length of life, X
a Find k. b Find P.X # 3/. (in years), of a computer component has an exponential
c Find ". d Find P.2 < X < "/. distribution with k D 25 . What is the probability that such a
aiting ime At a bus stop, the time X (in minutes) that a component will fail within three years of use What is the
randomly arriving person must wait for a bus is uniformly probability that it will last more than five years
1
distributed with density function f .x/ D 10 , where 0 ! x ! 10

Objective T N
o sc ss t e nor a str t on uite often, measured data in nature such as heights of individuals in a population
stan ar n ts an t e ta e of areas
n er t e stan ar nor a c r e are represented by a random variable whose density function may be approximated by
en C the bell-shaped curve in Figure 16. . The curve extends indefinitely to the right and left
and never touches the x-axis. This curve, called the normal curve, is the graph of the
most important of all density functions, the n rma density functi n.

A continuous random variable X is a n rma rand m ariab e, and equivalently, has

a normal (also called aussian1 ) distribution, if its density function is given by
1 2
f.x/ D p e!.1=2/Œ.x!!/="# "1<x<1
# 2$
called the normal density function. The parameters " and # are the mean and
standard deviation of X, respectively.

bserve in Figure 16. that f.x/ ! 0 as x ! ˙1. That is, the normal curve
has the x-axis as a horizontal asymptote. Also note that the normal curve is symmetric
about the vertical line x D ". That is, the height of a point on the curve d units to the
right of x D " is the same as the height of the point on the curve that is d units to
the left of x D ". ecause of this symmetry and the fact that the area under the normal
curve is 1, the area to the right (or left) of the mean must be 12 .
Each choice of values for " and # determines a different normal curve. The value
of " determines where the curve is centered , and # determines how spread out
the curve is. The smaller the value of #, the less spread out is the area near ". For
example, Figure 16.10 shows normal curves C1 ; C2 , and C3 , where C1 has mean "1
and standard deviation #1 ; C2 has mean "2 , and so on. Here C1 and C2 have the same
mean but different standard deviations: #1 > #2 . C1 and C3 have the same standard
deviation but different means: "1 < "3 . Curves C2 and C3 have different means and
different standard deviations.

1 After the erman mathematician Carl Friedrich auss (1777 1 55).

722 C Cont n o s an o ar a es

C2
f (x) o1 = o2 o1 Z o3
u1 Z u2 u1 = u3

C3

C1

o x o1 = o2 o3

FIGURE Normal curve. FIGURE Normal curves.

f(x)

x
o - 3u o - 2u o-u o o+u o + 2u o + 3u

68%
95%
99.7%
Areas under normal curve

FIGURE Probability and number of standard deviations from ".

The standard deviation plays a significant role in describing probabilities associated

with a normal random variable, X. ore precisely, the probability that X will lie within
one standard deviation of the mean is approximately 0.6 :
P." " # < X < " C # / D 0:6
In other words, approximately 6 of the area under a normal curve is within one
standard deviation of the mean (Figure 16.11). etween " ˙ 2# is about 5 of the
area, and between " ˙ 3# is about .7 :
P." " 2# < X < " C 2#/ D 0: 5
P." " 3# < X < " C 3#/ D 0: 7

The percentages in Figure 16.11 are Thus, it is highly li ely that X will lie within three standard deviations of the mean.
worth remembering.

E AM LE A T S

et X be a random variable whose values are the scores obtained on a nationwide test
given to high school seniors. Suppose, for modeling purposes, that X is normally dis-
tributed with mean 600 and standard deviation 0. Then the probability that X lies
within 2# D 2. 0/ D 1 0 points of 600 is 0. 5. In other words, 5 of the scores lie
between 420 and 7 0. Similarly, .7 of the scores are within 3# D 3. 0/ D 270
points of 600 that is, between 330 and 70.
Now ork Problem 17 G
If is a normally distributed random variable with " D 0 and # D 1, we obtain
the normal curve of Figure 16.12, called the standard normal curve.
 Section 6.2 e or a str t on 723

A continuous random variable is a standard normal random variable (equiva-

lently, has a standard normal distribution) if its density function is given by
1 2
f.z/ D p e!z =2
2$
called the standard normal density function. The variable has mean 0 and stan-
dard deviation 1.

f(z)

z
-3 -2 -1 0 1 2 3
= -3u = -2u = -u =o =u = 2u = 3u

FIGURE Standard normal curve: " D 0, # D 1.

ecause a standard normal random variable has mean 0 and standard deviation 1,
its values are in units of standard deviations from the mean, which are called standard
units. For example, if 0 < < 2:54, then lies within 2.54 standard deviations to the
right of 0, the mean. That is, 0 < < 2:54#. To find the probability P.0 < < 2:54/,
we have
Z 2:54 !z2 =2
e
P.0 < < 2:54/ D p dz
0 2$
The integral on the right cannot be evaluated by elementary functions. However, values
for integrals of this ind have been approximated and put in tabular form.
ne such table is given in Appendix C. The table there gives the area under a stan-
A(z0)
dard normal curve between z D 0 and z D z0 , where z0 # 0. This area is shaded
in Figure 16.13 and is denoted by A.z0 /. In the left-hand columns of the table are
z z-values to the nearest tenth. The numbers across the top are the hundredths val-
0 z0
ues. For example, the entry in the row for 2.5 and column under 0.04 corresponds
FIGURE to z D 2:54 and is 0.4 45. Thus, the area under a standard normal curve between z D 0
A.z0 / D P.0 < < z0 /. and z D 2:54 is (approximately) 0.4 45:

P.0 < < 2:54/ D A.2:54/ $ 0:4 45

The numbers in the table are necessarily approximate, but for the balance of this chapter
we will write A.2:54/ D 0:4 45 and the li e in the interest of improved readability.
Similarly, we can verify that A.2/ D 0:4772 and A.0:33/ D 0:12 3.
Using symmetry, we compute an area to the left of z D 0 by computing the corre-
sponding area to the right of z D 0. For example,

z P."z0 < < 0/ D P.0 < < z0 / D A.z0 /

-z0 0 z0
as shown in Figure 16.14. Hence, P."2:54 < < 0/ D A.2:54/ D 0:4 45.
FIGURE P."z0 < < 0/ D When computing probabilities for a standard normal variable, we may have to add
P.0 < < z0 /. or subtract areas. A useful aid for doing this properly is a rough s etch of a standard
normal curve in which we have shaded the entire area that we want to find, as Example 2
shows.
724 C Cont n o s an o ar a es

E AM LE S N
a Find P. > 1:5/.
z S This probability is the area to the right of z D 1:5 (Figure 16.15). That area
0 1.5
is equal to the difference between the total area to the right of z D 0, which is 0.5, and
the area between z D 0 and z D 1:5, which is A(1.5). Thus,
FIGURE P. > 1:5/.
P. > 1:5/ D 0:5 " A.1:5/
D 0:5 " 0:4332 D 0:066 from Appendix C
b Find P.0:5 < < 2/.
S This probability is the area between z D 0:5 and z D 2 (Figure 16.16). That
z area is the difference of two areas. It is the area between z D 0 and z D 2, which is
0 0.5 2
A.2/, minus the area between z D 0 and z D 0:5, which is A(0.5). Thus,
FIGURE P.0:5 < < 2/. P.0:5 < < 2/ D A.2/ " A.0:5/
D 0:4772 " 0:1 15 D 0:2 57
c Find P. ! 2/.
S This probability is the area to the left of z D 2 (Figure 16.17). That area is
equal to the sum of the area to the left of z D 0, which is 0.5, and the area between
z D 0 and z D 2, which is A(2). Thus,
z
0 2 P. ! 2/ D 0:5 C A.2/
FIGURE P. ! 2/. D 0:5 C 0:4772 D 0: 772
Now ork Problem 1 G

E AM LE S N

a Find P."2 < < "0:5/.

z
-2 -0.5 0 S This probability is the area between z D "2 and z D "0:5 (Figure 16.1 ).
y symmetry, that area is equal to the area between z D 0:5 and z D 2, which was
FIGURE P."2 < < "0:5/. computed in Example 2(b). We have
P."2 < < "0:5/ D P.0:5 < < 2/
D A.2/ " A.0:5/ D 0:2 57
b Find z0 such that P."z0 < < z0 / D 0: 642.
S Figure 16.1 shows the corresponding area. ecause the total area is 0. 642,
by symmetry the area between z D 0 and z D z0 is 12 .0: 642/ D 0:4 21, which is A.z0 /.
-z0 0 z0
z oo ing at the body of the table in Appendix C, we see that 0.4 21 corresponds to a
-value of 2.1. Thus, z0 D 2:1.
FIGURE
P."z0 < < z0 / D 0: 642.
Now ork Problem 3 G

T S N
If X is normally distributed with mean " and standard deviation #, one might thin
that a table of areas is needed for each pair of values of " and # . Fortunately, this is
not the case. Appendix C is still used. ut we must first express the area of a given
region as an equal area under a standard normal curve. This involves transforming X
into a standard variable (with mean 0 and standard deviation 1) by using the following
change-of-variable formula:
X""
D
#
 Section 6.2 e or a str t on 725

Here we convert a normal variable to a n the right side, subtracting " from X gives the distance from " to X. ividing by #
standard normal variable. expresses this distance in terms of units of standard deviation. Thus, is the number
of standard deviations that X is from ". That is, Formula (1) converts units of X into
standard units ( -values). For example, if X D ", then using Formula (1) gives D 0.
Hence, " is zero standard deviations from ".
Suppose X is normally distributed with " D 4 and # D 2. Then, to find for
example P.0 < X < 6/, we first use Formula (1) to convert the X-values 0 and 6 to
-values (standard units):
x1 " " 0"4
z1 D D D "2
# 2
x2 " " 6"4
z2 D D D1
# 2
It can be shown that

P.0 < X < 6/ D P."2 < < 1/

This means that the area under a normal curve with " D 4 and # D 2 between x D 0
and x D 6 is equal to the area under a standard normal curve between z D "2 and
z D 1 (Figure 16.20). This area is the sum of the area A1 between z D "2 and z D 0
and the area A2 between z D 0 and z D 1. Using symmetry for A1 , we have

P."2 < < 1/ D A1 C A2 D A.2/ C A.1/

-2 0 1
z D 0:4772 C 0:3413 D 0: 1 5

FIGURE P."2 < < 1/.

E AM LE E S

The wee ly salaries of 5000 employees of a large corporation are assumed to be nor-
mally distributed with mean 640 and standard deviation 56. How many employees
earn less than 570 per wee
S Converting to standard units, we have
! "
570 " 640
P.X < 570/ D P < D P. < "1:25/
56
This probability is the area shown in Figure 16.21(a). y symmetry, that area is equal to
the area in Figure 16.21(b) that corresponds to P. > 1:25/. This area is the difference
between the total area to the right of x D 0, which is 0.5, and the area between z D 0
and z D 1:25, which is A(1.25). Thus,

P.X < 570/ D P. < "1:25/ D P. > 1:25/

D 0:5 " A.1:25/ D 0:5 " 0:3 44 D 0:1056

z z
-1.25 0 0 1.25
(a) (b)

FIGURE iagram for Example 4.

That is, 10.56 of the employees have salaries less than 570. This corresponds to
0:1056.5000/ D 52 employees.
Now ork Problem 21 G
726 C Cont n o s an o ar a es

R BLEMS
If is a standard normal random variable, find each of the If X is normally distributed with " D and # D 1, find
following probabilities. P.X > " " #/.
a P.0 < < 2:3/ b P.0:35 < < 1:31/ If X is normally distributed such that " D 40 and
c P. > "0:57/ d P. ! 1:46/ P.X > 54/ D 0:0401, find #.
e P."2:3 < ! 1:70/ f P. > 0:1 / If X is normally distributed with " D 62 and # D 11,
find x0 such that the probability that X is between x0 and 62 is
If is a standard normal random variable, find each of the 0.4554.
following.
est Scores The scores on a national achievement test are
a P."1: 6 < < 1: 6/ b P."2:11 < < "1:35/
normally distributed with mean 500 and standard deviation 100.
c P. < "1:05/ d P. > 3#/ What percentage of those who too the test had a score between
e P.j j > 2/ f P.j j < 12 / 300 and 700
est Scores In a test given to a large group of people, the
In Pr b ems nd z0 such that the gi en statement is true
scores were normally distributed with mean 55 and standard
Assume that is a standard n rma rand m ariab e
deviation 10. What is the greatest whole-number score that a
P. < z0 / D 0:636 P. < z0 / D 0:066 person could get and yet score in about the bottom 10
P. > z0 / D 0: 5 P. > z0 / D 0:42 6 Adult eights The heights (in inches) of adults in a large
P."z0 < < z0 / D 0:2662 P.j j > z0 / D 0:0456 population are normally distributed with " D 6 and # D 3.
What percentage of the group is under 6 feet tall
If X is normally distributed with " D 16 and # D 4, find each
of the following probabilities. Income The yearly income for a group of 10,000
professional people is normally distributed with " D 60,000
a P.X < 27/ b P.X < 10/ and # D 5000.
c P.10: < X < 12:4/ a What is the probability that a person from this group has a
If X is normally distributed with " D 200 and # D 40, find yearly income less than 46,000
each of the following probabilities. b How many of these people have yearly incomes over
a P.X > 150/ b P.210 < X < 250/ 75,000
If X is normally distributed with " D 57 and # D 10, find IQ The I s of a large population of children are normally
P.X > 0/. distributed with mean 100.4 and standard deviation 11.6.
If X is normally distributed with " D 0 and # D 1:5, find a What percentage of the children have I s greater than 120
P.X < 3/. b About 5.05 of the children have I s greater than what
value
If X is normally distributed with " D 60 and # 2 D 100,
find P.50 < X ! 75/. Suppose X is a random variable with " D 10 and # D 2. If
P.4 < X < 16/ D 0:25, can X be normally distributed

Objective T N A
B
o s o t e tec n e of est at n We conclude this chapter by bringing together the notions of a discrete random variable
t e no a str t on sn t e
nor a str t on and a continuous random variable. Recall from Chapter that if X is a binomial random
variable (which is discrete), and if the probability of success on any trial is p, then for
n independent trials, the probability of x successes is given by

P.X D x/ D n Cx px qn!x

where q D 1 " p. Calculating probabilities for a binomial random variable can be time
consuming when the number of trials is large. For example, 100 C40 .0:3/40 .0:7/60 is a lot
of wor to compute by hand . Fortunately, we can approximate a binomial distribution
li e this by a normal distribution and then use a table of areas.
To show how this is done, let us ta e a simple example. Figure 16.22 gives a prob-
ability histogram for a binomial experiment with n D 10 and p D 0:5. The rectangles
centered at x D 0 and x D 10 are not shown because their heights are very close to 0.
Superimposed on the histogram is a normal curve, which approximates it. The approxi-
mation would be even better if n were larger. That is, as n gets larger, the width of each
unit interval appears to get smaller, and the outline of the histogram tends to ta e on
the appearance of a smooth curve. In fact, it is n t unusua t think f a density cur e as
 Section 6.3 e or a ro at on to t e no a str t on 727

the imiting case f a pr babi ity hist gram. In spite of the fact that in our case n is only
10, the approximation shown does not seem too bad. The question that now arises is,
Which normal distribution approximates the binomial distribution Since the mean
and standard deviation are measures of central tendency and dispersion of a random
variable, we choose the approximating normal distribution to have the same mean and
standard deviation as that of the binomial distribution. For this choice, we can estimate
the areas of rectangles in the histogram (that is, the binomial probabilities) by finding
the corresponding area under the normal curve. In summary, we have the following:

If X is a binomial random variable and n is su ciently large, then the distribution

of X can be approximated by a normal random variable whose mean and standard
p
deviation are the same as for X, which are np and npq, respectively.

A word of explanation is appropriate concerning the phrase n is su ciently large.

enerally spea ing, a normal approximation to a binomial distribution is not good if
n is small and p is near 0 or 1, because much of the area in the binomial histogram
would be concentrated at one end of the distribution (that is, at 0 or n). Thus, the dis-
tribution would not be fairly symmetric, and a normal curve would not fit well. A
general rule we can follow is that the normal approximation to the binomial distribu-
tion is reasonable if both np and nq are at least 5. This is the case in our example:
np D 10.0:5/ D 5 and nq D 10.0:5/ D 5.
et us now use the normal approximation to estimate a binomial probability for
n D 10 and p D 0:5. If X denotes the number of successes, then its mean is
np D 10.0:5/ D 5
and its standard deviation is
p p p
npq D 10.0:5/.0:5/ D 2:5 $ 1:5
The probability function for X is given by
x
f.x/ D 10 Cx .0:5/ .0:5/10!x
p
We approximate this distribution by the normal distribution with " D 5 and # D 2:5.

0.246
0.205

0.117

0.044
0.010 x
0 1 2 3 4 5 6 7 8 9 10

FIGURE Normal approximation to binomial distribution.

x
0 1 2 3 4 5 6 7 8 9 10

FIGURE Normal approximation to P.4 ! X ! 7/.

728 C Cont n o s an o ar a es

Suppose we estimate the probability that there are between 4 and 7 successes,
inclusive, which is given by
P.4 ! X ! 7/ D P.X D 4/ C P.X D 5/ C P.X D 6/ C P.X D 7/
X
7
x 10!x
D 10 Cx .0:5/ .0:5/
xD4

This probability is the sum of the areas of the rectang es for X D 4; 5; 6, and 7 in
Figure 16.23. Under the normal curve, we have shaded the corresponding area that we
will compute as an approximation to this probability. Note that the shading extends
not from 4 to 7, but from 4 " 12 to 7 C 12 that is, from 3.5 to 7.5. This continuity
correction of 0.5 on each end of the interval allows most of the area in the appropriate
rectangles to be included in the approximation, and such a c rrecti n must a ays be
made. The phrase continuity correction is used because X is treated as though it were
a continuous random variable. We now convert the X-values 3.5 and 7.5 to -values:
3:5 " 5
z1 D p $ "0: 5
2:5
7:5 " 5
z2 D p $ 1:5
2:5
Thus,
P.4 ! X ! 7/ $ P."0: 5 ! ! 1:5 /
which corresponds to the area under a standard normal curve between z D "0: 5 and
-0.95 0 1.58
z z D 1:5 (Figure 16.24). This area is the sum of the area between z D "0: 5 and z D 0,
which, by symmetry, is A.0: 5/, and the area between z D 0 and z D 1:5 , which is
FIGURE A.1:5 /. Hence,
P."0: 5 ! ! 1:5 /.
P.4 ! X ! 7/ $ P."0: 5 ! ! 1:5 /
D A.0: 5/ C A.1:5 /
D 0:32 C 0:442 D 0:771
This result is close to the true value, which to four decimal places is 0.7734.

A L IT I E AM LE N A B
n a game show, the grand prize Suppose X is a binomial random variable with n D 100 and p D 0:3. Estimate
is hidden behind one of four doors.
P.X D 40/ by using the normal approximation.
Assume that the probability of selecting
the grand prize is p D 14 . There were 20 S We have
40
winners among the last 60 contestants. P.X D 40/ D 100 C40 .0:3/ .0:7/60
Suppose that X is the number of con-
testants that win the grand prize, and X using the formula that was mentioned at the beginning of this section. We use a normal
is binomial with n D 60. Approximate distribution with
P.X D 20/ by using the normal approx-
imation.
" D np D 100.0:3/ D 30
and p p
p
#D npq D 100.0:3/.0:7/ D 21 $ 4:5
Converting the corrected X-values 3 .5 and 40.5 to -values gives
Remember the continuity correction. 3 :5 " 30
z1 D p $ 2:07
21
40:5 " 30
z2 D p $ 2:2
21
Therefore,
P.X D 40/ $ P.2:07 ! ! 2:2 /
 Section 6.3 e or a ro at on to t e no a str t on 729

This probability is the area under a standard normal curve between z D 2:07 and
z D 2:2 (Figure 16.25). That area is the difference of the area between z D 0 and
z D 2:2 , which is A.2:2 /, and the area between z D 0 and z D 2:07, which is
A.2:07/. Thus,
z
0 P.X D 40/ $ P.2:07 ! ! 2:2 /
2.07 2.29
D A.2:2 / " A.2:07/
FIGURE P.2:07 ! ! 2:2 /.
D 0:4 0 " 0:4 0 D 0:00 2 from Appendix C
Now ork Problem 3 G

E AM LE C

In a quality-control experiment, a sample of 500 items is ta en from an assembly line.

Customarily, of the items produced are defective. What is the probability that more
than 50 defective items appear in the sample
S If X is the number of defective items in the sample, then we will consider
X to be binomial with n D 500 and p D 0:0 . To find P.X # 51/, we use the normal
approximation to the binomial distribution with
" D np D 500.0:0 / D 40
and
p p p
#D npq D 500.0:0 /.0: 2/ D 36: $ 6:07
Converting the corrected value 50.5 to a -value gives
50:5 " 40
zD p $ 1:73
36:
Thus,
P.X # 51/ $ P. # 1:73/
This probability is the area under a standard normal curve to the right of z D 1:73
(Figure 16.26). That area is the difference of the area to the right of z D 0, which is
z 0.5, and the area between z D 0 and z D 1:73, which is A.1:73/. Hence,
0 1.73
P.X # 51/ $ P. # 1:73/
FIGURE P. # 1:73/.
D 0:5 " A.1:73/ D 0:5 " 0:45 2 D 0:041

Now ork Problem 7 G

R BLEMS
In Pr b ems X is a bin mia rand m ariab e ith the gi en axis out of service A taxi company has a eet of 100 cars.
a ues f n and p Ca cu ate the indicated pr babi ities by using At any given time, the probability of a car being out of service
the n rma appr ximati n due to factors such as brea downs and maintenance is 0.1.
n D 150; p D 0:4I P.X # 52/; P.X # 74/ What is the probability that 10 or more cars are out of service at
any time
n D 50, p D 0:3 P.X D 25/, P.X ! 20/
Quality Control In a manufacturing plant, a sample of 200
n D 200; p D 0:6I P.X D 125/; P.110 ! X ! 135/ items is ta en from the assembly line. For each item in the sample,
n D 50, p D 0:20I P.X # 10/ the probability of being defective is 0.05. What is the probability
Die ossing Suppose a fair die is tossed 300 times. What that there are 7 or more defective items in the sample
is the probability that a 5 turns up between 45 and 60 times, rue False Exam In a true false exam with 50 questions,
inclusive what is the probability of getting at least 25 correct answers by
Coin ossing For a biased coin, P. / D 0:4 and ust guessing on all the questions If there are 100 questions
P. / D 0:6. If the coin is tossed 200 times, what is the probability instead of 50, what is the probability of getting at least 50 correct
of getting between 0 and 100 heads, inclusive answers by ust guessing
730 C Cont n o s an o ar a es

Multiple Choice Exam In a multiple-choice test with 50 aste est An energy drin company sponsors a national
questions, each question has four answers, only one of which is taste test, in which sub ects sample its drin as well as the
correct. If a student guesses on the last 20 questions, what is the best-selling brand. Neither drin is identified by brand. The
probability of getting at least half of them correct sub ects are then as ed to choose the drin that tastes better. If
Po er In a po er game, the probability of being dealt a each of the 4 sub ects in a supermar et actually has no
hand consisting of three cards of one suit and two cards of another preference and arbitrarily chooses one of the drin s, what is the
suit (in any order) is about 0.1. In 100 dealt hands, what is the probability that 30 or more of the sub ects choose the drin from
probability that 16 or more of them will be as ust described the sponsoring company

Chapter 16 Review
I T S E
S Continuous Random Variables
continuous random variable uniform density function Ex. 1, p. 716
exponential density function exponential distribution Ex. 3, p. 717
cumulative distribution function (probability) density function Ex. 4, p. 71
mean, " variance, # 2 standard deviation, # Ex. 5, p. 71
S he ormal Distribution
normal distribution normal density function Ex. 1, p. 722
standard normal curve standard normal random variable Ex. 2, p. 724
standard normal distribution standard normal density function Ex. 4, p. 725
S he ormal Approximation to the Binomial Distribution
continuity correction Ex. 1, p. 72

S
A continuous random variable, X, can assume any value in The cumulative distribution function, F, for the continu-
an interval or intervals. A density function is a function that ous random variable X with density function f is given by
has the following properties:
Z
Z 1
x

f.x/ # 0 F.x/ D P.X ! x/ D f.t/dt

f.x/dx D 1 !1
!1

A density function is a density function f r the rand m eometrically, F.x/ represents the area under the density
ariab e X if curve to the left of x. y using F, we are able to find
Z b P.a ! x ! b/:
P.a ! X ! b/ D f.x/dx
a
P.a ! x ! b/ D F.b/ " F.a/
which means that the probability that X assumes a value in
the interval Œa; b! is to be given by the area under the graph of
The mean " of X (also called expectation of X) E.X/ is
f and above the x-axis from x D a to x D b. The probability
given by
that X assumes a particular value is 0.
The continuous random variable X has a uniform distri- Z 1
bution over Œa; b! if its density function is given by
" D E.X/ D xf.x/dx
8 !1
< 1
if a ! x ! b
f.x/ D b " a
: 0 otherwise provided that the integral is convergent. The variance is
given by
X has an exponential density function, f, if
Z 1
( !kx
ke if x # 0 # 2 D Var.X/ D .x " "/2 f.x/ dx
f.x/ D Z
!1
0 if x < 0 1
D x2 f.x/ dx " "2
where k is a positive constant. !1
 Chapter 6 e e 731

provided that the integral is convergent. The standard devia- z D 0 to z D z0 and is denoted A.z0 /. Values of A.z0 / appear
tion is given by in Appendix C.
p If X is normally distributed with mean " and standard
# D Var.X/ deviation # , then X can be transformed into a standard nor-
mal random variable by the change-of-variable formula
The graph of the normal density function
X""
1 D
f.x/ D p e!.1=2/..x!!/=" /
2 #
# 2$
With this formula, probabilities for X can be found by using
is called a normal curve and is bell shaped. If X has a normal areas under the standard normal curve.
distribution, then the probability that X lies within one stan- If X is a binomial random variable and the number, n, of
dard deviation of the mean " is (approximately) 0.6 within independent trials is large, then the distribution of X can be
two standard deviations, the probability is 0. 5 and within approximated by using a normal random variable with mean
p
three standard deviations, it is 0. 7. If is a normal random np and standard deviation npq, where p is the probability
variable with " D 0 and # D 1, then is called a standard of success on any trial and q D 1"p. It is important that con-
normal random variable. The probability P.0 < < z0 / is tinuity corrections be considered when we estimate binomial
the area under the graph of the standard normal curve from probabilities by a normal random variable.

R
Suppose X is a continuous random variable with density et X be uniformly distributed over the interval 2, 6 . Find
function given by P.X < 5/.
(1 et X be n rma y distributed ith mean 20 and standard
3
C kx2 if 0 ! x ! 1
f .x/ D de iati n 4 In Pr b ems determine the gi en pr babi ities
0 otherwise
P.X > 25/ P.X < 21/ P.14 < X < 1 /
P.X > 32/ P.X < 23/ P.21 < X < 31/
a Find k. b Find P. 12 < X < 34 /. c Find P.X # 12 /.
In Pr b ems and X is a bin mia rand m ariab e ith
d Find the cumulative distribution function. n D 100 and p D 0:35 Find the gi en pr babi ities by using the
Suppose X is exponentially distributed with k D 13 . Find n rma appr ximati n
P.X > 2/. P.25 ! X ! 47/ P.X D 4 /
Suppose X is a random variable with density function given by eights of Individuals The heights in meters of
(1 individuals in a certain group are normally distributed with mean
x if 0 ! x ! 4 1.73 and standard deviation 0.05. Find the probability that an
f .x/ D
0 otherwise individual from this group is taller than 1. 3.
Coin ossing If a fair coin is tossed 500 times, use the
normal approximation to the binomial distribution to estimate the
a Find ". b Find #.
probability that a head comes up at least 215 times.
 t ar a e
Ca c s

W
e now how to maximize a company s profit when both revenue and
17.1 Part a er at es cost are written as functions of a single quantity, namely, the number of
17.2 cat ons of Part a units produced. ut, of course, the production level is itself determined
er at es by many factors, and no single variable can represent all of them.
The amount of oil pumped from an oil field each wee , for example, depends on
17.3 er r er Part a both the number of pumps and the number of hours that the pumps are operated. The
er at es
number of pumps in the field will depend on the amount of capital originally available
17.4 a a an n a to build the pumps as well as the size and shape of the field. The number of hours
for nct ons of o that the pumps can be operated depends on the labor available to run and maintain the
ar a es pumps. In addition, the amount of oil that the owner will be willing to have pumped
17.5 a ran e t ers from the oil field will depend on the current demand for oil which is related to the
price of the oil.
17.6 t e nte ra s aximizing the wee ly profit from an oil field will require a balance between the
number of pumps and the amount of time each pump can be operated. The maximum
Chapter 17 e e
profit will not be achieved by building more pumps than can be operated or by running
a few pumps full time.
This is an example of the general problem of maximizing profit when production
depends on several factors. The solution involves an analysis of the production func-
tion, which relates production output to resources allocated for production. ecause
several variables are needed to describe the resource allocation, the most profitable
allocation cannot be found by differentiation with respect to a single variable, as in
preceding chapters. The more advanced techniques necessary to do the ob will be cov-
ered in this chapter. For the most part, we focus on functions of two variables because
the techniques involved in moving from one variable to two variables usually extend
unremar ably to introduction of further variables.

732
 Section 7. Part a er at es 733

Objective
o co te art a er at es Throughout this text we have encountered many examples of functions of several vari-
ables. We recall, from Section 2. , that the graph of a function of two variables is a
surface. Figure 17.1 shows the surface z D f.x; y/ and a plane that is parallel to the
x; z-plane and that passes through the point .a; b; f.a; b// on the surface. The equation
To review functions of several variables, of this plane is y D b. Hence, any point on the curve that is the intersection of the
see Section 2. . surface z D f.x; y/ with the plane y D b must have the form .x; b; f.x; b//. Thus, the
curve can be described by the equation z D f.x; b/. Since b is constant, z D f.x; b/ can
be considered a function of one variable, x. When the derivative of this function is eval-
uated at a, it gives the slope of the tangent line to this curve at the point .a; b; f.a; b//.
(See Figure 17.1.) This slope is called the partia deri ati e f f ith respect t x at
.a; b/ and is denoted fx .a; b/. In terms of limits,

f.a C h; b/ ! f.a; b/
fx .a; b/ D lim
h!0 h

z
z
Tangent line
Tangent line
(x, b, f (x, b))

(a, b, f (a, b)) (a, b, f (a, b))

z = f(x, b)
z = f(x, y)

z = f (a, y)
(a, y, f(a, y))

b
a b
y a
y
x (a, b, 0) (a, b, 0)
x

FIGURE eometric interpretation of fx .a; b/. FIGURE eometric interpretation of fy .a; b/.

n the other hand, in Figure 17.2, the plane x D a is parallel to the y; z-plane and
cuts the surface z D f.x; y/ in a curve given by z D f.a; y/, a function of y. When the
derivative of this function is evaluated at b, it gives the slope of the tangent line to this
curve at the point .a; b; f.a; b//. This slope is called the partia deri ati e f f ith
respect t y at .a; b/ and is denoted fy .a; b/. In terms of limits,

f.a; b C h/ ! f.a; b/
fy .a; b/ D lim
h!0 h
This gives us a geometric interpretation
of a partial derivative. We say that fx .a; b/ is the slope of the tangent line to the graph of f at .a; b; f.a; b//
in the x directi n similarly, fy .a; b/ is the slope of the tangent line in the y directi n.
For generality, by replacing a and b in Equations (1) and (2) by x and y, respectively,
we get the following definition.
734 C t ar a e Ca c s

If z D f.x; y/, the partial derivative of f with respect to x, denoted fx , is the function,
of two variables, given by
f.x C h; y/ ! f.x; y/
fx .x; y/ D lim
h!0 h
provided that the limit exists.
The partial eri ati e o f wit respe t to y, denoted fy , is the function, of two
variables, given by
f.x; y C h/ ! f.x; y/
fy .x; y/ D lim
h!0 h
provided that the limit exists.

y analyzing the foregoing definition, we can state the following procedure to find
fx and fy :

This gives us a mechanical way to find F fx .x; y/ fy .x; y/

partial derivatives. To find fx , treat y as a constant, and differentiate f with respect to x in the usual way.
To find fy , treat x as a constant, and differentiate f with respect to y in the
usual way.

E AM LE F

If f.x; y/ D xy2 C x2 y, find fx .x; y/ and fy .x; y/. Also, find fx .3; 4/ and fy .3; 4/.
S To find fx .x; y/, we treat y as a constant and differentiate f with respect to x:

fx .x; y/ D .1/y2 C .2x/y D y2 C 2xy

To find fy .x; y/, we treat x as a constant and differentiate with respect to y:

fy .x; y/ D x.2y/ C x2 .1/ D 2xy C x2

Note that fx .x; y/ and fy .x; y/ are each functions of the two variables x and y. To find
fx .3; 4/, we evaluate fx .x; y/ when x D 3 and y D 4:

fx .3; 4/ D 42 C 2.3/.4/ D 40

Similarly,

fy .3; 4/ D 2.3/.4/ C 32 D 33

Now ork Problem 1 G

Notations for partial derivatives of z D f.x; y/ are in Table 17.1. Table 17.2 gives
notations for partial derivatives evaluated at .a; b/. Note that the symbol @ (not d) is
used to denote a partial derivative. The symbol @z=@x is read the partial derivative of
z with respect to x.
 Section 7. Part a er at es 735

Table 1 .1 Table 1 .
Partial erivative of f Partial erivative of f Partial erivative of f Partial erivative of f
(or z) with Respect to x (or z) with Respect to y (or z) with Respect to x (or z) with Respect to y
Evaluated at .a; b/ Evaluated at .a; b/
fx .x; y/ fy .x; y/
fx .a; b/ fy .a; b/
@ @
. f .x; y// . f .x; y// ˇ ˇ
@x @y @f ˇˇ @f ˇˇ
@z @z @x ˇ.a;b/ @y ˇ.a;b/
@x @y ˇ ˇ
@z ˇˇ @z ˇˇ
@x ˇ xDa @y ˇ xDa
yDb yDb

E AM LE F

ˇ ˇ
3 3 2 @z @z @z ˇˇ
2 @z ˇˇ
a If z D 3x y ! x y C xy C 4y, find , , ˇ , and ˇ .
@x @y @x .1;0/ @y .1;0/
S To find @z=@x, we differentiate z with respect to x while treating y as a
constant:
@z
D 3.3x2 /y3 ! .2x/y C .1/y2 C 0
@x
D x2 y3 ! 1 xy C y2
Evaluating the latter equation at .1; 0/, we obtain
ˇ
@z ˇˇ
D .1/2 .0/3 ! 1 .1/.0/ C 02 D 0
@x ˇ.1;0/

To find @z=@y, we differentiate z with respect to y while treating x as a constant:

@z
D 3x3 .3y2 / ! x2 .1/ C x.2y/ C 4.1/
@y
D x3 y2 ! x2 C 2xy C 4

Thus,
ˇ
@z ˇˇ
D .1/3 .0/2 ! .1/2 C 2.1/.0/ C 4 D !5
@y ˇ.1;0/

b If D x2 e2xC3y , find @ =@x and @ =@y.

S To find @ =@x, we treat y as a constant and differentiate with respect to x.
Since x2 e2xC3y is a product of two functions, each involving x, we use the product rule:
@ @ @
D x2 .e2xC3y / C e2xC3y .x2 /
@x @x @x
D x2 .2e2xC3y / C e2xC3y .2x/
D 2x.x C 1/e2xC3y
To find @ =@y, we treat x as a constant and differentiate with respect to y:
@ @
D x2 .e2xC3y / D 3x2 e2xC3y
@y @y

Now ork Problem 27 G

736 C t ar a e Ca c s

We have seen that, for a function of two variables, two partial derivatives can be
considered. The concept of partial derivatives can be extended to functions of more
than two variables. For example, with D f.x; y; z/ we have three partial derivatives:
the partial with respect to x, denoted fx .x; y; z/, @ =@x, and so on
the partial with respect to y, denoted fy .x; y; z/, @ =@y, and so on
and
the partial with respect to z, denoted fz .x; yz/, @ =@z, and so on
To determine @ =@x, treat y and z as constants, and differentiate with respect to x.
For @ =@y, treat x and z as constants, and differentiate with respect to y. For @ =@z, treat
x and y as constants, and differentiate with respect to z. For a function of n variables,
we have n partial derivatives, which are determined in an analogous way.

E AM LE F T

If f.x; y; z/ D x2 C y2 z C z3 , find fx .x; y; z/, fy .x; y; z/, and fz .x; y; z/.

S To find fx .x; y; z/, we treat y and z as constants and differentiate f with respect
to x:
fx .x; y; z/ D 2x
Treating x and z as constants and differentiating with respect to y, we have
fy .x; y; z/ D 2yz
Treating x and y as constants and differentiating with respect to z, we have
fz .x; y; z/ D y2 C 3z2

Now ork Problem 23 G

E AM LE F F
ˇ
rsu @p @p @p ˇˇ
If p D g.r; s; t; u/ D 2 , find , , and .
rt C s2 t @s @t @t ˇ.0;1;1;1/
S To find @p=@s, first note that p is a quotient of two functions, each involving
the variable s. Thus, we use the quotient rule and treat r, t, and u as constants:
@ @
@p .rt2 C s2 t/ .rsu/ ! rsu .rt2 C s2 t/
D @s @s
@s .rt2 C s2 t/2
.rt2 C s2 t/.ru/ ! .rsu/.2st/
D
.rt2 C s2 t/2
Simplification gives
@p ru.rt ! s2 /
D a factor of t cancels
@s t.rt C s2 /2
To find @p=@t, we can first write p as
p D rsu.rt2 C s2 t/!1
Next, we use the power rule and treat r, s, and u as constants:
@p @
D rsu.!1/.rt2 C s2 t/!2 .rt2 C s2 t/
@t @t
D !rsu.rt2 C s2 t/!2 .2rt C s2 /
so that
@p rsu.2rt C s2 /
D!
@s .rt2 C s2 t/2
 Section 7. Part a er at es 737

etting r D 0, s D 1, t D 1, and u D 1 gives

ˇ
@p ˇˇ 0.1/.1/.2.0/.1/ C .1/2 /
ˇ D! D0
@t .0;1;1;1/ .0.1/2 C .1/2 .1//2

Now ork Problem 31 G

R BLEMS
In Pr b ems a functi n f t r m re ariab es is gi en f .r; s; t/ D rst.r2 C s3 C t4 /I fs .1; !1; 2/
Find the partia deri ati e f the functi n ith respect t each f ˇ ˇ
x2 ! y2 @z ˇˇ @z ˇˇ
the ariab es z D x2 !y2 ,
e @x ˇ x D 0 @y ˇ x D 1
yD1 yD0
f .x; y/ D 2x2 C 3xy C 4y2 C 5x C 6y ! 7
If z D xex!y C yey!x , show that
f .x; y/ D 2x2 C 3xy @z @z
f .x; y/ D 2y C 1 f .x; y/ D e! ln 2 C D ex!y C ey!x
@x @y
4 2
g.x; y/ D 3x y C 2xy ! 5xy C x ! y
Stoc Prices of a Dividend Cycle In a discussion of stoc
g.x; y/ D .x2 C 1/2 C .y3 ! 3/3 C 5xy3 ! 2x2 y2 prices of a dividend cycle, Palmon and aari1 consider the
p p
g.p; q/ D pq g. ; z/ D 3 2 C z2 function f given by
s2 C 1 u 2 .1 C r/1!z ln.1 C r/
h.s; t/ D h.u; / D 2 u D f .t; r; z/ D
2
t !1 u C 2 .1 C r/1!z ! t
p p where u is the instantaneous rate of as -price appreciation, r is an
u.q1 ; q2 / D ln q1 C 2 C ln 3 q2 C 5 annual opportunity rate of return, z is the fraction of a dividend
. ; k/ D 2 0:3 k1:7 ! 3 1:03 C 2k0:13 cycle over which a share of stoc is held by a midcycle seller, and
t is the effective rate of capital gains tax. They claim that
x2 C 3xy C y2 xC4
h.x; y/ D p h.x; y/ D @u t.1 C r/1!z ln2 .1 C r/
x2 C y2 xy2 ! x2 y D
@z Œ.1 C r/1!z ! t!2
z D e5xy z D .x3 C y3 /exyC3xC3y Verify this.
z D 5x ln.x2 C y/ z D ln.5x3 y2 C 2y4 /4 Money Demand In a discussion of inventory theory of
p money demand, Swanson2 considers the function
f .r; s/ D r ! s.r2 ! 2rs C s2 /
p b iC
f .r; s/ D rs e2Cr f .r; s/ D e3!r ln.7 ! s/ F.b; C; ; i/ D C
C 2
2 3
f .r; s/ D .5r C 3s /.2r ! 5s/ @F b i
and determines that D ! 2 C . Verify this partial derivative.
g.x; y; z/ D 2x3 y2 C 2xy3 z C 4z2 @C C 2
Interest Rate Deregulation In an article on interest rate
g.x; y; z/ D xy2 z3 C x3 yz2 C x2 y3 z deregulation, Christofi and Agapos3 arrive at the equation
g.r; s; t/ D esCt .r2 C 7s3 / g.r; s; t; u/ D rs ln.t/eu @r dC
r DrC C
In Pr b ems e a uate the gi en partia deri ati es @ d
where r is the deposit rate paid by commercial ban s, r is the rate
f .x; y/ D x y C 7x2 y2 I fx .1; !2/
3
ˇ earned by commercial ban s, C is the administrative cost of
p @z ˇˇ
z D 2x3 C 5xy C 2y2 I transforming deposits into return-earning assets, and is the
@x ˇ xD0 savings deposit level. Christofi and Agapos state that
yD1 ! "
p 1C" dC
g.x; y; z/ D exCyCz x2 C y2 C z2 gz .0; 3; 4/ r Dr C
" d
3x2 y2 C 2xy C x ! y r=
g.x; y; z/ D ; gy .1; 1; 5/ where " D is the deposit elasticity with respect to the
xy ! yz C xz @r=@
h.r; s; t; u/ D .rst2 u/ ln.1 C rstu/I ht .1; 1; 0; 1/ deposit rate. Express Equation (3) in terms of " to verify
7r C 3s2 u2 Equation (4).
h.r; s; t; u/ D I ht .4; 3; 2; 1/
s
1
. Palmon and U. aari, Taxation of Capital ains and the ehavior of
Stoc Prices over the ividend Cycle, he American Ec n mist VII, no. 1
(1 3), 13 22.
2 P.
E. Swanson, Integer Constraints on the Inventory Theory of oney
emand, uarter y urna f Business and Ec n mics 23, no. 1 (1 4),
32 37.
3 A.Christofi and A. Agapos, Interest Rate eregulation: An Empirical usti-
fication, Re ie f Business and Ec n mic Research (1 4), 3 4 .
738 C t ar a e Ca c s

Advertising and Profitability In an analysis of advertising where R is the ad usted rate of profit, r is the accounting rate of
and profitability, Swales4 considers a function R given by profit, a is a measure of advertising expenditures, and n is the
r number of years that advertising fully depreciates. In the analysis,
R D R.r; a; n/ D # $ Swales determines @R=@n. Find this partial derivative and the
n!1
1Ca other two partial derivatives.
2

Objective A
o e e o t e not ons of art a From Section 17.1, we now that if z D f.x; y/, then @z=@x and @z=@y can be geomet-
ar na cost ar na ro ct t
an co et t e an co e entar rically interpreted as giving the slopes of the tangent lines to the surface z D f.x; y/ in
ro cts the x- and y-directions, respectively. There are other interpretations: ecause @z=@x is
the derivative of z with respect to x when y is held fixed, and because a derivative is a
rate of change, we have

@z
Here we have rate of change is the rate of change of z with respect to x when y is held fixed.
interpretations of partial derivatives.
@x

Similarly,

@z
is the rate of change of z with respect to y when x is held fixed.
@y

We will now loo at some applications in which the rate of change notion of a partial
derivative is very useful.
Suppose a manufacturer produces x units of product and y units of product .
Then the total cost c of these units is a function of x and y and is called a oint cost
function. If such a function is c D f.x; y/, then @c=@x is called the partia margina
c st ith respect t x and is the rate of change of c with respect to x when y is held
fixed. Similarly, @c=@y is the partia margina c st ith respect t y and is the rate
of change of c with respect to y when x is held fixed. It also follows that @c=@x.x; y/ is
approximately the cost of producing one more unit of when x units of and y units
of are produced. Similarly, @c=@y.x; y/ is approximately the cost of producing one
more unit of when x units of and y units of are produced.
For example, if c is expressed in dollars and @c=@y D 2, then the cost of producing
an extra unit of when the level of production of is fixed is approximately two
dollars.
If a manufacturer produces n products, the oint-cost function is a function of
n variables, and there are n (partial) marginal-cost functions.

E AM LE M C

A company manufactures two types of s is, the ightning and the Alpine models.
Suppose the oint-cost function for producing x pairs of the ightning model and
y pairs of the Alpine model per wee is
c D f.x; y/ D 0:07x2 C 75x C 5y C 6000
where c is expressed in dollars. etermine the marginal costs @c=@x and @c=@y when
x D 100 and y D 50, and interpret the results.

4 . . Swales, Advertising as an Intangible Asset: Profitability and Entry arriers: A Comment on Ree ie and
hoyrub, App ied Ec n mics 17, no. 4 (1 5), 603 17.
 Section 7.2 cat ons of Part a er at es 739

S The marginal costs are

@c @c
D 0:14x C 75 and D 5
@x @y
Thus,
ˇ
@c ˇˇ
D 0:14.100/ C 75 D
@x ˇ.100;50/

and
ˇ
@c ˇˇ
D 5
@y ˇ.100;50/

Equation (1) means that increasing the output of the ightning model from 100 to 101
while maintaining production of the Alpine model at 50 increases costs by approxi-
mately . Equation (2) means that increasing the output of the Alpine model from
50 to 51 and holding production of the ightning model at 100 will increase costs by
approximately 5. In fact, since @c=@y is a constant function, the marginal cost with
respect to y is 5 at all levels of production.
Now ork Problem 1 G

E AM LE L B H

n a cold day, a person may feel colder when the wind is blowing than when the wind
is calm because the rate of heat loss is a function of both temperature and wind speed.
The equation
p
D .10:45 C 10 ! /.33 ! t/

indicates the rate of heat loss, (in ilocalories per square meter per hour), when the
air temperature is t (in degrees Celsius) and the wind speed is (in meters per second).
For D 2000, exposed esh will freeze in one minute.5

a Evaluate when t D 0 and D 4.

S When t D 0 and D 4,
p
D .10:45 C 10 4 ! 4/.33 ! 0/ D 72: 5

b Evaluate @ =@ and @ =@t when t D 0 and D 4, and interpret the results.

S
# $ ˇ
@ 5 ˇ @
ˇ
p ! 1 .33 ! t/;
@
D ˇ t D 0 D 4 :5
@
D4
ˇ
@ p @ ˇˇ
D .10:45 C 10 ! /.!1/; D !26:45
@t @t ˇ t D 0
D4

These equations mean that when t D 0 and D 4, increasing by a small amount

while eeping t fixed will ma e increase approximately 4 .5 times as much as
increases. Increasing t by a small amount while eeping fixed will ma e decrease
approximately 26.45 times as much as t increases.

5 . E. Fol , r., extb k f En ir nmenta Physi gy 2nd ed. (Philadelphia: ea Febiger, 1 74).
740 C t ar a e Ca c s

c When t D 0 and D 4, which has a greater effect on : a change in wind speed of

1m s or a change in temperature of 1ı C
S Since the partial derivative of with respect to is greater in magnitude
than the partial with respect to t when t D 0 and D 4, a change in wind speed of
1m s has a greater effect on .
Now ork Problem 13 G
The output of a product depends on many factors of production. Among these may
be labor, capital, land, machinery, and so on. For simplicity, let us suppose that output
depends only on labor and capital. If the function P D f. ; k/ gives the output P when
the producer uses units of labor and k units of capital, then this function is called a
production function. We define the marginal productivity with respect to to be
@P=@ . This is the rate of change of P with respect to when k is held fixed. i ewise,
the marginal productivity with respect to k is @P=@k and is the rate of change of P
with respect to k when is held fixed.

E AM LE M
p
A manufacturer of a popular toy determines that the production function is P D k,
where is the number of labor-hours per wee and k is the capital (expressed in hundreds
of dollars per wee ) required for a wee ly production of P gross of the toy. ( ne gross
is 144 units.) etermine the marginal productivity functions, and evaluate them when
D 400 and k D 16. Interpret the results.
S Since P D . k/1=2 ,
@P 1 k
D . k/!1=2 k D p
@ 2 2 k
and
@P 1
D . k/!1=2 D p
@k 2 2 k
Evaluating these equations when D 400 and k D 16, we obtain
ˇ
@P ˇˇ 16 1
ˇ D p D
@ D 400 2 400.16/ 10
k D 16
and
ˇ
@P ˇˇ 400 5
ˇ D p D
@k D 400 2 400.16/ 2
k D 16
Thus, if D 400 and k D 16, increasing to 401 and holding k at 16 will increase
1
output by approximately 10 gross. ut if k is increased to 17 while is held at 400, the
5
output increases by approximately gross.
2
Now ork Problem 5 G
C C
Sometimes, two products may be related such that changes in the price of one of them
affect the demand for the other. A typical example is that of butter and margarine. If
such a relationship exists between products A and , then the demand for each product
is dependent on the prices of both. Suppose qA and q are the quantities demanded for
A and , respectively, and pA and p are their respective prices. Then both qA and q
are functions of pA and p :
qA D f.pA ; p / demand function for A
q D g.pA ; p / demand function for
 Section 7.2 cat ons of Part a er at es 741

We can find four partial derivatives:

@qA
the margina demand f r A ith respect t pA
@pA
@qA
the margina demand f r A ith respect t p
@p
@q
the margina demand f r ith respect t pA
@pA
@q
the margina demand f r ith respect t p
@p
Under typical conditions, if the price of is fixed and the price of A increases, then
the quantity of A demanded will decrease. Thus, @qA =@pA < 0. Similarly, @q =@p < 0.
However, @qA =@p and @q =@pA may be either positive or negative. If
@qA @q
>0 and > 0
@p @pA
then A and are said to be competitive products, also nown as substitutes. In this
situation, an increase in the price of causes an increase in the demand for A, if it is
assumed that the price of A does not change. Similarly, an increase in the price of A
causes an increase in the demand for when the price of is held fixed. utter and
margarine are examples of substitutes.
Proceeding to a different situation, we say that if
@qA @q
<0 and < 0
@p @pA
then A and are complementary products. In this case, an increase in the price of
causes a decrease in the demand for A if the price of A does not change. Similarly, an
increase in the price of A causes a decrease in the demand for when the price of
is held fixed. For example, cars and gasoline are complementary products. An increase
in the price of gasoline will ma e driving more expensive. Hence, the demand for cars
will decrease. And an increase in the price of cars will reduce the demand for gasoline.

E AM LE A C
C

The demand functions for products A and are each a function of the prices of A and
and are given by
p
50 3 p 75pA
qA D p and q D q
pA 3
p2

respectively. Find the four marginal-demand functions, and determine whether A and
are competitive products, complementary products, or neither.
S Writing qA D 50p!1=2
A p1=3 and q D 75pA p!2=3 , we have
# $
@qA 1 !3=2 1=3
D 50 ! pA p D !25p!3=2 A p1=3
@pA 2
# $
@qA !1=2 1 50 !1=2 !2=3
D 50pA p!2=3 D p p
@p 3 3 A
@q
D 75.1/p!2=3 D 75p!2=3
@pA
# $
@q 2 !5=3
D 75pA ! p D !50pA p!5=3
@p 3
742 C t ar a e Ca c s

Since pA and p represent prices, they are both positive. Hence, @qA =@p > 0 and
@q =@pA > 0. We conclude that A and are competitive products.
Now ork Problem 19 G

R BLEMS
F r the j int c st functi ns in Pr b ems nd the indicated where P is product, A is land, B is labor, C is improvements, is
margina c st at the gi en pr ducti n e e liquid assets, E is wor ing assets, and F is cash operating
@c expenses. Find the marginal productivities for labor and
c D 7x C 0:3y2 C 2y C 00I ; x D 20; y D 30 improvements.
@y
p @c Production Function Suppose a production function is
c D 2x x C y C 6000I ; x D 70; y D 74
@x k
given by P D .
c D 0:03.x C y/3 ! 0:6.x C y/2 C :5.x C y/ C 7700 3k C 5
@c a etermine the marginal productivity functions.
; x D 50; y D 0 1
@x b Show that when k D , the marginal productivities sum to .
F r the pr ducti n functi ns in Pr b ems and nd the
margina pr ducti ity functi ns @P=@k and @P=@ MBA Compensation In a study of success among
graduates with master of business administration ( A) degrees,
P D 15 k ! 3 2 C 5k2 C 500 it was estimated that for staff managers (which include
P D 2:527 0:314 k0:6 6 accountants, analysts, etc.), current annual compensation (in
dollars) was given by
Cobb Douglas Production Function In economics, a
Cobb ouglas production function is a production function of the
z D 43; 60 C 44 0x C 34 2y
form P D A ˛ kˇ , where A, ˛, and ˇ are constants and ˛ C ˇ D 1.
For such a function, show that
where x and y are the number of years of wor experience before
a @P=@ D ˛P= b @P=@k D ˇP=k and after receiving the A degree, respectively. Find @z=@x
@P @P and interpret your result.
c Ck D P. This means that summing the products of the
@ @k Status A person s general status Sg is believed to be a
marginal productivity of each factor and the amount of that factor
function of status attributable to education, Se , and status
results in the total product P.
attributable to income, Si , where Sg ; Se , and Si are represented
In Pr b ems qA and q are demand functi ns f r pr ducts numerically. If
A and B respecti e y In each case nd @qA =@pA ; @qA =@p ; p p
@q =@pA ; and @q =@p and determine hether A and B are Sg D 7 3 Se Si
c mpetiti e c mp ementary r neither
qA D 1500 ! 40pA C 3p I q D 00 C 5pA ! 20p determine @Sg =@Se and @Sg =@Si when Se D 125 and Si D 100, and
interpret your results.
qA D 20 ! pA ! 2p I q D 50 ! 2pA ! 3p
Reading Ease Sometimes we want to evaluate the degree
100 500 of readability of a piece of writing. Rudolf Flesch10 developed a
qA D p I q D p
pA p p 3 pA function of two variables that will do this, namely,
Canadian Manufacturing The production function for the
Canadian manufacturing industries for 1 27 is estimated by6 R D f . ; s/ D 206: 35 ! .1:015 C 0: 46s/
P D 33:0 0:46 k0:52 , where P is product, is labor, and k is capital.
Find the marginal productivities for labor and capital, and where R is called the reading ease sc re is the average number
evaluate when D 1 and k D 1. of words per sentence in 100-word samples, and s is the average
number of syllables in such samples. Flesch says that an article
Dairy Farming An estimate of the production function for for which R D 0 is practically unreadable, but one with
dairy farming in Iowa (1 3 ) is given by7 R D 100 is easy for any literate person. a Find @R=@ and
@R=@s. b Which is easier to read: an article for which D 0
P D A0:27 B0:01 C0:01 0:23 0:0
E F0:27 and s D s0 , or one for which D 0 C 1 and s D s0

Adapted from A. . Weinstein and V. Srinivasen, Predicting anagerial Suc-

cess of aster of usiness Administration ( . .A.) raduates, urna f
App ied Psych gy 5 , no. 2 (1 74), 207 12.
6
P. aly and P. ouglas, The Production Function for Canadian anufac- Adapted from R. . ei and . F. ee er, athematica S ci gy (Engle-
tures, urna f the American Statistica Ass ciati n 3 (1 43), 17 6. wood Cliffs, N : Prentice-Hall, Inc., 1 75).
7 10 R. Flesch,
. Tintner and . H. rownlee, Production Functions erived from Farm he Art f Readab e riting (New or : Harper Row Publishers,
Records, American urna f Agricu tura Ec n mics 26 (1 44), 566 71. Inc., 1 4 ).
 Section 7.2 cat ons of Part a er at es 743

Model for Voice The study of frequency of vibrations of a a Classify A and as competitive, complementary, or neither.
taut wire is useful in considering such things as an individual s b If the unit prices of A and are 1000 and 2000,
voice. Suppose respectively, estimate the change in the demand for A when the
r price of is decreased by 20 and the price of A is held constant.
1 # Demand The demand equations for related products A and
!D
b $% are given by
r r
where ! (a ree letter read omega ) is frequency, b is diameter, p pA
is length, % (a ree letter read rho ) is density, and # (a ree qA D 10 and q D 3 3
pA p
letter read tau ) is tension.11 Find @!=@b; @!=@ ; @!=@%, and
@!=@#. where qA and q are the quantities of A and demanded and pA
ra c Flow Consider the following tra c- ow situation. and p are the corresponding prices (in dollars) per unit.
n a highway where two lanes of tra c ow in the same a Find the values of the two marginal demands for product A
direction, there is a maintenance vehicle bloc ing the left lane. when pA D and p D 16.
(See Figure 17.3.) Two vehicles ( ead and f ing) are in the b If p were reduced to 14 from 16, with pA fixed at , use part
right lane with a gap between them. The subject vehicle can (a) to estimate the corresponding change in demand for product A.
choose either to fill or not to fill the gap. That decision may be
based not only on the distance x shown in the diagram but also on oint Cost Function A manufacturer s oint-cost function
other factors (such as the velocity of the f ing vehicle). A gap for producing qA units of product A and q units of product is
index g has been used in analyzing such a decision.12;13 The given by
greater the g-value, the greater is the propensity for the subject
q2A .q3 C qA /1=2
vehicle to fill the gap. Suppose cD C q1=2
A q
1=3
C 500
16
# $
x F ! S
gD ! 0:75 C where c is in dollars.
F 1 :2
a Find the marginal-cost function with respect to qA .
where x (in feet) is as before, F is the velocity of the f ing b Evaluate the marginal-cost function with respect to qA
vehicle (in feet per second), and S is the velocity of the when qA D 1 and q D . Round your answer to two
subject vehicle (in feet per second). From the diagram, it seems decimal places.
reasonable that if both F and S are fixed and x increases, then g c Use your answer to part (a) to estimate the change in
should increase. Show that this is true by applying calculus to the cost if production of product A is decreased from 1 to
function g. Assume that x; F , and S are positive. 17 units, while production of product is held constant
at units.
Subject Maintenance Elections For the congressional elections of 1 74, the
vehicle vehicle
Republican percentage, R, of the Republican emocratic vote in
Left lane
a district is given (approximately) by14
x
R D f .Er ; Ed ; Ir ; Id ; /
Following Lead Right lane
vehicle vehicle D 15:4725 C 2:5 45Er ! 0:0 04E2r ! 2:364 Ed
C 0:06 7E2d C 2:1 14Ir ! 0:0 12I2r
FIGURE
! 0: 0 6Id C 0:00 1I2d ! 0:0277Er Ir
Demand Suppose the demand equations for related C 0:04 3Ed Id C 0: 57 ! 0:0061 2

products A and are

Here Er and Ed are the campaign expenditures (in units of
16
qA D e!.pA Cp /
and q D 2 2
10,000) by Republicans and emocrats, respectively Ir and Id
pA p are the number of terms served in Congress, p us ne for the
Republican and emocratic candidates, respectively and is the
where qA and q are the number of units of A and demanded percentage of the two-party presidential vote that Richard Nixon
when the unit prices (in thousands of dollars) are pA and p , received in the district for 1 6 . The variable gives a measure of
respectively. Republican strength in the district.
a In the Federal Election Campaign Act of 1 74, Congress set a
11 limit of 1 ,000 on campaign expenditures. y analyzing
R. . Thrall, . A. ortimer, . R. Rebman, and R. F. aum, eds., S me
athematica de s in Bi gy rev. ed., Report No. 40241-R-7. Prepared at
@R=@Er , would you have advised a Republican candidate who
University of ichigan, 1 67. served nine terms in Congress to spend 1 ,000 on his or her
12 campaign
P. . Hurst, . Perchono , and E. . Seguin, Vehicle inematics and ap
Acceptance, urna f App ied Psych gy 52, no. 4 (1 6 ), 321 24.
13 . Perchono and P. . Hurst, Effect of ane-Closure Signals upon river
14
ecision a ing and Tra c Flow, urna f App ied Psych gy 52, no. 5 . Silberman and . ochum, The Role of oney in etermining Election
(1 6 ), 410 13. utcomes, S cia Science uarter y 5 , no. 4 (1 7 ), 671 2.
744 C t ar a e Ca c s

b Find the percentage above which the Nixon vote had a et qA be demand f r pr duct A and supp se that
negative effect on R that is, find when @R=@ < 0. ive your qA D qA .pA ; p / s that qA is the quantity f A demanded hen
answer to the nearest percent. the price per unit f A is pA and the price per unit f pr duct is
Sales After a new product has been launched onto p he partia e asticity f demand f r A ith respect t pA
the mar et, its sales volume (in thousands of units) is given by den ted "pA is de ned as "pA D .pA =qA /.@qA =@pA / he partia
e asticity f demand f r A ith respect t p den ted "p is
A C 450 de ned as "p D .p =qA /.@qA =@p / R ugh y speaking "pA is the
SD p
AC 2 rati f a percentage change in the quantity f A demanded t a
where is the time (in months) since the product was first percentage change in the price f A hen the price f is xed
introduced and A is the amount (in hundreds of dollars) spent each Simi ar y "p can be r ugh y interpreted as the rati f a
month on advertising. percentage change in the quantity f A demanded t a percentage
change in the price f hen the price f A is xed In
a Verify that the partial derivative of sales volume with respect
Pr b ems nd "pA and "p f r the gi en a ues f pA
to time is given by
and p
@S A2 ! 450
D qA D 1000 ! 50pA C 2p I pA D 2; p D 10
@ .A C 2 /3=2
qA D 60 ! 3pA ! 2p I pA D 5; p D 3
b Use the result in part (a) to predict the number of months that p
will elapse before the sales volume begins to decrease if the qA D 1000=.p2A p / pA D 2, p D
amount allocated to advertising is held fixed at 000 per month.

Objective H
o co te er or er art a If z D f.x; y/, then not only is z a function of x and y, but also fx and fy are each functions
er at es
of x and y, which may themselves have partial derivatives. If we can differentiate fx and
fy , we obtain second order partial derivatives of f. Symbolically,
fxx means . fx /x fxy means . fx /y
fyx means . fy /x fyy means . fy /y
In terms of @-notation,
# $ # $
@2 z @ @z @2 z @ @z
means means
@x2 @x @x @y @x @y @x

# $ # $
@2 z @ @z @2 z @ @z
means means
@x @y @x @y @y2 @y @y

For z D f .x; y/, fxy D @2 z=@y@x. Note that to find fxy , we first differentiate f with respect to x. For @2 z=@x @y, we first
differentiate with respect to y.
We can extend our notation beyond second-order partial derivatives. For example,
fxxy ( D @3 z=@y@x2 ) is a third-order partial derivative of f, namely, the partial deriva-
tive of fxx (D @2 z=@x2 ) with respect to y. The generalization of higher-order partial
derivatives to functions of more than two variables should be obvious.

E AM LE S

Find the four second-order partial derivatives of f.x; y/ D x2 y C x2 y2 .

S Since

fx .x; y/ D 2xy C 2xy2

we have
@
fxx .x; y/ D .2xy C 2xy2 / D 2y C 2y2
@x
 S er r er Part a er at es 745

and
@
fxy .x; y/ D .2xy C 2xy2 / D 2x C 4xy
@y
Also, since
fy .x; y/ D x2 C 2x2 y
we have
@ 2
fyy .x; y/ D .x C 2x2 y/ D 2x2
@y
and
@ 2
fyx .x; y/ D .x C 2x2 y/ D 2x C 4xy
@x

Now ork Problem 1 G

The derivatives fxy and fyx are called mixed partial derivatives. bserve in Exam-
ple 1 that fxy .x; y/ D fyx .x; y/. Under suitable conditions, mixed partial derivatives of a
function are equal that is, the order of differentiation is of no concern. ou may assume
that this is the case for all the functions that we consider.

E AM LE M
ˇ
@3 ˇˇ
Find the value of if D .2x C 3y C 4z/3 .
@z@y@x ˇ.1;2;3/
S
@ @
D 3.2x C 3y C 4z/2 .2x C 3y C 4z/
@x @x
D 6.2x C 3y C 4z/2
@2 @
D 6 " 2.2x C 3y C 4z/ .2x C 3y C 4z/
@y@x @y
D 36.2x C 3y C 4z/
@3
D 36 " 4 D 144
@z@y@x
Thus,
ˇ
@3 ˇˇ
D 144
@z@y@x ˇ.1;2;3/

Now ork Problem 3 G

R BLEMS
In Pr b ems nd the indicated partia deri ati es f .x; y; z/ D x2 y3 z4 fx .x; y; z/, fxz .x; y; z/, fzx .x; y; z/
3
f .x; y/ D 5x y fx .x; y/, fxy .x; y/, fyx .x; y/ p @z @2 z
z D ln x2 C y2 ,
f .x; y/ D 2x y C 6x2 y3 ! 3xy
3 2
fx .x; y/, fxx .x; y/ @y @y2
2
f .x; y/ D 7x C 3y fy .x; y/, fyy .x; y/, fyyx .x; y/ ln.x2 C 5/ @z @2 z
zD ,
y @x @y @x
f .x; y/ D .x2 C xy C y2 /.xy C x C y/ fx .x; y/, fxy .x; y/
2xy In Pr b ems nd the indicated a ue
f .x; y/ D e fy .x; y/, fyx .x; y/, fyxy .x; y/
If f .x; y; z/ D 5, find fyxxz .4; 3; !2/.
f .x; y/ D ln.x2 C y3 / C 5 fx .x; y/, fxx .x; y/, fxy .x; y/
2 If f .x; y; z/ D z2 .3x2 ! 4xy3 /, find fxyz .1; 2; 3/.
f .x; y/ D .x C y/ .xy/ fx .x; y/, fy .x; y/, fxx .x; y/, fyy .x; y/
746 C t ar a e Ca c s

If f . ; k/ D 3 3 k6 ! 2 2 k7 , find fk k .2; 1/. For f .x; y/ D x4 y4 C 3x3 y2 ! 7x C 4, show that

If f .x; y/ D x3 y2 C x2 y ! x2 y2 , find fxxy .2; 3/ and fxyx .2; 3/. fxyx .x; y/ D fxxy .x; y/
2 x
If f .x; y/ D y e C ln.xy/, find fxyy .1; 1/. For f .x; y/ D e x2 CxyCy2
, show that
If f .x; y/ D 2x3 C 3x2 y C 5xy2 C 7y3 , find fxy .2; 3/.
fxy .x; y/ D fyx .x; y/
Cost Function Suppose the cost, c, of producing qA units xy
of product A and q units of product is given by For f .x; y/ D e , show that

c D .3q2A C q3 C 4/1=3 fxx .x; y/ C fxy .x; y/ C fyx .x; y/ C fyy .x; y/
and the coupled demand functions for the products are given by D f .x; y/..x C y/2 C 2/
qA D 10 ! pA C p2 @2 @2
For D ln.x2 C y2 /, show that C D 0. For
@x2 @y2
and D ln.x2 C y2 C z2 /, show that
q D 20 C pA ! 11p @2 @2 @2 2
C C D 2
Find the value of @x2 @y2 @z2 x C y2 C z2
@2 c
@qA @q
when pA D 25 and p D 4.

Objective M M F
o sc ss re at e a a an re at e T
na to n cr t ca o nts an to
a t e secon er at e test for a We now extend the notion of relative maxima and minima (relative extrema) to
f nct on of t o ar a es functions of two variables.

A function z D f.x; y/ is said to have a relative maximum at the point .a; b/ if, for
all points .x; y/ in the plane that are su ciently close to .a; b/, we have
f.a; b/ # f.x; y/
For a relative minimum, we replace # by $ in Inequality (1).

To say that z D f.x; y/ has a relative maximum at .a; b/ means, geometrically,

that the point .a; b; f.a; b// on the graph of f is higher than or is as high as all other
points on the surface that are near .a; b; f.a; b//. In Figure 17.4(a), f has a relative
maximum at .a; b/. Similarly, the function f in Figure 17.4(b) has a relative minimum
when x D y D 0, which corresponds to a low point on the surface.

z z

Relative
maximum
point

Graph of f
Graph of f

a b (0, 0, 0)
y y
x x Relative
minimum
(a, b, 0) point
(a) (b)

FIGURE Relative extrema.

 Section 7.4 a a an n a for nct ons of o ar a es 747

z z

Surface
Tangent (a, b, f (a, b))
Tangent
Surface

Plane y = b

a a
b b
y y
x (a, b, 0) x Plane x = a (a, b, 0)

(a) (b)

FIGURE At relative extremum, fx .x; y/ D 0 and fy .x; y/ D 0.

Recall that in locating extrema for a function y D f.x/ of one variable, we examine
those values of x in the domain of f for which f 0 .x/ D 0 or f 0 .x/ does not exist. For
functions of two (or more) variables, a similar procedure is followed. However, for the
functions that concern us, extrema will not occur where a derivative does not exist, and
such situations will be excluded from our consideration.
Suppose z D f.x; y/ has a relative maximum at .a; b/, as indicated in Figure 17.5(a).
Then the curve where the plane y D b intersects the surface must have a relative maxi-
mum when x D a. Hence, the slope of the tangent line to the surface in the x-direction
must be 0 at .a; b/. Equivalently, fx .x; y/ D 0 at .a; b/. Similarly, on the curve where
the plane x D a intersects the surface Figure 17.5(b) , there must be a relative maxi-
mum when y D b. Thus, in the y-direction, the slope of the tangent to the surface must
be 0 at .a; b/. Equivalently, fy .x; y/ D 0 at .a; b/. Since a similar discussion applies to
a relative minimum, we can combine these results as follows:

R
If z D f.x; y/ has a relative maximum or minimum at .a; b/, and if both fx and fy are
defined for all points close to .a; b/, it is necessary that .a; b/ be a solution of the
system
%
fx .x; y/ D 0
fy .x; y/ D 0

Rule 1 does not imply that there must be A point, .a; b/, for which fx .a; b/ D fy .a; b/ D 0 is called a critical point of f. Thus,
an extremum at a critical point. ust as in from Rule 1, we infer that, to locate relative extrema for a function, we should examine
the case of functions of one variable, a its critical points.
critical point can give rise to a relative Two additional comments are in order: First, Rule 1, as well as the notion of a
maximum, a relative minimum, or critical point, can be extended to functions of more than two variables. For example,
neither. A critical point is only a
candidate for a relative extremum. to locate possible extrema for D f.x; y; z/, we would examine those points for which
x D y D z D 0. Second, for a function whose domain is restricted, a thorough
examination for absolute extrema would include a consideration of boundary points.

E AM LE F C

Find the critical points of the following functions.

a f.x; y/ D 2x2 C y2 ! 2xy C 5x ! 3y C 1.
S Since fx .x; y/ D 4x ! 2y C 5 and fy .x; y/ D 2y ! 2x ! 3, we solve the system
%
4x ! 2y C 5 D 0
!2x C 2y ! 3 D 0
This gives x D !1 and y D 12 . Thus, .!1; 12 / is the only critical point.
748 C t ar a e Ca c s

3
b f. ; k/ D C k3 ! k.
S
(
f . ; k/ D 3 2 ! k D 0
fk . ; k/ D 3k2 ! D 0

From Equation (2), k D 3 2 . Substituting for k in Equation (3) gives

0 D 27 4 ! D .27 3 ! 1/

Hence, either D 0 or D 13 . If D 0, then k D 0 if D 13 , then k D 13 . The critical

points are therefore .0; 0/ and . 13 ; 13 /.

c f.x; y; z/ D 2x2 C xy C y2 C 100 ! z.x C y ! 100/.

S Solving the system
8
< fx .x; y; z/ D 4x C y ! z D 0
fy .x; y; z/ D x C 2y ! z D 0
: f .x; y; z/ D !x ! y C 100 D 0
z

gives the critical point .25; 75; 175/, which the reader should verify.

Now ork Problem 1 G

E AM LE F C

Find the critical points of

f.x; y/ D x2 ! 4x C 2y2 C 4y C 7

S We have fx .x; y/ D 2x ! 4 and fy .x; y/ D 4y C 4. The system

%
2x ! 4 D 0
4y C 4 D 0

gives the critical point .2; !1/. bserve that we can write the given function as

f.x; y/ D x2 ! 4x C 4 C 2.y2 C 2y C 1/ C 1
D .x ! 2/2 C 2.y C 1/2 C 1

and f.2; !1/ D 1. Clearly, if .x; y/ ¤ .2; !1/, then f.x; y/ > 1. Hence, a relative
minimum occurs at .2; !1/. oreover, there is an abs ute minimum at .2; !1/, since
f.x; y/ > f.2; !1/ for a .x; y/ ¤ .2; !1/.
Now ork Problem 3 G
Although in Example 2 we were able to show that the critical point gave rise to a
relative extremum, in many cases this is not so easy to do. There is, however, a second-
derivative test that gives conditions under which a critical point will be a relative max-
imum or minimum. We state it now, omitting the proof.
 Section 7.4 a a an n a for nct ons of o ar a es 749

R S T F T
Suppose z D f.x; y/ has continuous partial derivatives fxx , fyy , and fxy at all points
.x; y/ near a critical point, .a; b/. et be the function defined by
.x; y/ D fxx .x; y/fyy .x; y/ ! . fxy .x; y//2
Then
if .a; b/ > 0 and fxx .a; b/ < 0, then f has a relative maximum at .a; b/
if .a; b/ > 0 and fxx .a; b/ > 0, then f has a relative minimum at .a; b/
if .a; b/ < 0, then f has a sadd e p int at .a; b/ (see Example 4)
if .a; b/ D 0, then no conclusion about an extremum at .a; b/ can be drawn,
and further analysis is required.

We remar that when .a; b/ > 0, the sign of fxx .a; b/ is necessarily the same as
the sign of fyy .a; b/. Thus, when .a; b/ > 0 we can test either fxx .a; b/ or fyy .a; b/,
whichever is easiest, to ma e the determination required in parts 1 and 2 of the second
derivative test.

E AM LE A S T

Examine f.x; y/ D x3 C y3 ! xy for relative maxima or minima by using the second-

derivative test.
S First we find critical points:
fx .x; y/ D 3x2 ! y fy .x; y/ D 3y2 ! x
In the same manner as in Example 1(b), solving fx .x; y/ D fy .x; y/ D 0 gives the critical
points .0; 0/ and . 13 ; 13 /. Now,
fxx .x; y/ D 6x fyy .x; y/ D 6y fxy .x; y/ D !1
Thus,
.x; y/ D .6x/.6y/ ! .!1/2 D 36xy ! 1
Since .0; 0/ D 36.0/.0/ ! 1 D !1 < 0, there is no relative extremum at .0; 0/.
Also, since . 13 ; 13 / D 36. 13 /. 13 / ! 1 D 3 > 0 and fxx . 13 ; 13 / D 6. 13 / D 2 > 0, there is a
& '
relative minimum at 13 ; 13 . At this point, the value of the function is

f. 13 ; 13 / D . 13 /3 C . 13 /3 ! . 13 /. 13 / D ! 27
1

Now ork Problem 7 G

E AM LE AS

Examine f.x; y/ D y2 ! x2 for relative extrema.

S Solving
fx .x; y/ D !2x D 0 and fy .x; y/ D 2y D 0
we get the critical point (0, 0). Now we apply the second-derivative test. At (0, 0), and
indeed at any point,
fxx .x; y/ D !2 fyy .x; y/ D 2 fxy .x; y/ D 0
750 C t ar a e Ca c s

The surface in Figure 17.6 is called a ecause .0; 0/ D .!2/.2/ ! .0/2 D !4 < 0, no relative extremum exists at .0; 0/. A
hyperbolic paraboloid. s etch of z D f.x; y/ D y2 ! x2 appears in Figure 17.6. Note that, for the surface curve
cut by the plane y D 0, there is a maximum at (0,0) but for the surface curve cut by the
plane x D 0, there is a minimum at (0,0). Thus, on the surface no relative extremum
can exist at the origin, although (0, 0) is a critical point. Around the origin the curve is
saddle shaped, and .0; 0/ is called a sadd e p int of f.
Now ork Problem 11 G

z = f(x, y) = y2 - x2

Saddle point
at (0, 0)

(0, 0)

x fx(0, 0) = fy(0, 0) = 0

FIGURE Saddle point.

E AM LE F R E

Examine f.x; y/ D x4 C .x ! y/4 for relative extrema.

S If we set

fx .x; y/ D 4x3 C 4.x ! y/3 D 0

and

fy .x; y/ D !4.x ! y/3 D 0

then, from Equation (5), we have x ! y D 0 equivalently, x D y. Substituting into

Equation (4) gives 4x3 D 0 equivalently, x D 0. Thus, x D y D 0, and (0,0) is the only
critical point. At (0,0),

fxx .x; y/ D 12x2 C 12.x ! y/2 D 0

fyy .x; y/ D 12.x ! y/2 D 0

and

fxy .x; y/ D !12.x ! y/2 D 0

Hence, .0; 0/ D 0, and the second-derivative test gives no information. However, for
all .x; y/ ¤ .0; 0/, we have f.x; y/ > 0, whereas f.0; 0/ D 0. Therefore, at (0, 0) the
graph of f has a low point, and we conclude that f has a relative (and absolute) minimum
at (0,0).
Now ork Problem 13 G
 Section 7.4 a a an n a for nct ons of o ar a es 751

A
In many situations involving functions of two variables, and especially in their appli-
cations, the nature of the given problem is an indicator of whether a critical point is
in fact a relative (or absolute) maximum or a relative (or absolute) minimum. In such
cases, the second-derivative test is not needed. ften, in mathematical studies of applied
problems, the appropriate second-order conditions are assumed to hold.

E AM LE M

et P be a production function given by

P D f. ; k/ D 0:54 2 ! 0:02 3 C 1: k2 ! 0:0 k3
where and k are the amounts of labor and capital, respectively, and P is the quantity
of output produced. Find the values of and k that maximize P.
S To find the critical points, we solve the system P D 0 and Pk D 0:
P D 1:0 ! 0:06 2 Pk D 3:7 k ! 0:27k2
D 0:06 .1 ! / D 0 D 0:27k.14 ! k/ D 0
D 0; D 1 k D 0; k D 14
There are four critical points: (0,0), (0,14), (1 ,0), and (1 ,14).
Now we apply the second-derivative test to each critical point. We have
P D 1:0 ! 0:12 Pkk D 3:7 ! 0:54k Pk D 0
Thus,
. ; k/ D P Pkk ! .P k /2
D .1:0 ! 0:12 /.3:7 ! 0:54k/
At (0,0),
.0; 0/ D 1:0 .3:7 / > 0

Since .0; 0/ > 0 and P D 1:0 > 0, there is a relative minimum at (0,0). At (0,14),
.0; 14/ D 1:0 .!3:7 / < 0

ecause .0; 14/ < 0, there is no relative extremum at (0,14). At (1 ,0),

.1 ; 0/ D .!1:0 /.3:7 / < 0

Since .1 ; 0/ < 0, there is no relative extremum at (1 ,0). At (1 ,14),

.1 ; 14/ D .!1:0 /.!3:7 / > 0

ecause .1 ; 14/ > 0 and P D !1:0 < 0, there is a relative maximum at (1 , 14).
Hence, the maximum output is obtained when D 1 and k D 14.
Now ork Problem 21 G

E AM LE M

A candy company produces two types of candy, A and , for which the average costs
of production are constant at 2 and 3 per pound, respectively. The quantities qA ; q
(in pounds) of A and that can be sold each wee are given by the oint-demand
functions
qA D 400.p ! pA /
752 C t ar a e Ca c s

and
q D 400. C pA ! 2p /
where pA and p are the selling prices (in dollars per pound) of A and , respectively.
etermine the selling prices that will maximize the company s profit, P.
S The total profit is given by
0 10 1 0 10 1
profit pounds profit pounds
P D @ per pound A @ of A A C @ per pound A @ of A
of A sold of sold

For A and , the profits per pound are pA ! 2 and p ! 3, respectively. Thus,
P D .pA ! 2/qA C .p ! 3/q
D .pA ! 2/Œ400.p ! pA /! C .p ! 3/Œ400. C pA ! 2p /!

Notice that P is expressed as a function of two variables, pA and p . To maximize P,

we set its partial derivatives equal to 0:
@P
D .pA ! 2/Œ400.!1/! C Œ400.p ! pA /!.1/ C .p ! 3/Œ400.1/!
@pA
D0
@P
D .pA ! 2/Œ400.1/! C .p ! 3/Œ400.!2/! C 400. C pA ! 2p /!.1/
@p
D0

Simplifying the preceding two equations gives

%
!2pA C 2p ! 1 D 0
2pA ! 4p C 13 D 0
whose solution is pA D 5:5 and p D 6. oreover, we find that
@2 P @2 P @2 P
D ! 00 D !1600 D 00
@p2A @p2 @p @pA
Therefore,
.5:5; 6/ D .! 00/.!1600/ ! . 00/2 > 0
Since @2 P=@p2A < 0, we indeed have a maximum, and the company should sell candy
A at 5.50 per pound and at 6.00 per pound.
Now ork Problem 23 G

R BLEMS
In Pr b ems nd the critica p ints f the functi ns In Pr b ems nd the critica p ints f the functi ns F r
2 2
f .x; y/ D x ! 3y ! x C y C 3xy each critica p int determine by the sec nd deri ati e test
hether it c rresp nds t a re ati e maximum t a re ati e
f .x; y/ D x2 C 3y2 ! 4x ! 30y minimum r t neither r hether the test gi es n inf rmati n
5 2 15 f .x; y/ D x2 C 4y2 ! 6x ! 32y C 1
f .x; y/ D x3 C y3 ! x2 C y2 ! 4y C 7
3 3 2
f .x; y/ D !2x2 C x ! 3y2 C 24y C 7
f .x; y/ D xy ! x C y
f .x; y/ D y ! y2 ! 3x ! 6x2
f .x; y; z/ D 2x2 C xy C y2 C 100 ! z.x C y ! 200/
3
f .x; y; z; / D x2 C y2 C z2 C .x C y C z ! 3/ f .x; y/ D 2x2 C y2 C 3xy ! 10x ! y C 2
2
 Section 7.4 a a an n a for nct ons of o ar a es 753

f .x; y/ D x2 C 3xy C y2 ! x ! 11y C 3 Profit A monopolist sells two competitive products, A and
, for which the demand functions are
f .x; y/ D 2x3 C 3y2 C 6xy C 6x C 6y
1 3 qA D 16 ! pA C p and q D 24 C 2pA ! 4p
f .x; y/ D .x C y3 / ! 2.x2 C y2 / C 1
3 If the constant average cost of producing a unit of A is 2 and a
f .x; y/ D x2 C y2 ! xy C x3 unit of is 4, how many units of A and should be sold to
2 maximize the monopolist s profit
f . ; k/ D C 2 k C 3k2 ! 6 ! 164k C 17 Profit For products A and , the oint-cost function for a
2
2 1 1 manufacturer is
f . ; k/ D C 4k2 ! 4 k f .x; y/ D xy ! !
x y
c D q2A C 6q2
f .x; y/ D .x ! 3/.y ! 3/.x C y ! 3/
and the demand functions are pA D 1 ! q2A and p D 0 ! 2q2 .
f .x; y/ D .y2 ! 4/.ex ! 1/ Find the level of production that maximizes profit.
f .x; y/ D ln.xy/ C 2x2 ! xy ! 6x Profit For a monopolist s products A and , the oint-cost
Maximi ing Output Suppose function is c D 2.qA C q C qA q /, and the demand functions are
qA D 20 ! 2pA and q D 10 ! p . Find the values of pA and p
2
P D f . ; k/ D 2:1 ! 0:02 3 C 1: 7k2 ! 0:03k3 that maximize profit. What are the quantities of A and that
correspond to these prices What is the total profit
is a production function for a firm. Find the quantities of inputs
and k that maximize output P. Cost An open-top rectangular box is to have a volume of
6 ft3 . The cost per square foot of materials is 3 for the bottom, 1
Maximi ing Output In a certain o ce, computers C and for the front and bac , and 0.50 for the other two sides. Find the
are utilized for c and d hours, respectively. If daily output dimensions of the box so that the cost of materials is minimized.
is a function of c and d, namely, (See Figure 17.7.)
D 10c C 20d ! 3c2 ! 4d2 ! cd
find the values of c and d that maximize .
In Pr b ems un ess ther ise indicated the ariab es pA
and p den te se ing prices f pr ducts A and respecti e y x = width
y = length
Simi ar y qA and q den te quantities f A and that are y z = height
pr duced and s d during s me time peri d In a cases the z
ariab es emp yed i be assumed t be units f utput input
m ney and s n x
Front
Profit A candy company produces two varieties of candy,
A and , for which the constant average costs of production are FIGURE
60 and 70 (cents per lb), respectively. The demand functions for
A and are given by Collusion Suppose A and are the only two firms in the
mar et selling the same product. (We say that they are du p ists.)
qA D 5.p ! pA / and q D 500 C 5.pA ! 2p /
The industry demand function for the product is
Find the selling prices pA and p that maximize the company s
profit. p D 2 ! qA ! q
Profit Repeat Problem 23 if the constant costs of where qA and q denote the output produced and sold by A and ,
production of A and are a and b (cents per lb), respectively. respectively. For A, the cost function is cA D 10qA for , it is
Price Discrimination Suppose a monopolist is practicing c D 0:5q2 . Suppose the firms decide to enter into an agreement
price discrimination in the sale of a product by charging different on output and price control by ointly acting as a monopoly. In
prices in two separate mar ets. In mar et A the demand function is this case, we say they enter into c usi n. Show that the profit
function for the monopoly is given by
pA D 100 ! qA
P D pqA ! cA C pq ! c
and in it is
Express P as a function of qA and q , and determine how output
p D 4!q should be allocated so as to maximize the profit of the monopoly.
where qA and q are the quantities sold per wee in A and , and Suppose f .x; y/ D x2 C 3y2 C , where x and y must satisfy
pA and p are the respective prices per unit. If the monopolist s the equation x C y D 2. Find the relative extrema of f, sub ect to
cost function is the given condition on x and y, by first solving the second
equation for y (or x). Substitute the result in the first equation.
c D 600 C 4.qA C q / Thus, f is expressed as a function of one variable. Now find where
how much should be sold in each mar et to maximize profit relative extrema for f occur.
What selling prices give this maximum profit Find the maximum Repeat Problem 31 if f .x; y/ D x2 C 4y2 C 11, sub ect to the
profit. condition that x ! y D 1.
754 C t ar a e Ca c s

Suppose the oint-cost function where x and y represent his wee ly expenditures (in dollars) on
newspaper and radio advertising, respectively. The profit is 300
c D q2A C 3q2 C 2qA q C aqA C bq C d
per sale, less the cost of advertising, so the wee ly profit is given
has a relative minimum value of 15 when qA D 3 and q D 1. by the formula
etermine the values of the constants a, b, and d. # $
7x 4y
P D 300 C !x!y
Suppose that the function f .x; y/ has continuous partial 2Cx 5Cy
derivatives fxx , fyy , and fxy at all points .x; y/ near a critical point
.a; b/. et .x; y/ D fxx .x; y/fyy .x; y/ ! . fxy .x; y//2 and suppose Find the values of x and y for which the profit is a relative
that .a; b/ > 0. maximum. Use the second-derivative test to verify that your
answer corresponds to a relative maximum profit.
a Show that fxx .a; b/ < 0 if and only if fyy .a; b/ < 0.
b Show that fxx .a; b/ > 0 if and only if fyy .a; b/ > 0. Profit from omato Crop The revenue (in dollars per
square meter of ground) obtained from the sale of a crop of
Profit from Competitive Products A monopolist
tomatoes grown in an artificially heated greenhouse is given by
sells two competitive products, A and , for which the demand
equations are r D 5 .1 ! e!x /
pA D 35 ! 2q2A C q
and where is the temperature (in ı C) maintained in the greenhouse
p D 20 ! q C qA and x is the amount of fertilizer applied per square meter. The cost
of fertilizer is 20x dollars per square meter, and the cost of heating
The oint-cost function is is given by 0:1 2 dollars per square meter.
1
c D ! ! 2q3A C 3qA q C 30qA C 12q C q2A a Find an expression, in terms of and x, for the profit per
2 square meter obtained from the sale of the crop of tomatoes.
a How many units of A and should be sold to obtain a relative b Verify that the pairs
maximum profit for the monopolist Use the second-derivative
test to ustify your answer. . ; x/ D .20; ln 5/ and . ; x/ D .5; ln 54 /
b etermine the selling prices required to realize the relative
maximum profit. Also, find this relative maximum profit. are critical points of the profit function in part (a). ( te ou
need not derive the pairs.)
Profit and Advertising A retailer has determined that the c The points in part (b) are the only critical points of the profit
number of TV sets he can sell per wee is function in part (a). Use the second-derivative test to determine
7x 4y whether either of these points corresponds to a relative maximum
C profit per square meter.
2Cx 5Cy

Objective L M
o n cr t ca o nts for a f nct on We will now find relative maxima and minima for a function on which certain c n
s ect to constra nts a n t e
et o of a ran e t ers straints are imposed. Such a situation could arise if a manufacturer wished to minimize
a oint-cost function and yet obtain a particular production level.
Suppose we want to find the relative extrema of
D x2 C y2 C z2
sub ect to the constraint that x, y, and z must satisfy
x ! y C 2z D 6
We can transform , which is a function of three variables, into a function of two
variables such that the new function re ects constraint (2). Solving Equation (2) for x,
we get
x D y ! 2z C 6
which, when substituted for x in Equation (1), gives
D .y ! 2z C 6/2 C y2 C z2
Since is now expressed as a function of two variables, to find relative extrema we
follow the usual procedure of setting the partial derivatives of equal to 0:
@
D 2.y ! 2z C 6/ C 2y D 4y ! 4z C 12 D 0
@y
@
D !4.y ! 2z C 6/ C 2z D !4y C 10z ! 24 D 0
@z
 Section 7.5 a ran e t ers 755

Solving Equations (5) and (6) simultaneously gives y D !1 and z D 2. Substituting

into Equation (3), we get x D 1. Hence, the only critical point of Equation (1) sub ect to
the constraint represented by Equation (2) is .1; !1; 2/. y using the second-derivative
test on Equation (4) when y D !1 and z D 2, we have
@2 @2 @2
D4 D 10 D !4
@y2 @z2 @z @y
.!1; 2/ D 4.10/ ! .!4/2 D 24 > 0

Thus , sub ect to the constraint, has a relative minimum at .1; !1; 2/.
This solution was found by using the constraint to express one of the variables in the
original function in terms of the other variables. ften this is not practical, but there is
another technique, called the method of Lagrange multipliers, after the French math-
ematician oseph- ouis agrange (1736 1 13), that avoids this step and yet allows us
to obtain critical points.
The method is as follows. Suppose we have a function f.x; y; z/ sub ect to the con-
straint g.x; y; z/ D 0. We construct a new function, F, of f ur variables defined by the
following (where & is a ree letter read lambda ):

F.x; y; z; &/ D f.x; y; z/ ! &g.x; y; z/

It can be shown that if .a; b; c/ is a critical point of f, sub ect to the constraint
g.x; y; z/ D 0, there exists a value of &, say, &0 , such that .a; b; c; &0 / is a critical
point of F. The number &0 is called a Lagrange multiplier. Also, if .a; b; c; &0 / is a
critical point of F, then .a; b; c/ is a critical point of f, sub ect to the constraint. Thus,
to find critical points of f, sub ect to g.x; y; z/ D 0, we instead find critical points of F.
These are obtained by solving the simultaneous equations
8̂
F .x; y; z; &/ D 0
< x
Fy .x; y; z; &/ D 0
:̂Fz .x; y; z; &/ D 0
F! .x; y; z; &/ D 0
At times, ingenuity must be used to solve the equations. nce we obtain a critical point
.a; b; c; &0 / of F, we can conclude that .a; b; c/ is a critical point of f, sub ect to the
constraint g.x; y; z/ D 0. Although f and g are functions of three variables, the method
of agrange multipliers can be extended to n variables.
et us illustrate the method of agrange multipliers for the original situation, namely,

f.x; y; z/ D x2 C y2 C z2 sub ect to x ! y C 2z D 6

First, we write the constraint as g.x; y; z/ D x ! y C 2z ! 6 D 0. Second, we form the

function

F.x; y; z; &/ D f.x; y; z/ ! &g.x; y; z/

D x2 C y2 C z2 ! &.x ! y C 2z ! 6/

Next, we set each partial derivative of F equal to 0. For convenience, we will write
Fx .x; y; z; &/ as Fx , and so on:
8̂
ˆ Fx D 2x ! & D 0
ˆ
ˆ
ˆ
< Fy D 2y C & D 0
ˆ
ˆ Fz D 2z ! 2& D 0
ˆ
ˆ
:̂F! D !x C y ! 2z C 6 D 0

From Equations (7) ( ), we see immediately that

& &
xD yD! zD&
2 2
756 C t ar a e Ca c s

Substituting these values into Equation (10), we obtain

& &
! ! ! 2& C 6 D 0
2 2
!3& C 6 D 0
&D2
Thus, from Equation (11),
x D 1 y D !1 z D 2
Hence, the only critical point of f, sub ect to the constraint, is .1; !1; 2/, at which there
may exist a relative maximum, a relative minimum, or neither of these. The method
of agrange multipliers does not directly indicate which of these possibilities occurs,
although from our previous wor , we saw that .1; !1; 2/ is indeed a relative mini-
mum. In applied problems, the nature of the problem itself may give a clue as to how
a critical point is to be regarded. ften the existence of either a relative minimum or
a relative maximum is assumed, and a critical point is treated accordingly. Actually,
su cient second-order conditions for relative extrema are available, but we will not
consider them.

E AM LE M L M

Find the critical points for z D f.x; y/ D 3x!yC6, sub ect to the constraint x2 Cy2 D 4.
S We write the constraint as g.x; y/ D x2 C y2 ! 4 D 0 and construct the
function
F.x; y; &/ D f.x; y/ ! &g.x; y/ D 3x ! y C 6 ! &.x2 C y2 ! 4/
Setting Fx D Fy D F! D 0, we have
8̂
ˆ
ˆ 3 ! 2x& D 0
<
! 1 ! 2y& D 0
ˆ
ˆ 2
:̂!x ! y2 C 4 D 0
From Equations (12) and (13), we can express x and y in terms of &. Then we will
substitute for x and y in Equation (14) and solve for &. nowing &, we can find x and
y. To begin, from Equations (12) and (13), we have
3 1
xD and y D !
2& 2&
Substituting into Equation (14), we obtain
1
! ! C4D0
4&2 4&2
10
! 2 C4D0
4&
p
10
&D˙
4
p
With these &-values, we can find x and y. If & D 10=4, then
p p
3 3 10 1 10
xD p !D 5 yD! p ! D! 5
10 10
2 2
4 4
p
Similarly, if & D ! 10=4,
p p
3 10 10
xD! yD
5 5
 Section 7.5 a ran e t ers 757
p p
Thus, the critical points of f, sub ect to the constraint, are .3 10=5; ! 10=5/ and
p p
.!3 10=5; 10=5/. Note that the values of & do not appear in the answer they are
simply a means to obtain the solution.
Now ork Problem 1 G

E AM LE M L M

Find critical points for f.x; y; z/ D xyz, where xyz ¤ 0, sub ect to the constraint x C
2y C 3z D 36.
S We have
F.x; y; z; &/ D xyz ! &.x C 2y C 3z ! 36/
Setting Fx D Fy D Fz D F! D 0 gives, respectively,
8̂
yz ! & D 0
<
xz ! 2& D 0
:̂xy ! 3& D 0
!x ! 2y ! 3z C 36 D 0
ecause we cannot directly express x, y, and z in terms of & only, we cannot follow the
procedure in Example 1. However, observe that we can express the products yz, xz, and
xy as multiples of &. This suggests that, by loo ing at quotients of equations, we can
obtain a relation between two variables that does not involve &. (The & s will cancel.)
Proceeding to do this, we write the foregoing system as
8̂ yz D &
ˆ
ˆ
ˆ
< xz D 2&
ˆ
ˆ xy D 3&
ˆ
:̂
x C 2y C 3z ! 36 D 0
ividing each side of Equation (15) by the corresponding side of Equation (16), we get
yz & x
D so y D
xz 2& 2
This division is valid, since xyz ¤ 0. Similarly, from Equations (15) and (17), we get
yz & x
D so z D
xy 3& 3
Now that we have y and z expressed in terms of x only, we can substitute into
Equation (1 ) and solve for x:
(x) (x)
xC2 C3 ! 36 D 0
2 3
x D 12
Thus, y D 6 and z D 4. Hence, (12, 6, 4) is the only critical point satisfying the given
conditions. Note that in this situation, we found the critical point without having to find
the value for &.
Now ork Problem 7 G

E AM LE M C

Suppose a firm has an order for 200 units of its product and wishes to distribute its
manufacture between two of its plants, plant 1 and plant 2. et q1 and q2 denote the
outputs of plants 1 and 2, respectively, and suppose the total-cost function is given by
758 C t ar a e Ca c s

c D f.q1 ; q2 / D 2q21 C q1 q2 C q22 C 200. How should the output be distributed in order
to minimize costs
S We minimize c D f.q1 ; q2 /, given the constraint q1 C q2 D 200. We have

F.q1 ; q2 ; &/ D 2q21 C q1 q2 C q22 C 200 ! &.q1 C q2 ! 200/

8̂
@F
ˆ
ˆ D 4q1 C q2 ! & D 0
ˆ
ˆ @q
ˆ
ˆ
1
ˆ
ˆ
< @F
D q1 C 2q2 ! & D 0
ˆ
ˆ @q2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ @F
:̂ D !q1 ! q2 C 200 D 0
@&
We can eliminate & from Equations (1 ) and (20) and obtain a relation between q1 and
q2 . Then, solving this equation for q2 in terms of q1 and substituting into Equation (21),
we can find q1 . We begin by subtracting Equation (20) from Equation (1 ), which gives

3q1 ! q2 D 0 so q2 D 3q1

Substituting into Equation (21), we have

!q1 ! 3q1 C 200 D 0

!4q1 D !200
q1 D 50

Thus, q2 D 150. Accordingly, plant 1 should produce 50 units and plant 2 should
produce 150 units in order to minimize costs.
Now ork Problem 13 G
An interesting observation can be made concerning Example 3. From Equation (1 ),
& D 4q1 C q2 D @c=@q1 , the marginal cost of plant 1. From Equation (20),
& D q1 C 2q2 D @c=@q2 , the marginal cost of plant 2. Hence, @c=@q1 D @c=@q2 ,
and we conclude that, to minimize cost, it is necessary that the marginal costs of each
plant be equal to each other.

E AM LE L C I C

Suppose a firm must produce a given quantity, P0 , of output in the cheapest possible
manner. If there are two input factors, and k, and their prices per unit are fixed at p
and pk , respectively, discuss the economic significance of combining input to achieve
least cost. That is, describe the least-cost input combination.
S et P D f. ; k/ be the production function. Then we must minimize the cost
function

c D p C kpk

sub ect to

P0 D f. ; k/

We construct

F. ; k; &/ D p C kpk ! &. f. ; k/ ! P0 /

 Section 7.5 a ran e t ers 759

We have
8̂
ˆ @F @
ˆ
ˆ D p ! & . f. ; k// D 0
ˆ
ˆ @ @
ˆ
< @F @
D pk ! & . f. ; k// D 0
ˆ
ˆ @k @k
ˆ
ˆ
ˆ
ˆ @F
:̂ D !f. ; k/ C P0 D 0
@&
From Equations (22) and (23),
p pk
&D D
@ @
. f. ; k// . f. ; k//
@ @k
Hence,

@
. f. ; k//
p @
D
pk @
. f. ; k//
@k
We conclude that when the least-cost combination of factors is used, the ratio of the
marginal productivities of the input factors must be equal to the ratio of their corre-
sponding unit prices.
Now ork Problem 15 G

M C
The method of agrange multipliers is by no means restricted to problems involv-
ing a single constraint. For example, suppose f.x; y; z; / were sub ect to constraints
g1 .x; y; z; / D 0 and g2 .x; y; z; / D 0. Then there would be two lambdas, &1 and &2
(one corresponding to each constraint), and we would construct the function
F D f ! &1 g1 ! &2 g2 . We would then solve the system

Fx D Fy D Fz D F D F!1 D F!2 D 0

E AM LE M L M T C

Find critical points for f.x; y; z/ D xy C yz, sub ect to the constraints x2 C y2 D and
yz D .
S Set

F.x; y; z; &1 ; &2 / D xy C yz ! &1 .x2 C y2 ! / ! &2 .yz ! /

Then
8̂
ˆ
ˆ Fx D y ! 2x&1 D 0
ˆ
ˆ
ˆ
ˆ F D x C z ! 2y&1 ! z&2 D 0
ˆ
< y
Fz D y ! y&2 D 0
ˆ
ˆ
ˆ
ˆ F! D !x2 ! y2 C D 0
ˆ
ˆ
ˆ 1
:̂F D !yz C D 0
!2
760 C t ar a e Ca c s

This appears to be a challenging system to solve. Some ingenuity will come into play.
Here is one sequence of operations that will allow us to find the critical points. We can
write the system as
8̂ y
ˆ
ˆ D &1
ˆ
ˆ 2x
ˆ
ˆ
ˆ x C z ! 2y&1 ! z&2 D 0
<
&2 D 1
ˆ
ˆ
ˆ
ˆ x2 C y2 D
ˆ
ˆ
ˆ
:̂
zD
y
In deriving Equation (30) we assumed x ¤ 0. This is permissible because if x D 0,
then by Equation (25) we have also y D 0, which is impossible because the second con-
straint, yz D , provides y ¤ 0. We also used y ¤ 0 to derive Equations (32) and (34).
Substituting &2 D 1 from Equation (32) into Equation (31) and simplifying gives
the equation x ! 2y&1 D 0, so
x
&1 D
2y
Substituting into Equation (30) gives
y x
D
2x 2y
y2 D x2
Substituting into Equation (33) gives x2 Cx2 D , from which it follows that x D ˙2. If
x D 2, then, from Equation (35), we have y D ˙2. Similarly, if x D !2, then y D ˙2.
Thus, if x D 2 and y D 2, then, from Equation (34), we obtain z D 4. Continuing in
this manner, we obtain four critical points:
.2; 2; 4/ .2; !2; !4/ .!2; 2; 4/ .!2; !2; !4/

Now ork Problem 9 G

R BLEMS
In Pr b ems nd by the meth d f agrange mu tip iers the Production Allocation To fill an order for 100 units of its
critica p ints f the functi ns subject t the gi en c nstraints product, a firm wishes to distribute production between its two
plants, plant 1 and plant 2. The total-cost function is given by
f .x; y/ D x2 C 4y2 C 6I 2x ! y D 20
c D f .q1 ; q2 / D q21 C 3q1 C 25q2 C 1000
f .x; y/ D 3x2 ! 2y2 C xCyD1
where q1 and q2 are the numbers of units produced at plants 1
f .x; y; z/ D x2 C y2 C z2 xCyCzD1 and 2, respectively. How should the output be distributed in order
to minimize costs (Assume that the critical point obtained
f .x; y; z/ D x C y C zI xyz D corresponds to the minimum cost.)
f .x; y; z/ D 2x2 C xy C y2 C zI x C 2y C 4z D 3 Production Allocation Repeat Problem 13 if the cost
function is
f .x; y; z/ D xyz2 I x ! y C z D 20 .xyz2 ¤ 0/ c D 3q21 C q1 q2 C 2q22
f .x; y; z/ D xyz x C y C z D 1 (xyz ¤ 0) and a total of 200 units are to be produced.
2
f .x; y; z/ D x C 4y C z 2 2
xCyCzD3 Maximi ing Output The production function for a firm is
2
f .x; y; z/ D x2 C 2y ! z2 I 2x ! y D 0; y C z D 0 f . ; k/ D 12 C 20k ! ! 2k2
The cost to the firm of and k is 4 and per unit, respectively. If
f .x; y; z/ D x2 C y2 C z2 I x C y C z D 4; x ! y C z D 4
the firm wants the total cost of input to be , find the greatest
f .x; y; z/ D xy2 zI x C y C z D 1; x ! y C z D 0 .xyz ¤ 0/ output possible, sub ect to this budget constraint. ( ou may
assume that the critical point obtained does correspond to the
f .x; y; z; / D x2 C 2y2 C 3z2 ! 2
4x C 3y C 2z C D 10 maximum output.)
 Section 7.6 t e nte ra s 761

Maximi ing Output Repeat Problem 15, given that c The profit may be considered a function of , k, and q (that is,
2 2 P D 64q ! ! 16k), sub ect to the constraint
f . ; k/ D 20 C 25k ! ! 3k
and the budget constraint is 2 C 4k D 50. 16q D 65 ! 4. ! 4/2 ! 2.k ! 5/2
Advertising Budget A computer company has a monthly
advertising budget of 20,000. Its mar eting department estimates Use the method of agrange multipliers to find all critical points
that if x dollars are spent each month on advertising in newspapers of P D 64q ! ! 16k, sub ect to the constraint.
and y dollars per month on television advertising, then the Pr b ems refer t the f ing de niti n A utility function
monthly sales will be given by S D 0x1=4 y3=4 dollars. If the is a functi n that attaches a measure t the satisfacti n r uti ity a
profit is 10 of sales, less the advertising cost, determine how to c nsumer gets fr m the c nsumpti n f pr ducts per unit f time
allocate the advertising budget in order to maximize the monthly Supp se D f .x; y/ is such a functi n here x and y are the
profit. ( ou may assume that the critical point obtained does am unts f t pr ducts and he marginal utility f is
correspond to the maximum profit.) @ =@x and appr ximate y represents the change in t ta uti ity
Maximi ing Production When units of labor and k units resu ting fr m a ne unit change in c nsumpti n f pr duct per
of capital are invested, a manufacturer s total production, q, is unit f time e de ne the margina uti ity f simi ar y If the
given by the Cobb- ouglas production function, P D 3=5 k2=5 . prices f and are p and p respecti e y and the c nsumer
Each unit of labor costs 50, and each unit of capital costs 3 . If has an inc me r budget f I t spend then the budget
exactly 30,750 is to be spent on production, determine the c nstraint is
numbers of units of labor and capital that should be invested to
maximize production. (Assume that the maximum occurs at the xp C yp D I
critical point obtained.)
Political Advertising Newspaper advertisements for In Pr b ems nd the quantities f each pr duct that the
political parties always have some negative effects. The recently c nsumer sh u d buy subject t the budget that i a
elected party assumed that the three most important election maximum satisfacti n hat is in Pr b ems and nd a ues
issues, X, , and , had to be mentioned in each ad, with space x, f x and y that maximize D f .x; y/ subject t xp C yp D I
y, and z units, respectively, allotted to each. The combined bad Perf rm a simi ar pr cedure f r Pr b em Assume that such a
effect of this coverage was estimated by the party s bac room maximum exists
operative as D x3 y3 I pX D 2; p D 3; I D 4 .x3 y3 ¤ 0/
2 2 2
B.x; y; z/ D x C y C 2z D 40x ! x2 C 2y ! y2 pX D 4, p D 6, I D 100
Aesthetics dictated that the total space for X and together must D f .x; y; z/ D xyz p D 1 p D 2 p D 3 I D 100
be 20, and realism suggested that the total space allotted to and .xyz ¤ 0/
together must also be 20 units. What values of x, y, and z in each
et D f .x; y/ be a utility function sub ect to the budget
ad would produce the lowest negative effect ( ou may assume
constraint xp C yp D I, where pX , p , and I are constants. Show
that any critical point obtained provides the minimum effect.)
that, to maximize satisfaction, it is necessary that
Maximi ing Profit Suppose a manufacturer s production
function is given by fx .x; y/ fy .x; y/
&D D
16q D 65 ! 4. ! 4/2 ! 2.k ! 5/2 p p
and the cost to the manufacturer is per unit of labor and 16 where fx .x; y/ and fy .x; y/ are the marginal utilities of and ,
per unit of capital, so that the total cost (in dollars) is C 16k. respectively. Show that fx .x; y/=p is the marginal utility of one
The selling price of the product is 64 per unit. dollar s worth of X. Hence, maximum satisfaction is obtained
a Express the profit as a function of and k. ive your answer in when the consumer allocates the budget so that the marginal
expanded form. utility of a dollar s worth of X is equal to the marginal utility per
b Find all critical points of the profit function obtained in dollar s worth of . Performing the same procedure as that for
part (a). Apply the second-derivative test at each critical point. If D f .x; y/, verify that this is true for D f .x; y; z; /, sub ect to
the profit is a relative maximum at a critical point, compute the the corresponding budget equation. In each case, & is called the
corresponding relative maximum profit. margina uti ity f inc me.

Objective M I
o co te o e an tr e nte ra s Recall that the definite integral of a function of one variable is concerned with inte-
gration over an inter a . There are also definite integrals of functions of two variables,
called (definite) double integrals. These involve integration over a regi n in the plane.
For example, the symbol
Z 2 Z 4 Z 2 #Z 4 $
xydxdy D xydx dy
0 3 0 3
762 C t ar a e Ca c s

y is the double integral of f.x; y/ D xy over a region determined by the bounds of inte-
gration. That region consists of all points .x; y/ in the x; y-plane such that 3 $ x $ 4
and 0 $ y $ 2. (See Figure 17. .) P
2
A double integral is a limit of a sum of the form f.x; y/dxdy, where, in this
example, the points .x; y/ are in the shaded region. A geometric interpretation of a
double integral will be given later.
x To evaluate
3 4 Z 2Z 4 Z 2 #Z 4 $
FIGURE
R2R4 Region over which xydxdy D xydx dy
0 3 0 3
0 3 xydxdy is evaluated.
we use successive integrations starting with the innermost integral. First, we evaluate
Z 4
xydx
3

by treating y as a constant and integrating with respect to x between the bounds 3 and 4:
Z 4 ˇ4
x2 y ˇˇ
xydx D
3 2 ˇ3
Substituting the limits for the variable x, we have
42 " y 32 " y 16y y 7
! D ! D y
2 2 2 2 2
Now we integrate this result with respect to y between the bounds 0 and 2:
Z 2 ˇ2
7 7y2 ˇˇ 7 " 22
ydy D D !0D7
0 2 4 ˇ0 4
Thus,
y
Z 2 Z 4
xydxdy D 7
0 3
1
y = x2
Now consider the double integral
Z 1 Z x2 Z Z !
1 x2
3
y = x3 .x ! xy/dydx D .x3 ! xy/dy dx
0 x3 0 x3

1
x Here, we integrate first with respect to y and then with respect to x. The region over
which the integration ta es places is all points .x; y/ such that x3 $ y $ x2 and 0 $ x $ 1.
FIGURE Region over which
R 1 R x2 3 (See Figure 17. .) This double integral is evaluated by first treating x as a constant and
0 x3 .x ! xy/dydx is evaluated. integrating x3 ! xy with respect to y between x3 and x2 , and then integrating the result
with respect to x between 0 and 1:

Z Z Z Z ! Z # $ˇx2
1 x2
3
1 x2
3
1
3 xy2 ˇˇ
.x ! xy/dydx D .x ! xy/dy dx D x y! dx
0 x3 0 x3 0 2 ˇx3

Z 1 ## $ # $$
3 2 x.x2 /2 3 3 x.x3 /2
D x .x / ! ! x .x / ! dx
0 2 2

Z 1 # $ Z 1# 5 $
5 x5 6 x7 x 6 x7
D x ! !x C dx D !x C dx
0 2 2 0 2 2

# $ˇ1 # $
x6 x7 x ˇˇ 1 1 1 1
D ! C D ! C !0D
12 7 16 ˇ0 12 7 16 336
 Section 7.6 t e nte ra s 763

E AM LE E I
Z 1 Z 1!x
Find .2x C 1/dydx
!1 0
S Here we first integrate with respect to y and then integrate the result with
respect to x:
Z 1 Z 1!x Z 1 #Z 1!x $
.2x C 1/dydx D .2x C 1/dy dx
!1 0 !1 0
Z ˇ1!x Z 1
1 ˇ
D ˇ
.2xy C y/ˇ dx D ..2x.1 ! x/ C .1 ! x// ! 0/dx
!1 0 !1
Z # $ˇ1
1
2x3 x2 ˇ
D 2
.!2x C x C 1/dx D ! C C x ˇˇ
3 2
#!1 $ # $ !1
2 1 2 1 2
D ! C C1 ! C !1 D
3 2 3 2 3
Now ork Problem 9 G
E AM LE E I
Z ln 2 Z 2
Find dxdy
1 ey
S Here we first integrate with respect to x and then integrate the result with
respect to y:
Z ln 2 Z 2 Z ln 2 #Z 2 $ Z ln 2 ˇ2
ˇ
dxdy D dx dy D xˇˇ dy
1 ey 1 ey 1 ey
Z ln 2 ˇln 2
y
ˇ
y ˇ
D .2 ! e /dy D .2y ! e /ˇ
1 1
D .2 ln 2 ! 2/ ! .2 ! e/ D 2 ln 2 ! 4 C e
D ln 4 ! 4 C e
Now ork Problem 13 G
z

z = f(x, y)

a c
d
b
y
dx
dy

x
RbRd
FIGURE Interpreting a c f .x; y/dydx in terms of volume, where f .x; y/ # 0.
764 C t ar a e Ca c s

A double integral can be interpreted in terms of the volume of a region between

the x; y-plane and a surface z D f.x; y/ if z # 0. In Figure 17.10 is a region whose vol-
ume we will consider. The element of volume for this region is a vertical column with
height approximately x D f.x; y/ and base area dyx. Thus, its volume is approximately
f.x; y/dydx. The volume of the entire region can be found by summing the volumes of
all such elements for a $ x $ b and c $ y $ d via a double integral:
Z bZ d
volume D f.x; y/dydx
a c
riple integrals are handled by successively evaluating three integrals, as the next
example shows.

E AM LE E T I
Z 1Z xZ x!y
Find xdzdydx.
0 0 0
S
Z 1 Z xZ x!y Z 1 Z x #Z x!y $
xdzdydx D xdz dydx
0 0 0 0 0 0
Z Z ˇx!y Z Z
1 x ˇ 1 x
D ˇ
.xz/ˇ dydx D .x.x ! y/ ! 0/dydx
0 0 0 0 0
Z 1 Z x Z 1 #Z x $
2 2
D .x ! xy/dydx D .x ! xy/dy dx
0 0 0 0
Z # $ˇx Z 1 ## $ "
1
2 xy2 ˇˇ 3 x3
D x y! dx D x ! ! 0 dx
0 2 ˇ0 0 2
Z ˇ1
1
x3 x4 ˇˇ 1
D dx D ˇ D
0 2 0

Now ork Problem 21 G

R BLEMS
Z Z Z Z
In Pr b ems e a uate the mu tip e integra s 1 1!x 3 3y
Z 3Z 4 Z 4Z 3 3.x C y/ dy dx 5x dx dy
!1 x 0 y2
x dy dx y dy dx
0 0 1 0 Z 1 Z y Z 1 Z 1
Z 1 Z 1 Z 1 Z 2 exCy dx dy ey!x dx dy
2
xy dx dy xy dydx 0 0 0 0
0 0 0 0 Z 1 Z 2 Z 3 Z 1 Z xZ xCy
Z 3 Z 2 Z 3 Z 2 x3 y2 zdxdydz x2 dz dy dx
2 2
.x ! y/ dx dy .y ! 2xy/ dy dx 0 0 0 0 0 0
1 1 !2 0 Z 1 Z xZ xy
Z eZ x Z y
Z 1 Z 2 Z 3 Z x dz dy dx dz dy dx
1 ln x 0
.x C y/ dy dx .x2 C y2 / dy dx 0 x2 0
0 0 0 0
Z Z Z Statistics In the study of statistics, a oint density function
2Z x2 2 x!1
2y dy dx z D f .x; y/ defined on a region in the x y-plane is represented by a
ydydx
1 0 1 0 surface in space. The probability that
Z 1 Z x2 Z 2 Z x2
a$x$b and c $ y $ d
14x2 y dy dx xy dy dx
0 3x 0 0
is given by
Z 3 Z p
!x2
Z 1 Z y2
dxdy Z Z
y dy dx d b
0 0 0 y3 P.a $ x $ b; c $ y $ d/ D f .x; y/ dx dy
c a
 Chapter 7 e e 765

and is represented by the volume between the graph of f and the Statistics In Problem 23, let f .x; y/ D 6e!.2xC3y/ for
rectangular region given by x; y # 0. Find

a $ x $ b and c $ y $ d P.1 $ x $ 3; 2 $ y $ 4/
and give your answer in terms of e.
If f .x; y/ D e!.xCy/ is a oint density function, where x # 0 and
Statistics In Problem 23, let f .x; y/ D 1, where 0 $ x $ 1
y # 0, find
and 0 $ y $ 1. Find P.x # 1=2; y # 1=3/.
P.0 $ x $ 2; 1 $ y $ 2/ Statistics In Problem 23, let f be the uniform density
function f .x; y/ D 1= defined over the rectangle
0 $ x $ 4; 0 $ y $ 2. etermine the probability that 0 $ x $ 1
and give your answer in terms of e.
and 0 $ y $ 1.

Chapter 17 Review
I T S E
S Partial Derivatives ˇ
@z @z ˇˇ
partial derivative D fx .x; y/ D fx .a; b/ Ex. 2, p. 735
@x @x ˇ.a;b/
S Applications of Partial Derivatives
oint-cost function production function marginal productivity Ex. 3, p. 740
competitive products complementary products Ex. 4, p. 741
S igher Order Partial Derivatives
@2 z @2 z @2 z @2 z
D fxy D fyx D f xx D fyy Ex. 1, p. 744
@y@x @x@y @x2 @y2
S Maxima and Minima for Functions of wo Variables
relative maximum and minimum critical point Ex. 1, p. 747
second-derivative test for functions of two variables Ex. 3, p. 74
S Lagrange Multipliers
agrange multipliers Ex. 1, p. 756
S Multiple Integrals
double integral triple integral Ex. 3, p. 764

S
For a function of n variables, we can consider n partial deriva- If units of labor and k units of capital are used to pro-
tives. For example, if D f.x; y; z/, we have the partial duce P units of a product, then the function P D f. ; k/ is
derivatives of f with respect to x, with respect to y, and with called a production function. The partial derivatives of P are
respect to z, denoted either fx , fy , and fz , or @ =@x, @ =@y, called marginal productivity functions.
and @ =@z, respectively. To find fx .x; y; z/, we treat y and z Suppose two products, A and , are such that the quan-
as constants and differentiate f with respect to x in the usual tity demanded of each is dependent on the prices of both. If
way. The other partial derivatives are found similarly. We can qA and q are the quantities of A and demanded when the
interpret fx .x; y; z/ as the approximate change in that results prices of A and are pA and p , respectively, then qA and
from a one-unit change in x when y and z are held fixed. There q are each functions of pA and p . When @qA =@p > 0 and
are similar interpretations for the other partial derivatives. @q =@pA > 0, then A and are called competitive products
Functions of several variables occur frequently in busi- (or substitutes). When @qA =@p < 0 and @q =@pA < 0, then
ness and economic analysis, as well as in other areas of study. A and are called complementary products.
If a manufacturer produces x units of product and y units A partial derivative of a function of n variables is itself a
of product , then the total cost, c, of these units is a func- function of n variables. y successively ta ing partial deriva-
tion of x and y and is called a oint-cost function. The partial tives of partial derivatives, we obtain higher-order partial
derivatives @c=@x and @c=@y are called the marginal costs derivatives. For example, if f is a function of x and y, then
with respect to x and y, respectively. We can interpret, for fxy denotes the partial derivative of fx with respect to y fxy is
example, @c=@x as the approximate cost of producing an extra called the second-partial derivative of f, first with respect to
unit of while the level of production of is held fixed. x and then with respect to y.
766 C t ar a e Ca c s

If the function f.x; y/ has a relative extremum at .a; b/, y solving the system
then .a; b/ must be a solution of the system 8̂
F D0
< x
% Fy D0
fx .x; y/ D 0
:̂ Fz D0
fy .x; y/ D 0 F! D0

we obtain the critical points of F. If .a; b; c; &0 / is such a

Any solution of this system is called a critical point of f. Thus, critical point, then .a; b; c/ is a critical point of f, sub ect
critical points are the candidates at which a relative extremum to the constraint. It is important to write the constraint in
may occur. The second-derivative test for functions of two the form g.x; y; z/ D 0. For example, if the constraint is
variables gives conditions under which a critical point cor- 2xC3y!z D 4, then g.x; y; z/ D 2xC3y!z!4. If f.x; y; z/ is
responds to a relative maximum or a relative minimum. The sub ect to two constraints, g1 .x; y; z/ D 0 and g2 .x; y; z/ D 0,
test states that if .a; b/ is a critical point of f and then we would form the function F D f ! &1 g1 ! &2 g2 and
solve the system
.x; y/ D fxx .x; y/fyy .x; y/ ! Œfxy .x; y/!2 8̂
Fx D 0
ˆ
ˆ
< Fy D 0
then Fz D 0
ˆ
ˆF!1 D 0
:̂
F!2 D 0
if .a; b/ > 0 and fxx .a; b/ < 0, then f has a relative
maximum at .a; b/ When wor ing with functions of several variables, we
if .a; b/ > 0 and fxx .a; b/ > 0, then f has a relative can consider their multiple integrals. These are determined
minimum at .a; b/ by successive integration. For example, the double integral
Z 2Z y
if .a; b/ < 0, then f has a saddle point at .a; b/
.x C y/dxdy
if .a; b/ D 0, no conclusion about an extremum at 1 0
.a; b/ can yet be drawn, and further analysis is required.
is determined by first treating y as a constant and integrating
x C y with respect to x. After evaluating between the bounds
To find critical points of a function of several variables,
0 and y, we integrate that result with respect to y from y D 1
sub ect to a constraint, we can sometimes use the method of
to y D 2. Thus,
agrange multipliers. For example, to find the critical points Z 2Z y Z 2 #Z y $
of f.x; y; z/, sub ect to the constraint g.x; y; z/ D 0, we first
.x C y/dxdy D .x C y/dx dy
form the function 1 0 1 0

Triple integrals involve functions of three variables and are

F.x; y; z; &/ D f.x; y; z/ ! &g.x; y; z/
also evaluated by successive integration.

R
In Pr b ems nd the indicated partia deri ati es D exCyCz ln.xyz/ @3 =@z@y@x
f .x; y/ D ln.x2 C y2 / fx .x; y/, fy .x; y/ P D 100 0:11 k0: I @2 P=@k@
3 3
PD C k ! kI @P=@ ; @P=@k xCy
If f .x; y; z/ D , find fxyz .2; 7; 4/.
x @z @z xz
zD I ;
xCy @x @y If f .x; y; z/ D .6x C 1/ey
2 ln.zC1/
, find fxyz .0; 1; 0/.
f .pA ; p / D 4.pA ! 10/ C 5.p ! 15/I fp .pA ; p /
Production Function If a manufacturer s production
p @ function is defined by P D 100 0: k0:2 , determine the marginal
f .x; y/ D e x Cy
2 2
. f .x; y// productivity functions.
@y
p @ 2
D ex yz I oint Cost Function A manufacturer s cost for producing
D x2 C y2 xy .x; y; z/
@y x units of product and y units of product is given by
f .x; y/ D xy ln.xy/I fxy .x; y/
c D 3x C 0:05xy C y C 500
2 2 2 @2
f .x; y; z/ D .x C y C z/.x C y C z /I . f .x; y; z//
@z2 etermine the (partial) marginal cost with respect to x when
z D .x2 ! y2 /2 @2 z=@y@x x D 50 and y D 100.
 Chapter 7 e e 767

Competitive Complementary Products If Maximi ing Profit A dairy produces two types of cheese,
qA D 100 ! pA C 2p and q D 150 ! 3pA ! 2p , where qA and A and , at constant average costs of 50 cents and 60 cents per
q are the number of units demanded of products A and , pound, respectively. When the selling price per pound of A is pA
respectively, and pA and p are their respective prices per unit, cents and of is p cents, the demands (in pounds) for A and ,
determine whether A and are competitive products or are, respectively,
complementary products or neither.
Innovation For industry, the following model describes qA D 250.p ! pA /
the rate ˛ (a ree letter read alpha ) at which an innovation
substitutes for an established process:15 and

˛D C 0:530P ! 0:027S q D 32;000 C 250.pA ! 2p /

Here, is a constant that depends on the particular industry, P is
Find the selling prices that yield a relative maximum profit. Verify
an index of profitability of the innovation, and S is an index of the
that the profit has a relative maximum at these prices.
extent of the investment necessary to ma e use of the innovation.
Find @˛=@P and @˛=@S. Find all critical points of f .x; y; z/ D xy2 z, sub ect to the
condition that
Examine f .x; y/ D x2 C 2y2 ! 2xy ! 4y C 3 for relative
extrema. x C y C z ! 1 D 0 .xyz ¤ 0/
3
Examine f . ; z/ D C z3 ! 3 z C 5 for relative extrema. p
Minimi ing Material An open-top rectangular cardboard Find all critical points of f .x; y/ D x2 C y2 , sub ect to the
box is to have a volume of 32 cubic feet. Find the dimensions of constraint 5x C y D 1. Explain the answer geometrically.
the box so that the amount of cardboard used is minimized.
In Pr b ems e a uate the d ub e integra s
The function Z 2Z y Z 1 Z y2
x2 y2 dx dy xy dx dy
f .x; y/ D ax2 C by2 C cxy ! x C y 1 0 0 0
has a critical point at .x; y/ D .0; 1/, and the second-derivative test Z 4 Z 2x Z 1 Z x2
is inconclusive at this point. etermine the values of the constants y dy dx 7.x2 C 2xy ! 3y2 / dy dx
p
1 x2 0 x
a, b, and c.

15
A. P. Hurter, r., A. H. Rubenstein, et al., ar et Penetration by New Inno-
vations: The Technological iterature, echn gica F recasting and S cia
Change 11 (1 7 ), 1 7 221.
This page intentionally left blank
PP Co o n nterest
a es

769
770 A A Co o n nterest a es

r D 0:005 r D 0:0075
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.005000 0. 5025 0. 5025 1.000000 1 1.007500 0. 2556 0. 2556 1.000000

2 1.010025 0. 0075 1. 50 2.005000 2 1.015056 0. 5167 1. 77723 2.007500
3 1.015075 0. 514 2. 7024 3.015025 3 1.02266 0. 77 33 2. 55556 3.022556
4 1.020151 0. 024 3. 504 6 4.030100 4 1.03033 0. 70554 3. 26110 4.045225
5 1.025251 0. 75371 4. 25 66 5.050251 5 1.03 067 0. 6332 4. 440 5.075565
6 1.03037 0. 7051 5. 63 4 6.075502 6 1.045 52 0. 5615 5. 455 6.113631
7 1.03552 0. 656 0 6. 62074 7.105 7 7 1.0536 6 0. 4 040 6.7 463 7.15 4 4
1.040707 0. 60 5 7. 22 5 .14140 1.0615 0. 41 75 7.736613 .2131 0
1.045 11 0. 56105 .77 064 .1 2116 1.06 561 0. 34 63 .671576 .27477
10 1.051140 0. 5134 .730412 10.22 026 10 1.0775 3 0. 2 003 .5 5 0 10.34433
11 1.0563 6 0. 46615 10.677027 11.27 167 11 1.0 5664 0. 210 5 10.520675 11.421 22
12 1.06167 0. 41 05 11.61 32 12.335562 12 1.0 3 07 0. 1423 11.434 13 12.5075 6
13 1.066 6 0. 3721 12.556151 13.3 7240 13 1.102010 0. 07432 12.342345 13.6013 3
14 1.072321 0. 32556 13.4 70 14.464226 14 1.110276 0. 00677 13.243022 14.703404
15 1.0776 3 0. 27 17 14.416625 15.53654 15 1.11 603 0. 3 73 14.136 5 15. 1367
16 1.0 3071 0. 23300 15.33 25 16.614230 16 1.126 2 0. 731 15.024313 16. 322 2
17 1.0 4 7 0. 1 707 16.25 632 17.6 7301 17 1.135445 0. 0712 15. 05025 1 .05 274
1 1.0 3 2 0. 14136 17.17276 1 .7 57 1 1.143 60 0. 74156 16.77 1 1 1 .1 471
1 1.0 3 0. 0 5 1 .0 2356 1 . 7 717 1 1.152540 0. 6764 17.646 30 20.33 67
20 1.104 6 0. 05063 1 . 741 20. 7 115 20 1.1611 4 0. 611 0 1 .50 020 21.4 121
21 1.110420 0. 00560 1 . 7 7 22.0 4011 21 1.16 3 0. 5477 1 .3627 22.652403
22 1.115 72 0. 60 0 20.7 405 23.1 4431 22 1.17 667 0. 4 416 20.211215 23. 222 6
23 1.121552 0. 1622 21.6756 1 24.310403 23 1.1 7507 0. 42100 21.053315 25.000 63
24 1.127160 0. 71 6 22.562 66 25.431 55 24 1.1 6414 0. 35 31 21. 146 26.1 471
25 1.1327 6 0. 2772 23.44563 26.55 115 25 1.2053 7 0. 2 60 22.71 755 27.3 4 4
26 1.13 460 0. 7 3 0 24.32401 27.6 1 11 26 1.214427 0. 23434 23.5421 2 .5 0271
27 1.144152 0. 74010 25.1 02 2 . 30370 27 1.223535 0. 17304 24.35 4 3 2 . 046
2 1.14 73 0. 6 662 26.0676 2 . 74522 2 1.232712 0. 11220 25.170713 31.02 233
2 1.155622 0. 65335 26. 33024 31.1243 5 2 1.241 57 0. 051 1 25. 75 3 32.260 45
30 1.161400 0. 61030 27.7 4054 32.2 0017 30 1.251272 0.7 1 7 26.7750 0 33.502 02
31 1.167207 0. 56746 2 .650 00 33.441417 31 1.260656 0.7 323 27.56 31 34.754174
32 1.173043 0. 524 4 2 .5032 4 34.60 624 32 1.270111 0.7 7333 2 .355650 36.014 30
33 1.17 0 0. 4 242 30.351526 35.7 1667 33 1.27 637 0.7 1472 2 .137122 37.2 4 41
34 1.1 4 03 0. 44022 31.1 554 36. 60575 34 1.2 234 0.775654 2 . 12776 3 .56457
35 1.1 0727 0. 3 23 32.035371 3 .14537 35 1.2 04 0.76 0 30.6 2656 3 . 53 13
36 1.1 66 1 0. 35645 32. 71016 3 .336105 36 1.30 645 0.76414 31.446 05 41.152716
37 1.202664 0. 314 7 33.702504 40.5327 5 37 1.31 460 0.75 461 32.205266 42.461361
3 1.20 677 0. 27351 34.52 54 41.73544 3 1.32 34 0.752 14 32. 5 0 0 43.77 22
3 1.214721 0. 23235 35.3530 42. 44127 3 1.33 311 0.747210 33.7052 0 45.10 170
40 1.2207 4 0. 1 13 36.17222 44.15 47 40 1.34 34 0.74164 34.446 3 46.4464 2
41 1.226 0. 15064 36. 72 1 45.37 642 41 1.35 461 0.736127 35.1 3065 47.7 4 30
42 1.233033 0. 1100 37.7 300 46.606540 42 1.36 650 0.730647 35. 13713 4 .1532 1
43 1.23 1 0. 06 74 3 .605274 47. 3 572 43 1.37 15 0.72520 36.63 21 50.521 41
44 1.2453 4 0. 02 5 3 .40 232 4 .07 770 44 1.3 256 0.71 10 37.35 730 51. 00 56
45 1.251621 0.7 64 40.2071 6 50.324164 45 1.3 676 0.714451 3 .0731 1 53.2 0112
46 1.257 7 0.7 4 41.0021 5 51.5757 5 46 1.410173 0.70 133 3 .7 2314 54.6 7
47 1.26416 0.7 1034 41.7 321 52. 33664 47 1.420750 0.703 54 3 .4 616 56.0 61
4 1.2704 0.7 70 42.5 031 54.0 7 32 4 1.431405 0.6 614 40.1 47 2 57.520711
4 1.276 42 0.7 31 2 43.363500 55.36 321 4 1.442141 0.6 3414 40. 7 1 5 5 . 52116
50 1.2 3226 0.77 2 6 44.1427 6 56.645163 50 1.452 57 0.6 252 41.566447 60.3 4257
 ppen i Co o n nterest a es 771

r D 0:01 r D 0:0125
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.010000 0. 00 0. 00 1.000000 1 1.012500 0. 7654 0. 7654 1.000000

2 1.020100 0. 02 6 1. 703 5 2.010000 2 1.025156 0. 75461 1. 63115 2.012500
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4 1.040604 0. 60 0 3. 01 66 4.060401 4 1.050 45 0. 51524 3. 7 05 4.075627
5 1.051010 0. 51466 4. 53431 5.101005 5 1.0640 2 0. 3 777 4. 17 35 5.126572
6 1.061520 0. 42045 5.7 5476 6.152015 6 1.0773 3 0. 2 175 5.746010 6.1 0654
7 1.072135 0. 3271 6.72 1 5 7.213535 7 1.0 0 50 0. 16716 6.662726 7.26 03
1.0 2 57 0. 234 3 7.65167 .2 5671 1.1044 6 0. 053 7.56 124 .35
1.0 36 5 0. 14340 .56601 .36 527 1.11 2 2 0. 4221 .462345 .463374
10 1.104622 0. 052 7 .471305 10.462213 10 1.132271 0. 31 1 .345526 10.5 1666
11 1.11566 0. 6324 10.36762 11.566 35 11 1.146424 0. 72277 10.217 03 11.713 37
12 1.126 25 0. 744 11.255077 12.6 2503 12 1.160755 0. 6150 11.07 312 12. 60361
13 1.13 0 3 0. 7 663 12.133740 13. 0 32 13 1.175264 0. 50 73 11. 301 5 14.021116
14 1.14 474 0. 6 63 13.003703 14. 47421 14 1.1 55 0. 4036 12.770553 15.1 63 0
15 1.160 6 0. 6134 13. 65053 16.0 6 6 15 1.204 2 0. 2 3 13.600546 16.3 6335
16 1.17257 0. 52 21 14.717 74 17.257 64 16 1.21 0 0. 1 746 14.4202 2 17.5 1164
17 1.1 4304 0. 44377 15.562251 1 .430443 17 1.23513 0. 0 626 15.22 1 1 . 11053
1 1.1 6147 0. 36017 16.3 26 1 .61474 1 1.250577 0.7 631 16.02 54 20.0461 2
1 1.20 10 0. 27740 17.22600 20. 10 5 1 1.266210 0.7 75 16. 1 30 21.2 676
20 1.2201 0 0. 1 544 1 .045553 22.01 004 20 1.2 2037 0.7 000 17.5 316 22.562 7
21 1.2323 2 0. 11430 1 . 56 3 23.23 1 4 21 1.2 063 0.77037 1 .36 6 5 23. 45016
22 1.244716 0. 033 6 1 .66037 24.4715 6 22 1.3142 0.760 6 1 .130563 25.14307
23 1.257163 0.7 5442 20.455 21 25.716302 23 1.330717 0.751475 1 . 2037 26.457367
24 1.26 735 0.7 7566 21.2433 7 26. 73465 24 1.347351 0.7421 7 20.624235 27.7 0 4
25 1.2 2432 0.77 76 22.023156 2 .243200 25 1.3641 3 0.733034 21.35726 2 .135435
26 1.2 5256 0.77204 22.7 5204 2 .525631 26 1.3 1245 0.723 4 22.0 1253 30.4 62
27 1.30 20 0.764404 23.55 60 30. 20 27 1.3 511 0.715046 22.7 62 31. 0 73
2 1.3212 1 0.756 36 24.316443 32.12 0 7 2 1.415 2 0.70621 23.50251 33.27 3 4
2 1.334504 0.74 342 25.0657 5 33.4503 2 1.4336 2 0.6 7500 24.20001 34.6 5377
30 1.347 4 0.741 23 25. 0770 34.7 4 2 30 1.451613 0.6 24. 06 36.12 06
31 1.361327 0.734577 26.5422 5 36.132740 31 1.46 75 0.6 03 4 25.56 2 0 37.5 06 2
32 1.374 41 0.727304 27.26 5 37.4 406 32 1.4 131 0.671 4 26.241274 3 .050441
33 1.3 6 0 0.720103 27. 6 3 3 . 6 00 33 1.506732 0.6636 26. 04 62 40.53 571
34 1.402577 0.712 73 2 .702666 40.2576 34 1.525566 0.6554 4 27.560456 42.045303
35 1.416603 0.705 14 2 .40 5 0 41.660276 35 1.544636 0.647402 2 .207 5 43.570 70
36 1.43076 0.6 25 30.107505 43.076 7 36 1.563 44 0.63 40 2 . 47267 45.115505
37 1.445076 0.6 2005 30.7 510 44.507647 37 1.5 34 3 0.631515 2 .47 7 3 46.67 44
3 1.45 527 0.6 5153 31.4 4663 45. 52724 3 1.6032 7 0.62371 30.102501 4 .262 42
3 1.474123 0.67 370 32.163033 47.412251 3 1.62332 0.61601 30.71 520 4 . 6622
40 1.4 64 0.671653 32. 346 6 4 . 6373 40 1.64361 0.60 413 31.326 33 51.4 557
41 1.503752 0.665003 33.4 6 50.375237 41 1.664165 0.600 02 31. 27 35 53.133177
42 1.51 7 0 0.65 41 34.15 10 51. 7 42 1.6 4 67 0.5 34 4 32.52131 54.7 7341
43 1.533 7 0.651 00 34. 1000 53.3 777 43 1.70602 0.5 6157 33.107475 56.4 230
44 1.54 31 0.645445 35.455454 54. 31757 44 1.727354 0.57 20 33.6 63 5 5 .1 337
45 1.564 11 0.63 055 36.0 450 56.4 1075 45 1.74 46 0.571773 34.25 16 5 . 156 1
46 1.5 045 0.63272 36.727236 5 .045 5 46 1.770 0 0.564714 34. 22 2 61.664637
47 1.5 6263 0.626463 37.3536 5 .626344 47 1.7 2 43 0.557742 35.3 0624 63.435445
4 1.612226 0.620260 37. 73 5 61.22260 4 1. 15355 0.550 56 35. 314 1 65.22 3
4 1.62 34 0.61411 3 .5 07 62. 34 34 4 1. 3 047 0.544056 36.475537 67.043743
50 1.644632 0.60 03 3 .1 611 64.4631 2 50 1. 61022 0.53733 37.012 76 6 . 17 0
772 A A Co o n nterest a es

r D 0:015 r D 0:02
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.015000 0. 5222 0. 5222 1.000000 1 1.020000 0. 03 2 0. 03 2 1.000000

2 1.030225 0. 70662 1. 55 3 2.015000 2 1.040400 0. 6116 1. 41561 2.020000
3 1.04567 0. 56317 2. 12200 3.045225 3 1.06120 0. 42322 2. 3 3 3.060400
4 1.061364 0. 421 4 3. 543 5 4.0 0 03 4 1.0 2432 0. 23 45 3. 0772 4.12160
5 1.0772 4 0. 2 260 4.7 2645 5.152267 5 1.1040 1 0. 05731 4.713460 5.204040
6 1.0 3443 0. 14542 5.6 71 7 6.22 551 6 1.126162 0. 7 71 5.601431 6.30 121
7 1.10 45 0. 01027 6.5 214 7.322 4 7 1.14 6 6 0. 70560 6.471 1 7.4342 3
1.1264 3 0. 7711 7.4 5 25 .432 3 1.17165 0. 534 0 7.3254 1 .5 2 6
1.1433 0 0. 745 2 .360517 .55 332 1.1 50 3 0. 36755 .162237 .75462
10 1.160541 0. 61667 .2221 5 10.702722 10 1.21 4 0. 2034 . 25 5 10. 4 721
11 1.177 4 0. 4 33 10.07111 11. 63262 11 1.243374 0. 04263 .7 6 4 12.16 715
12 1.1 561 0. 363 7 10. 07505 13.041211 12 1.26 242 0.7 4 3 10.575341 13.4120 0
13 1.213552 0. 24027 11.731532 14.236 30 13 1.2 3607 0.773033 11.34 374 14.6 0332
14 1.231756 0. 11 4 12.5433 2 15.4503 2 14 1.31 47 0.757 75 12.10624 15. 73 3
15 1.250232 0.7 52 13.343233 16.6 213 15 1.345 6 0.743015 12. 4 264 17.2 3417
16 1.26 6 0.7 031 14.131264 17. 32370 16 1.3727 6 0.72 446 13.57770 1 .63 2 5
17 1.2 020 0.7763 5 14. 0764 1 .201355 17 1.400241 0.714163 14.2 1 72 20.012071
1 1.307341 0.764 12 15.672561 20.4 376 1 1.42 246 0.70015 14. 2031 21.412312
1 1.326 51 0.753607 16.42616 21.7 6716 1 1.456 11 0.6 6431 15.67 462 22. 4055
20 1.346 55 0.742470 17.16 63 23.123667 20 1.4 5 47 0.672 71 16.351433 24.2 7370
21 1.36705 0.7314 17. 00137 24.470522 21 1.515666 0.65 776 17.01120 25.7 3317
22 1.3 7564 0.7206 1 .620 24 25. 375 0 22 1.545 0 0.646 3 17.65 04 27.2 4
23 1.40 377 0.710037 1 .330 61 27.225144 23 1.576 0.634156 1 .2 2204 2 . 44 63
24 1.42 503 0.6 544 20.030405 2 .633521 24 1.60 437 0.621721 1 . 13 26 30.421 62
25 1.450 45 0.6 206 20.71 611 30.063024 25 1.640606 0.60 531 1 .523456 32.030300
26 1.472710 0.67 021 21.3 632 31.513 6 26 1.67341 0.5 757 20.121036 33.670 06
27 1.4 4 00 0.66 6 22.067617 32. 667 27 1.706 6 0.5 5 62 20.706 35.344324
2 1.517222 0.65 0 22.726717 34.4 147 2 1.741024 0.574375 21.2 1272 37.051210
2 1.53 1 0.64 35 23.376076 35. 701 2 1.775 45 0.563112 21. 443 5 3 .7 2235
30 1.5630 0 0.63 762 24.015 3 37.53 6 1 30 1. 11362 0.552071 22.3 6456 40.56 07
31 1.5 6526 0.63030 24.646146 3 .101762 31 1. 475 0.541246 22. 37702 42.37 441
32 1.610324 0.620 3 25.26713 40.6 2 32 1. 4541 0.530633 23.46 335 44.227030
33 1.63447 0.611 16 25. 7 54 42.2 612 33 1. 22231 0.52022 23. 564 46.111570
34 1.65 6 0.602774 26.4 172 43. 330 2 34 1. 60676 0.51002 24.4 5 2 4 .033 02
35 1.6 3 1 0.5 3 66 27.0755 5 45.5 20 35 1. 0 0.50002 24. 61 4 . 447
36 1.70 140 0.5 50 0 27.6606 4 47.275 6 36 2.03 7 0.4 0223 25.4 42 51. 4367
37 1.734777 0.576443 2 .237127 4 . 510 37 2.0 06 5 0.4 0611 25. 6 453 54.034255
3 1.7607 0.567 24 2 . 05052 50.71 5 3 2.1222 0.4711 7 26.440641 56.114 40
3 1.7 7210 0.55 531 2 .3645 3 52.4 06 4 3 2.164745 0.461 4 26. 025 5 .23723
40 1. 1401 0.551262 2 . 15 45 54.267 4 40 2.20 040 0.452 0 27.35547 60.401 3
41 1. 4122 0.543116 30.45 61 56.0 1 12 41 2.252200 0.444010 27.7 4 62.610023
42 1. 6 47 0.5350 30. 4050 57. 23141 42 2.2 7244 0.435304 2 .2347 4 64. 62223
43 1. 6 0 0.5271 2 31.521232 5 .7 1 43 2.3431 0.42676 2 .661562 67.15 46
44 1. 25333 0.51 3 1 32.040622 61.6 6 44 2.3 0053 0.41 401 2 .07 63 6 .502657
45 1. 54213 0.511715 32.552337 63.614201 45 2.437 54 0.4101 7 2 .4 0160 71. 2710
46 1. 3526 0.504153 33.0564 0 65.56 414 46 2.4 6611 0.402154 2 . 2314 74.330564
47 2.01327 0.4 6702 33.5531 2 67.551 40 47 2.536344 0.3 426 30.2 65 2 76. 17176
4 2.04347 0.4 362 34.042554 6 .56521 4 2.5 7070 0.3 653 30.673120 7 .35351
4 2.074130 0.4 2130 34.5246 3 71.60 6 4 2.63 12 0.37 5 31.05207 1. 405 0
50 2.105242 0.475005 34. 6 73.6 2 2 50 2.6 15 0.37152 31.423606 4.57 401
 ppen i Co o n nterest a es 773

r D 0:025 r D 0:03
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.025000 0. 75610 0. 75610 1.000000 1 1.030000 0. 70 74 0. 70 74 1.000000

2 1.050625 0. 51 14 1. 27424 2.025000 2 1.060 00 0. 425 6 1. 13470 2.030000
3 1.076 1 0. 2 5 2. 56024 3.075625 3 1.0 2727 0. 15142 2. 2 611 3.0 0 00
4 1.103 13 0. 05 51 3.761 74 4.152516 4 1.12550 0. 4 7 3.7170 4.1 3627
5 1.13140 0. 3 54 4.645 2 5.25632 5 1.15 274 0. 6260 4.57 707 5.30 136
6 1.15 6 3 0. 622 7 5.50 125 6.3 7737 6 1.1 4052 0. 374 4 5.4171 1 6.46 410
7 1.1 6 6 0. 41265 6.34 3 1 7.547430 7 1.22 74 0. 130 2 6.2302 3 7.662462
1.21 403 0. 20747 7.170137 .736116 1.266770 0.7 40 7.01 6 2 . 2336
1.24 63 0. 0072 7. 70 66 . 5451 1.304773 0.766417 7.7 610 10.15 106
10 1.2 00 5 0.7 11 .752064 11.2033 2 10 1.343 16 0.7440 4 .530203 11.463 7
11 1.3120 7 0.762145 .51420 12.4 3466 11 1.3 4234 0.722421 .252624 12. 077 6
12 1.344 0.743556 10.257765 13.7 5553 12 1.425761 0.7013 0 . 54004 14.1 2030
13 1.37 511 0.725420 10. 31 5 15.140442 13 1.46 534 0.6 0 51 10.634 55 15.6177 0
14 1.412 74 0.707727 11.6 0 12 16.51 53 14 1.5125 0 0.66111 11.2 6073 17.0 6324
15 1.44 2 0.6 0466 12.3 137 17. 31 27 15 1.557 67 0.641 62 11. 37 35 1 .5 14
16 1.4 4506 0.673625 13.055003 1 .3 0225 16 1.604706 0.623167 12.561102 20.156 1
17 1.52161 0.6571 5 13.7121 20. 64730 17 1.652 4 0.605016 13.16611 21.7615
1 1.55 65 0.641166 14.353364 22.3 634 1 1.702433 0.5 73 5 13.753513 23.414435
1 1.5 650 0.62552 14. 7 1 23. 46007 1 1.753506 0.5702 6 14.3237 25.116 6
20 1.63 616 0.610271 15.5 162 25.54465 20 1. 06111 0.553676 14. 77475 26. 70374
21 1.67 5 2 0.5 53 6 16.1 454 27.1 3274 21 1. 602 5 0.53754 15.415024 2 .6764 6
22 1.721571 0.5 0 65 16.765413 2 . 62 56 22 1. 16103 0.521 3 15. 36 17 30.5367 0
23 1.764611 0.5666 7 17.332110 30.5 4427 23 1. 735 7 0.5066 2 16.44360 32.452 4
24 1. 0 726 0.552 75 17. 4 6 32.34 03 24 2.0327 4 0.4 1 34 16. 35542 34.426470
25 1. 53 44 0.53 3 1 1 .424376 34.157764 25 2.0 377 0.477606 17.41314 36.45 264
26 1. 002 3 0.526235 1 . 50611 36.01170 26 2.1565 1 0.4636 5 17. 76 42 3 .553042
27 1. 47 00 0.513400 1 .464011 37. 12001 27 2.2212 0.4501 1 .327031 40.70 634
2 1. 64 5 0.500 7 1 . 64 3 . 5 01 2 2.2 7 2 0.437077 1 .76410 42. 30 23
2 2.046407 0.4 661 20.453550 41. 562 6 2 2.356566 0.424346 1 .1 455 45.21 50
30 2.0 756 0.476743 20. 302 3 43. 02703 30 2.427262 0.411 7 1 .600441 47.575416
31 2.150007 0.465115 21.3 5407 46.000271 31 2.5000 0 0.3 7 20.00042 50.00267
32 2.203757 0.453771 21. 4 17 4 .15027 32 2.5750 3 0.3 337 20.3 766 52.50275
33 2.25 51 0.442703 22.2 1 1 50.354034 33 2.652335 0.377026 20.7657 2 55.077 41
34 2.315322 0.431 05 22.7237 6 52.612 5 34 2.731 05 0.366045 21.131 37 57.730177
35 2.373205 0.421371 23.145157 54. 2 207 35 2. 13 62 0.3553 3 21.4 7220 60.4620 2
36 2.432535 0.4110 4 23.556251 57.301413 36 2. 27 0.345032 21. 32252 63.275 44
37 2.4 334 0.401067 23. 5731 5 .733 4 37 2. 5227 0.334 3 22.167235 66.174223
3 2.5556 2 0.3 12 5 24.34 603 62.2272 7 3 3.0747 3 0.325226 22.4 2462 6 .15 44
3 2.61 574 0.3 1741 24.730344 64.7 2 7 3 3.167027 0.315754 22. 0 215 72.234233
40 2.6 5064 0.372431 25.102775 67.402554 40 3.26203 0.306557 23.114772 75.401260
41 2.7521 0 0.363347 25.466122 70.0 7617 41 3.35 0.2 762 23.412400 7 .6632
42 2. 20 5 0.3544 5 25. 20607 72. 3 0 42 3.4606 6 0.2 5 23.70135 2.0231 6
43 2. 1520 0.345 3 26.166446 75.660 03 43 3.564517 0.2 0543 23. 1 02 5.4 3 2
44 2. 63 0 0.337404 26.503 4 7 .552323 44 3.671452 0.272372 24.254274 .04 40
45 3.037 03 0.32 174 26. 33024 1.516131 45 3.7 15 6 0.26443 24.51 713 2.71 61
46 3.113 51 0.321146 27.154170 4.554034 46 3. 5044 0.256737 24.77544 6.501457
47 3.1 16 7 0.313313 27.4674 3 7.667 5 47 4.011 5 0.24 25 25.02470 100.3 6501
4 3.2714 0 0.305671 27.773154 0. 5 5 2 4 4.132252 0.241 25.266707 104.40 3 6
4 3.353277 0.2 216 2 .07136 4.131072 4 4.25621 0.234 50 25.501657 10 .54064
50 3.43710 0.2 0 42 2 .362312 7.4 434 50 4.3 3 06 0.22 107 25.72 764 112.7 6 67
774 A A Co o n nterest a es

r D 0:035 r D 0:04
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.035000 0. 661 4 0. 661 4 1.000000 1 1.040000 0. 6153 0. 6153 1.000000

2 1.071225 0. 33511 1. 6 4 2.035000 2 1.0 1600 0. 24556 1. 60 5 2.040000
3 1.10 71 0. 01 43 2. 01637 3.106225 3 1.124 64 0. 6 2.7750 1 3.121600
4 1.147523 0. 71442 3.67307 4.214 43 4 1.16 5 0. 54 04 3.62 5 4.246464
5 1.1 76 6 0. 41 73 4.515052 5.362466 5 1.216653 0. 21 27 4.451 22 5.416323
6 1.22 255 0. 13501 5.32 553 6.550152 6 1.26531 0.7 0315 5.242137 6.632 75
7 1.27227 0.7 5 1 6.114544 7.77 40 7 1.315 32 0.75 1 6.002055 7. 2 4
1.316 0 0.75 412 6. 73 56 .0516 7 1.36 56 0.7306 0 6.732745 .214226
1.362 7 0.733731 7.6076 7 10.36 4 6 1.423312 0.7025 7 7.435332 10.5 27 5
10 1.4105 0.70 1 .316605 11.7313 3 10 1.4 0244 0.675564 .110 6 12.006107
11 1.45 70 0.6 4 46 .001551 13.141 2 11 1.53 454 0.64 5 1 .760477 13.4 6351
12 1.51106 0.6617 3 .663334 14.601 62 12 1.601032 0.6245 7 .3 5074 15.025 05
13 1.563 56 0.63 404 10.30273 16.113030 13 1.665074 0.600574 . 564 16.626 3
14 1.61 6 5 0.6177 2 10. 20520 17.676 6 14 1.731676 0.577475 10.563123 1 .2 1 11
15 1.67534 0.5 6 1 11.517411 1 .2 56 1 15 1. 00 44 0.555265 11.11 3 7 20.0235
16 1.733 6 0.576706 12.0 4117 20. 71030 16 1. 72 1 0.533 0 11.6522 6 21. 24531
17 1.7 4676 0.557204 12.651321 22.705016 17 1. 47 00 0.513373 12.16566 23.6 7512
1 1. 574 0.53 361 13.1 6 2 24.4 6 1 1 2.025 17 0.4 362 12.65 2 7 25.645413
1 1. 22501 0.520156 13.70 37 26.3571 0 1 2.106 4 0.474642 13.133 3 27.67122
20 1. 7 0.502566 14.212403 2 .27 6 2 20 2.1 1123 0.4563 7 13.5 0326 2 .77 07
21 2.05 431 0.4 5571 14.6 7 74 30.26 471 21 2.27 76 0.43 34 14.02 160 31. 6 202
22 2.131512 0.46 151 15.167125 32.32 02 22 2.36 1 0.421 55 14.451115 34.247 70
23 2.206114 0.4532 6 15.620410 34.460414 23 2.464716 0.405726 14. 56 42 36.617
24 2.2 332 0.437 57 16.05 36 36.66652 24 2.563304 0.3 0121 15.246 63 3 .0 2604
25 2.363245 0.423147 16.4 1515 3 . 4 57 25 2.665 36 0.375117 15.6220 0 41.645 0
26 2.445 5 0.40 3 16. 0352 41.313102 26 2.772470 0.3606 15. 276 44.311745
27 2.531567 0.3 5012 17.2 5365 43.75 060 27 2. 336 0.346 17 16.32 5 6 47.0 4214
2 2.620172 0.3 1654 17.66701 46.2 0627 2 2. 703 0.333477 16.663063 4 . 675 3
2 2.711 7 0.36 74 1 .035767 4 . 107 2 3.11 651 0.320651 16. 3715 52. 662 6
30 2. 067 4 0.35627 1 .3 2045 51.622677 30 3.2433 0.30 31 17.2 2033 56.0 4 3
31 2. 05031 0.344230 1 .736276 54.42 471 31 3.373133 0.2 6460 17.5 4 4 5 .32 335
32 3.00670 0.3325 0 1 .06 65 57.334502 32 3.50 05 0.2 505 17. 73551 62.70146
33 3.111 42 0.321343 1 .3 020 60.341210 33 3.64 3 1 0.2740 4 1 .147646 66.20 527
34 3.220 60 0.310476 1 .7006 4 63.453152 34 3.7 4316 0.263552 1 .4111 6 . 57 0
35 3.3335 0 0.2 77 20.000661 66.674013 35 3. 460 0.253415 1 .664613 73.652225
36 3.450266 0.2 33 20.2 04 4 70.007603 36 4.103 33 0.24366 1 . 0 2 2 77.5 314
37 3.571025 0.2 0032 20.570525 73.457 6 37 4.26 0 0 0.2342 7 1 .14257 1.702246
3 3.6 6011 0.270562 20. 410 7 77.02 5 3 4.43 13 0.2252 5 1 .367 64 5. 70336
3 3. 25372 0.261413 21.102500 0.724 06 3 4.616366 0.216621 1 .5 44 5 0.40 150
40 3. 5 260 0.252572 21.355072 4.55027 40 4. 01021 0.20 2 1 .7 2774 5.025516
41 4.0 7 34 0.244031 21.5 104 .50 537 41 4. 3061 0.20027 1 . 3052 . 26536
42 4.24125 0.23577 21. 34 3 2.607371 42 5.1 27 4 0.1 2575 20.1 5627 104. 1 5
43 4.3 702 0.227 06 22.0626 6. 4 62 43 5.4004 5 0.1 516 20.3707 5 110.0123 2
44 4.543342 0.220102 22.2 27 1 101.23 331 44 5.616515 0.17 046 20.54 41 115.412 77
45 4.70235 0.21265 22.4 5450 105.7 1673 45 5. 41176 0.1711 20.720040 121.02 3 2
46 4. 66 41 0.20546 22.700 1 110.4 4031 46 6.074 23 0.164614 20. 4654 126. 7056
47 5.0372 4 0.1 520 22. 43 115.350 73 47 6.317 16 0.15 2 3 21.042 36 132. 453 0
4 5.2135 0.1 1 06 23.0 1244 120.3 257 4 6.57052 0.1521 5 21.1 5131 13 .263206
4 5.3 6065 0.1 5320 23.276564 125.601 46 4 6. 3334 0.146341 21.341472 145. 33734
50 5.5 4 27 0.17 053 23.45561 130. 7 10 50 7.1066 3 0.140713 21.4 21 5 152.6670 4
 ppen i Co o n nterest a es 775

r D 0:05 r D 0:06
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.050000 0. 523 1 0. 523 1 1.000000 1 1.060000 0. 433 6 0. 433 6 1.000000

2 1.102500 0. 0702 1. 5 410 2.050000 2 1.123600 0. 6 1. 333 3 2.060000
3 1.157625 0. 63 3 2.72324 3.152500 3 1.1 1016 0. 3 61 2.673012 3.1 3600
4 1.215506 0. 22702 3.545 51 4.310125 4 1.262477 0.7 20 4 3.465106 4.374616
5 1.2762 2 0.7 3526 4.32 477 5.525631 5 1.33 226 0.74725 4.212364 5.6370 3
6 1.3400 6 0.746215 5.0756 2 6. 01 13 6 1.41 51 0.704 61 4. 17324 6. 7531
7 1.407100 0.7106 1 5.7 6373 .14200 7 1.503630 0.665057 5.5 23 1 .3 3 3
1.477455 0.676 3 6.463213 .54 10 1.5 3 4 0.627412 6.20 7 4 . 746
1.55132 0.64460 7.107 22 11.026564 1.6 47 0.5 1 6. 016 2 11.4 1316
10 1.62 5 0.613 13 7.721735 12.577 3 10 1.7 0 4 0.55 3 5 7.3600 7 13.1 07 5
11 1.71033 0.5 467 .306414 14.2067 7 11 1. 2 0.5267 7. 6 75 14. 71643
12 1.7 5 56 0.556 37 . 63252 15. 17127 12 2.0121 6 0.4 6 6 .3 3 44 16. 6 41
13 1. 564 0.530321 .3 3573 17.712 3 13 2.132 2 0.46 3 . 526 3 1 . 213
14 1. 7 32 0.50506 . 641 1 .5 632 14 2.260 04 0.442301 .2 4 4 21.015066
15 2.07 2 0.4 1017 10.37 65 21.57 564 15 2.3 655 0.417265 .71224 23.275 70
16 2.1 2 75 0.45 112 10. 37770 23.6574 2 16 2.540352 0.3 3646 10.105 5 25.67252
17 2.2 201 0.4362 7 11.274066 25. 40366 17 2.6 2773 0.371364 10.477260 2 .212 0
1 2.40661 0.415521 11.6 5 7 2 .1323 5 1 2. 5433 0.350344 10. 27603 30. 05653
1 2.526 50 0.3 5734 12.0 5321 30.53 004 1 3.025600 0.330513 11.15 116 33.75 2
20 2.6532 0.376 12.462210 33.065 54 20 3.207135 0.311 05 11.46 21 36.7 55 1
21 2.7 5 63 0.35 42 12. 21153 35.71 252 21 3.3 564 0.2 4155 11.764077 3 . 2727
22 2. 25261 0.341 50 13.163003 3 .505214 22 3.603537 0.277505 12.0415 2 43.3 22 0
23 3.071524 0.325571 13.4 574 41.430475 23 3. 1 750 0.2617 7 12.30337 46. 5 2
24 3.225100 0.31006 13.7 642 44.501 24 4.04 35 0.246 7 12.55035 50. 15577
25 3.3 6355 0.2 5303 14.0 3 45 47.7270 25 4.2 1 71 0.232 12.7 3356 54. 64512
26 3.555673 0.2 1241 14.3751 5 51.113454 26 4.54 3 3 0.21 10 13.003166 5 .1563 3
27 3.733456 0.267 4 14.643034 54.66 126 27 4. 22346 0.20736 13.210534 63.705766
2 3. 2012 0.2550 4 14. 127 5 .4025 3 2 5.1116 7 0.1 5630 13.406164 6 .52 112
2 4.116136 0.242 46 15.141074 62.322712 2 5.41 3 0.1 4557 13.5 0721 73.63 7
30 4.321 42 0.231377 15.372451 66.43 4 30 5.7434 1 0.174110 13.764 31 7 .05 1 6
31 4.53 03 0.22035 15.5 2 11 70.7607 0 31 6.0 101 0.164255 13. 2 0 6 4. 01677
32 4.764 41 0.20 66 15. 02677 75.2 2 32 6.4533 7 0.154 57 14.0 4043 0. 77
33 5.0031 0.1 73 16.00254 0.063771 33 6. 405 0 0.1461 6 14.230230 7.343165
34 5.25334 0.1 0355 16.1 2 04 5.066 5 34 7.251025 0.137 12 14.36 141 104.1 3755
35 5.516015 0.1 12 0 16.3741 4 0.320307 35 7.6 60 7 0.130105 14.4 246 111.4347 0
36 5.7 1 16 0.172657 16.546 52 5. 36323 36 .147252 0.122741 14.620 7 11 .120 67
37 6.0 1407 0.164436 16.7112 7 101.62 13 37 .6360 7 0.1157 3 14.7367 0 127.26 11
3 6.3 5477 0.156605 16. 67 3 107.70 546 3 .154252 0.10 23 14. 4601 135. 04206
3 6.704751 0.14 14 17.017041 114.0 5023 3 .703507 0.103056 14. 4 075 145.05 45
40 7.03 0.142046 17.15 0 6 120.7 774 40 10.2 571 0.0 7222 15.0462 7 154.761 66
41 7.3 1 0.1352 2 17.2 436 127. 3 763 41 10. 02 61 0.0 171 15.13 016 165.0476 4
42 7.7615 0.12 40 17.42320 135.231751 42 11.557033 0.0 6527 15.224543 175. 50545
43 .14 667 0.122704 17.545 12 142. 333 43 12.250455 0.0 1630 15.306173 1 7.507577
44 .557150 0.116 61 17.662773 151.143006 44 12. 54 2 0.07700 15.3 31 2 1 .75 032
45 . 500 0.1112 7 17.774070 15 .700156 45 13.764611 0.072650 15.455 32 212.743514
46 .43425 0.105 7 17. 0066 16 .6 5164 46 14.5 04 7 0.06 53 15.524370 226.50 125
47 . 05 71 0.100 4 17. 1016 17 .11 422 47 15.465 17 0.06465 15.5 02 241.0 612
4 10.401270 0.0 6142 1 .07715 1 .0253 3 4 16.3 3 72 0.060 15.650027 256.56452
4 10. 21333 0.0 1564 1 .16 722 1 .426663 4 17.377504 0.057546 15.707572 272. 5 401
50 11.467400 0.0 7204 1 .255 25 20 .347 6 50 1 .420154 0.0542 15.761 61 2 0.335 05
776 A A Co o n nterest a es

r D 0:07 r D 0:0
n .1 C r/n .1 C r/!n an r sn r n .1 C r/n .1 C r/!n an r sn r

1 1.070000 0. 3457 0. 3457 1.000000 1 1.0 0000 0. 25 26 0. 25 26 1.000000

2 1.144 00 0. 7343 1. 0 01 2.070000 2 1.166400 0. 5733 1.7 3265 2.0 0000
3 1.225043 0. 162 2.624316 3.214 00 3 1.25 712 0.7 3 32 2.5770 7 3.246400
4 1.3107 6 0.762 5 3.3 7211 4.43 43 4 1.3604 0.735030 3.312127 4.506112
5 1.402552 0.712 6 4.1001 7 5.75073 5 1.46 32 0.6 05 3 3. 2710 5. 66601
6 1.500730 0.666342 4.766540 7.1532 1 6 1.5 6 74 0.630170 4.622 0 7.335 2
7 1.6057 1 0.622750 5.3 2 .654021 7 1.713 24 0.5 34 0 5.206370 . 22 03
1.71 1 6 0.5 200 5. 712 10.25 03 1. 50 30 0.54026 5.74663 10.63662
1. 3 45 0.543 34 6.515232 11. 77 1. 005 0.50024 6.246 12.4 755
10 1. 67151 0.50 34 7.0235 2 13. 1644 10 2.15 25 0.4631 3 6.7100 1 14.4 6562
11 2.104 52 0.4750 3 7.4 674 15.7 35 11 2.33163 0.42 3 7.13 64 16.6454 7
12 2.2521 2 0.444012 7. 426 6 17. 451 12 2.51 170 0.3 7114 7.53607 1 . 77126
13 2.40 45 0.414 64 .357651 20.140643 13 2.71 624 0.3676 7. 03776 21.4 52 7
14 2.57 534 0.3 7 17 .74546 22.5504 14 2. 371 4 0.340461 .244237 24.214 20
15 2.75 032 0.362446 .107 14 25.12 022 15 3.17216 0.315242 .55 47 27.152114
16 2. 52164 0.33 735 .44664 27. 054 16 3.425 43 0.2 1 0 . 5136 30.3242 3
17 3.15 15 0.316574 .763223 30. 40217 17 3.70001 0.27026 .12163 33.750226
1 3.37 32 0.2 5 64 10.05 0 7 33. 033 1 3. 601 0.25024 .371 7 37.450244
1 3.61652 0.27650 10.3355 5 37.37 65 1 4.315701 0.231712 .6035 41.446263
20 3. 6 6 4 0.25 41 10.5 4014 40. 54 2 20 4.660 57 0.21454 . 1 147 45.761 64
21 4.140562 0.241513 10. 35527 44. 65177 21 5.033 34 0.1 656 10.016 03 50.422 21
22 4.430402 0.225713 11.061240 4 .00573 22 5.436540 0.1 3 41 10.200744 55.456755
23 4.740530 0.210 47 11.2721 7 53.436141 23 5. 71464 0.170315 10.37105 60. 32 6
24 5.072367 0.1 7147 11.46 334 5 .176671 24 6.3411 1 0.1576 10.52 75 66.76475
25 5.427433 0.1 424 11.6535 3 63.24 03 25 6. 4 475 0.14601 10.674776 73.105 40
26 5. 07353 0.1721 5 11. 2577 6 .676470 26 7.3 6353 0.135202 10. 0 7 7 . 54415
27 6.213 6 0.160 30 11. 670 74.4 3 23 27 7. 061 0.1251 7 10. 35165 7.35076
2 6.64 3 0.150402 12.137111 0.6 76 1 2 .627106 0.115 14 11.05107 5.33 30
2 7.114257 0.140563 12.277674 7.34652 2 .317275 0.10732 11.15 406 103. 65 36
30 7.612255 0.131367 12.40 041 4.4607 6 30 10.062657 0.0 377 11.2577 3 113.2 3211
31 .145113 0.122773 12.531 14 102.073041 31 10. 6766 0.0 2016 11.34 7 123.345 6
32 .715271 0.114741 12.646555 110.21 154 32 11.7370 3 0.0 5200 11.434 134.213537
33 .325340 0.107235 12.7537 0 11 . 33425 33 12.676050 0.07 11.513 145. 50620
34 . 7 114 0.10021 12. 5400 12 .25 765 34 13.6 0134 0.073045 11.5 6 34 15 .626670
35 10.6765 1 0.0 3663 12. 47672 13 .236 7 35 14.7 5344 0.067635 11.65456 172.316 04
36 11.423 42 0.0 7535 13.03520 14 . 13460 36 15. 6 172 0.062625 11.7171 3 1 7.10214
37 12.22361 0.0 1 0 13.117017 160.337402 37 17.245626 0.057 6 11.77517 203.070320
3 13.07 271 0.076457 13.1 3473 172.561020 3 1 .625276 0.0536 0 11. 2 6 220.315 45
3 13. 4 20 0.071455 13.264 2 1 5.6402 2 3 20.1152 0.04 713 11. 7 5 2 23 . 41221
40 14. 7445 0.0667 0 13.33170 1 .635112 40 21.724521 0.046031 11. 24613 25 .05651
41 16.022670 0.062412 13.3 4120 214.60 570 41 23.4624 3 0.042621 11. 67235 2 0.7 1040
42 17.144257 0.05 32 13.45244 230.632240 42 25.33 4 2 0.03 464 12.0066 304.243523
43 1 .344355 0.054513 13.506 62 247.7764 6 43 27.366640 0.036541 12.043240 32 .5 3005
44 1 .62 460 0.050 46 13.557 0 266.120 51 44 2 .555 72 0.033 34 12.077074 356. 4 646
45 21.002452 0.047613 13.605522 2 5.74 311 45 31. 2044 0.03132 12.10 402 3 6.505617
46 22.472623 0.0444 13.650020 306.751763 46 34.4740 5 0.02 007 12.13740 41 .426067
47 24.045707 0.0415 7 13.6 160 32 .2243 6 47 37.232012 0.026 5 12.164267 452. 00152
4 25.72 07 0.03 67 13.730474 353.2700 3 4 40.210573 0.024 6 12.1 136 4 0.132164
4 27.52 30 0.036324 13.7667 37 . 000 4 43.42741 0.023027 12.212163 530.342737
50 2 .457025 0.033 4 13. 00746 406.52 2 50 46. 01613 0.021321 12.2334 5 573.770156
PP a e of e ecte
nte ra s
R F C .a C bu/
Z nC1
u
un du D C C; n ¤ !1
nC1
Z
du 1
D ln ja C buj C C
a C bu b
Z
u du u a
D ! 2 ln ja C buj C C
a C bu b b
Z 2
u du u2 au a2
D ! 2 C 3 ln ja C buj C C
a C bu 2b b b
Z ˇ ˇ
du 1 ˇˇ u ˇˇ
D ln ˇ CC
u.a C bu/ a a C bu ˇ
Z ˇ ˇ
du 1 b ˇˇ a C bu ˇˇ
D ! C 2 ln ˇ CC
u2 .a C bu/ au a u ˇ
Z ! "
u du 1 a
D 2 ln ja C buj C CC
.a C bu/2 b a C bu
Z
u2 du u a2 2a
D ! ! ln ja C buj C C
.a C bu/2 b2 b3 .a C bu/ b3
Z ˇ ˇ
du 1 1 ˇˇ u ˇˇ
D C 2 ln ˇ CC
u.a C bu/2 a.a C bu/ a a C bu ˇ
Z ˇ ˇ
du a C 2bu 2b ˇˇ a C bu ˇˇ
D! 2 C 3 ln ˇ CC
u2 .a C bu/2 a u.a C bu/ a u ˇ
Z ˇ ˇ
du 1 ˇ a C bu ˇ
D ln ˇ ˇCC
.a C bu/.c C ku/ bc ! ak ˇ c C ku ˇ
Z hc i
u du 1 a
D ln jc C kuj ! ln ja C buj C C
.a C bu/.c C ku/ bc ! ak k b
p
F C a C bu
Z
p 2.3bu ! 2a/.a C bu/3=2
u a C bu du D CC
15b2
Z
p 2. a2 ! 12abu C 15b2 u2 /.a C bu/3=2
u2 a C bu du D CC
105b3
Z p
u du 2.bu ! 2a/ a C bu
p D CC
a C bu 3b2
Z p
u2 du 2.3b2 u2 ! 4abu C a2 / a C bu
p D CC
a C bu 15b3

777
778 A B a e of e ecte nte ra s

Z ˇp ˇ
ˇ a C bu ! pa ˇ
du 1 ˇ ˇ
p D p ln ˇ p p ˇ C C; a>0
u a C bu a ˇ a C bu C a ˇ
Z p Z
a C bu du p du
D 2 a C bu C a p
u u a C bu

p
F C a2 ! u2
Z
du u
D p CC
.a2
!u /2 3=2
a aˇ2 ! u2
2
Z p ˇ
du 1 ˇˇ a C a2 ! u2 ˇˇ
p D ! ln ˇ ˇCC
u a2 ! u2 a ˇ u ˇ
Z p
du a2 ! u2
p D! CC
u2 a2 ! u2 a2 u
Z p 2 ˇ ˇ
p ˇ a C pa2 ! u2 ˇ
a ! u2 du ˇ ˇ
D a2 ! u2 ! a ln ˇ ˇ C C; a>0
u ˇ u ˇ

p
F C u2 ˙ a2
Z p ˇ ˇ$
1# p 2 ˇ p ˇ
u2 ˙ a2 du D u u ˙ a2 ˙ a2 ln ˇu C u2 ˙ a2 ˇ C C
2
Z p u p a4 ˇˇ p ˇ
ˇ
u2 u2 ˙ a2 du D .2u2 ˙ a2 / u2 ˙ a2 ! ln ˇu C u2 ˙ a2 ˇ C C
Z p 2 ˇ ˇ
p ˇ a C pu2 C a2 ˇ
u C a2 du ˇ ˇ
D u C a ! a ln ˇ
2 2 ˇCC
u ˇ u ˇ
Z p 2 p ˇ p ˇ
u ˙ a2 du u2 ˙ a2 ˇ 2 ˙ a2 ˇˇ C C
D ! C ln ˇ u C u
Z u2 u
ˇ p ˇ
du ˇ ˇ
p D ln ˇu C u2 ˙ a2 ˇ C C
u2 ˙ a2 ˇp ˇ
Z
du 1 ˇˇ u2 C a2 ! a ˇˇ
p D ln ˇ ˇCC
u u2 C a2 a ˇ u ˇ
Z ! p ˇ ˇ"
2
u du 1 ˇ p ˇ
p D u u ˙ a " a ln ˇu C u ˙ a ˇˇ C C
2 2 2 ˇ 2 2
u2 ˙ a2 2
Z p
du ˙ u2 ˙ a2
p D! CC
u2 u2 ˙ a2 a2 u
Z ˇ ˇ
u 2 p 3a4 ˇˇ p ˇ
2 2 3=2
.u ˙ a / du D .2u ˙ 5a / u ˙ a C 2 2 2 ln ˇu C u ˙ a ˇˇ C C
2 2

Z
du ˙u
2 2 3=2
D p CC
.u ˙ a / a u2 ˙ a2 ˇ
2
Z ˇ
u2 du !u ˇ p ˇ
2 2 3=2
Dp C ln ˇu C u ˙ a ˇˇ C C
ˇ 2 2
.u ˙ a / u2 ˙ a2

R F C a2 ! u2 u2 ! a2
Z ˇ ˇ
du 1 ˇˇ a C u ˇˇ
D ln CC
a2 ! u2 2a ˇ a ! u ˇ
Z ˇ ˇ
du 1 ˇˇ u ! a ˇˇ
D ln CC
u2 ! a2 2a ˇ u C a ˇ
 ppen i a e of e ecte nte ra s 779

E L F
Z
eu du D eu C C
Z
au
au du D C C; a > 0; a ¤ 1
Z ln a
au
e
ueau du D 2 .au ! 1/ C C
Z a Z
n au un eau n
u e du D ! un!1 eau du
Z a a Z au
eau du eau a e du
D ! C ; n¤1
Z un .n ! 1/un!1 n!1 un!1
ln u du D u ln u ! u C C
Z
unC1 ln u unC1
un ln u du D ! C C; n ¤ !1
nC1 .n C 1/2
Z Z
unC1 m m
un lnm u du D ln u ! un lnm!1 u du; m; n ¤ !1
ˇ n Cˇ 1 n C 1
Z
du ˇ ˇ
D ln ˇ ln uˇˇ C C
ˇ
u ln u
Z ! ˇ ˇ"
du 1 ˇ ˇ
D ˇ cu ˇ
cu ! ln ˇa C be ˇ C C
a C be cu ac

M F
Z r p
aCu p p
du D .a C u/.b C u/ C .a ! b/ ln. a C u C b C u/ C C
bCu ˇ ˇ
Z p
du ˇa C b ˇ
p ˇ
D ln ˇ C u C .a C u/.b C u/ˇˇ C C
.a C u/.b C u/ 2
Z p
2cu C b p
a C bu C cu2 du D a C bu C cu2
4c
ˇ ˇ
b2 ! 4ac ˇˇ p p ˇ
! 3=2
ln ˇ2cuCbC2 c a C bu C cu ˇˇ CC; c > 0
2
c
 PP reas n er
t e tan ar
or a C r e

780
 ppen i C reas n er t e tan ar or a C r e 781

z
A(z) = 3 1 e-x2/2 dx
0 2r

z A(-z) = A(z)
0

z .00 .01 .02 .03 .04 .05 .06 .07 .0 .0

0.0 .0000 .0040 .00 0 .0120 .0160 .01 .023 .027 .031 .035
0.1 .03 .043 .047 .0517 .0557 .05 6 .0636 .0675 .0714 .0753
0.2 .07 3 .0 32 .0 71 .0 10 .0 4 .0 7 .1026 .1064 .1103 .1141
0.3 .117 .1217 .1255 .12 3 .1331 .136 .1406 .1443 .14 0 .1517
0.4 .1554 .15 1 .162 .1664 .1700 .1736 .1772 .1 0 .1 44 .1 7
0.5 .1 15 .1 50 .1 5 .201 .2054 .20 .2123 .2157 .21 0 .2224
0.6 .2257 .22 1 .2324 .2357 .23 .2422 .2454 .24 6 .2517 .254
0.7 .25 0 .2611 .2642 .2673 .2704 .2734 .2764 .27 4 .2 23 .2 52
0. .2 1 .2 10 .2 3 .2 67 .2 5 .3023 .3051 .307 .3106 .3133
0. .315 .31 6 .3212 .323 .3264 .32 .3315 .3340 .3365 .33
1.0 .3413 .343 .3461 .34 5 .350 .3531 .3554 .3577 .35 .3621
1.1 .3643 .3665 .36 6 .370 .372 .374 .3770 .37 0 .3 10 .3 30
1.2 .3 4 .3 6 .3 .3 07 .3 25 .3 44 .3 62 .3 0 .3 7 .4015
1.3 .4032 .404 .4066 .40 2 .40 .4115 .4131 .4147 .4162 .4177
1.4 .41 2 .4207 .4222 .4236 .4251 .4265 .427 .42 2 .4306 .431
1.5 .4332 .4345 .4357 .4370 .43 2 .43 4 .4406 .441 .442 .4441
1.6 .4452 .4463 .4474 .44 4 .44 5 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .45 2 .45 1 .45 .460 .4616 .4625 .4633
1. .4641 .464 .4656 .4664 .4671 .467 .46 6 .46 3 .46 .4706
1. .4713 .471 .4726 .4732 .473 .4744 .4750 .4756 .4761 .4767
2.0 .4772 .477 .47 3 .47 .47 3 .47 .4 03 .4 0 .4 12 .4 17
2.1 .4 21 .4 26 .4 30 .4 34 .4 3 .4 42 .4 46 .4 50 .4 54 .4 57
2.2 .4 61 .4 64 .4 6 .4 71 .4 75 .4 7 .4 1 .4 4 .4 7 .4 0
2.3 .4 3 .4 6 .4 .4 01 .4 04 .4 06 .4 0 .4 11 .4 13 .4 16
2.4 .4 1 .4 20 .4 22 .4 25 .4 27 .4 2 .4 31 .4 32 .4 34 .4 36
2.5 .4 3 .4 40 .4 41 .4 43 .4 45 .4 46 .4 4 .4 4 .4 51 .4 52
2.6 .4 53 .4 55 .4 56 .4 57 .4 5 .4 60 .4 61 .4 62 .4 63 .4 64
2.7 .4 65 .4 66 .4 67 .4 6 .4 6 .4 70 .4 71 .4 72 .4 73 .4 74
2. .4 74 .4 75 .4 76 .4 77 .4 77 .4 7 .4 7 .4 7 .4 0 .4 1
2. .4 1 .4 2 .4 2 .4 3 .4 4 .4 4 .4 5 .4 5 .4 6 .4 6
3.0 .4 7 .4 7 .4 7 .4 .4 .4 .4 .4 .4 0 .4 0
3.1 .4 0 .4 1 .4 1 .4 1 .4 2 .4 2 .4 2 .4 2 .4 3 .4 3
3.2 .4 3 .4 3 .4 4 .4 4 .4 4 .4 4 .4 4 .4 5 .4 5 .4 5
3.3 .4 5 .4 5 .4 5 .4 6 .4 6 .4 6 .4 6 .4 6 .4 6 .4 7
3.4 .4 7 .4 7 .4 7 .4 7 .4 7 .4 7 .4 7 .4 7 .4 7 .4
3.5 .4 .4 .4 .4 .4 .4 .4 .4 .4 .4
This page intentionally left blank
 ns ers to ere
Pro e s
p p p p
false !13 is not a real number 11x ! 2y ! 3 6t2 ! 2s2 C 6 a C 5 3b ! c
p p
false the natural numbers are 1, 2, 3, and so on 7x2 C 7xy ! 2z 3
3y ! 4 4z !15x C 15y ! 27
p
false 3 is not rational 2x2 ! 33y2 ! 7xy 6x2 C 6
p
false 25 D 5; a positive integer !12u3 ! u2 C u ! 20 6x2 C 11x ! 10
false we cannot divide by 0 2
! 3 ! 10 10x2 C 1 x C 6 X2 C 4X C 4 2
p
true see Figure 0.1 in the text 4 ! 14X C X2 5x ! 4 5x C 4 4s2 ! 1
true we can regard a terminating sequence as being the same as x3 C 4x2 ! 3x ! 12 3x4 C 2x3 ! 13x2 ! x C 4
the sequence obtained from it by appending infinitely many zeros 1 t3 ! 111t2 ! 24t 3s2 ! st ! 2t2 C 11s C t ! 4
a3 C 36a2 C 54a C 27 x3 ! 36x2 C 54x ! 27
1 15
false false false false false z!1 2u3 C 3u ! 2 xC2!
3u xC5
distributive associative !37
3x2 ! x C 17 C x2 ! 2x C 4 !
commutative and distributive the definition of division xC2 xC2
distributive and commutative b!a 7
x!2C
3x C 2
5 !1 0: a
1
!7x 6Cy 6 !ab
3 5b.x C 1/ 5x.2y C z/
X 20 C 4x 0 X !x abcd2 .3a2 ! 4b2 c C 2a2 c3 d/ .z C 7/.z ! 7/

3ab by 10 x C 3y aCc .p C 3/.p C 1/ .5y C 2/.5y ! 2/ .a C 7/.a C 5/

c x xy 3a b .y C 3/.y C 5/ 5.x C 3/.x C 2/ 3.x C 1/.x ! 1/

17 6y .5x C 1/.x C 3/ 2s.3s C 4/.2s ! 1/

x2 0; for X ¤ 0
12 ! a2=3 b.a3 ! 4b2 / 2x.x C 3/.x ! 2/
x yz
4.2x C 1/2 x.xy ! /2 .x C 2/.x ! 2/2
.a C 3b/.2x C y/.2x ! y/ .b C 4/.b2 ! 4b C 16/
x a21
25 .D32/ 177 x6 y .x C 1/.x2 ! x C 1/.x ! 1/.x2 C x C 1/
y14 b20
1 2.x C 4/2 .x C 1/.x ! 2/ P.1 C r/2
x4 y4 5 !2
2 .4u ! /.4u C /
1 1 p .y4 C 1/.y2 C 1/.y C 1/.y ! 1/
7 27 5 2
4 16 .X 2 C 5/.X C 1/.X ! 1/ .a C 2/2 .a ! 2/2 b
p p p p
x32 7u4 4 2 ! 15 3 C 4 3 2
t2 a5 3 1
3z2 x2 ! 3x C x!5
4 b3 c2 ab2 c3 t2 for x ¤ !3
x xC5
x =4 z3=4 p
51=5 x2=5 x1=2 ! y1=2 5
.a ! b C c/3 5x C 2 1 y2 b ! ax
y1=2 for x ¤ !
p p xC7 3 .y ! 3/.y C 2/ ax C b
1 3 1 6 5 2 2x
p
5
p
5
!p
5 3
x4 3 27 3 5 x for x ¤ !2 and x ¤ !1 and x ¤ 1
p p
20
x
5
.3b/4 16a10 b15 2x6 X 2
2 5 for u; ¤ 0
3b ab y3 2 3
312 y6 4y4 2x2
xyz !27x2 1 for x ¤ !1=2 and x ¤ 3
x2 4 x2 x!1
a10 c1 .2x C 3/.1 C x/
x2 y5=2 x 1 for x ¤ !6; !3; !2; 5 ! xC2
b24 xC4

AN-1
AN-2 ns ers to ere Pro e s
p
20x C 3 1 3x2 C 1 5 1
˙ ;˙ 3; 0 =2; 21=5
5x2 1 ! x3 .x C 1/.3x ! 1/ 5 2
2.x C 2/ 35 ! x x 3 1
; !1 6; !2 ! ;1
.x ! 3/.x C 1/.x C 3/ .x ! 1/.x C 5/ xC1 2 2
5; !2 No real roots !2
x 5x C 2 .x C 2/.6x ! 1/ p
1 ! xy 3x 2x2 .x C 3/ 5 ! 21
6 4;
p p p p p 2
3. 3 x ! 3 x C h/ a! b 6C2 3
p p ! 0 1 " 64:15; 3:35
3
xCh 3 x a2 ! b 3
p 6 inches by inches
p 3t ! 3 7
15 ! 3 1 year and 10 years age 23 never
t2 ! 7
a s b 5.4 s or 2.6 s
D 1:065=..0:75/.1:15// " 1:2347 26
R C
.bc/17=5 1
17 p p for h ¤ 0
0 !2 a xChC x
4 r
adding 3 equivalence guaranteed !.2x C h/ S
for h ¤ 0 rD n !1 34 :2
raising to fourth power equivalence not guaranteed .x C h/2 x2 P
dividing by x equivalence not guaranteed
multiplying by x ! 1 equivalence not guaranteed
4 5
multiplying by .2x ! 3/=2x equivalence not guaranteed 120 ft 64 oz of A, 0 oz of B
12 p
3:14=! 1 !1 1600 ml D5! 60=! " 0:63 m.
5
27 26 1
! 5=6 126 ! " 13;077 tons 4000 at 6 , 16;000 at 7
4 2
37 25 1 4.25 r # 7:7217345 0 000
! tD 2
1 52 5 120 to brea approximately even
2 ad ! bc 7
; if a ¤ c 116.25 30 0,000
14 a!c 2
7 43 either 440 or 460 100
tD! 3 1 11
4 16 42 115 m by 67:5 m
13 10 4
xD ! 2 23 11.51 cm long, 6.51 cm wide
2 36
100E
I pC1 S!P 232,000 60 acres
a D !2 rD qD tD 100 ! p
Pt r Pr
Ai S 2mI 125 of A and 100 of B or 150 of A and 125 of B.
RD rD n !1 nD !1
1 ! .1 C i/!n P rB
A I
170 m c D 1:065x 3 years
5375
2172
" 46:2 hours 20 150 ! x4 # 0 3x4 ! 210 # 0 x4 C 60 # 0 x4 # 0
47
d d
tD Ic D r ! " 602 ft " 13
r!c t ! "
1
.7; 1/ .!1; 4" !1; !
2
2 t D !7 or t D 3 3; !1 4; ˙2
7 4 1
2 3 -
2
0; 5 !1=2; 3=2 1; 5; !2 0; ! #
3 2 2
1 4 !1; .0; 1/ Œ1; 1"
0; 1; !4 0; 1; !1; !3 0; ; ! 3; ˙2 7
2 3
7 p 2 0 1
3; 4 4; !6 1˙2 2 7
3 ! # p !
p 2 3!2
!5 ˙ 57 ! ;1 ; !1;
no real roots 40; !25 7 2

p p 1 2
no real roots ˙ 3; ˙ 2 3; -
7
3- 2
2 2
 ns ers to ere Pro e s AN-3

.!1; 4 / .3=7; 1/ .!1; 1/ no, second is 1 5 times first

1
256
48 3
7
120
! # 22:5; 23:4; 24:3; 25:2; 26:1
17
;1 Œ!12; 1/ .0; 1/ 6; 4:5; 3; 1:5; 0
1=2; !1=22 ; 1=23 ; !1=24 ; 1=25
17 -12 0
9 100; 105; 110:25; 115:7625; 121:550625
.!1; 0/ .!1; !2" 600 < S < 1 00 55 1024 6
50.1 ! .1:07/!10 /
600 1800 "1 : 0 6
0 -2 1 ! .1:07/!1
x < 70 degrees not possible, jrj D 17 > 1 2000
33 0
50; 000.1:0 /11 " 116;5 2
120,001 17,000 214,2 6 .12;000 C 1 ;000/ D 124;000
2
x # 2 ; 000 1000 t > 36:7 1 500.1:05/!1
" 6 77:00 D 10;000
1 ! .1:05/!1
A I no, differences are not common
j ! 22j $ 0:3 a 2; 4; 6; ; 10; : : :
b 2; 4; ; 16; 32; : : :
c 2; 4; 16; 256; 65; 536; : : :
13 2 7 !4 < x < 4
p d 2; 4; 16; 65; 536; 265;536 ; : : :
10 ! 3
a jx ! 7j < 3 b jx ! 2j < 3 c jx ! 7j $ 5 R C
! # ! "
d jx ! 7j D 4 e jx C 4j < 2 f jxj < 3 2 5
g jxj > 6 h jx ! 105j < 3 i jx ! 50j < 100 x # !4 ;1 ; !1;
3 2
jp1 ! p2 j # 5 ˙7 ˙35
2 .!1; 1/ !5=3; 3 .!1; 4/
f!4; 14g x D 1=5; 1 ! " $ #
5 1 7
.! ; / .!1; ! / [ . ; 1/ !1; ! [ ;1 4320
2 2
.!10; !4/ .!1; 0/ [ .1; 1/
$ # 542 10; 000 c < 212; 14
16
Œ1=2; 5=2" .!1; 0" [ ;1 100; 102; 104:04; 106:120 ; 10 :243216
3
100.1 ! .1:02/5 /
jd ! 35:2j $ 0:2 " 520:40
!0:02
.!1; # ! h$/ [ .# C h$; 1/
A I

P60 a a.r/ D !r2 b .!1; 1/ c r # 0

12, 17, t 45 532 iD36 i 300
P6 P a t.r/ D b .!1; 1/ ! f0g c r > 0
kD2 3
k
iD1 2
i ; 750 5 r
300 % x& 600 % x & 1200
4 3 d t.x/ D t D t D
37,750 030 2 5,425 x 2 x 4 x
200 %x&
.n C 1/.2n C 1/ 300c
15 ! e time scaled by a factor of c t D
2n2 c x
a 300 b 21.00 per pizza c 16.00 per pizza
A I 5500 400 11, 00 16,000
1 3, 201, 21 , 237, 255, 273
. :57.1:06/k!1 /4kD1
1225, 1213, 1201, 11 , 1177, 1165, 1153 f¤g h¤k
21620, 1 0, 1 2 , 16 35 .!1; 1/ ! f1g .!1; !1/ [ Œ2; 1/
' (
220.5 44, 65.1 7
.!1; 1/ .!1; 1/ !
2
' (
1
.!1; 1/ ! f2g .!1; 1/ ! ! ; 2
2:3 1 23 3
.!1 C .k ! 1/3/4kD1 ..!1/kC1 2k /4kD1 3; !7; 13 !62, 2 ! u2 , 2 ! u4
no, first term of first is 64 that of second is !26 10; 2 ! 2 ; 2x2 C 4ax C 2a2 ! x ! a
AN-4 ns ers to ere Pro e s

4; 0; x2 C 2xh C h2 C 2x C 2h C 1 1
a 2x2 C x ! 1 b !x ! 1 c ! d x4 C x3 ! x2 ! x
2x ! 5 xCh!5 2
0, 2 , 0; ; 1=4 x!1
4x C 1 x2 C 2xh C h2 C 1 e for x ¤ !1 f 3 g x4 C 2x3 C x2 ! 1
x
a 4x C 4h ! 5 b 4 h x4 ! x2 i 72
a x2 C 2hx C h2 C 2x C 2h b 2x C h C 2 4 14 2
6 !32 2
C C1 2
a 3 ! 2x ! 2h C 4x2 C xh C 4h2
b !2 C x C 4h .t ! 1/ t!1 t C 7t
s
1 !1 2 2 C3
a b for h ¤ 0 f.x/ D x ! 7, g.x/ D 11x
xCh!1 .x ! 1/.x C h ! 1/ 3 !2 2 !3
3 for h ¤ 0
3
y is a function of x x is a function of y g.x/ D x2 C x C 1, f.x/ D is a possibility
x
y is a function of x x is not a function of y 3
g.x/ D x2 C x f.x/ D is another
es .t/ D 50; 000 C .2300/t P is a function of q xC1
402.72 35.52 supply increases as price increases p x2 ! 1
p p f.x/ D 4
x, g.x/ D
a 4 b 3 2 c f.2I0 / D 2 3 2f.I0 / xC3
p a r.x/ D :75x b e.x/ D 4:25x C 4500
doubling intensity increases response by a factor of 2 3 2
c .r ! e/.x/ D 5:5x ! 4500
a 3000, 2 00, 2300, 2000 12, 10
b 10, 12, 17, 20 3000, 2300 12.20m ! m2 / revenue from output of m employees
a !5:13 b 2.64 c !17:43 a 14.05 b 116 .64 a 1 4.47 b 0.2
a 6: 4 b 40:2 c 0:67

A I x 7
f!1 .x/ D ! F!1 .x/ D 2x C 14
a p.n/ D 125 b premiums do not change 3 3
c constant function r
A
a quadratic function b 2 c 3 r.A/ D
8 4! ! # ! #
< 3:50n if n $ 5 1 7
c.n/ D 3:00n if 5 < n $ 10 not one-to-one for example g ! D Dg !
: 3 3
2:75n if n > 10
5
7Š D 5040 h.x/ D .5x C 12/2 , for x # ! , is one-to-one
p 12
23 5 1;200;000
xD C qDs ,p>0
4 4 p
yes no yes no
yes, is one-to-one
.!1; 1/ .!1; 1/ a 4 b 5
a 7 b 1 , , 2, !1, 0, 2
7, 2, 2, 2 362; 0 2 n A I
! #
f.I/ D 2:50, where I is income constant function 145
y D !600x C 7250 x-intercept ;0
a C D 50 C 3q b 250 12
8̂ y-intercept .0; 7250/
ˆ 0:075 if 0 $ j $ 44; 701
<
0:11 if 44; 701 < j $ ; 401 y D 24: 5 horizontal line no x-intercept
c.j/ D y-intercept (0,24. 5)
ˆ 0:13 if ; 401 < j $ 13 ; 5 6 64
:̂
0:145 if 13 ; 5 6 < j
y
17 33
a all such that 30 $ $ 3 b 4, ,
4 4 36 (2.5, 30)
a 11 2.74 b 4 5.27 c 252.15
Miles

24
a 2.21 b . c !14:52
12
(5, 0)
x
A I 1 2 3 4 5 hours
(0, 0)
c.s.x// D c.x C 3/ D 2.x C 3/ D 2x C 6
let side length be .x/ D x C 3 y
let area of square with side length x be a.x/ D x2
Cost (dollars)

(100, 59.3)
then g.x/ D .x C 3/2 D . .x//2 D a. .x// 60

40 (70, 37.1)

20
a 2x C b c !2 d x2 C x C 15 x
xC3 (0, 0) 20 40 60 80 100 therms
e 3 f g xC h 11 i xC 11
xC5
 ns ers to ere Pro e s AN-5

(0,0) function one-to-one .!1; 1/ .!1; 1/

y
3 rd 4 th 2 nd on positive x-axis

Q.II Q.I
6 x
(- 52 , 4)
2 (6, 0)
x
-4 2 6
(-1, -3) -4 (4, -2)
(0, 0) not a function of x
Q.III Q.IV
y
a 1, 2, 3, 0 b .!1; 1/ c .!1; 1/ d !2
a 0, 1, 1 b .!1; 1/ c Œ0; 1/ d 0
(0,0) function one-to-one .!1; 1/ .!1; 1/
x
y

(0,2), (1,0) function one-to-one .!1; 1/ .!1; 1/

x
y

! # x
5 1
.0; !5/, ;0 function one-to-one .!1; 1/ .!1; 1/
3

y
.!1; 1/ 3 (0, 3)
u y

5
x
3
3
-5 v
1
1 x

.0; 0/ is only intercept

domain .!1; 1/ range .!1; 1/
y intercepts are .0; !1/ and .1; 0/
p
.!1; 1/ Œ!3; 1/ (0, 1), .2 ˙ 3; 0/
y
x

2- 3
1
x
2+ 3
y is a function of x is one-to-one both are .!1; 1/ (2, -3)
every point on y-axis not a function of x

y .!1; 1/ .!1; 1/ (0,0)

f(t)

x
t
AN-6 ns ers to ere Pro e s

.!1; !3" [ Œ!3; 1/ Œ0; 1/ .!3; 0/, (3,0) .n/ D 700 ! 300n .0; 700/ initial debt .2 ; 0/ number of
months to clear debt
s as price increases, quantity increases p is a function of q
p

r 50
-3 3

30

10
f(x) q
30 90 150 210

y
2

x 1000
( 27 , 0)

Œ0; 1/ .0; 2/, .2=7; 0/

300
.!1; 1/ ! f0g .0; 1/ no intercepts

F(t) x
7 14 21

0.3 !0:61; !0:04

f(x) !1:70; 0

(-3.59, 0)
x
Œ0; 1/ Œ1; /

a 1 .60 b !10: 6 a 5 b 4
8
a 2 b .!1; 2 " c !4:02, 0:60
5
f(x)

p 30
7
20
10
.!1; 1/ Œ0; 1/
0 x
1 2 3 4 5
g(x)

(0,0) sym about origin .˙2; 0/, .0; / sym. about y-axis
! # ! #
13 13
x ˙ ;0 0; ˙ sym about x-axis, y-axis, and origin not
3 5 12
sym about y D x
.!7; 0/ symmetric about x-axis sym. about x-axis
(a), (b), (d)
.!21; 0/; .0; !7/; .0; 3/ (1, 0), (0, 0)
! #
2
0; no sym of the given inds
27
 ns ers to ere Pro e s AN-7

.0; !2/, .0; 2/, . =5; 0/ symmetric about x-axis

y
y

-5
y = x3 - 1
x
-5 5 x
5 f (x) = x3

.˙2; 0/, .0; 0/ sym about origin translate 2 units left stretch result vertically from x-axis by factor
of 3
y
y

5
-5
x x
-2 2 5
-5

y y
(0,0) sym about x-axis, y-axis, origin, y D x
f(x) = x
y
2
f(x) = 1x x
1 -1 y = x + 1 -2
x
x –1 1 -2
-1
2
y=
-2 3x

! # ! #
5 5
˙ ; 0 , 0; ˙ sym about x-axis, y-axis, origin shift y D x3 three units left, two units up
3 2
y
y
5 y = 2 + (x+3)3
2 y = x3

x
5 5
-3 3
y

5
-2 y= -x f(x) = x

a .˙0: ; 0/, .0; 5/ b 5 c .!1; 5"

x
y

translate 5 units right and 1 unit up shrin result by a factor of 1 2

vertically towards x-axis and re ect about x-axis
(0, 3)
re ect about y-axis move 5 units down

x A I
(0, - 3 ) (3, 0)
2
(- 3 , 0) a 3260 b 4410
2

3 !6 !1 17
a2 C 2ab C b2 C 2ah C 2bh C h2
AN-8 ns ers to ere Pro e s

00 yD2 zD6 (0, 4) only intercept, symmetry y

z
about y-axis
z
6
1
3
4
1 1 y
2
x
x y
2
4
x
.0; 2/, .!4; 0/ Œ!4; 1/ Œ0; 1/ G(v)
z z

4 2
v
-4

y 2
x
3 6
y ! #
x 1
0; .!1; 1/ ! f4g Œ0; 1/ g(t)
2
z
1
2
3
t
4

3 y
.!1; 1/ .!1; 2"
3 y
x

2
y

x+y=3 x

x+y=2
3
shrin by a factor of 1 2 towards the x-axis re ect in the x-axis
x+y=1
2 and translate up by 2
y
1
y = x2
2
x
1 2 3
x

1
y = - 2 x2 + 2
R C
.!1; 1/ ! f1; 5g .!1; 1/ (a) and (c)
!0:67; 0:34; 1:73 !1:50; !0: ; !0:11; 1:0 ; 1:40
Œ2; 1/ ! f3g. 5, 1 , 40, 2! 2 ! 3! C 5
p p a .!1; 1/ b (1. 2,0), (0,7)
p
4
p
4 3 xC4 u
0, 2, t ! 2, x3 ! 3 , 0, , f(x)
5 x u!4
20, !3, !3, undefined a 1 ! 3x ! 3h b !3 for h ¤ 0
a 3.x C h/2 C .x C h/ ! 2 b 6x C 1 C 3h for h ¤ 0
a 5x C 2 b 22 c x ! 4 d 6x2 C 7x ! 3 e 10
3x ! 1 x
f g 6x C h 3 i 6x C 1
2x C 3
1 1 1 C x2 p
2
, 2 C1D 2
x3 C 2, .x C 2/3=2
.x C 1/ x x if k is even then symmetry about the y-axis if k is odd then no axial or
only intercept .0; 0/ symmetric about the origin original symmetries
 ns ers to ere Pro e s AN-9

z y 7 .0; 0/ 0I 3
2 5
10 2x C 3y ! 5 D 0 y D ! x C
3 3
4 5
10
1.5 4x C y ! 5 D 0 y D x C
3
10 y 1 3 57
6x ! y ! 57 D 0 y D x!
x 4
0.5 parallel parallel neither
perpendicular perpendicular y D 2x C 7
x
2.5 5 7.5 1
yD1 yD xC5 xD5
3
2 2
A I yD! x! no such point
3 3
depreciating at rate of 4000 per year !2: price dropped an average of 2. 0 per year
y y D 2 ;000x ! 100;000 !t C d ! 1 4 D 0
C D .61:34/ C 3:14 the slope is 7:1.
60
A I
x D number of s is y D number of boots x C 14y D 1000
3
x p D q C 1025
20
answers may vary, two possibilities: .0; 60/ and .2; 140/
S D 14 C FD C C 32 f(x)
5
125 125 1000
slope D I y-intercept D
3 3
C ! 5F C 160 D 0
F 500

100
x
10 20
C f.t/ D 2:3t C 32:2 f.x/ D 70x C 150
-100 100

-100
!4 0 5I !7
slopes of sides are 0, 7, and 1 no pair of which y h(t )
are negative reciprocals no sides perpendicular so
not a right triangle

x t
1
4=3 ! undefined 0
2
5x C y ! 2 D 0 -7
2x ! 3y C 10 D 0
y 1 5 p(q)
!
3 3
10 5
3 3

x q
-5

3x ! 7y C 25 D 0 4x C y C 16 D 0 f.x/ D 4x C 3 f.x/ D !2x C 4

2x ! y C 4 D 0 x C 2y C 6 D 0 2 10
f.x/ D ! x ! f.x/ D x C 1
y D !2 x!2D0 3
3 4
4I !6 ! p D ! q C 24: 0 1 .50
5 5 25
slope undefined no y-intercept q D .1=2/p ! 3=2 c D 3q C 10 115
AN-10 ns ers to ere Pro e s

f.x/ D 0:125x C 4:15 vertex: .!3; 0/ .!3; 0/, .0; / range: Œ0; 1/
D !1 0t C 1 00 slope D !1 0 v
s

1800 9

t
10 t
y D 53x C 65 -3
C D 500 C 3n 500 3
x C 10y D 100
35 225 ! # ! "
a yD xC b 52:2 1 17 17
44 11 vertex: ;! .0; !5/ range !1; !
2 4 4
a p D 0:05 t C 0:025 b 0:556
1 y
a t D c C 37 b add 37 to # of chirps in 15 seconds
4
A I -1 1 1
x
2
vertex: .1; 400/ intercepts: vertex: .1; 24/ intercepts:
p ! ( 12 , - 174 )
.0; 3 /, .!1 ; 0/, .21; 0/ 6
y .0; /, 1 ˙ ;0 -5
2
400 y
30
p
vertex: .4; !2/ .0; 14/ .4 ˙ 2; 0/ range: Œ!2; 1/
100
x t
-25 25
x
-5 5

14

1000 units 3000 maximum revenue

4- 2 4+ 2
s
quadratic not quadratic (4, -2)
quadratic quadratic
! #
5 13
a ! ; b lowest point minimum 1 4=23 maximum !10
6 12
! # p
1 25 g!1 .x/ D 1 C x ! 3, x # 3
a !6 b !3; 2 c ! ; !
2 4
y
f(x)
g(x)

x
-1 7

g-1(x)
(3, -16)

vertex .3; !16/ intercepts .!1; 0/, .7; 0/, .0; !7/ range Œ!16; 1/ x
! # ! "
3
vertex: ! ; .0; 0/, .!3; 0/ range: !1;
2 2 2
y

9
2

-3
x
3
-2
max revenue 250 when q D 5
200 units 240,000 maximum revenue
 ns ers to ere Pro e s AN-11

maximum of 51; 250 at 225

P(x) p p
.1 ˙ 13; 5 ' !2 13/ .!3; !4/ .2; 1/
400 p p !
1˙ 7 6˙ 7
no solution ;
2 2
p p
.0; 0/ .1; 1/ .˙ 17; 2/ .˙ 14; !1/
.7; 6/
x
-20 30 y ! 4 D 4.x ! 2/ or y ! 4 D 4x ! or y D 4x ! 4
two .!1:3; 5:1/
70 grams " 134. 6 ft " 2.7 sec x D 1:76 x D !1:46
! #
45 224
vertex ; h-intercept .0; 14/
16 16
p !
45 C 13 % 173 equilibrium .75; 7:75/ (5,212.50)
for t # 0, t-intercept ;0
16 p
h(t )

11 (75, 7.75)

5
q
75

t ( ,3 ) (15,5)
45
16 brea -even quantity is 2500 units

125 ft & 250 ft p

TR
15,000 TC
A I
120,000 at and 0,000 at (2500, 10000)
5000
500 of species A and 1000 of species
q
20;000 4 2000 6000
infinitely many solutions of form A D ! r,
3 3
B D r, where 0 $ r $ 5000
1 1 1 cannot brea even cannot brea even
lb of A lb of lb of C
6 3 2
a 12 b 12.1
5 40 units 40 units 1 40 units 4
x D !1, y D 1 .2; !1/ a .1; 1/, .4; 2/
u D 6, D !1 x D !3, y D 2 b
no solution x D 12; y D !12
y
1 1 1
; xD ,yD ,zD
2 2 4 10
x D 2, y D !1, z D 4
x D 0 ! 2p, y D 3 C p, z D 0 C p, p in .!1; 1/
1 5 q
x D ! r, y D r, z D r r in .!1; 1/ -10 10
3 3
f.5 C 3r ! s; r; s/ j r; s in .!1; 1/g c maximum profit for q in .1; 4/
1 2
533 gal of 20 soln, 266 gal of 35 soln decreases by 0.70 PA D PB D 10
3 3
0.5 lb of cotton 0.25 lb of polyester 0.25 lb of nylon 2.4 and 11.3
675 m per h in still air, 75 m per h speed of wind y
240 units early American, 200 units contemporary
40
550 at Exton, 450 at Whyton
4 on first 100,000, 6 on remainder
1 0 boxes, 760 clamshells
100 chairs, 50 side tables, 0 coffee tables q
12
10 semis illed wor ers, 5 s illed wor ers, 55 shipping cler s
AN-12 ns ers to ere Pro e s

R C y x D 0:75, y D 1:43
y D !2x ! 1 2x C y C 1 D 0
(13.26, 0.17)
y D 3x ! 21 3x ! y ! 21 D 0 yD7 y!7D0
2 x
y D x ! 3 2x ! 5y ! 15 D 0 perpendicular
5
neither parallel, both lines have slope 5
4
y D .5=3/x ! 7=3 slope 7=3 yD 0
3
A I
y
graph shapes are the same
A scales second coordinate by A
ultiplicative
( 57 , 0) ear Increase Expression
x

(0,- 73 ) 0 1 1:10
1 1.1 1:11
2 1.21 1:12
!5 .0; 17/ (3,0), .!3; 0/, (0, ) (0, ) 3 1.33 1:13
y y 4 1.46 1:14

17 9 1.1 investment increases by 10 every year

.1 C 1.0:1/ D 1 C 0:1 D 1:1/ y
17
2
5 -3 3
x x
1

x
intercept (0,3) vertex .!1; 2/ 2 4 6 8 years
y between 7 and years

ultiplicative
ear ecrease Expression
3
0 1 0: 50
t 1 0. 5 0: 51
-1 1
2 0.72 0: 52
3 0.61 0: 53
.0; !3/ .!1; !2/
s y 0. 5 car depreciates by 15 every year
.1 ! 1.0:15/ D 1 ! 0:15 D 0: 5/ y
x
-1 -2 2
t -3
1

x
1 2 3 4 5 years
!5 .0; 0/
between 4 and 5 years
! # ! # y D 1:0 t!3 shift graph 3 units right
17
;! 2; ! .4; 0/ .1; 2; 3/ 36 4. 7 16 4. 7 117 employees
7 7 5
p p ! P
!5 ˙ 65 !21 ˙ 5 65
;
4 1
.!2 ! 2r; 7 C r; r/ r in .!1; 1/
.r; r; 0/ r in .!1; 1/ 2a C 3b C D 0 a D !
f.x/ D .!3=2/x ! =2 50 units 5000 " 6:55 t
10 20 years
1250 units 20,000 2.36 tons per sq m
 ns ers to ere Pro e s AN-13

y
y y

(-1, 5) x
1
4 -1
x

1
x " 0:16 0 .ek /t , where b D ek
a 12 b . c 3.1 d 22 hr
y y
27 yrs 0.1465 y

8 9
5

x
-5 5
2 -5
1
x x
-1 1 -2

y 3.17 4.2 min yrs

y

3
9

1 8 A I
x 7 t D log2 16 number of times the bacteria have doubled
1
I
6 D 10 :3 y
I0
5
6 y = log1.5x
4
3 3

2 x
5 10 multiplicative
1
increase
x
-2 -1 1 2 3 y
-1
-2 8
y = log0.8x
1 3 7 4
A 13 ,750 ; ;
2 4
x
a " 231 .55 b " 31 .55 1 multiplicative
a 1 64.76 b 1264.76 decrease

a 1 ,30 .16 b 15,30 .16 "13. " .2

a 6256.36 b 1256.36 a 64 .6 b 164 .6

" 6 00. 1 a D 400.1:05/t b 420 c 4 6

ear Factor
log 10;000 D 4 210 D 1024
0 1 ln 20:0 55 D 3 e1:0 61 D 3
1 1.32 y y
2 1.7424
3 2.2 6 1
x 1 4
4 3.035 5776 1 5 x
1
5 4.007464243 -1

334,4 5 4.4 17 0.44 3 -5

AN-14 ns ers to ere Pro e s

y y ln x
log x D ln 3
ln 10

1 1
x
A I
x
4 6 1 e
-1 1 day 20 67.5 times as intense
-2

2 4 1 !4 0 !3
1 1=2 2.75 !3 2
343 125 e!3 2 4
1000 0.125 .1=3/ ln 11 " 0:7 0.02 5.140
1 5 5 C ln 3
3 4 .ln 7/=5 !0:073 2.322
27 3 2
2.3 7 0 2.00013 y D log1:10 x 3 .1=5/..ln 11= ln 3/ ! 7/ " !0: 63 0.4 3
a 2 0 b k is the doubling time 2.4 6 1003 2.222 " !2:072
" 72.2 minutes zD y3=2
1.353 0.5 S D 12:4A0:26 a 100 b 46

a .0; 1/ b Œ!0:37; 1/ .ln 1: ! ln 1:5/= ln 1:02 " 11:

log. 0 ! q/
y 1.41 3.06 pD 4.32 3.33
log 2
5
(0.69, 2)

x R C
-5 5
log3 243 D 5 11=4 D 3 ln 10 6:63 " 7
1
-5
5 !4 !2 4
1024
! !
A I 73 x2 y
! # !6=5 3.a C 1/ log ln
00; 000 52 z3
log. 00; 000/ ! log. 000/ D log D log.100/ D 2
000 !
x1 =3
log.10; 000/ D log.104 / D 4 ln 7 ln x C 5 ln y C 3 ln z
.x ! 1/2 .x ! 2/3
1 1
.ln x C ln y C ln z/ .ln y ! ln z/ ! ln x
3 2
aCbCc a!b 3a ! c 2.a C c/
b ln.x C 5/ 5:20 45
4 !7 2:77 " " 1: 5565 n
a ln 3 2: 0735
1 2 C2
! 2 .1=5/x ! 4y 2x C 1 y D ex
2
ln x C 2 ln.x C 1/ 2 ln x ! 3 ln.x C 1/ y
3.2 ln x C ln.x C 2/ ! ln.x C 1//
ln x C ln.x C 1/ ! ln.x C 2/
8
1
ln x ! 2 ln.x C 1/ ! 3 ln.x C 2/
2
2 1
ln x ! ln.x C 1/ ! ln.x C 2/
5 5
.2x/3 1
log 24 log2 .xC2/5 log3 .57 % 174 / x
-3
1
log.100.1:05/10 / 1 no solution
64 1
%z& 4 !4 10 0. 0
ln.2x C 1/ ln.x2 C 1/ 3e
2
f!3; 1g y D ln
ln 2 ln 3 7 ln ! 2 " 0:07 44 " 0:4 0
amounts to ln.B C E/ D ln.B C E/ C ln B ! ln B a 3 2 .04 b 122 .04 14
a P D 6000.0: 5/t b " 5707
7
a 10 b 10e!0:41 " 6:6 c 10e!0:41.5/ " 1:3
ln 2 ln.100/
d " 1:7 e " 11:2
0:41 0:41
-10 10 a " 7 b " 60
1
.!1; 0:37" 2.53 g.x/ D
-2 x
 ns ers to ere Pro e s AN-15

y
42 .73 502. 4
5
a 221.43 b 25 c 1 6.43

-5 5
x multiply all entries by A D 10; 000:
-5 Prin. ut. Principal utstanding at eginning
Prin. Repd. Interest Repaid at End

Period Prin. ut. Interest Payment Prin. Repd.

A I 1 ! .0:05/a3 0:05
1
4. 7 years, 16 days 7.720 1) 1 0.05
a3 0:05 a3 0:05
11.25 compounded quarterly is better the 10,000 investment a2 0:05 a2 0:05 1 ! .0:05/a2 0:05
1
is better over 20 years 2) .0:05/
a3 0:05 a3 0:05 a3 0:05 a3 0:05
a1 0:05 a1 0:05 a 1 ! .0:05/a1 0:05
3) .0:05/ 1 0:05 .1:05/
a3 0:05 a3 0:05 a3 0:05 a3 0:05
a 11,105.5 b 5105.5
" 2:7 5 " 3.562 3 3
Total !1 1
a 10 b 10.25 c 10.3 1 a3 0:05 a3 0:05
d 10.471 e 10.516
.0 .0 years 41 :41 Prin. uts. Interest Pmt. Prin.
3044 .33 a 1 b 1 .56 at for at Repaid
Period eginning Period End at End
31 .54 compounded annually
a " 3:35 b " 3:30 10.757 6.2 1 00.00 22.50 1 3.72 171.22
2 72 .7 1 .22 1 3.72 175.50
3 553.2 13. 3 1 3.72 17 .
2261.34 1751. 3 5 21.55 4 373.3 .33 1 3.72 1 4.3
4 62.31 " 702 :03 11,3 1. 5 1 .00 4.73 1 3.73 1 .00
14,0 1.10 123 .5 3244.63 Total 6 .61 6 .61 00.00
a " ! 15; 35 :1635 2 b not profitable
savings account 226.25 .55
13 1606
a 20 .6 b 1 7 .33 c 211.36 d 3 1, 07
54 113,302 25.64
5 1 . 7 1 1 . 7 2217.30 " 2:0201
3.05 10 .42 77 , 00.7
a 43,24 .06 b 20,737.6 4000 1; 00;000 4 00 1 e2
" 2: 55 16 years
R C
a 1072.51 b 10 3.30 c 1072.1
ln 2
a 45 .51 b this is better by 26. 0 .5 compounded annually
ln.1 C r/
1005.41 a 1 7.13 b 3325.37
A I
" 66 :62 6. 314.00
6.20 101, 25 121, 25 723.03
13, 62.01 45,502.06 4 ,0 5.67 Prin. uts. Interest Pmt. Prin.
at for at Repaid
Period eginning Period End at End

" 23:0 1244 .2131 0 2 50.3 1 15,000.00 112.50 3067. 4 2 55.34

2 , 4.06 7.0 " 172; 562:13 2 12,044.66 0.33 3067. 4 2 77.51
3 067.15 6 .00 3067. 4 2 . 4
204, 77.46 24,5 4.36 5106.27
4 6067.31 45.50 3067. 4 3022.34
1332.73 a " 77 6:23 b 43 6:23
5 3044. 7 22. 4 3067. 1 3044. 7
3474.12 1725 102. 1305
Total 33 .17 15,33 .17 15,000.00
10,475.72 66.30 1, 72, 4.02
205,073 142,146 1 1,26 .25 127 .36
AN-16 ns ers to ere Pro e s

A I 0 24 4
xD ;yD! x D 6; y D
3 & 2 or 2 & 3 2 2 2 33
2 3 45 105
1 2 4 16 x D !6; y D !14; z D 1 4 1750 1400 5
41 2 4 165 50 60
1 2 4 16 $ " $ "
15 !4 26 !10 22 12
1.16
4 7 30 24 36 !44

a 2 & 3I 3 & 3I 3 & 2I 2 & 2I 4 & 4I 1 & 2I 3 & 1I 3 & 3I 1 & 1 A I

b B, , E, ,
c , upper triangular , lower triangular 57 0 22, 43.75
2 3 2 3
d F e G $ "
61 57 6 7
6 A24 D !2 0 5410 6 7 y D657
4 1 5 x 455
$ " $ " 1
135 4 16 25 3 3
a AD b CD
024 16 25 36

120 entries, 1, 0, 1, 0
2 3 !12 1 1 2 & 2, 4
000 $ "
0000
a 40 0 05 b 3 & 5 15
2 3
2&1 2 3&1 3 3&1 3
0000
000 1000 $ " $ "
2 3 60 1 0 07 12 !12 23
2 0 7 6 7
$ " 40 0 1 05 10 6 50
6 2 6 5 3 7
6 7 0001
!3 4 4 !3 6 !2 5 2 3
0 2 1 1 !4 2
4 2 2 45 Œ!4 5 !1 !1 "
a A and C b all of them
!3 !2 3
x D 2, y D 7, z D !5 x D 0; y D 0 $ " $ " $ "
a 4 b 4 c February d none for either 63 70 7 4 !5 !
e anuary f anuary g 37 72 0 !21 !12 !5 !20
2 3 2 3 2 3
3 1 1 z $ " 500
4y5 2x1 C x2 C 3x3 40 7 05
61 7 4 7
!2001 6 7 4x1 C x2 C 7x3
44 3 1 5 x 005
27 3
2 6 2
0 0 2 3
$ " 63 7 !1 5
6 7
!1 !20 60 7 07 4 2 175
!2 23 6 3 7
A I 4 75 1 31
$ " 0 0
230 220 3
x1 D 670; x2 D 35; x3 D 1405 2 3
1 0 255 0 0 !4 $ "
4 2 !1 !2 5 !1 2
undefined
1 0
0 0
2 3
2 3 $ " 2 0 0 $ "
$ " !3 !4 0 3 0 1 !1 0
10 1 40 2 0 5
4 !4 ! 5 Œ!4 !2 10" !1 !1 2 0 1 1
33 0 0 2
!2 6
$ " $ " $ "$ " $ "
!12 36 !42 !6 6 !7 3 1 x 6
undefined D
!42 !6 !36 12 !7 2 ! y 5
2 3 2 32 3 2 3
0 13 $ "
4 2 2 !1 3 r
4 5 !4 5 0 45 !1 25 4 s 5 D 4 5 5
6 !6
1 6 !5 3 !2 2 t 11
$ " $ " 2075 2 , 50
66 51
undefined
0 !10 12 a 1 0,000 520,000 400,000 270,000 3 0,000 640,000
2 3 2 3 b 3 0,000 100,000 00,000 c 2,3 0,000
1 6 134 4 7
4! 3 ! 3 5 4 2 !35 110 12
d
!32 26 20 2 23 23
$ " $ " $ "
7 16 72: 2 ! : 15:606 64:0
undefined 51:32 !36:32 !73 :42 373:056
5
 ns ers to ere Pro e s AN-17
8̂2 3 2 3 2 3 9
A I ˆ 0 0 !1 >
>
<6 7 6 7 6 7 =
5 bloc s of A 2 bloc s of 1 bloc of C 6 2 7 C r 6 !3 7 C s 6 !4 7 j r; s in .!1; 1/
ˆ4 0 5 4 15 4 05 >
>
3 of 4 of 2 of :̂ ;
0 0 1
A D 3 I D 1000 ! 2 I C D 500 ! I D any amount 8̂2 3 2 3 2 3 9
ˆ 3 3 !1 >
>
between 0 and 500 <6 7 6 7 6 7 =
0 1
6 7 C r6 7 C s6 0 7 j r; s in .!1; 1/
ˆ 24 5 4 0 5 4 2 5 >
:̂ >
;
0 0 1
8̂2 3 2 3 2 3 2 3 9
not reduced reduced not reduced ˆ 1 !2 1 !2 >
>
ˆ
ˆ >
2 3 <6 4
6 7
7 6 !1 7
6 7
6 !2 7
6 7
6 17
6 7
>
=
2 3 1000 6 0 7 C r 6 1 7 C s 6 0 7 C t 6 0 7 j r; s; t in .!1; 1/
$ " 1 2 3 6 7 6 7 6 7 6 7
60 1 0 07 ˆ
ˆ >
>
1 0 40 0 05 6 7 ˆ4 0 5 4 05 4 15 4 05 >
>
0 1 40 0 1 05 :̂ ;
0 0 0 0 0 0 1
0001
infinitely many trivial solution only
x D 5, y D 2 no solution $ "
0
2 5 1 7 infinitely many
x D ! r C ; y D ! r C ; z D r, for all r in .!1; 1/ 0
3 3 6 6 8̂ 2 3 9
2 3 6 >
ˆ
ˆ ! 7 >
>
2 ˆ
< 6 57 >
=
6
no solution X D 405 6
r6 7 j r in 1/
! 7
.!1;
ˆ
ˆ >
>
3 ˆ 4 15 5 >
>
:̂ ;
x D 3, y D !2, z D 1 1
8 2 3 9
x1 D r; x2 D 0; x3 D 0; x4 D 0; x5 D r for all r in .!1; 1/ < 1 =
federal, 72,000 state, 24,000 x D 0, y D 0 r 4 !2 5 j r in .!1; 1/
: ;
1
A, 2000 , 4000 C, 5000 8̂ 2 3 9
ˆ !2 >
>
a 3 of , 4 of 2 of , 1 of , 5 of 1 of , 2 of , 6 of < 6 =
6 !3 7
7
3 of , 7 of b 3 of , 4 of c 3 of , 4 of 3 of , 7 of r4 5 j r in .!1; 1/
8 ˆ 1 >
>
:̂ ;
< 12s C 20d C 32g D 220 stoc A 1
a Solve 16s C 12d C 2 g D 176 stoc
:
s C 2 d C 36g D 264 stoc C
2 3 A I
1015
b 40 1 1 5 yes EET AT N N FRI A
0000 2 3
s D 5 ! r, d D ! r, g D r, r in .!1; 1/ 2 1 1
6 3 ! ! 7
c .s; d; g/ in f.5; ; 0/; .4; 7; 1/; .3; 6; 2/; .2; 5; 3/; .1; 4; 4/; .0; 3; 5/g 6 6 37
6 1 5 17
d C D C.s; d; g/ D 300s C 400d C 600g. E!1 D6
6! 3 ! 7 F is not invertible
s d g 300.s/ C 400.d/ C 600.g/ D C 6 6 377
4 1 1 25
! !
5 0 300.5/ C 400. / C 600.0/ D 4700 3 6 3
4 7 1 300.4/ C 400.7/ C 600.1/ D 4600 A: 5000 shares : 2500 shares C: 2500 shares
3 6 2 300.3/ C 400.6/ C 600.2/ D 4500
2 5 3 300.2/ C 400.5/ C 600.3/ D 4400
1 4 4 300.1/ C 400.4/ C 600.4/ D 4300
0 3 5 300.0/ C 400.3/ C 600.5/ D 4200 2 3
minimum C of 4200 for .s; d; g/ D .0; 3; 5/ $ " 1 0 0
!1 1 4 1=2 !1=4 0 5
not invertible
7 !6
!1=4 1= 1=2
A I
not invertible not invertible
infinitely many solutions 2 3 2 3
8̂ 2 3 9 1 !2 1 1 0 2
1 > 40
ˆ
ˆ
ˆ ! >
> 1 !25 40 1 0 5
< 6 6 2 7
7
>
= 0 0 1 3 0 7
r6
6
1 7 j r in .!1; 1/
7
ˆ
ˆ ! >
> 2 3
ˆ 4 25 >
> 11 1
:̂ ; 6 3 !3
1 2
!1=5 2=5 !6=5
3
6 377
6 7 27
4 3=5 !1=5 3=5 5 6! ! 7
6 3 7
!1=5 2=5 !1=5 6 3 37
8̂2 3 2 3 9 4 2 15
ˆ !1 7 >
> !1
< 6 27 6 7 = 3 3
6 7 C r 6 !5 7 j r in .!1; 1/ $ " $ " $ "
ˆ4 4 5 4 !7 5 >
> 27 17 2
:̂ ; XD XD XD
0 1 3 !20 1
AN-18 ns ers to ere Pro e s
$ " $ "
coe
$ cient invertible
"$ " $ " 215 7 3 :7
1=10 3=10 7 1 141 35:1
D
3=10 !1=10 1 2
2 3 2 3 A I
0
1
X D 415 X D 4 !1 5 y > !1:375x C 62:5
2
2 !1 x C y # 50, x # 2y, y # 0
2 3
1 $ "
6 37 3 1
no solution XD6 7
4 !2 5 1=14
2 !4 y y
7
a 40 of model A, 60 of model 1
2
2
b 45 of model A, 50 of model 3
$ "$ " $ " x x
11 13 3 7
b .AB/!1 D D
15 24 11 23
yes
: 5000 shares E: 1000 shares F: 4000 shares
y y

1152 agriculture, 6 6 forestry 2

2 3 x x
6 0 !135 !4 2 ; 00 4
6 7
reduce 4 ! 0 540 !240 21; 600 5
! 0 !135 600 0
$ " $ "
12:5 220 y y
a XD b XD
1125 2 0
2 3 2 3
155 : 1 1073
X D 4 1112:44 5 X D 4 1016 5
173 :04 52 x x

R C
2 3
$ " 1 42 5
3 42 !1 !75
!16 !10 y
1 0 !2 y
$ " $ " $ "
11 !4 36 !1 !2
11 24 2 1
$ " $ " $ "
2 0 3 1 0 x x
XD
0 21 0 1
2 3
101 $ "
40 1 25 0
XD no solution
0
000
2 3 y y
3 5
6! 2 67
6 7 no inverse exists
4 1 15
!
2 6 x
x
2 32 3 2 3
1=2 !1=2 1=2 3 2
4 1=2 1=2 !1=2 5 4 4 5 D 4 1 5
!1=2 1=2 1=2 5 3
2 3
0 0 1 y y
A2 D 40 0 05 ; A3 D 0; A1000 D 0, no inverse
0 0 0
a x, y, z represent wee ly doses for I, II, III
x 4 3 2 1 x
there are four possibilities: y 6 3 0
z 0 1 2 3 x
b x D 1; y D 0; z D 3
 ns ers to ere Pro e s AN-19

y y

x + y … 100 .7; 0/ D 14 .1 ; 0; 0/ D 216 .0; 0; 4/ D 4

5
xÚ0
yÚ0 C.3; 0; 1/ D 0 .3; 0; 5/ D 2
100
10
x put A on 700,000 on 2,600,000 at cost 1,215,000
x
3 100
C.AC; SC; A ; S / D C.150; 0; 0; 150/ D 1050
x: number of lb from A a column 3:1,3,3: column 4:0,4,
y: number of lb from B
b x1 D 10, x2 D 0, x3 D 20, x4 D 0 c 0 in

x # 0, y # 0, 3x C 2y $ 240, 0:5x C y $ 0
A I

8̂ inimize D 60;000y1 C 2000y2 C 120y3 sub ect to

ˆ 300y1 C 20y2 C 3y3 # 300
<
1 45 220y1 C 40y2 C y3 # 200
P D 112 when x D ,yD0
2 2 ˆ 1 0y1 C 20y2 C 2y3 # 200
:̂
D !10 when x D 2, y D 3 y1 ; y2 ; y3 # 0
8
No optimum solution (empty feasible region) < 20y1 C y2 $ 6
inimize D y1 C 0y2 sub ect to 6y1 C 16y2 $ 2
C D 1 at .0; 1/ :
y1 ; y2 # 0
minimum =3, at .4=3; 4=3/
produce 5 of device 1, 0 of device 2 and 15 of device 3
No optimum solution (unbounded)
10 truc s, 20 spinning tops 110
4 units of food A, 4 units of food
! # minimize D 4y1 C 5y2 sub ect to
6500 25 125
CD at ;
3 3 6
minimum cost 4,600,000 using 6 A and 7 3y1 C 2y2 # 2

c x D 0, y D 200 !y1 C 3y2 # 3

D 15:54 when x D 2:56, y D 6:74 y1 ; y2 # 0
D !75: when x D :4 , y D 16:67
maximize D 4y1 ! 3y2 sub ect to

A I y1 C y2 $ 2
0 of Type 1, 72 of Type 2, 12 of Type 3 for max 20,400 !y1 C 4y2 $ !3
2y1 ! 3y2 $ 5
y1 ; y2 # 0
.0; 4/ D . =3; 5=3/ D 7 P.3; 2/ D 2
minimize D 13y1 ! 3y2 ! 11y3 sub ect to
.0; 5; 0/ D 20 .1; 0; 0/ D 2
.22=13; 50=13/ D 72=13 .3; 0; 2; 0/ D 310 !y1 C y2 ! y3 # 1
P.4; 1; 4; 0/ D 600 P.0; 2400/ D 1200 2y1 ! y2 ! y3 # !1
P.0; 300; 100/ D 10; 00
y1 ; y2 ; y3 # 0
A I
maximize D !3y1 C 3y2 sub ect to
plant I: 500 standard, 700 deluxe
plant II: 500 standard, 100 deluxe
maximum profit ,500 !y1 C y2 $ 4
y1 ! y2 $ 4
y1 C y2 $ 6
P.10=3; =3/ D 3 =3 . =3; 0; 5=3/ D 23=3
y1 ; y2 # 0
. ; 2; 0/ D 2 .3; 2/ D !17
no solution (empty feasible region) P.2; 5/ D 2 .0; 4=5; 7=5/ D 43=5 .3; 0/ D 15
D 000 at .200; 0/ .6; 7/ D 0 C.n; r/ D C.250; 1400/ D 1650

30 in A, 0 in AA, 70 in AAA 6.6 .s; ss; sc; sca/ D .0; 0; 40; 20/ D 1200
AN-20 ns ers to ere Pro e s

R C Green
Die Result
y y Red 1 1, 1
Die 2 1, 2
1
3 1, 3
4 1, 4
x x
2 5
-
3
5 1, 5
-3 6 1, 6

1 2, 1
y y 2 2, 2
2
3 2, 3
4 2, 4
5 2, 5
x x
6 2, 6

1 3, 1
2 3, 2
y 3
3 3, 3
4 3, 4
5 3, 5
6 3, 6
Start
x 1 4, 1
2 4, 2
3 4, 3
4 4 4, 4
5 4, 5
D 3 when x D 3, y D 0
! # 6 4, 6
70 20 10
D !2 when x D 0, y D 2 D at ;
1 5, 1
D 32 on .1 ! t/.0; 4/ C t.2; 3/ for 0 $ t $ 1 2 5, 2

. ; 0/ D 16 3 5, 3
! # 4 5, 4
5 5 5
D at 0; 0; D 24 when x1 D 0, x2 D 12 5 5, 5
3 3
6 5, 6
7 5
D when x1 D , x2 D 0, x3 D
2 4 4 1 6, 1
no solution ( has no upper bound) 2 6, 2
a minimum is .5; 0; 0/ D 5 there may be others 3 6, 3
0 units of , 6 units of , 14 units of 3 6 4 6, 4
500,000 gal from A to , 100,000 gal from A to C, 400,000 gal 5 6, 5
from to C 1 ,000 6 6, 6
10 g of food A only 36 possible results
D 12 : 3 when x D :3 , y D 1:63 60 6 1024 120 720 10;0 0
1000 displayed error message 6 336 12 6
11; 0 360 720 2520 5040 624
24 a 11, 0 b 1 ,00 4 4320
Assembly Finishing Production
line line route

D AD
A
E AE 35 1 360 11,62
D BD 74Š
Start B 4 5 56 415, 00
E BE 10Š % 64Š
D CD 15 40;320 16 0 252
C
E CE 756,756 40 3 4 712
6 possible production routes a 1 b 1 c 1 3744 5,250, 60
 ns ers to ere Pro e s AN-21

A I 1
1 .34
10,5 6, 00 12
a 45=100 D =20 4:1
b 45=100 D =20
c 5=100 D 1=20
f , H, C, Sg 7 3
f1HH, 1HT, 1TH, 1TT, 2HH, 2HT, 2TH, 2TT, 3HH, 3HT, 3TH, 7:3
12 10
3TT, 4HH, 4HT, 4TH, 4TT, 5HH, 5HT, 5TH, 5TT, 6HH, 6HT, 6TH, 0:7
D 3 W 11 " 56.
6TTg 0:22
f64, 6 , 60, 61, 46, 4 , 40, 41, 6, 4, 0, 1, 06, 04, 0 , 01, 16, 14,
1 , 10g
a
1 4 1 1
a b c d e 0
f.R; R; R/; .R; R; /; .R; R; B/; .R; ; R/; .R; ; /; .R; ; B/; 5 5 4 2
.R; B; R/; .R; B; /; .R; B; B/; . ; R; R/; . ; R; /; . ; R; B/; 1 2
1 0.3 a b
. ; ; R/; . ; ; /; . ; ; B/; . ; B; R/; . ; B; /; . ; B; B/; 2 3
.B; R; R/; .B; R; /; .B; R; B/; .B; ; R/; .B; ; /; .B; ; B/; 1 1 1 3
.B; B; R/; .B; B; /; .B; B; B/g a b c d
4 3 4
b 5 35 11 10 25
a b c d e f
5 3 25 47 6
f.R; ; B/; .R; B; /; . ; R; B/; . ; B; R/; .B; R; /; .B; ; R/g
1 4 2
a b
set of ordered 6-tuples of elements of fH,Tg 64 2 3
f.c; i/jc is a card; i D 1; 2; 3; 4; 5; 6g 312 1 1 1
a b
combinations of 52 cards ta en 4 at a time 270,725 2 4 3
f1, 3, 5, 7, g f2; 4; 6; g 1 1 5 1
f1, 2, 4, 6, , 10g S 11 6 12 13
E1 and E4 , E2 and E3 , E2 and E4 , E3 and E4 1 1
3 =51
E and G, F and I, G and , G and I 16; 575 5525 51
47 27
a SDf ; ; ; ; ; ; ; g a b a 2=3 b 4=7
b EDf ; ; ; g 100 47 20
c FDf ; ; ; ; ; ; g 1 11
" 1:4 0.04
d E[FDS e E\FDf ; ; g 4 00
f .E [ F/0 D S0 D ; 4
a 0:10 b 0:2 5
g .E \ F/0 D f ; ; ; ; g 31
a fA C, AC , AC, CA, CA , C Ag b fA C, AC g
c f AC, CA, CA , C Ag

1 5 1 2 1 1 1
a b c d e f g
4 6 3 3 12 2 3
500 a 0.5 b 0. no
7
5 1 1 1 1 1 5 independent independent
a b c d e f g 1
36 12 4 36 2 2 6
dependent dependent
1 1 1 1 1 1 4
a b c d e f g a independent
b dependent c dependent d no
52 4 13 2 2 52 13
1 1 1 3
h i 0 dependent
26 1 16 676
1 1 1 1 3 1 1
a b c d a b c
624 156 7 16 10 40 10
a 4=22100 D 0:0001 0 5 b 2 6=22100 D 0:012 41176 a 3=5 b 1=5 c 7=20 d 1 =20 e 1=20
1 3 1 7 4 1
a b c d a b 7 35 3 3
5 5 a b
54 162 11 200
a 0.1 b 0.35 c 0.7 d 0. 5 e 0.1, 0.35, 0.7, 0. 5 1 3
1 a b
a 1=510 D 0:000000102 172
7
120 & 47
13 % 4 C4 % 12 % 4 C1 13 % 12 % 4 a " 0:2013265 2 b " 0: 7 12611
D 510
52 C5 52 C5 c " 0:120 73 2
6545 4140
a " 0:040 b " 0:026 0.0106
161;700 161;700
AN-22 ns ers to ere Pro e s

a 0.1 b 5. c 1.56
a =17 b =23 "4 3 3
E.X/ D D 1:5 $ 2 D D 0:75 $ " 0: 7
2 4
25 14
a " 0:275 b " 0:005 6 3
37 3021 E.X/ D D 1:2 $ 2 D D 0:36 $ D D 0:6
5 6 27 5 25 5
1= " 0: 2
31 62 3 15 10
f.0/ D f.1/ D f.2/ D
3 24 4 2 2 2
" 0: 2
4 2 5 a !2:2222 b !4:4444
4 101.43 3.00
5 410 ! 0.50, 2.00
23 17
a b c
0 23 0
A I
R C
336 36 17,576,000 x P(x)

1024 210 462 2401

0
a 2024 b 253 34,650 1; 6 1; 600 10,000
4116
a f1, 2, 3, 4, 5, 6, 7g b f4, 5, 6g c f4, 5, 6, 7, g 1
d ; e f4, 5, 6, 7, g f no 10,000
2646
a fR1 R2 R3 ; R1 R2 G3 ; R1 G2 R3 ; R1 G2 G3 ; G1 R2 R3 ; G1 R2 G3 ; 2
G1 G2 R3 ; G1 G2 G3 g 10,000
b fR1 R2 G3 ; R1 G2 R3 ; G1 R2 R3 g 756
3
c fR1 R2 R3 ; G1 G2 G3 g 10,000
45 1
0.2 4
512 10,000

62 15 1 3
a D 1=4 b D 5=22 a b
122 66 16 p
6 10 16 1 2 2 2
3W5 0:33 f.0/ D f.1/ D f.2/ D #D $D
7 13 25 25 25 5 5
p
2 1 1 1 2 4 6
a b f.0/ D f.1/ D f.2/ D f.3/ D #D2 $ D
11 1 4 27 27 3
1 23 6 165
a b independent dependent 0:0307
3 34 625 204
1225 33 % 72 5
a .0:7/4 D 0:2401 b 0:4116 c 0:6517 a b
3456 2 %5 64 32
22 1
13 36 % 7
45 4 " 0: 0653 6 0.75
16 214
3
a 0.01625 b " 0:23
13

1 3
no no yes aD bD
# D 1:5 Var.X/ D 1:05 $ " 1:02 3 4
f(x) a D 0:7 b D 0:1 c D 0:2 yes no
$ "
0
X1 D D X2 D X3
0.4 1
0.3 $ " $ " $ "
0:42 0:416 0:416
0.2 X1 D X2 D X3 D
0:5 0:5 4 0:5 32
0.1
x 1 1
0 1 2 3 Π33 21 46 "T Π271 230 4 "T
100 1000
1
p Π276 241 4 13 "T
# D 25=12 Var.X/ D 3=144 $ D 3=12 10000
 ns ers to ere Pro e s AN-23
2 3 2 3 2
3 2 3 2 3
5 3 7 0:10 0:130 0:1310
6 7 6 16 16 7
7 b 3 c X1 D 4 0:15 5 X2 D 4 0:155 5 X3 D 4 0:15 5 5
a 2 D6
4
7 3 6
D4
3 55 75 16 0:75 0:715 0:70 5
16 16 $ " $ "
2 =25 =25 3 41=125 42=125
2 3 2 3 a D D
16=25 17=25 4=125 3=125
0:50 0:23 0:27 0:230 0:36 0:327
a 2 D 40:40 0:6 0:545 3 D 40:6 0 0:530 0:5435 b =25 c 4=125
0:10 0:0 0:1 0:0 0 0:101 0:130 1
Œ4 "T
D
b 0.40 c 0.36 13
2 3
3 a 76 b 74.4 apanese, 25.6 non- apanese
$ "
2=5 677 1 c 75 apanese, 25 non- apanese
6 7 Π22 20 3 "T
3=5 445 1
7 A I
exists if and only if a not an integer
$flu no "u
4
a flu 0:1 0:2 b 37 36 ! 3616 20 2
no flu 0: 0: 3

$A "
a A 0:7 0:4 b 0.61 a 1 b 0 c 1
0:3 0:6
2 3 a 1 b does not exist c 3
0:7 0:1 0:3
4 0 0:7 0:1 5. f.!0: / D !3:7 f.!0: / D !3: 7
a rows, columns labelled in order , C,
f.!0: / D !3: 7 f.!1:001/ D !4:003
0:3 0:2 0:6
b 20 c 24 f.!1:01/ D !4:03 f.!1:1/ D !4:3 !4

A Comp
$ "
a A 0: 0:3 b 65 c 60 x !0:001 !0:0001 0.0001 0.001 0.01 0.1
Comp 0:2 0:7 ln 2
2 3 f.x/ 0:6 2 0:6 31 0.6 32 0.6 34 0.6 56 0.7177
3 3
65 57 5
a 6 7
4 2 2 5 b Œ0:6 0:4"
T c Œ0:6 0:4"T 16 20 !47 ! 0
2
5 5 5 !2 3 5 !1=4
2 3
2 1 11
637 1 ! 4 2x 7
a 6 7
4 1 5 b 33 3
5
1
2x 3x2 ! x a 1 b 0
3 4
1:56 23076 4.00 does not exist

R C A I
f(x) limx!1 p.x/ D 0 graph decreases rapidly to 0
demand is a decreasing function of price
limx!1 y.x/ D 500 even with unlimited advertising sales
bounded by 500,000
0.5 limx!1 C.x/ D 1 cost increases without bound as production
0.3 increases without bound
0.2
x does not exist 250
1 2 3

# D 2:1 Var.X/ D 0:4 $ D 0:7

1 1 a 1 b 1 c !1 d does not exist e 0 f 0 g 0
a f.1/ D D f.7/ f.2/ D f.3/ D f.4/ D f.5/ D f.6/ D h 1 i 2 does not exist 2
12 6
b E.X/ D 4 12 !1 !1 1 0
! 0:10 a 176 b 704,000 1
C1 0 1 0
f.0/ D 0:522 f.1/ D 0:36 f.2/ D 0:0 2
f.3/ D 0:011, f.4/ D 0:0005 # D 0:6 $ " 0:71 2
0 1 0 1
53 2072 5
256=72 3125 16 1
24 % 35 3=7 !1 1
3 2
a D 0:3 b D 0:2 c D 0:5 C1 1 does not exist 1
AN-24 ns ers to ere Pro e s

0 1 R C
a 1 b 2 c does not exist d 1 e 2
5 2 x ! 0 3=11
a 0 b 0 c 0 d !1 e !1 3
1
6 c does not exist !1 !1
lim c = 6 C1 !1 1 !1 21
q q
continuous everywhere f is a polynomial function

6
x D !3 none x D !2; 3 xD2
q .!1; !6/ [ .2; 1/ .!1; 7"
5000
.!1; !5/ [ .!1; 1/ .!1; !1/ [ Œ0; 4/ [ Œ6; 1/
40; 000 20 1.00 0 Œ2:00; 1/
1, 0.5, 0.525, 0.631, 0. 12, 0. 6, 0. 1
0 a 11 b c does not exist A I
d
D 40 ! 32t
dt
continuous at !2 and 0 discontinuous at ˙5
continuous at 2 and 0 f is a polynomial function a
composite of continuous functions
x-value of !3 !2:5 !2:2 !2:1 !2:01 !2:001
none x D 23 none x D !5, 3
0, ˙3 none xD0 none xD2 mP 1 15.25 13.24 12.61 12.0601 12.0060
1, 2, 3 yes, no, no b estimate mtan D 12
y y
1 3 !7 0
6 5
600 2x C 4 6q C 2 p
0.20 x2 2 5x
0.16 !4 0 yDxC4
0.12 100
0.08
x y D 4x C 2 y ! 1 D .!1=5/.x ! 2/
5 10 15
r
t !3:000, 13.445
1 2 3 3 12 dC
r !r!
d
!5:120, 0.03
A I
if tangent at .a; f.a// horizontal then f0 .a/ D 0
0<x<4
D 10x4 ! 15x2

A I
.!1; !1/ [ .4; 1/ Œ!2; 5" .2=3; 5/
50 ! 0:6q
no solution .!1; !7" [ Œ!1; 2"
.!1; !4/ [ .0; 5/ Œ0; 1/
.!1; 0" [ Œ1; 2" .!1; !3/ [ .0; 3/ 0 17x16 0x7 1 x
.!1; 1/ .!1; !5/ [ Œ!2; 1/ [ Œ3; 1/ 6 1 5 s4
56 x 1
.!5; !1/ .!1; !3" [ Œ1=2; 1/ 5 6
integers in Œ37; 103" 14in. by 14in. 1
x!2 4p3 ! p2 4x3 ! x!2=3
.!1; !7:72" .!1; !0:5/ [ .0:667; 1/ 3
4 3
f(x) 55x4 C 36x2 ! 5 ! x3 x
3
45 21
5 16x3 C 3x2 ! x C x C x6
7 5
e
x 2 3 !1=4 10 2=3 11 !1=2
5 x C x x
-5 7x5=7 4 3 2
!6x!7 !6x!7 ! 4x!5 ! 2x!3
-5
2r!2=3
!
yes, appears to be and is continuous on .0; 1/, supports conclusions !x!2 !40x!6 !4x!4
5t4
of Example 5, appears to have and has a minimum of !1 at 1, 4
appears and can be sh n lim f.x/ D 0 x! !3x!2=3 ! 2x!7=5
x!0C x3
 ns ers to ere Pro e s AN-25

1 10 7=3 1 20
! x!5=3 !x!3=2 x 1 :2
3 3 235
1 !2=3 10 !5=3
30x4 ! 27x2 45x4 x ! x 4
3 3 a dr=dq D 30 ! 0:6q b c
2 45
3C 2 2x C 1 0 .x/ D 1 for x ¤ 0 0:432
q 4150 5.07 unit
t
4, 16, !14 0, 0, 0 y D 13x C 2
A I
1 1 dR
y ! D ! .x ! 2/ y ! 7 D 14.x ! 1/ D 6:25 ! 6x
4 ! 4 # dx
5 125 0 .x/ D 2x ! x2 0 .1/
.0; 0/, ; .3; !3/ D1
3 54
0 yDx!1

.4x C 1/.6/ C .6x C 3/.4/ D 4 x C 1 D 6. x C 3/

A I .5 ! 3t/.3t2 ! 4t/ C .t3 ! 2t2 /.!3/ D !12t3 C 33t2 ! 20t
2.5 units ˇ .14r6 ! 15r4 /.5r2 ! 2r C 7/ C .2r7 ! 3r5 /.10r ! 2/
dy dy ˇˇ
D 16 ! 32t D0 x3 ! 10x
dt dt ˇtD0:5
.2x C 5/.6x2 ! 5x C 4/ C .x2 C 5x ! 7/.12x ! 5/
when t D 0:5 ob ect at maximum height
. 2 C 3 ! 7/.6 2 / C .2 3 ! 4/.2 C 3/
1.2 and 120
.x2 ! 1/. x2 ! 6/ C .3x3 ! 6x C 5/.2x/ ! 4. x C 2/
5 !1=2 /.11p C 2/ C .2pp ! 3/.11//
7 ..p
0
&t 1 0.5 0.2 0.1 0.01 0.001 .5/.2x ! 5/.7x C / C .5x C 3/.2/.7x C / C .5x C 3/.2x ! 5/.7/
.x ! 1/.5/ ! .5x/.1/ 65
&s=&t 7.4 7.2 7.02 7.002 .x ! 1/2 3x6
ad ! bc .z2 ! 4/.!2/ ! .6 ! 2z/.2z/
estimate and confirm 7 m s
.cx C d/2 .z2 ! 4/2
a 70 m b 25 m s c 24 m s
a 32 b 1 :1505 c 1 .3x2 ! 2x C 1/. x C 3/ ! .4x2 C 3x C 2/.6x ! 2/
a 2m b 10.261 m s c ms .3x2 ! 2x C 1/2

dy .2x2 ! 3x C 2/.2x ! 4/ ! .x2 ! 4x C 3/.4x ! 3/

D .27=2/x5=4 432
dx .2x2 ! 3x C 2/2
0.27 dc=dq D 10 10 100x 2 3 !1
!
.x100 C 7/2 2
dc=dq D .0:4/q C 4
15x2 ! 2x C 1 10 .3x C 1/.2/ ! .2x/.3/
dc=dq D 2q C 50 0, 2, 4 C
3x4=3 .2x C 5/2 .3x C 1/2
dc
D 0:04q C 3 4:6 6:2 Œ.x C 2/.x ! 4/".1/ ! .x ! 5/.2x ! 2/
dq
Œ.x C 2/.x ! 4/"2
dc=dq D 0:00006q2 ! 0:02q C 6 4.6, 11
Œ.t2 ! 1/.t3 C 7/".2t C 3/ ! .t2 C 3t/.5t4 ! 3t2 C 14t/
dr=dq D 0: 0. , 0. , 0.
Œ.t2 ! 1/.t3 C 7/"2
dr=dq D 240 C 0q ! 6q2 440 0 !560
!4x3 ! 3x2 C 1 x ! 6 2a
dc=dq D 6:750 ! 0:000656q 5.43 2C 25
.x3 ! 3x2 C 2x/2 .a ! x/2
!10; 4 4:6
cN D C 6:750 ! 0:00032 q 0. 51655 3 15
q yD! xC y D 16x C 24 1=2
2 2
PD 5; 000; 000R!0: 3 dP=dR D !4; 650; 000R!1: 3 216
1 m, !1:5 m s 0 ! 0:04q !3
a !7:5 b 4.5 .q C 2/2
1 1 dC
a 1 b c 1 d e 11.1 D 0:672 7=24 17=24 0.615 0.3 5
xC4 dI
4x 40 dc 6q.q C 4/
a 4x b c 40 d e 1 .51 a 0:23 b 0:02 D
2x2 C 5 205 dq .q C 2/2
3x2 3 0:7355 1
a !3x2 b ! c !3 d ! e !42: !
! x3 7 10 .1 C 0:02744x/2 120
AN-26 ns ers to ere Pro e s

A I R C
p
2 t 3
!2x p 0
2 x
p
3ex2 C 2 3 3x C 14x C 5 4s3 C 4s D 4s.s2 C 1/

2x
!3 5
.3.x2 C 1/2 C 6.x2 C 1//2x
.3x ! 5/2
6x5 C 30x4 ! 2 x3 C 15x2 C 70x
0 0 1 .3x C 2/5 7.2 C 3x5 /6 .15x4 /
400.x C 1/.2x2 C 4x/
500.x3 ! 3x2 C 2x/ .3x2 ! 6x C 2/ !6x.x2 ! 2/!4
2.ax C b/..cx C d/a ! .ax C b/c/
10 .cx C d/3
! .2x C 5/.x2 C 5x ! 2/!12=7
7
10z
1 2.x2 C 1/3 . x2 C 32x C 1/
.10x ! 1/.5x2 ! x/!1=2 .1=3/.5x C 7/!2=3 .5/ .z2 C 4/2
2
12 2 4
.x C 1/!4=7 .2x/ !6.4x ! 1/.2x2 ! x C 1/!2 .4x ! 1/!2=3 x.1 ! x2 /!3=2
7 3
!2.2x ! 3/.x2 ! 3x/!3 !36x. x2 C 1/!3=2 Am!1 Bn!1 Cp!1 .mBC C AnC C ABp/
where A D x C a, B D x C b, C D x C c
p
.1=5/.5x/!4=5 .5/ C 5
5.1/ 34 3
! .1 C 2!11= /x!11=
.x C 6/2 4
3x2 .2x C 3/7 C x3 .7/.2x C 3/6 .2/
p x.x2 C 4/
10x2 .5x C 1/!1=2 C x 5x C 1 x.x C 4/.x3 C 6x2 C /!2=5
.x2 C 5/3=2 5
5.x2 C 2x ! 1/2 .7x2 C x ! 1/
.2z C 3/.3z C 5/2 .30z C 47/ y D !4x C 3
16. x ! 1/2 .2x C 1/3 .7x C 1/
! # ! # 1 4 5
ax C b 10 .cx C d/a ! .ax C b/c yD xC " 0:714 71:4
11 12 3 7
cx C d .cx C d/2
dr=dq D 20 ! 0:2q 0:56 0:44
! #
1 x C 1 !1=2 !6 !2.5x2 ! 15x ! 4/
dr=dq D 500 ! 0:2q
2 x!5 .x ! 5/2 .x2 C 4/4
dc=dq D 0:125 C 0:00 7 q 0:73 6
. x ! 1/4 .4 x ! 31/ 4 1
.3x ! 1/4 4 eggs mm a b 400!
3 24
12x.x4 C 5/!1=2 .10x4 C 2x2 C 25/ 10; 000
4q !
! # q2
2 2t C 3 6
2C ! .14=5/
.t C 3/2 5 a !315:456 b !0:00025 c no, dr=dmjmD240 < 0
.x2 ! 7/3 ..3/.3x C 2/2 .3/.x C 1/4 C .3x C 2/3 .4/.x C 1/3 / 0.305 !0:32
!.3x C 2/3 .x C 1/4 .3/.x2 ! 7/2 .2x/
.x2 ! 7/6
A I
0 0 y D 4x ! 11 dq 12p dR 1
D 2 D
y ! 1 D .!1=10/.x ! 4/ 400 130 dp 3p C 4 dI I ln 10

" 13:
q q
a !p b ! p
q2 C 20 100 q2 C 20 ! q2 ! 20 a a 2 2x 3.4X3 C 1/
!
q x ax C b x 1 ! x2 X.2X3 C 1/
q2
c 100 ! p ! q2 C 20
q2 C 20 ax2
ln t 2x ln.ax C b/ C
ax C b .ln 3/. x ! 1/
4q3 C 16q
!340 4 !.10/!1 $ "
.q2 C 2/3=2 1 1 ! ln z
2x 1 C
.ln 2/.x2 C 4/ z2
a !0:001416x3 C 0:01356x2 C 1:6 6x ! 34: , !256:23
b !0:016 !1:57 .ln x/.4x3 C 6x C 1/ ! .x3 C 3x C 1/
!4 10 4:03 .ln x/2
 ns ers to ere Pro e s AN-27

d.2ax C b/ x 2 x A I
ax2 C bx C c 1 C x2 1! 2 1 ! x4 dP
D 0:5.P ! P2 /
p.2ax C b/ q.2hx C k/ f0 .x/ g0 .x/ dt
C 2 C ˇ
ax2 C bx C c hx C kx C f.x/ g.x/ d dr d ˇˇ
D 4!r2 D2 0! in3 =min
dt dt dt ˇrD12
2.x2 C 1/ 3.1 C ln2 x/ 4 ln3 .ax/
C2x ln.2xC1/
2x C 1 x x ft sec
f0 .x/ f0 .x/ C 1=x 4
p y D 5x ! 20
2f.x/ 2 .f.x/ C ln x/
ln.3/ ! 1 25 22
p
ln2 3 7 2p C 1 x 7x 3 2
y y1=4
for y ¤ 0 !p !
6a y 3y2 3 2
x x1=4
!1:65, 1.65
. ! a2 C a /.a ! / y 1Cy 4y ! 2x2
! for x ¤ 0 ! for x ¤ !1
x 1Cx y2 ! 4x
A I
d 4y3=4 1 ! 15x2 y4
D Ckekt 20x5 y3 C 2y
dt 2y1=4 C 1

1=x ! yexy
for 1=y ! xexy ¤ 0
5ex 4xe 2x2 C3 !5e !5x 1=y ! xexy
ey C yex
.12r2 C 10r C 2/e 4r3 C5r2 C2rC6 xe!1 ex .e C x/ ! for xey C ex ¤ 0 6e3x .1 C e3x /.x C y/ ! 1
xey C ex
2 ex ! e!x 3
2xe!x .1 ! x2 / .6x2 /52x ln 5 3 4x0
3 ! 0 !
5 y0
. 2 C C 1/aea ! ea .2 C 1/
. 2 C C 1/2 y C 1 D !.x C 1/ y D 3.x C 1/ y ! 1 D !2.x C 1/
p
e1! x 2ex !1 dq .q C 5/3
! p 5x4 ! 5x ln 5 for q ¤ 0 D!
2 x .e C 1/2
x 3q2 dp 40
f 3
1 xx .ln x C 1/ eaCbCc .2a C b/ !(I !
(

y D e!2 x C 3e!2 dp=dq D !0:015e!0:001q !0:015e!0:5

dc=dq D 10eq=700 10e0:5 I 10e $ "

2 1 2x
!12e 100e!2 !b.10A!b / ln 10 .x C 1/2 .x ! 2/.x2 C 3/ C C 2
xC1 x!2 x C3
0.0036 !0: , 0.56 " #
1 x2 6
.3x3 ! 1/2 .2x C 5/3 C
3x3 ! 1 2x C 5
101 p ! #
!3 elastic !1 unit elasticity ! , elastic 1p p 1 1 2x
! # 100 x C 1 x ! 1 x2 C 1 C C 2
150 2 xC1 x!1 x C1
! ! 1 elastic !1 unit elasticity p ! #
e 3
1 C x2 2x 1
1 !
!2 elastic ! inelastic 1Cx 3.1 C x2 / 1 C x
ˇ ˇ 2 $ "
ˇ ˇ .2x2 C 2/2 4x 2 3
'ˇˇ D !2 elastic 'ˇˇ D !0:5 inelastic ! !
pD10 pD5 .x C 1/2 .3x C 2/ x2 C 1 x C 1 3x C 2
ˇ
ˇ r $ "
'ˇˇ D !1 unit elasticity 1 .x C 3/.x ! 2/ 1 1 2
pD7:5 C !
2 2x ! 1 xC3 x ! 2 2x ! 1
!1:2; 0:6 decrease ! ! #
r r # r x2C1 p
x 2 C ln x
cp2 b b b
2
xx C1 C 2x ln x x p
a 'D! b ; c x 2 x
b ! cp2 2c c 2c
! #
207 10x
a 'D! D !13: elastic b 27.6 .2x C 3/5x 5 ln.2x C 3/ C
15 2x C 3
c increase, since demand is elastic
dr 4ex x3x .4 C 3 ln x/ 12
' D !1:6 D 30 y D 6x C 36 y ! ee D 2ee .x ! e/
dq
p
maximum at q D 5 minimum at q D 5 x D 1= e 0.1 decrease
AN-28 ns ers to ere Pro e s

A I 0:02 unit elasticity ' D !0:5 inelastic

43 1 5 3
! increase 1.76 3
16

" 0:2016 " !0:6 2327 04 !0:6 233 A I

0.33767 1. 07 5 4:17 " 1:5052. rel max q D 2 rel min q D 5 2 hours after in ection
13.33 2. 0 3.45

A I dec on .!1; !1/, .3; 1/ inc on .!1; 3/ rel min .!1; !1/ rel
d2 h max .3; 4/
D !32 ft sec2 c00 .3/ D 14 dollars unit2
dt2 dec on .!1; !2/, .0; 2/ inc on .!2; 0/, .2; 1/ rel min .!2; 1/,
.2; 1/ no rel max
inc on .!3; 1/, .2; 1/ dec on .!1; !3/, .1; 2/
24 0 ex 6 ln x C 11 rel max x D 1 rel min x D !3, x D 2
20 1 dec on .!1; !1/ inc on .!1; 3/, .3; 1/ rel min x D !1
! for q ¤ 0 !
q6 4. ! r/3=2 .2x C 3/3 dec on .!1; 0/, .0; 1/ no rel ext
! # inc .!1; !1/ dec .!1; 1/ max (at) !1
4 1 1
! 2 C
.x ! 1/3 x .x C a/2 dec on .!1; !5/, .1; 1/ inc on .!5; 1/ rel min x D !5
rel max x D 1
1
ez .6 C 1 z C z2 C z3 / 275 ! dec on .!1; !1/, .0; 1/ inc on .!1; 0/, .1; 1/ rel max x D 0
y3
r rel min x D!˙1 # ! #
4 a y 1 1 1
! inc on !1; , .2; 1/ dec on ; 2 rel max x D rel min
y3 b x 3 3 3
xD2 ! # ! # ! #
2.y C 1/ 2 5 2 5 2
for x ¤ !1 inc on !1; , ; 1 dec on rel max x D rel
.x C 1/2 3 2
;
3 2 3
y 25 300.5x ! 3/2 5
.1 ! y/3 32 min x D
2
0:04 x D 1; 2 !4: 1. 4 dec .!1; 1/, .1; 1/ no rel ext
inc on .!1; !1/, .1; 1/ dec on .!1; 0/, .0; 1/ rel max x D !1
R C rel min x D 1
x2 e2 !1 14r C 4 dec on .!1; !4/, .0; 1/ inc on .!4; 0/ rel min x D !4 rel
3ex C 2xe C e2 x max x D 0
7r2 C 4r C 5 p p p p
incpon .!1; ! 2/, .0; 2/ dec on .! 2; 0/, . 2; 1/ rel max
.6x C 5/e3x
2 C5xC7
ex .x2 C 2x C 2/
x D ˙ 2 rel min x D 0
p $ " inc on .!2; 0/, .2; 1/ dec on .!1; !2/, .0; 2/ rel max x D 0
.x ! 6/.x C 5/. ! x/ 1 1 1
C C rel min x D ˙2
2 x!6 xC5 x!
m n dec on .!1; !2/, .!2; 1/ no rel ext
1 ! x ln x C
2
ln.3/.10x C 3/35x C3xC2
xCa xCb dec on .0; 1/ no rel ext
xex
dec on .!1; 0/, .4; 1/ inc on (0,2), (2,4) rel min x D 0 rel max
4e2xC1 .2x ! 1/ 16 1C2 C32 xD4
x2 . x C 5/ ln 2 1C C 2C 3
inc on .!1; !3/, .!1; 1/ dec on .!3; !2/, .!2; !1/ rel max
! #
2x.x C 1/ 2 !3t2 x D !3 rel min x D !1
.x2 C1/xC1 ln.x2 C 1/ C 2 C r ! r ! r !
x C1 t 2.5 ! t3 / d d d
" # a inc on 0; ! , ! ; 1 dec on !1; ! ! ,
c c c
x 2x 6.x2 C 1/ r !
y 2 C ! .xx /x .x C 2x ln x/ d
x C1 3.x2 C 2/ 5.x3 C 3x/ ! ! ; 0 rel min x D 0 b as for a with inc and dec
c
4 !2e!e y ! 4 D 4.x ! ln 2/ .0; 4 ln 2/ interchanged min replaced by max
y dec .!1; !1/ inc .!1; 1/ min (at) !1
2 2 !1 ! for x ¤ 0
x ! # ! #
1 1
inc on .!1; 0/, 0; , .6; 1/ dec on ; 6 rel max
4 dy y C 1 d2 y yC1 7 7
D D! 3 1
dx y dx2 y xD rel min x D 6
7
f0 .t/ D 0:00 e!0:01t C 0:00004e!0:0002t dec on .!1; 1/ no rel ext
 ns ers to ere Pro e s AN-29
p ! p ! p
3 2 3 2 3 2 int .1; 0/, .2; 0/, .0; !4/ no app(arent) sym(metry)
dec on 0; inc on ; 1 rel min x D inc .!1; 1/, .1; =5/, .2; 1/ dec . =5; 2/ max =5 min 2
2 2 2
. =5; 10 =3125/, .2; 0/
inc on .!1; 1/ no rel ext
p p p y
dec .0; 1= e/ inc .1= e; 1/ min at 1= e
! # ! #
3 3 3
dec on !1; inc on ; 1 rel min x D int .!2; 0/, 8 , 108
2 2 2 5 3125
(5,0), .0; !10/
x
(2, 0)
(1, 0)
y

x
-2 3 5 dec on .1; 1/ inc on y
2
(0,1) rel max x D 1 int (0,0), (4,0)
-10
49 y
-
4 6

4
1
dec on .!1; !1/, .1; 1/ inc onp.!1; 1/ rel min x D !1 rel 2
x
max x D 1 sym about origin int (˙ 3; 0), (0,0) 1 4
x
2 4
y

2 q < 50 q D 30
a 25,300 b 4 c 17,200
x rel min .!3: 3; 0:6 /
-1 1
rel max (2.74,3.74) rel min .!2:74; !3:74/
-2
rel min 0, 1.50, 2.00 rel max 0.57, 1.77
a f0 .x/ D 4 ! 6x ! 3x2 c dec on .!1; !2:53/, .0:53; 1/ inc
.!2:53; 0:53/
inc on .!1; 1/, .2; 1/ dec on .1; 2/ rel max x D 1 rel min
x D 2 int (0,0)

y ! #
1
max .3; 6/ min .1; 2/ max !1; min .0; 1/
6
5
max .0; 50/ min .4; 2/
4
max .0; 0/ min .!1; !31=12/, min .1; !31=12/
p
max . 2; 4/ min .2; !16/
p !
3 2 73
x max .0; 2/, .3; 2/ min ;!
1 2 2 4
max .!26; /, .2 ; / min .1; 0/
a !3:22, !0:7 b 2.75 c d 14,2 3
inc on .!1; 0/, .1; 1/ dec on .!1; !1/,
p .0; 1/ abs minpx D ˙1,
relative max x D 0 sym about x D 0 int .! 2; 0/, .0; 0/, . 2; 0/

! # ! #
y 1 1 1
conc up !1; ! , .2; 1/ conc down ! ; 2 inf pt x D ! ,
2 2 2
xD2
cdn (concave down) .!1; !1/, .!1; 2/ cup (concave up) .2; 1/
1 i (in ection point at) 2
-1 p p p p
x conc up .!1; ! 2/, . 2; 1/ conc down .! 2; 2/ no inf pt
- 2 2
conc down .!1; 1/
conc down .!1; !1/ conc up .!1; 1/ inf pt x D !1
AN-30 ns ers to ere Pro e s
! # ! #
b b int .0; !1 / inc .!1; 2/, .4; 1/ dec (2,4) rel max x D 2 rel
conc up down ! ; 1 conc down up !1; for
3a 3a min x D 4 conc down .!1; 3/ conc up .3; 1/ inf pt x D 3
b
a > 0 a < 0 inf pt x D
3a y
cdn .!1; 1/, .2; 3/ cup .1; 2/, .3; 1/, i 1; 2; 3
conc up .!1; 0/ conc down .0; 1/ inf pt x D 0 x
! # ! # ! #
7 1 7 1
conc up !1; ! , ; 1 conc down ! ; inf pt
2 3 2 3
7 1
xD! ,xD
2 3
p p !
3! 5 3C 5
conc down .!1; 0/, ; conc up
2 2
p ! p ! p
3! 5 3C 5 3˙ 5
0; , ; 1 inf pt x D 0, x D p p p p
2 2 2 inc on .!1; ! 5/, . 5; 1/ dec on .! 5; 5/ conc down
! #
p p p p 10 p
conc up .!1; !2/, .! 3; 3/, .2; 1/ conc down .!2; ! 3/, .!1; 0/ conc up .0; 1/ rel max ! 5; . 5/ , rel min
p p p ! # 3
. 3; 2/ inf pt x D !2, x D ! 3, x D 3, x D 2 p 10 p p
5; ! . 5/ inf pt .0; 0/ sym about .0; 0/ int .˙ 15; 0/, .0; 0/
cup .!1; !1/ cdn .!1; 1/ no i 3
p p
conc down .!1; !1= 3/, .1= 3; 1/ conc up
p p p
.!1= 3; 1= 3/ inf pt x D ˙1= 3 y
! # ! #
2 2 2
conc down .!1; !3/, !3; conc up ; 1 inf pt x D ( - 5, 10
5)
7 7 7 3

conc up .!1; 1/
(- 15, 0)
conc down up .!1; !2/ conc up down on .!2; 1/ for a > 0 x
(0, 0) ( 15, 0)
a < 0 inf pt x D !2
cdn .0; 1/ no i 10
! # ! # ( 5, - 3 5)
1 1
int .!2; 0/, (3,0), .0; !6/ dec !1; inc ; 1 rel min
2 2
1
xD conc up .!1; 1/ int .!1; 0/, .0; 1/ no app sym inc .!1; !1/, .!1; 1/ no rel
2
ext cdn .!1; !1/ cup .!1; 1/ i at x D !1

y
y
10

5
x

x
-4 -2 2 4

-5

!# ! # ! #
5 5 5 5 -10
int (0,0), ; 0 inc !1; dec ; 1 rel max x D
2 4 4 4
conc down .!1; 1/
int (0,0), .4=3; 0/ inc .!1; 0/, (0,1) dec .1; 1/ rel max x D 1
conc up .0; 2=3/ conc down .!1; 0/, .2=3; 1/ inf pt x D 0, x D 2=3
y

x
 ns ers to ere Pro e s AN-31
p p p
int .0; !2/ dec .!1; !2/, .2; 1/ inc .!2; 2/ rel min x D !2 int (0,0), .˙2; 0/ inc .!1; ! 2/, .0; 2/ dec .! 2; 0/,
rel max x D 2 conc up .!1; 0/ conc down .0; 1/ inf pt x D 0 p p
. 2; 1/ rel max x D ˙ 2 rel min x D 0 conc down
p p p p
y .!1; ! 2=3/, . 2=3; 1/ conc up .! 2=3; 2=3/ inf pt
p
x D ˙ 2=3 sym about y-axis
y

int (0,0), ( ,0) dec .!1; 0/, (0,2) inc .2; 1/ rel min x D 2
conc up .!1; !4/, .0; 1/ conc down .!4; 0/ inf pt x D !4, x D 0
int .0; !2/, (1,0) inc on .!1; 1/, .1; 1/ conc down .!1; 1/ y
conc up .1; 1/ inf pt x D 1

y
12 3 4

x
-4 2 8

! # ! # ! #
2 2 2 2 3
2
dec on !1; ! p4
, p
4
; 1 inc on ! p4
; p conc up -6
5 5 5 45
! #
2 12
.!1; 0/ conc down .0; 1/ rel min ! p 4
;! .5/3=4 rel max int (0,0), .!4; 0/ dec .!1; !1/ inc on .!1; 0/, .0; 1/ rel min
5 25
! # x D !1 conc up .!1; 0/, .2; 1/ conc down .0; 2/ inf pt x D 0,
2 12 xD2
p
4
; .5/3=4 inf pt .0; 0/ sym about .0; 0/ int .˙2; 0/,.0; 0/
5 25
y
y

63 2

x
(-2, 0) (2, 0)
x
-4 -1 2
-3

int .0; 3/ no app sym dec on .!1; 0/, .0; 1/ inc on .1; 1/, min ! ! ##
! # ! # 64 64
2 2 dec on .!1; 0/, inc on 0;;1 conc down .!1; 0/,
at x D 1 cup .!1; 0/, ; 1 cdn 0; i at x D 0, x D 2=3 27 27
3 3 ! #
64 32
.1; 1/, .2=3; 147= 1/ .0; 1/ rel min .0; 0/ rel max ; no inf pt vertical tangent at
27 27
y .0; 0/ no sym .0; 0/, . ; 0/

5
2 (8, 0)
x
(0, 0) 2
x
-3 -1 3
AN-32 ns ers to ere Pro e s

f.x/ D x.x ! 3/2 an example y int (0,0) dec on .!1; 1/, .1; 1/ conc up .1; 1/ conc down
.!1; 1/ asymp x D 1, y D 1
y
y
5 1
x
1
x 1
-5 5 x
1

-5

S b f (r )
dec on .!1; !1/, (0,1) inc on .!1; 0/, .1; 1/ rel min x D ˙1
60 conc up .!1; 0/, .0; 1/ sym about x D 0 asymp x D 0
6.2
y

r
A 1 10 2
625
c 0.26
x
two above tangent line concave up -1 1

!2:61, !0:26
int .0; !1/ inc on .!1; !1/, .!1; 0/ dec (0,1), .1; 1/ rel max
x D 0 conc up .!1; !1/, .1; 1/ conc down .!1; 1/ asymp x D 1,
x D !1, y D 0 sym about y-axis
y
5
rel min x D abs min
2
1
rel max x D abs max
4 x
rel max x D !5 rel min x D 1 -1 -1 1

rel max x D !2 rel min x D 3

test fails, rel and abs min .0; 3/ ! #
1 1 2
rel max x D ! rel min x D asymp x D 3, y D !1 int 0; , .!2; 0/ inc on .!1; 3/,
3 3 3
7 .3; 1/ conc up on .!1; 3/ conc down on .3; 1/
rel min x D !5, x D !2 rel max x D !
2
y

!7 1 (0, 23 )
xD1 yD1 xD yD 3
2 2 x
xD0 yD0 y D 0 x D 1, x D !1 (-2, 0) -1

none y D 2 x D 2, x D !3
p p
x D ! 7, x D 7 y D 15 xD5 yD7 no app sym ver ast x D 1 non-ver ast y D x C 1 int .0; 0/ inc
1 1 .!1; 1/, .2; 1/ dec .0; 1/, .1; 2/ max 0 min 2 cdn .!1; 1/ cup
y D !x C 1 x D 0, x D !1 yD xD!
4 2 .1; 1/ min pt .2; 4/
1 4
yD xD! y D !2 y
2 3
sym about .0; 0/ ver ast x D 0 hor ast y D 0 no int dec
.!1; 0/, .0; 1/ no ext cdn .!1; 0/ cup .0; 1/

x
 ns ers to ere Pro e s AN-33
! # ! # ! # ! # ! #
2 2 1 1 4 1
int 0; ! inc on !1; ! , ! ; dec on ; , asymp x D 1, y D 2x C 1 int .0; 0/, ; 0 inc on
3 3 3 3 3 p ! ! 2 !
! # ! # ! # p p
4 1 2 4 2 2 2
; 1 rel max x D conc up !1; ! , ; 1 conc down !1; 1 ! , 1C ; 1 dec on 1 ! ;1 ,
3 3 3 3 2 2 2
! # p !
2 4 2 4 2
! ; asymp y D 0, x D ! , x D 1; 1 C conc down .!1; 1/ conc up .1; 1/, rel max
3 3 3 3 2
p ! p !
2 p 2 p
1! ; 3 ! 2 2 rel min 1 C ; 3 C 2 2 sym .1; 3/ a
y 2 2
concept not covered in HPW

x
1
- 23 4
3 x
(0, 0) ( 12 , 0)

1
3
, -1

y D !1 for x ¤ !1 and x ¤ 1
! # ! # ! # ! #
1 1 4 2 y
int ! ; 0 , 0; dec on !1; ! , ; 1 inc on
! # 3 4 ! 3 3#
4 2 4 7 5
! ; rel min x D ! conc down !1; ! conc up
3 3 3 3
! # ! #
7 2 2 7 2
! ; , ; 1 inf pt x D ! asymp x D , y D 0
3 3 3 3 3
x
-5 5

y (-1, -1) (1, -1)

(0, -1)

-5

y
2
x
3
7 2
- 3 , - 27 4 1
- 3 , - 12 1
x
p p p 2
p int .!1; 0/, (1,0) inc
p on .! 3; 0/, .0;p 3/ dec on .!1; ! 3/,
. 3; 1/prel maxpx D 3 rel minpx D ! p3 conc down p
.!1; ! 6/, .0; 6/ conc up .! 6; 0/, . 6; 1/ inf pt x D ˙ 6
asympx D 0, y D 0 sym about origin
y

x
-1 2
- 3
x
3

a 1
xD! yD
b b
limt!1 250 ! 3e!t D 250 since limt!1 e!t D 0
AN-34 ns ers to ere Pro e s

hor ast y D 2 ver ast x D 1, x D 2, x D 3 int .0; 20/, inc on .!1; !2/, .2; 1/ dec on .!2; 2/ rel max
x D !2 rel min x D 2 conc up .0; 1/ conc down .!1; 0/ inf pt
y " 0:4
xD0
y
(-2, 36)

4 ;4 300 ft by 250 ft
100 units 15
12 0
maximum 105 minimum " 56:1
23
quantity 350, profit 5625, price 57:50
(2, 4)
22 120 units 6,000
x
625 units 4 22 154, 0
2 3 3 p ! p !
3 ft & 3 ft & 3=2 ft in. in
6 27 3 3
int .0; 0/, .!1; 0/, .1; 0/ inc on !1; ! , ; 1 dec
130 units, p D 340, P D 36; 0 125 units, p D 350, 3 3
p p !
P D 34;175 3 3
on ! ; conc down .!1; 0/ conc up .0; 1/ inf pt x D 0
375 lot ( lots) 60 attendees 3 3
sym about origin
60 mi h 3400
p p
5 ! 3 tons 5 ! 3 tons 10 cases 50.55 y

R C (1, 0)
x
5 2 (-1, 0) (0, 0)
y D 3, x D 4, x D !4 yD ,xD!
3
0
7=6, 1 critical, f.2=3/ not defined
inc on .!1; 7/ dec on .!1; !1/, .7; 1/ int .!5; 0/ inc .!10; 0/ dec on .!1; !10/, .0; 1/ rel min
p p p p p p x D !10 conc up .!15; 0/, .0; 1/ conc down .!1; !15/ inf pt
pdec on p.!1; ! 6/, .0; 3/, . 3; 6/ inc on .! 6; ! 3/,
x D !15 asymp y D 0 x D 0
.! 3; 0/, . 6; 1/
! # ! #
1 1
conc up .!1; 0/, ; 1 conc down 0; f(x)
2 2
! # ! #
2 2
conc down 1; ! , conc up on ! ; 1
3 3
cup .!1; 17=12/, .5=2; 1/ cdn .17=12; 5=2/ x
rel max x D 1 rel min x D 2
rel min x D !1
2
rel max x D ! rel min x D 0
5 p ! # ! #
1 1
2 i =5 0, 3 ˙ 3 inc on 1; ! dec on ! ; 1 , .1; 1/ conc up .!1; !1/,
! # 2 2
6 1 ! # ! #
max .2; 16/ min .1; !1/ max .0; 0/ min ! ; ! 1 4 1
5 120 .1; 1/ conc down .!1; 1/ rel max ! ; inf pt !1;
2 27
a e!1 b .0; 1/, none asymp y D 0, x D 1 no symmetry int .0; 0/
int .0; !21/, .!3; 0/, .7; 0/ no app sym no ast dec .!1; 2/ inc
y
.2; 1/ rel min pt .2; !25/ cup .!1; 1/

y
4
(- 12 , 27 )

(-1, 18 )
x
(7, 0) (0, 0) 1
x
(-3, 0)

(0, -21)
(2, -25)
 ns ers to ere Pro e s AN-35

int .0; 0/ no ast sym origin x > 0 implies y.x/ > 0, (x < 0
implies y.x/ < 0) inc .!1; 1/ cdn .!1; 0/ cup .0; 1/ i pt .0; 0/ x 5
7x C C CC ! CC
y 6x6
5 2 t2
! CC ! CC 4t C CC
6x6 3t3=2 2
y6 5y2
x ! CC t3 ! 2t2 C 5t C C
6 2
p x2 2x6
. 2 C e/x C C ! CC
10 1
x :3 x7 1 1
a false b false c true d false e false !ex C C ! ! 3 ! 2 CC
:3 7 x 2x
q>2 x5
4x3=2 p 4
max .!1:34; 12:33/ min .0:45; 1:26/ ! CC 53 xCC C x!3 C C
20 3
3 2 3 2 4
y
C CC u ! uCC
2 3 10 5
14
ueC1 p p
3 4
(-1.38, 12.33) C eu C C 6 x! x CC
eC1
1 =5 4 p 711 2 x4 5x2
x ! xC x CC ! x3 C ! 15x C C
5 2 4 2
2x5=2
C 2x3=2 C C
5
(0.45, 1.26) 27 4
x u C 1 u3 C 1 u2 C u C C
4 4
-4 5 1
x3 C 4x C C C ..1=5/z5 C .7=2/z2 / C C
x 3
x " !0:60 20 200
2 ,000 =10 & =4 x C ex C C

a 200 stands at 120 per stand b 300 stands no, F.x/ ! G.x/ ¤ 0 is possible
1
p CC
2
x C1
3x2 2
a dx p dx ! dx
2 x3 ! 27 x3
2x 2 C3
A I
dx 3e2x .12x2 C 4x C 3/ dx
x2 C7 .t/ D 00t C 200et C 6317:37
&y D a dx D dy &y D 0:0701 dy D 0:07 y.t/ D 14t3 C 12t2 C 11t C 3
&y " 0:04 , dy D 0:050 a !1 b 2.
577 25
" 16: 7 " !0:002
34 12
1 1 31
1.001 3x2
2 6p.p2 C 5/2 yD ! 4x C 1 4
2
p3 !3 4 5 1 2 x4
! ! y D ! x4 C x3 C x C 3 yD C x2 ! 5x C 13
2 2 5 12 3 3 12
3
!17 !1 :3 2.04 " 0:1 p D 0:7 p D 275 ! 0:5q ! 0:1q2
40
2:47q C 15 5 4
" 1: 3 & 10!11 c 42 units
P2
GD! C 2P C 20
A I 50
R R 0 (dc=dqjqD50 D 27:50 is not relevant)
2 :3 dq D 2 :3q C C 0:12t2 dt D 0:04t3 C C
R 4 0 240
! dt D 2 C C
t3 t
R p A I
.500 C 300 t/dt D 500t C 200t3=2 C C
S.t/ D 0:7t3 ! 32:7t2 C 4 1:6t C C .t/ D 10e!0:5t C C 35 ln jt C 1j C C
AN-36 ns ers to ere Pro e s

.x C 3/6 .x2 C 3/6 1 5 4 3 1 3

CC CC x C x ! 2 ln jxj C C .2x C 4x C 1/3=2 C C
6 6 5 3 3
3 3 6p 1 5x
.y C 3y2 C 1/5=3 C C 5.3x ! 1/!2 ! 4 ! 5x C C 2 CC
5 ! CC 5 5 ln 2
6
2 2
2 1 7x2 ! 4e.1=4/x C C x2 ! 3x C ln j3x ! 1j C C
.7x C 3/3=2 C C .5x ! 2/6 C C 3
21 30
5 5 13=x
.5u2 ! /15 3 ln.7e2x C 4/ C C ! e CC
CC .27 C x5 /4=3 C C 14 13
150 5
2 p
1 3t2 C2tC1 x2 ! ln.x2 C 1/ C C . x C 2/3 C C
e3x CC e CC
2 1 2
3 5x2 4 3.x1=3 C 2/5 C C .ln x/ C C
e CC ! e!3x C C 2
10 3 1 1 ln 11
.ln.r2 C 1//3=2 C C x CC
3 ln 11
ln jx C 5j C C ln jx3 C x4 j C C
2 .x3 C1/=2
e CC ln j ln.x C 3/j C C
2 3
! 2 CC 7 ln jxj C C 2
3.z ! 5/6 x2 .ln.x4 C 1/3 /3=2 C C
C x C ln jx2 ! 3j C C 3
1 5 2
3
ln js3 C 5j C C ! ln j4 ! 2xj C C
2 1p 4 2
x ! 4x ! .ln 2/x C C x2 ! x ! 6 ln jxj ! CC
p 2 x2
2 2 5 3=2 p
.5x/3=2 C C D x CC x ! ln jx C 1j C C
2
ex C 2 C C
15 3
1p 2 1 y4 C1 1 1 2 p 5=2
ax C b C C e CC ! .e!x C 5/4 C C .x C 2/ C C
a 2 4 5
1 1 1 2 p3
3
! e!2 C1 C C ! e!5x C 2ex C C p Œ. x/3=2 C 3"3=2 C C ! e! s C C
6 5 36 2 3
1 2 ln jx3 C 4xj C C x3 1
! .7 ! 2x2 ! 5x/4 C C CxCC x ! .ln x/2 C C
2 3 2
ln..1 C 4s C 3s2 /4 / C C 1 100
ln.2x2 C 1/ C C pD c D 20 ln j.q C 5/=5j C 2000
4 qC3
1 3 1 4 p 3 1p 71
.x ! x6 /! C C .x C x2 /2 C C C D 2. I C 1/ CD I! IC
27 4 4 3 12
1 3x3 C5x2 C2 a 140 per unit b 14,000 c 14,2 0
3.5 ! x ! x2 /!3 C C e CC
2
p 1504 ID3
1 2 2 3=2 p 1=2
! . ! 5x2 /5=2 C C x ! 2x C C
25 3
A I
x5 2x3
C CxCC 5 75
5 3
1
.ln.x2 C 1/ ! .x2 C 1/!1 C C
2
13 25
5 3
3 ln j5x C 2j C .x C x6 /!4 C C 64
12 $ ! # ! #
1 1 2 % n &" 2.n C 1/
2 1 Sn D 4 C4 C %%% C 4 D
.3x C 1/3=2 ! ln.x2 C 3/ C C n n n n n
2
p 1 1 nC1 3 3
2e x CC ! e!x C ex C C a Sn D C1 b
4 4 2n 2 2
1 !1 6 1 1 5
.ln.2x2 C 3x//2 C C .5 ! 7x/4 C 12 20 !1
2 2 2 3 6
5 1
y D ln j1=xj C x ! 160e0:05t C 1 0 0 11=2 14.7 2.4 !25:5
2 2
Rr2
C B1 ln jrj C B2 A I
4
32, 30 2 ,750
 ns ers to ere Pro e s AN-37

2 1 1
CS D 166 , PS D 53
15 3 3 2
15 !20 63
2 CS " 114 , PS " 251
4 76 5
74
3 7 3
244 p
3=20 4 ln e5 x
5 p CC 16x2 C 3
! x2 ! CC
5 1 15 3x
.e ! 1/ 3 =3 ˇ ˇ ˇp ˇ
3 14 2 1 ˇˇ x ˇˇ 1 ˇˇ x2 C ! 3 ˇˇ
ln ˇ CC
1 1
!
1
# 6 6 C 7x ˇ ln ˇ
3 ˇ x
ˇCC
ˇ
ln 3 eC !2
2 2 e ! #
1 4 2
p ln j4 C 5xj ! ln j2 C 3xj C C
5 2C3 3 3 1 2 5 3
! C C 2 ! 3
4 e 2e e 1% &
2x ! ln j3 C e2x j C C
.1=6/.e24 ! e / " 4; 414; 52; 33 6
! #
1 1 ˇˇ x ˇˇ 4
47 7 C ln ˇ ˇ CC 1 C ln
6 C ln 1 5.5 C 2x/ 25 5 C 2x
12
% p ˇ p ˇ&
1 ˇ ˇ
6 ! 3e !7 0 ˛ 5=2 x x2 ! 3 ! 3 ln ˇx C x2 ! 3ˇ C C
2
Rb !
.!Ax!B /dx 63 1 0; 667 76 ; 126 1 x2 e2x e2x
a
144 ! .2x ! 1/ C C
1 3:15 1367. 6 64 2 2Ri 2 4
p p !
1=2 3.52 14.34 5x2 C 1 ˇp p ˇ
5 ˇ ˇ
! p C ln ˇ 5x C 5x C 1ˇ C C
2
2 5x
! #
1 1
ln j1 C 3xj C CC
R C 1 C 3x
x4 !
C x2 ! 7x C C 2160
!1
CC ˇ p7 C p5x ˇ
.x C 2/3
1 1 ˇ ˇ
4 p p ln ˇ p p ˇ CC
p 5 2 7 7 ! 5x
11 3 11
2 ln jx3 ! 6x C 1j C C !4 x6 .6 ln.3x/ ! 1/ C C
4
!
y4 2y3 y2 21 17=21 6 7=6 2. x ! 2/.1 C 3x/3=2
C C CC t ! t CC 2 CC
4 3 2 17 7 135
2 p
ln.21/ ! ln.5/ " 1:435 .3x3 C 2/3=2 C C 1 ˇˇ p ˇ
ˇ 16 ! x2
27 ln ˇ2x C 4x2 ! 13ˇ C C ! CC
2 x
1 2y 2 272
.e C e!2y / C C ln jxj ! CC 1 p p
2 x 15 .4 x ! ln j! C 7e4 x j/ C C
2!
4 p p 3 2
. 125 ! / 4!33 2 ! p CC 1
3 t t ln.x2 C 1/ C C
2
3 p 1 p
! 5 ln 2 e x C x xCC 1=2
2 3 .1=3/.ln x/3 C C
ˇ ˇ $ "
p ˇx ! 3ˇ x4 1
.1 C e2x /4 2 103x ˇ
ln ˇ ˇCC ln.x/ ! CC
CC CC x ! 2ˇ 4 4
ln 10
1 2x 2 2
yD e C 3x ! 1 4 2=3 e3x .3x2 ! 1/ C C x.ln x/2 ! 2x ln.x/ C 2x C C
2
125 2 Formula 15 with u D x, a D 5, and b D 2 so that
6 C ln 4 Z Z 1 p !ˇ
6 3 1 2xdx xdx 2.2x ! 10/ 5 C 2x ˇˇ1
p D2 p D2 ˇ D
a3 !1 5 C 2x !1 5 C 2x 12 !1
55:07 e!1 p
6 ! p
7C4 3
p 3
p D 100 ! 2q 14 3.33
p p 7 3
2.2 2 ! 7/ ln.2/ !
1 ! e!0:7 " 0:5034 15 2 4
AN-38 ns ers to ere Pro e s
ˇ ˇ
ˇ qn .1 ! q0 / ˇ
ln ˇˇ ˇ a 755 .0 b 16, 30.75
q0 .1 ! qn / ˇ 2
yD!
3x2 C C
a !2000e.e!1
! 1/ " 3436
b !1666:67e1:5 .e!1:5 ! 1/ " 5 02 y D .x2 C 1/ ln.x2 C 1/ ! .x2 C 1/ C C

A I y D Cex , C > 0 y D Cx; C > 0

!
p x4 C 4e
76. 0 ft yD 3
3x ! 2 y D ln
4
5.77 gm
v !2
u
4 .3x2 C 2/2 u 3x2 3
yD yD t C !1
1 4 C 23.3x2 C 2/2 2 2
413 0:26 " 0:767 0.750 ! p #
4 1
" 0:7377 2,115,215 3.0 y D ln x2 C 3 c D .q C 1/e1=.qC1/
2
0.771 120 wee s
3
1
P.t/ D 60;000e 10 .4 ln 2!ln 3!ln 5/t 6 ,266
yes, " 30; 34 tiles
2e1:14 2 billion 0:01204 57:57 sec
2 00 years D 0e
k.t!t0 / , t # t0
7
5=3.20 / 13=12 13
2 12.21=3 / " 15:1 A D 400.1 ! e!t=2 / 157 g m2
ln 16 e3 !e 1 16 t
ln.3 =600/
3 3 a V D 60; 000e :5 b une 202
16 e C 2 ln 2 D C ln 4
2 2
1 3 7
2=3.125/ a b c
16 4 16
6 ,200 500 1 0
1
b 375 c " 4:1322 days 2:20 . .
7
a ln b ln 5 ! 1 c 2 ! ln 4 155,555.56 D !. ! 0 /e
!kt
3
1. "7 :11
R3 2 2
R4 A I
0 .2x ! .x ! x//dx C 3 ..x ! x/ ! 2x/ dx
R1 p 20 ml
0 ..y C 1/ ! 1 ! y/ dy
R2 2 2
1 ..7 ! 2x / ! .x ! 5// dx
4
36 40 1 1
3 divergent e!37 divgt
1 2
125 44
1=2 0 a k D 500 b 1=2
12 3
1
255 14 25,000,000 20,000 increase
! 4 ln 2 12 3
32 45
3 R C
" 0:62 6052 4.76 6.17
2m3 ! # p !
1 3 1 p 3C 13
x ln x ! 2 13 C ln
3 3 3 2
CS D 25:6 PS D 3 :4 CS D 50 ln 2 ! 25 PS D 1:25 ln j3x C 1j C 4 ln jx ! 2j C C
p
CS D 225 PS D 450 426:67 1 1 x ! 4 ! x2
C ln j jCC CC
" 45; 432 CS " 11 7 PS " 477 2.x C 2/ 4 xC2 4x
1
.ln jx ! aj ! ln jx C aj/ C C
2a
7 2 1
!1 33=5 11,050 e7x .7x ! 1/ C C ln j ln x2 j C C
3 3 4
3155.13 3 a3
x! ln j3 C 2xj C C
2 6
A I 243
e!1 " 0:6
I D I0 e!0:00 5x
 ns ers to ere Pro e s AN-39

2 1 a 0. 70 b 0.066 c 0.0 73
CS D 166 , PS D 53 12:25
3 3 0:0107 0. 351 5
3 2
yD Cex Cx , C>0 2=3 0: 2 a 4:55 b 1:26

divgt 2;000;000 A I
is fine 450
D 0.03 6
1 C 224e!1:02t
4:16 . . 0: 5

0: 207I 0:0122 0:0430I 0: 232

A I 0:7507 0: 525
1 0:607 0.5557 0.53 0:0336
3
mean 5 years, standard deviation 5 years
R C
8̂
ˆ 0 if x < 0
3 <1 2
a 2 b c d F.x/ D x C x3 if 0 $ x $ 1
p 32 4 ˆ 3 3
5 11 13 :̂
a b D 0:6 75 c D 0: 125 d !1 C 10 1 if x > 1
12 16 16 p
8 a #D = b $ D .2 2/=3
<1
if 1 $ x $ 4 0.1056 0.2417
a f.x/ D 3
:0 otherwise
0:7734 0: 17 0:022
f(x)

fx .x; y/ D 4x C 3y C 5 fy .x; y/ D 3x C y C 6
1
4 fx .x; y/ D 0I fy .x; y/ D 2
x gx .x; y/ D 12x3 y C 2y2 ! 5y C
1 4 gy .x; y/ D 3x4 C 4xy ! 5x !
p
2 5 1 5 3 q p
b c 0 d e
g 1 f 0 h i gp .p; q/ D p gq .p; q/ D p
3 6 3 2 2 2 pq 2 pq
8̂
<
0 if x < 1 2s .s2 C 1/.2t/
1 @h=@s D @h=@t D ! 2
F.x/ D .x ! 1/ if 1 $ x $ 4 !1t2 .t ! 1/2
:̂ 3 @u 1 @u 1
1 if x > 4 D D
@q1 2.q1 C 2/ @q2 3.q2 C 5/
1 2
P.X < 2/ D P.1 < X < 3/ D
3 3 hx .x; y/ D .x3 C xy2 C 3y3 /.x2 C y2 /!3=2
F (x)
hy .x; y/ D .3x3 C x2 y C y3 /.x2 C y2 /!3=2
@z @z
D 5ye5xy D 5xe5xy
@x @y
1
@z 2x2 @z 5x
D5 2 C ln.x2 C y/ D 2
2 6
x @x x Cy @y x Cy
@f=@r D .5=2/.r ! s/3=2 @f=@s D !.5=2/.r ! s/3=2
a e!2 ! e!4 " 0:11702 b 1 ! e!6 " 0: 752 @f @f e3!r
c e!10 " 0:00005 d 1 ! e!3 " 0: 5021 D !e3!r ln.7 ! s/ D!
' @r @s 7!s
0 if x < 0 gx .x; y; z/ D 6x2 y2 C 2y3 z gy .x; y; z/ D 4x3 y C 6xy2 z
e F.x/ D
1 ! e!2x if x # 0 gz .x; y; z/ D 2xy3 C z
p
1 5 3 2 2 gr .r; s; t/ D 2resCt gs .r; s; t/ D .7s3 C 21s2 C r2 /esCt
a b c " 0:60 d 1 e f
16 64 3 3 gt .r; s; t/ D esCt .r2 C 7s3 /
p 7
g 2 2 h 50 2 =5.e7 / 0 26
16
@R 2 @R !2r.n ! 1/
7 D D
5 min e!3 " 0:050 @r 2 C a.n ! 1/ @a .2 C a.n ! 1//2
10
@R !2ra
D
@n .2 C a.n ! 1//2

a 0:4 3 b 0:26 1 c 0:7157 d 0: 27 e 0: 467

f 0:4247 20 1374.5
0.35 !1:0 0:34 @P=@ D 0:77347 .k= /0:6 6 @P=@k D 1:733522. =k/0:314
AN-40 ns ers to ere Pro e s

@qA =@pA D !40 @qA =@p D 3 @q =@pA D 5 @q =@p D !20 .0; !2/, .0; 2/ not rel ext D 72:67, k D 43:7
competitive
pA D 0, p D 5
@qA 100 @qA 50 @q 500
D! D! D! qA D 4 , q D 40, pA D 52, p D 44, profit D 3304
@pA 2 1=2 @p 3=2 @pA 4=3
p p A pA p 3p p A .3; 3/ rel max 1 ft by 2 ft by 3 ft
! #
@q 500 3 1
D! complementary ; rel min a D ! , b D !12, d D 33
@p 2
p p
1=3 2 2
A
@P a 2 of A, 3 of b selling price for A is 30, for is 1 , rel
D 0:01A0:27 B!0: C0:01 0:23 0:0
E F0:27 max profit is 25
@B
@P a P D 5 .1 ! e!x / ! 20x ! 0:1 2 c .20; ln 5/ rel max,
D 0:01A0:27 B0:01 C !0: 0:23 0:0
E F0:27 ! #
@C 5
5; ln not rel ext
44 0 4
a !1:015I !0: 46 b one with D 0 and s D s0
@g 1 ! #
D > 0 for F > 0 if F and s fixed and x increases then 1 5
@x F .2; !2/ .1=3; 1=3; 1=3/ 0; ;
4
g increases. ! # ! # ! #
1 1 1 2 4 4 1 1 1
@qA ˇˇ 20 @qA ˇˇ 5 ; ;
3 3 3
; ;!
3 3 3
; ;
4 2 4
a ˇ D! ˇ D
@pA pA D ;p D16 27 @p pA D ;p D16 12 .11; / 74 when D , k D 7
5
b demand for A decreases by " x D 5; 000 newspaper, y D 15; 000 TV
6
a no b 70 x D 5, y D 15, z D 5 x D 12, y D
5 1 .100=3; 50=3; 100= /
'pA D ! ; 'p D
46 46
'pA D !2 for all prices 'p D !1=2 for all prices
1 1=4 2=3
3 31=10 !5 =5
!1 e2 =2 ! e C 1=2
fx .x; y/ D 15x2 y fxy .x; y/ D 15x2 (fy D 5x3 ) fyx .x; y/ D 15x2
27 1=4
3 0 0
1 xe2xy 1 e2xy .2xy C 1/ 72x.1 C xy/e2xy e!4 ! e!2 ! e!3 C e!1
3x2 y C 4xy2 C y3 3xy2 C 4x2 y C x3 6xy C 4y2 6xy C 4x2 1=3

y x2 ! y2
@z=@y D @2 z=@y2 D 2 R C
x2 C y2 .x C y2 /2 y x
2x 2y !
fy D 0 fyx D 0 fyxx D 0 fyxxz D 0 fyxxz .4; 3; !2/ D 0 fx D 2 2
fy D 2 2 .x C y/2 .x C y/2
x Cy x Cy
1 y p 2
744 2e ! p
2
e x Cy
2
2xzex yz .1 C x2 yz/
x2 C y2
2x C 2y C 6z exCyCz .ln.xyz/ C 1=x C 1=y C 1=z/
1
@P=@ D 0.k= /0:2 @P=@k D 20. =k/0:
.1; 2/ .0; !2/, .0; 1/, .3; !2/, .3; 1/ 64
neither rel min at .2; 2/
.50; 150; 350/ .3; 4/ rel min
! # 4 ft by 4 ft by 2 ft
1 1 max for A at cents lb and at 4 cents lb
! ; rel max .3; 1/ .3; 1/ < 0 no rel ext
4 2
! # ! # .5=26; 1=26/ (is point on 5x C y D 5 closest to the origin)
1 1
.0; 0/ rel max 4; rel min 0; , .4; 0/, not rel ext 1=12
2 2
.43; 13/ rel min .!1; !1/ rel min of 3 1=30
 n e
A Applications of inequalities, 5 62 inomials, 15
Absolute extrema, on closed interval, Applied maxima and minima. See axima inomial theorem, 432 433, 4 2
5 1 5 3, 5 2 and minima ond redemption, 50 51
Absolute value, 62 66 Approximate integration, 672 67 orrower, 212
equations, 62 63 Simpson s rule, 673, 674 676 ounded feasible region, 301
inequalities, 63 64 trapezoidal rule, 672 674 ounds of integration, 650
notation, 64 Approximations of e, 1 2 ounds of summation, 67
properties of, 65 APR (annual percentage rate), 1 1, 20 rea -even chart, 16
Absolute-value functions, 3, 10 10 Area, computing using right-hand endpoints, rea -even point, 167 16
Accumulated amount, 1 0 650 651 rea -even quantity, 16 , 16 170
Accumulated amount of continuous annuity, Area, rectangular, 4 4 rea -even revenue, 16
670 Area between curves, 67 6 7 SF (basic feasible solution), 30 , 316.
Accumulated amount of interest, 20 Arithmetic sequences, 74 0 See a s egeneracy
Acid test ratio, 62 Artificial ob ective function, 321 udget equation, 2 5
Addition Artificial problem, 320 udget line, 2 5
of algebraic expressions, 16 Artificial variables, 320 32
associative properties of, 4 defined, 320 C
closure properties of, 4 and equality constraints, 325 327 Calculus, 61 , 653 661. See a s
commutative properties of, 4, 5 example of, 323 325 ultivariable calculus
elimination-by-addition method, 154 155 Assets, current, 60 Cancelled factors, 23
of fractions, 24 26 Associative properties, of addition and Case-defined function, 10 110, 466 467,
matrix, 246 24 multiplication, 4, 6 471 472
Algebra Associative property, matrix multiplication, Cash ows, 217
real numbers. See Real numbers 257 Cells, 363 365
review of, 1 46 Asymptotes, 5 3 603 Central tendency, 42
Algebraic expressions horizontal, 5 5 5 Certain event, 370
addition, 16 nonvertical, 5 6 Chain rule, 51 527
defined, 15 oblique, 5 6 5 defined, 51
multiplication, 1 vertical, 5 3 5 4 example of, 520 522
operations with, 15 1 Augmented coe cient matrix, 265 266 marginal-revenue product, 524 525
special products, 17 1 reducing, 26 271 power rule, 522 524
subtraction, 16 Average cost per unit, 505 Change-of-base formula, 1
Algebraic manipulation, 457 45 , 62 630 Average rate of change, 500 Chebyshev s inequality theorem, 42
Amortization Average value of function, 6 0 6 2 Clearance rate, 60
defined, 230 Average velocity, 500 Closed half-plane, 2 5
finance charge, 230 Axis of symmetry, 145 Closed intervals, 57
formulas, 231 Closure properties, of addition and
of loans, 230 234 B multiplication, 4
periods of loan, 233 ase, 10 Coe cient matrix, 262, 265
schedule, 230 asic Counting Principle, 34 351 augmented, 265 266
Annual percentage rate (APR), 1 1, 20 , 220 asic feasible solution ( SF), 30 , 316. not invertible, 2 4
Annual rate of interest, 20 See a s egeneracy reduced, 276
Annuities, 222 232 asic variables, 30 Coe cient of inequality, 6 6
continuous, 230, 670 671 ayes formula, 411 41 Coe cients, 15
defined, 222 defined, 414 Column vector, 242
future value of, 226 22 elly beans in bag, 416 417 Combinations, 355 361
integration applied to, 66 671 partition, 413 basic combinational identity, 360 361
ordinary, 222 posterior probability, 413 committee selection, 357
payment period, 222 prior probabilities, 413 defined, 356
periodic payment of, 225 quality control, 414 416 Pascal s Triangle, 361
present value of, 223 226 ayes probability tree, 416 permutations s., 356 357
sin ing fund, 227 22 ernoulli trials, 434 po er hand, 357 35
term, 222 inomial coe cients, 433 and sets, 35 361
Annuity due, 222, 225 226, 227 inomial distribution, 432 437 sum of, ma ority decision and, 35
Antiderivative, 625 626 ernoulli trials, 434 Common difference, 74 75
Apartment rent, determination of, 51 52 experiment, 434 Common factors, 20
Applications, and linear functions, 13 145 independent trials, 433 Common logarithms, 1 1
Applications, of systems of equations. See random variable, 435 Common ratio, 75
Systems of equations, applications of theorem, 432 433 Commutative properties, of addition and
Applications of equations, 4 54. See a s inomial distribution, normal approximation multiplication, 4, 5
Equations to, 726 72 Compensating balance, 54

I-1
I-2 n e

Competitive products, 740 742 of derivatives, 4 0 enominators, rationalizing, 12, 24

Complementary products, 740 742 discontinuous, 46 ensity function, 714 71
Complement of events, probability of, 370, infinite discontinuity, 471 defined, 67
371 372 point of discontinuity, 46 example of, 70
Completing the square, 146 of polynomial functions, 470 exponential, 717 71
Composition functions, 100 post-o ce function, 472 474 normal, 721
Compound amount, 1 0 1 1, 20 rational function, 471 normal-distribution, 53
Compounded continuously. See Interest Continuity correction, 72 probability, 715
compounded continuously Continuous, 46 uniform, 716
Compound experiment, 3 4 Continuous annuity, 230, 670 671 eparting variable, 310
Compound interest, 20 214 Continuous random variables, 714 721 ependent events, 402
accumulated amount, 1 0, 20 defined, 425 ependent variable, 5
annual percentage rate, 1 1, 20 density functions, 714 71 epreciation, 3 , 144, 206
comparing interest rates, 212 mean, 71 720 erivative(s), 4 3 4 1. See a s
compound amount, 1 0 1 1, 20 standard deviation, 71 720 ifferentiation Rate of change Tangent
defined, 1 0, 20 Contour lines, 126 line
doubling money, 210, 211 212 Convergent, 706 antiderivatives, 625 626
effective rate, 210 211 Coordinate axes, 104 continuity of, 4 0
example of, 1 0 1 1, 20 , 210 Coordinate planes, 124 defined, 4 6
formula, 20 Coordinates, 3 definite integral of, 65
interest periods, 1 1 Corner points, 300, 301 difference quotient, 4 6
negative interest rates, 212 213 Corollary, 277 differentiability of, 4 0
nominal rate, 1 1, 20 Critical points, 573, 747 74 differentiable, 4 6
periodic rate, 1 1, 20 Critical value, 573 differentiation, 4 6
rate per period, 20 Cube roots, 10 of exponential functions. See erivatives of
solving, 210 Cumulative distribution function, 71 71 exponential functions
tables, 76 776 Current assets, 60 finding, 4 6 4 7, 4 4 0
Concavity, 5 3 5 0 Current liabilities, 60 higher-order, 562 566
criteria for, 5 5 Current ratio, 60 higher-order partial derivatives, 744 746
curve s etching, 5 7 5 Curves, 621 of logarithmic functions. See erivatives of
defined, 5 4 area between, 67 6 7 logarithmic functions
in ection point, 5 5 5 7 normal, 721, 722 mixed partial derivatives, 745
testing for, 5 5 5 6 Curve s etching, 56 61 partial derivatives, 733 73
Conditional probability, 3 3 4. See a s absolute extrema, on closed interval, as rate of change. See Rate of change
Probability 5 1 5 3 secant line, 4 3
advertising, 3 3 3 4 applied maxima and minima. See axima second order, 562, 563, 744 745
defined, 3 and minima slope of a curve, 4 4, 4 7
of event, 3 2 concavity. See Concavity tangent line. See Tangent line
formula for, 3 0 relative extrema, 570 5 1 third order, 562
genders of offspring, 3 3 erivatives of exponential functions, 537 542
general multiplication law, 3 3 with base 4, 540
elly beans in bag, 3 0 3 1 ecay constant, 1 5, 6 6 6 7 to base b, 540 541
quality control, 3 2 ecimal numbers, nonterminating, 2 different forms, 540 541
reduced sample space, 3 ecision variables, 30 functions involving eu , 53
subspace, 3 eclining-balance depreciation, 206 functions involving ex , 53
survey, 3 1 3 2 de Fermat, Pierre, 34 inverse function rule, 53
Venn diagram for, 3 3 0 efinite integral, 647 653, 672 normal-distribution density function, 53
Constant factor rule, 4 5 area requiring two, 67 power functions, 541
Constant functions, 1 2 computing area by using right-hand erivatives of logarithmic functions, 532 537
derivatives of, 4 2 endpoints, 650 651 to base 2, 536
Constant of integration, 626 defined, 650, 654 to base 10, 536
Constants, 15, 2 of derivative, 65 to base b, 535 536
Constraints evaluating, 651, 657 derivative of ln x, 533
defined, 2 finding, by using tables, 66 functions involving ln u, 534
equality, 325 327 finding and interpreting, 657 65 functions involving ln x, 533
multiple, 75 760 integrand, 650 functions involving logarithms, 535
Consumers surplus, 6 7 6 0 properties of, 656 65 overview of, 532 533
Consumption function, 516 517, 645 variable of integration, 650 rewriting before differentiating, 534 535
Continuity, 46 47 egeneracy, 316 iagonal function, 127
applied to inequalities, 474 47 egree, of polynomial, 15 iagonal matrix, 245
applying definition of, 46 470 emand, elasticity of, 543 54 ifference quotient, , 4 6
case-defined functions, 471 472 emand curves, 112, 13 140 ifference rule, 4 6
continuous, 46 emand equation, , 140, 141, 164, 201, 543 ifferentiability, of derivatives, 4 0
continuous on an interval, 470 emand function, , 543 ifferential equations, 6 2 6
continuous on its domain, 470 emand schedule, 1, 112 ancient tool, estimating age of, 6 7
defined, 46 emand vectors for an economy, 24 applications of, 6 705
 n e I-3

decay constant, finding, 6 6 6 7 simplex method and, 340 342 Equiprobable spaces, 374 377
defined, 6 2 slac variable, 33 defined, 375
exponential growth and decay, 6 4 6 7 uality, 335 equally li ely outcomes, 374
first-order, 6 2 probability of simple event, 375
general solution, 6 3 E relative frequency, 375
half-life, finding, 6 6 6 7 Economic lot size, 607 trial, 374
particular solution, 6 3 Effective rate, 210 211 Equivalent equations, 2 30
population growth, 6 5 6 6 Elastic, 545 Equivalent inequalities, 56
separation of variables, 6 3 6 4 Elasticity, and revenue, 546 547 Equivalent matrices, 266
ifferentials, 620 625 Elasticity of demand, 543 54 Events, 36 373
computing, 620 Element, 2 certain, 370
curve, 621 Elementary row operations, 265, 266 complement, 370, 371 372
defined, 620 Elimination-by-addition method, 154 155 defined, 370
estimate change in quantity, 621 622 Elimination by substitution, 155 156 dependent, 402
estimate function value, 622 623 Ellipsoid, 127 example of, 370 371
finding dp=dq from dq=dp, 623 Empirical probability, 3 4 3 5 impossible, 370
finding in terms of dx, 621 Empty feasible region, 301, 327 32 independent. See Independent events
tangent line, 621 Empty set, 2, 34 intersection, 371 372
ifferentiation, logarithmic. See ogarithmic Endpoints, 57 mutually exclusive dis oint, 372 373, 37 ,
differentiation Entering variable, 310 3 3, 40
iscontinuous, 46 . See a s Continuity Entries, matrix, 240, 241 properties of, 372
iscount rate, 21 Environmental demand equation, 174 simple, 370
iscrete random variable, 425 432 Environmental supply equation, 174 union, 370, 371 372
central tendency, 42 Equality Venn diagram, 370, 3 3 0
Chebyshev s inequality theorem, 42 of functions, 7 0 Expected value, 425 432
continuous, 425 of matrix, 243 Exponential density function, 717 71
defined, 425 transitive property of, 3 Exponential distribution, 717
dispersion, 42 Equality constraints, 325 327 Exponential equations, 200 204
expected gain, 427 42 Equality sign, 2 , 461 defined, 200
expected value, 427 Equally li ely outcomes, 374 demand equation, 201
mean, 42 430 Equation of degree two. See uadratic oxygen composition, 200
probability function, 425 equations solving, 200 201
probability histogram, 426 Equation of motion, 500 Exponential form, 1
probability table, 426 Equation of the first degree, 135 Exponential functions, 176 1
random variables, 425 427 Equation of value, 215 217 with base e, 1 3 1 5
standard deviation, 42 430 Equations compound interest. See Compound interest
variance, 42 430 absolute-value, 62 63 with constant base, graph of, 17
is oint events, 372 373, 3 0 applications of, 4 54 defined, 176
ispersion, 42 basic, 2 6 2 7 derivatives of. See erivatives of
isplacement, 500 defined, 2 exponential functions
istance, 62 demand, graphs involving e, 1 3 1 4
istribution, of random variable, 425 427 differential. See ifferential equations graphs of, 177 17
istributive properties, 4 5, 6, 257 25 equivalent, 2 30 hemocytometer and cells, 1 4 1 5
ivergent, 706 examples of, 2 2 natural, 1 3, 63 63
ividend, 1 exponential. See Exponential equations number e, 1 2 1 3
ivision fractional, 34 35 population growth, 1 1 1 2, 1 4
defined, 5 graphs, 11 properties of, 17
fractions, 23 24 linear. See inear equations radioactive decay, 1 5 1 6
before integration, 642 of lines, 133 136 rules for exponents, 176
long, 1 1 literal, 32 34, 35 transformations of, 17
multinomial by monomial, 1 logarithmic. See ogarithmic equations Exponential growth, and decay, 6 4 6 7
by zero, 5 matrix, 251, 262 Exponential law of decay, 1 5, 6 5
ivisor, 1 mixture, 4 Exponential law of growth, 6 5
omain of present value, 215 217 Exponents, 10 15
of function, 5, , 103, 10 quadratic. See uadratic equations basic laws of, 11
of sequence, 71 radical, 35 36 examples of, 12 13
ouble-declining-balance depreciation, 206 roots, 107 rules for, 176
ouble integrals, 761 764 systems of, applications of. See Systems of External demand, 2 6, 2
ouble subscripts, 241 equations, applications of Extrema. See Relative extrema
oubling money, 210, 211 212 of tangent line, 4 7 Extreme values, 5 1
oubling period, 712 terminology for, 2 Extreme-value theorem, 5 1 5 3
ual, linear programming, 335 344 Equilibrium, 164 167
defined, 337 with nonlinear demand, 167 F
of maximization problem, 33 340 tax effect on, 165 167 Factored expressions, 21
of minimization problem, 340 Equilibrium price, 164 Factorial notation, 3
primal, 337 33 Equilibrium quantity, 164 Factorials, 3
I-4 n e

Factoring, 20 22 Fundamental theorem of integral calculus, I

Factors, 15 653 661 Identity function,
Feasible points, 2 Future value, 214, 223 Identity matrix, 260, 2 1
Feasible region, 300, 301 of annuities, 226 22 Identity properties, 4
Finance charge, 230 Imaginary unit, 42
Finite sample space, 367 Implicit differentiation, 54 554
Finite sequence, 71 G higher-order, 564 565
First-derivative test for relative extrema, ap index, 743 Impossible event, 370
574 57 auss, Carl Friedrich, 721 Improper integrals, 706 70
First octant, 124 aussian distribution. See Normal distribution Income tax, 4 5
First-order differential equation, 6 2 eneral linear equation, 135, 136, 15 Indefinite integral, 625 631, 643
Fixed costs, 4 , 16 eneral multiplication law, 3 3 of constant and of power of x, 627
Fractional equations, 34 35 eneral solution, of differential equation, constant of integration, 626
Fractions, 22 2 6 3 of constant times a function, 627 62
addition, 24 26 enetics, 3 4 defined, 655
combined operations with, 26 eometric manipulation, 11 finding, 62
division, 23 24 eometric sequences, 74 0 integral sign, 626
least common denominator, 25 ompertz equation, 204 integrand, 626
multiplication, 23 raphs graphing of sum, 62 62
as percentages, 27 absolute-value function, 10 10 of sum and difference, 62
rationalizing denominator, 24 case-defined function, 10 110 using algebraic manipulation to find,
simplifying, 22 23 defined, 105 62 630
subtraction, 24 26 equations, 11 variable of integration, 626
Function, increasing decreasing nature of, of exponential functions, 177 17 Independent events, 401 411. See a s Events
570 571 of function, 121 aptitude test, 40 40
Function of four variables, partial derivatives of functions involving e, 1 3 1 4 cards, 404 405, 407 40
of, 736 737 general linear equation, 136 defined, 402, 407
Function of three variables, partial derivatives horizontal-line test, 110 dice, 406
of, 736 intercepts of, 106 10 example of, 402
Functions of inverse, 14 150 sex of offspring, 406 407
average value of, 6 0 6 2 limit estimation from, 452 453 smo ing and sinusitis, 402 403
combinations of, 6 101 of linear function, 141 142 special multiplication law, 403
composition, 100 of logarithmic functions, 1 1 1 survival rates, 403 404
defined, 4, 5 plane, 125 Independent trials, 433
demand, quadratic functions, 147 14 Index of summation, 67
dependent variable for, 5 square-root function, 10 Indicators, 310
difference quotient, raphs in rectangular coordinates, 104 112 Inelastic, 545
domain of, 5, , 10 absolute-value function, 10 10 Inequalities
equality of, 7 0 coordinate axes, 104 absolute value, 63 64
exponential. See Exponential functions function-value axis, 107 applications of, 5 62
independent variable for, 5 intercepts and, 106 10 , 125 continuity and, 474 47
inverse, 101 104 origin, 104 defined, 47, 55
inverse of, 14 150 points, 104 equivalent, 56
linear. See inear functions quadrants, 104 105 linear. See inear inequalities
logarithmic. See ogarithmic functions rectangular coordinate system, 104 rules for, 55 56
probability, 3 0 3 2 square-root function, 10 triangle, 65
quadratic. See uadratic functions vertical-line test, 10 , 10 usage, example of, 47
range of, 5, 10 x-intercept, 105, 106 Inequality symbol, 55
rational. See Rational functions x; y-plane, 104 Infinite geometric sequence, 7
scalar product, 7 y-intercept, 105, 106 Infinite limits, 461 467
of several variables. See Functions, of rouping symbols, removing, 16 17 Infinite sequence, 71
several variables Infinity, limits at. See imits at infinity
special. See Special functions H Infinity symbol, 4 0
supply, 0 Half-life, 1 5, 1 2 1 3, 6 6 6 7 In ection point, 5 5 5 7
values, 6, , 10 Half-plane, 2 5 2 6 Initial amount, 1 5
Functions, of several variables, 120 127 Higher-order derivative, 562 566 Initial conditions, integration with, 631 635
graphing plane, 125 Higher-order implicit differentiation, 564 565 cost from marginal cost, 634
level curves, 126 127 Higher-order partial derivatives, 744 746 defined, 631
temperature humidity index, Histogram, probability, 426 demand function from marginal revenue,
123 125 Homogeneous system, 275 27 633 634
3-dimensional rectangular coordinate Horizontal asymptotes, 5 5 5 income and education, 633
system, 123 Horizontal line equations, 135 problem, 631 632
of two variables, 122 123 Horizontal-line test, 110 problem involving yn , 632
Functions of two variables, 746 754 Horizontal strips, 6 3 6 4, 6 6 Initial distribution, 43
Function-value axis, 107 Horizontal translation, 11 Initial simplex table, 30
Fundamental principle of fractions, Horner s method, 5 Initial state vector, 43
 n e I-5

Input output analysis, 2 6 2 2 to solve equations, 103 first-degree, 31

application of inverses, 2 symmetry and, 117 solving, 31 32
basic equation, 2 6 2 7 Inverse properties, 4 inear equations systems, 152 162
example of, 2 0 2 1 Inverse rule, integration of, 666 three-variable systems, 15 160
external demand, 2 6 Inverses, 27 2 5 two-variable systems, 152 15
internal demand, 2 7 application of, 2 inear functions
eontief matrix, 2 7 defined, 27 2 0 applications and, 13 145
Input output matrix, 2 6 determination of, 2 1 2 2 defined, 141
Input variable, 5 example of, 2 0 demand curves, 13 140
Instantaneous rate of change, 501 finding, 2 2 2 3 demand equation, 140, 141
Instantaneous velocity, 501 to solve a system, 2 0 2 1, 2 3 2 4 determination of, 142 143
Integers, 2 Investment, modeling example, 50 diet for hens, 143
Integral, definite. See efinite integral Irrational numbers, 2 graphing, 141 142
Integral calculus, 61 Isocost line, 144 inverses of, 102
Integral sign, 626 Isoprofit line, 126, 144, 301 production levels, 13
Integrand, 626, 650 Isotherms, 126 supply curves, 13 140
Integration supply equation, 140
application of, 644 645 in x and y, 2
approximate. See Approximate integration oint-cost function, 73 inear inequalities, 55 5
consumption function, 645 defined, 56, 2 5, 2
definite integral, finding by using tables, L rules for, 55 56
66 affer, Arthur, 56 solving, 57 5 , 2 6 2 7
differential equations. See ifferential affer curve, 56 system of, 2 7 2
equations agrange multipliers, 754 761 in two variables, 2 5 2
formulas. See Integration formulas defined, 755 inearly related variables, 135
improper integrals, 706 70 least-cost input combination, 75 75 inear programming, 2 306
indefinite integrals, 643 method of, 756 757 artificial variables, 320 32
integrating bu , 643 644 minimizing costs, 757 75 bounded feasible region, 301
of inverse rule, 666 multiple constraints, 75 760 corner points, 300
natural exponential functions and, 63 63 atency, 542 dual. See ual, linear programming
power rule for, 635 63 aw of compound probability, 3 4 empty feasible region, 301,
preliminary division before, 642 eading coe cient, 2 327 32
techniques of, 642 647 eading entry, matrix, 267 equality constraints, 325 327
Integration applied to annuities, 66 671 earning equation, 204 feasible points, 2
Integration by tables, 666 672 east common denominator ( C ), , 25 feasible region, 300, 301
annuities, applied to, 66 671 eibniz, ottfried Wilhelm, 4 isoprofit line, 301
examples of, 667 66 eibniz notation, 4 6 linear function in x and y, 2
Integration formulas, 635 642 ender, 212 maximization, 2
involving logarithmic functions, 63 640 eontief, Wassily W., 2 6 minimization, 2 , 330 335
natural exponential functions and, eontief matrix, 2 7, 2 2 0 nonempty feasible region, 301
63 63 evel curves, 126 nonnegativity conditions, 300
power rule for integration, 635 63 iabilities, current, 60 ob ective function, 2 , 300
Integration with initial conditions. See Initial ife-table function, 660, 675 overview, 2 4
conditions, integration with imits, 451 46 problems, 2 , 303
Intercepts, 106 10 , 114 116, 125 and algebraic manipulation, 457 45 profit function, 300
testing, 577 57 for case-defined function, 466 467 and simplex method. See Simplex method
Interest, compound. See Compound interest defined, 451, 452 standard maximum linear programming
Interest compounded continuously, 21 222 estimation from graph, 452 453 problem, 307 30
compound amount, 21 finding, 457 45 unbounded feasible region, 301
defined, 21 form 0=0, 45 inear supply curve, 140
effective rate under, 21 220 infinite, 461 467 ines, 132 13
present value under, 220 nonexistent, 453 454 equations of, 133 136
trust fund, 220 one-sided, 461 forms of equations of lines, 136
Interest periods, 1 1 overview, 451 452 general linear equation, 135, 136
Internal demand, 2 7 perpetuities, 235 236 horizontal line equations, 135
Intersection of events, probability of, polynomial function, 456 457 isocost, 144
371 372 properties of, 454 457 isoprofit, 144
Interval notations, 57 of sequence, 236 linearly related variables, 135
Intervals, 57 special, 45 45 parallel, 137
Inverse function rule, 53 imits at infinity, 463 467 perpendicular, 137
Inverse functions, 101 104 for polynomial functions, 466 point-slope form, 134
defined, 101 for rational functions, 464 466 slope-intercept form, 134 135
graphing, 14 150 ine, symmetry about, 116 117 slope of. See Slope of line
linear, 102 inear demand curve, 140 vertical line equations, 135
one-to-one function, 102 inear equations iteral constants, 32
restricting domain of, 103 defined, 30 iteral equations, 32 34, 35
I-6 n e

oans, amortization of, 230 234 atrix addition, 246 24 maximization of profit, 60 610
ogarithm, integral of, 666 defined, 247 maximizing output, 751
ogarithmic differentiation, 554 55 example of, 247 maximizing recipients of health-care
defined, 554 properties of, 247 24 benefits, 60
differentiating form u , 555 55 scalar multiplication, 24 250 maximizing revenue, 605
example of, 555 atrix equations, 251 maximizing TV cable company revenue,
relative rate of change of a product, 556 atrix matrices, 241 27 607 60
ogarithmic equations, 200 204 addition. See atrix addition minimizing average cost, 606
defined, 200 applications of, 240 minimizing cost of fence, 603 605
demand equation, 201 coe cient. See Coe cient matrix minimizing costs, 757 75
oxygen composition, 200 column vector, 242 profit maximization, 751 752
predator prey relation, 202 constructing, 243 relative, 746
solving, 202 203 defined, 242 relative extrema, finding, 572, 750
ogarithmic form, 1 diagonal, 245 saddle point, 74 750
ogarithmic functions, 1 1 4 double subscripts, 241 second-derivative test, 74
base, 1 1 elementary row operations, 265, 266 ean, 1 4, 427, 42 430, 71 720
common logarithms, 1 1 entries, 240, 241 ethod of reduction, 264 27
defined, 1 equality of, 243 example of, 267 26
derivatives of. See erivatives of equations, 251, 262 homogeneous system, 275 27
logarithmic functions equivalent, 266 nonhomogeneous system, 275 277
graph of, 1 1 1 identity, 260, 2 1 parametric form of solution, 271 273
integrals involving, 63 640 input-output, 2 6 solving system by, 26 271
natural logarithms, 1 1 inverses of. See Inverses two-parameter family of solutions,
properties of. See ogarithms, properties of leading entry, 267 274 275
radioactive decay and half-life, 1 2 1 3 eontief matrix, 2 7, 2 2 0 inimization, 2 , 330 335
solving, 1 1 1 2 lower triangular, 245 ixed partial derivatives, 745
ogarithms, properties of, 1 4 1 main diagonal square, 245 ixture equations, 4 , 157 15
base 5, evaluating, 1 multiplication. See atrix multiplication odeling, 4 , 702 703
change-of-base formula, 1 nonzero-row, 267, 277 onomials, 15, 1
combining logarithms, 1 6 operations involving I and 0, 260 261 onopolist, 60
expressions, rewriting, 1 5 power of, 261 ultinomials, 15, 1
expressions, simplifying, 1 7 rectangular arrays, 241, 242 ultiple constraints, 75 760
finding, 1 4 1 5 reduced, 267 26 ultiple integrals, 761 764
writing in terms of simpler logarithms, row vector, 242 ultiple optimal solutions, 301
1 6 size of, 242 ultiplication
ogistic curve, 700 solving systems by reducing. See ethod associative properties of, 4
ogistic function, 700, 701 of reduction closure properties of, 4
ogistic growth, 6 702 special, 244 245 commutative properties of, 4, 5
ong division, 1 1 square, 244 245 fractions, 23
orenz curve, 6 6 subtraction of, 250 251 matrix. See atrix multiplication
oss, profit and, 16 16 transition, 43 , 440 multinomials, 1
ower bound, 650 transpose of, 244 scalar, 24 250
ower triangular matrix, 245 triangular, 245 ultiplication law, 403
upper triangular, 245 ultivariable calculus, 732 767
M zero, 244 applications of partial derivatives, 73 744
ain diagonal square matrix, 245 zero-row, 267 higher-order partial derivatives, 744 746
arginal cost, 504 505, 634, 73 73 atrix multiplication, 253 264 agrange multipliers, 754 761
arginal productivity, 740 associative property, 257 maxima minima for functions of two
arginal propensity to save, 516 517 cost vector, 256 variables, 746 754
arginal revenue, 505 506, 515 516, defined, 253 multiple integrals, 761 764
633 634 distributive property, 257 25 partial derivatives, 733 73
arginal-revenue product, 524 525 matrix product, 255 256 utually exclusive events, 372 373, 37 ,
arginal utility of income, 761 power of matrix, 261 3 3, 40
argin of profit, 54 properties of, 257 25
ar ov, Andrei, 437 raw materials and cost, 25 25 N
ar ov chains, 437 446 sizes of matrices and their product, 255 Natural exponential function, 1 3, 63 63
defined, 437 transpose of product, 25 260 Natural logarithms, 1 1
demography, 441 axima and minima, 603 614 Negative, 4, 250
initial state vector, 43 critical points, finding, 747 74 Negative integers, 2
k-state, 43 dual linear programming, 33 340 Negative interest rates, 212 213
regular, 443 economic lot size, 607 Net present value (NPV), 217
state vector, 43 for functions of two variables, 746 754 Newton s law of cooling, 703 704
steady-state vectors, 442 445 guide for solving applied max min Newton s method, 55 562
transition matrix, 43 , 440 problems, 605 defined, 55 , 55
transition probabilities, 43 linear programming, 2 , 330 335 recursion formula, 560
two-state, 43 , 43 maximization applied to enzymes, 606 root approximation by, 560 561
 n e I-7

Nominal rate, 1 1, 20 , 220 defined, 734 equations of, 215 217

Nonbasic variables, 30 finding, 734 735 example of, 214 215
Nonempty feasible region, 301 of function of four variables, 736 737 investments, comparing, 216
Nonexistent limits, 453 454 of function of three variables, 736 of perpetuities, 234 235
Nonhomogeneous system, 275 277 higher-order, 744 746 single-payment trust fund, 215
Nonlinear demand, 167 oint-cost function, 73 Present value of a continuous annuity, 670
Nonlinear inequality, 47 loss of body heat, 73 740 Present value of continuous income stream,
Nonlinear systems, 162 164 marginal cost, 73 73 670
Nonnegativity conditions, 300 marginal productivity, 740 Price quantity relationship, 133
Nonrational function inequality, 47 mixed, 745 Pricing, 4 50
Nonterminating decimal numbers, 2 production function, 740 Primal, 337 33
Nonvertical asymptote, 5 6 second-order, 744 745 Principal, 1 0, 20
Nonzero constant, 15 Particular solution, of differential equation, Principal nth root, 10
Nonzero-row, of matrix, 267 6 3 Principal square root, 11
Normal approximation to binomial Partition, 413 Prior probabilities, 413
distribution, 726 72 Pascal, laise, 34 Probability, 34 , 374 3 . See a s
Normal curve, 721, 722 Pascal s Triangle, 361 Conditional probability
Normal density function, 721 Payment period, annuities, 222 birthday surprise, 3 0
Normal distribution, 721 726 Percentage rate of change, 506 507 of complement of events, 370
analysis of test scores, 722 723 Percentages, 27 conditional. See Conditional probability
defined, 721 Periodic payment, of annuities, 225 dice, 3 1 3 2
standard normal variable, 723 725 Periodic rate, 1 1, 20 empirical, 3 4 3 5
Normal-distribution density function, 53 Permutations, 351 355 equiprobable spaces. See Equiprobable
Normal random variable, 721 club o cers, 352 353 spaces
Notations combinations s., 356 357 interrupted gambling, 3 2
absolute value, 64 defined, 351 of intersection of events, 371
factorial, 3 legal firm, 353 odds, 3 5 3 6
interval, 57 political questionnaire, 353 opinion poll, 3 4 3 5
eibniz, 4 6 with repeated ob ects, 361 363 properties of, 377 3 2
summation. See Summation notation Perpendicular lines, 137 quality control, 37 3 0
Not equal to symbol, 2 Perpetuities, 234 237 of union of events, 370, 37
NPV (net present value), 217 defined, 234 of winning prize, 3 6
Number e, 1 2 1 3 limits, 235 236 Probability density function. See ensity
Numerators, rationalizing, 12 present value of, 234 235 function
Numerical coe cient, 15 Pivot column, 311 Probability functions, 3 0 3 2
Pivot entry, 311 Probability histogram, 426
Pivot row, 311 Probability of E, 376
b ective function, 2 , 300 Planes, 124, 125 Probability of simple event, 375
artificial, 321 Point elasticity of demand, 544, 545 546 Probability table, 426
b ective row, 30 Point of discontinuity, 46 Probability tree, 3 4, 416
blique asymptotes, 5 6 5 Point of equilibrium, 164, 6 7 Producers surplus, 6 7 6 0
ctants, 124 Point of in ection, 5 5 5 7 Product, finding, 6 7
dds, 3 5 3 6 Points, 104 Production function, 740
ne-sided limits, 461 Point-slope form, 134 Product rule, 50 51
ne-to-one correspondence, 3 Poiseuille s law, 664 applying, 510 511
ne-to-one function, 102 Poisson distribution function, 1 4 consumption function, 516 517
pen half-plane, 2 5 Polynomial, 15 differentiating product of three factors,
pen intervals, 57 Polynomial functions 511 512
perating ratio, 2 asymptotes and, 5 overview, 510
perations with algebraic expressions, 15 1 continuity of, 470 usage to find slope, 512 513
rdered n-tuple, 121 defined, 2 Profit, 4 , 5 , 16
rdered pair, 4, 104, 121 limits at infinity for, 466 Profit function, 300
rdered triple, 120 121 limits of, 456 457 Profit maximization, 60 610, 751 752
rdinary annuity, 222 Polynomial inequality, 476 477 Prolate spheroid, 127
rigin, 2, 104, 113 114 Population growth, 1 1 1 2, 1 4 Promissory note, 21
utput variable, 5 Position function, 500 Properties
Positive integers, 2 absolute value, 65
Posterior probability, 413 definite integral, 656 65
Paired-associate learning, 1 6 Power function, 4 2, 541 demand vectors for an economy, 24
Parabola, 145 Power rule, 522 524, 53 events, 372
Parallel lines, 137 Power rule for integration, 635 63 exponential function, 17
Parameters, 157, 717 Preliminary division, before integration, 642 limits, 454 457
Partial derivatives, 733 73 Present value, 214 21 logarithmic functions, 1 4 1
applications of, 73 744 of annuities, 223 226 matrix addition, 247 24
competitive and complementary products, under continuous interest, 220 matrix multiplication, 257 25
740 742 defined, 214 probability, 377 3 2
I-8 n e

Properties (c ntinued) total-revenue function, 505 differentiating a constant times a function,

real numbers, 3 velocity, 501 502 4 5 4 6
scalar multiplication, 250 of volume, 503 differentiating sums and differences of
summation notation, 6 70 Rational functions functions, 4 6 4 7
Proportional change, 624 defined, 2 equation of tangent line, finding,
Purchasing, renting s., 5 60 discontinuities in, 471 4 7 4
inequality, 477 47 rewriting functions in form xa , 4 4 4 5
limits at infinity for, 464 466 sum difference rule, 4 6
uadrants, 104 105 Rationalizing denominators, 12, 24
uadratic equations, 3 45 Rational numbers, 2 S
defined, 3 Real-number line, 3 Saddle point, 74 750
factoring, solution by, 3 40, 41 Real numbers, 2 Sample points, 367
fractional equation leading to, 40 41 properties of, 3 Sample spaces, 367 36
higher-degree, solving by factoring, 40 sets of, 2 3 defined, 367
with no real roots, 43 Recall interval, 542 finite, 367
with one real roots, 42 Reciprocal, of number, 4, 5 elly beans in bag, 36 36
quadratic-form, 43 Rectangular area, 4 4 outcome, 367
quadratic formula, 41 43 Rectangular arrays, 241, 242 po er hand, 36
with two real roots, 42 Rectangular coordinate plane, 104 reduced, 3
uadratic formula, 41 43 Rectangular coordinate system, 104 replacement, 36
uadratic functions, 145 152 Rectangular hyperbola, 107 roll of two dice, 36
axis of symmetry, 145 Recursion formula, 560 sample points, 367
defined, 2, 145 Recursively defined sequences, 73 74 three tosses of coin, 36
graphing, 147 14 Reduced coe cient matrix, 276 toss of two coins, 36
inverse, graphing, 14 150 Reduced matrix, 267 26 Scalar multiplication, 24 250
maximum revenue, 150 151, 605 Reduced sample space, 3 Scalar product, 7
parabola, 145 Reduction methods, matrices. See ethod of Scalars, 24
vertex, 145, 146 reduction Secant line, 4 3, 4 4
uadratic inequality, 476 Re ections Second-degree equation. See uadratic
uic ratio, 62 shrin ing and, 11 120 equations
uotient rule, 50 51 translations and, 11 120 Second-derivative test
applying, 514 Regular ar ov chains, 443 for functions of two variables, 74
differentiating quotients without using, 515 Regular transition matrix, 443 for relative extrema, 5 1 5 2
marginal revenue, 515 516, 633 634 Relation, defined, 4 Second-order derivative, 562, 563
overview, 513 Relative extrema, 570 5 1 Second-order partial derivatives, 744 745
rewriting before differentiating, 514 515 absolute maximum minimum, 572 Separation of variables, 6 3 6 4
condition for, 572 Sequence of length, 70, 71
R criteria for, 573 Sequences, 70 0
Radical equations, 35 36 critical point, 573 arithmetic, 74 0
Radicals, 10 15 critical value, 573 domain of, 71
basic laws of, 11 curve s etching, 577 57 equality of, 73
examples of, 13 14 finding, 577 finite, 71
Radioactive decay, 1 5 1 6, 1 2 1 3 first-derivative test for, 574 57 formula for, 72 73
Random variables, 425 427, 435 relative maximum minimum, 572 geometric, 74 0
Range second-derivative test for, 5 1 5 2 infinite, 71
of function, 5, 10 where f 0 .x/ does not exist, 576 introduction, 70 73
of sequence, 72 Relative frequency, 375 limits of, 236
Rate of change, 4 50 . See a s Relative maximum minimum, 746 range of, 72
erivative(s) Relative rate of change, 506 507 recursively defined, 73 74
applications to economics, 504 505 Remainder, 1 sums of, 76 7
average, 500 Rent, apartment, determination of, 51 52 term of, 71, 72, 75 76
average cost per unit, 505 Renting s purchasing, 5 60 of trials, 3 4
average velocity, 500 Repeated ob ects, permutations with, 361 363 Sets, 4
of enrollment, 504 Replacement, sample spaces, 36 combinations and, 35 361
equation of motion, 500 Response magnitude, 206 defined, 2
estimating !y by using dy=dx, 502 Revenue, elasticity and, 546 547 empty, 2, 34
finding, 503 Right-hand endpoint, 650 651 of real numbers, 2 3
instantaneous, 501 Root, 10, 2 , 107 solution, 2
instantaneous velocity, 501 Row vector, 242 union, 64
marginal cost, 504 505 Rules for differentiation, 4 1 4 . See a s Shadow price, 337
marginal revenue, 505 506 erivative(s) Shrin ing, and re ections, 11 120
percentage, 506 507 constant factor rule, 4 5 Sides, 2
position function, 500 derivative, finding, 4 7 Sigmoid function, 705
price with respect to quantity, 503 derivative of constant, 4 2 Sign chart, 475
relative, 506 507 derivative of xa , 4 3 4 4 Signs. See Symbols
total-cost function, 504 derivatives of powers of x, 4 4 Simple events, 370
 n e I-9

Simplex method, 306 31 Standard normal density function, 723 Systems of equations, applications of,
basic feasible solution, 30 Standard normal distribution, 723 164 171
basic variables, 30 Standard normal random variable, brea -even points, 167 16
decision variables, 30 723 725 equilibrium, 164 167
defined, 306 Standard units, 723
departing variable, 310 State vector, 43
dual linear programming and, Statistics, 34 . See a s Probability T
340 342 Statistics application, 67 6 0 Tangent line
entering variable, 310 Steady-state vectors, 442 444 defined, 4 3
example of, 314 31 Step function, 472 equation of, 4 7, 4 7 4
indicators, 310 Stochastic processes, 3 4 3 . See a s example of, 621
initial simplex table, 30 ar ov chains slope of, 4 5 4 6
nonbasic variables, 30 cards, 3 4 3 6 vertical, function with, 4
ob ective row, 30 compound experiment, 3 4 Temperature humidity index (THI),
pivot column, 311 defective computer chips, 3 6 3 7 123 125
pivot entry, 311 defined, 3 4 Term
pivot row, 311 elly beans in bag, 3 7 3 of annuities, 222
slac variable, 30 law of compound probability, 3 4 of sequence, 71, 75 76
solving, 313 probability tree, 3 4 3 5 Theorem of integral calculus, 653 661
standard maximum linear programming stages, 3 4 Theoretical probabilities, 3 4
problem, 307 30 trials, 3 4 Third-degree equation, 40
Simpson s rule, 673, 674 676 Straight-line depreciation, 3 , 144 Third-order derivative, 562
Single declining-balance depreciation, Subset, 2 3-dimensional rectangular coordinate system,
206 Substitutes products, 741 123
Single-payment trust fund, 215 Substitution, elimination by, 155 156 Three-variable linear equations systems,
Sin ing fund, 227 22 Subtraction 15 160
S etching, of surface, 125 126 of algebraic expressions, 16 general linear equation, 15
Slac variable, 30 , 33 defined, 5 one-parameter family of solutions,
Slope-intercept form, 134 135 of fractions, 24 26 15 160
Slope of curve, 4 4, 4 7 of matrices, 250 251 solving, 15 15
Slope of line Summation notation, 66 70 two-parameter family of solutions, 160
defined, 132 bounds of summation, 67 Topographic map, 126
equations of lines, 133 136 defined, 66, 67 Total cost, 4 , 16
general linear equation, 135, 136 index of summation, 67 Total-cost function, 504
horizontal line equations, 135 properties of, 6 70 Total revenue, 4 , 16
parallel line, 137 sums, evaluating, 67 Total-revenue function, 505
perpendicular line, 137 sums, writing, 67 6 Traces, 125
point-slope form, 134 Sum rule, 4 6 Transformations, 11 11
price quantity relationship, 133 Supply curves, 112, 13 140 of exponential functions, 17
slope-intercept form, 134 135 Supply equation, 140, 164 Transition matrix, 43 , 440, 443
from two points, 134 Supply function, 0 Transitive property of equality, 3, 6
vertical line equations, 135 Surface, s etching, 125 126 Translations, and re ections, 11 120
Solutions, of equation, 2 Surplus, consumers and producers , Transpose of matrix, 244
Solution set, 2 6 7 6 0 Transpose of product, 25 260
Special functions, 1 6 Surplus variable, 321 Trapezoidal rule, 672 674
absolute-value functions, 3 Symbols Tree diagram, 34
case-defined function, 2 3 equality, 2 , 461 Trend equation, 130
constant functions, 1 2 grouping, removing, 16 17 Trials, 3 4
factorials, 3 inequality, 55 Triangle inequality, 65
genetics, 3 4 infinity, 4 0 Triangular matrix, 245
Horner s method, 5 integral sign, 626 Trinomials, 15, 20 21
polynomial function. See Polynomial not equal to, 2 Triple integrals, 764
functions radical, 11 Trivial solution, homogeneous system,
rational functions. See Rational union, 64 276
functions for variables, 2 Two-level tree diagram, 34
Special limits, 45 45 Symmetry, 113 11 Two-point form, 135
Special matrices, 244 245 about line, 116 117 Two-state ar ov chains, 43 , 43
Special multiplication law, 403 axis of, 145 Two-variable functions, 122 123
Special products, 17 1 intercepts and, 114 116 Two-variable linear equations systems,
Square matrix, 244 245 and inverse functions, 117 152 15
Square-root function, 10 of origin, 113 114 elimination-by-addition method,
Square roots, 10 testing, 57 154 155
Standard deviation, 42 430, 71 720 tests for, 114 elimination by substitution, 155 156
Standard maximum linear programming x-Axis, 113 with infinitely many solutions,
problem, 307 30 y-Axis, 113 114 156 157
Standard normal curve, 722, 7 1 System of inequalities, 2 7 2 mixture, 157 15
I-10 n e

U Variables
Unbounded feasible region, 301 defined, 2 x-Axis symmetry, 113
Unbounded solution, 316 dependent, 5 x-intercept, 105, 106, 147
Unconditional probability, 3 , independent, 5 x; y-plane, 104
413 restrictions on, 2
Uniform density function, 716 symbols for, 2
Uniform distribution, 716 Variance, 42 430 y-Axis symmetry, 113 114
Union, of sets, 64 Vehicle inspection pit, 4 4 y-intercept, 105, 106
Union of events, probability of, 370, Velocity, 501 502
371 372, 37 Venn diagram, 370, 3 3 0
Union symbol, 64 Verhulst Pearl logistic function, eno of Elea, 450
Unit distance, 2 700 ero
Unit elasticity, 545 Vertex, 145, 146 division by, 5
Upper bound, 650 Vertical asymptotes, 5 3 5 4 exponent, 10
Upper triangular matrix, 245 Vertical bars, 241 ero-coupon bond, 213
Vertical-line equations, 135 ero function, 2
Vertical-line test, 10 , 10 ero matrix, 244
Variable costs, 4 , 16 Vertical strips, 67 , 6 0 6 3 ero-row, of matrix, 267
Variable of integration, 626, 650 Volume, rate of change, 503 eros, of function, 107
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 n e of cat ons
B arginal revenue, 4 6, 506, 515 Supply schedule, 0
aximizing output, 751 Supply-price, changes in, 534
Advertising, 3 3, 465 aximizing revenue, 607 Taxes and, 1, 33, 165, 436, 60
rea -even point, 16 , 16 inimizing costs, 603, 606, 757 Total revenue, 650
rea -even quantities, 560 ultiple ventures, 50 Value of product, 142
Cashiers ma ing change, 453 Naming of, 353, 363 Value-added-tax, 1, 33
Cellular phone networ , 713 il production, 732 Wor place, time-analysis matrix, 243
Competition, 43 , 62 Postal rates, 472
Competitive pricing, 251 Price reductions, 511
Competitive complementary products,
741
Price-quantity relationship, 133 E
Pricing, 57, , 3,
Consumer goods sales, 1, 47, 3, 246, Product warranty, 717 Amortization of loans, 230, 231
24 , 2 6 Production, assembled parts, 3 6 Annuities, 223, 224, 66 , 670
Cost hours of production, 33 Production, component mixing, 157 ond redemption, 50
Costs, 107, 122, 466 Production levels, 131, 13 , 247, 323 Consumers producers surplus, 6 7,
Costs, trac ing, 242 Production quantities, 316, 34 6
emand equation, 141 Production run, units in, 607 Consumption of a country, 645
emand function, 463 Productivity, 457, 740 Current ratio, 60
epreciation, 17 , 1 0 Profit, 4 , 5 , 147, 340, 502, 60 , emand vectors, 24 , 256
istribution, 2 751 epreciation, 17 , 1 0
Employees, 1 1, 435, 524 Profit and distribution, 335 oubling money, 210, 211
Equipment, regulation of, 47 Promotional campaign, 200 Effective rate of interest, 211, 212
Equipment, renting vs. purchasing, 5 Purchasing power, 1 3 Financing a car, 20
Equipment rental, 22 , 336 uality control, 37 , 3 2, 414, 72 Interest, effective rate of, 211, 212
Expansion of, 72 Real estate, 51, 224, 225, 257, 65 Interest, nominal rate of, 20 , 211, 224
Fast-food chains, 200 Rentals businesses, 10 Interest compounded annually, 77,
Food sales, , 511, 575 Resource allocation, 2 4 1 0, 227
Health insurance premiums, 2 Revenue, rate of change in, 627 Interest compounded continuously,
Income pro ections, 655 Revenue function, 456, 511 1 1, 21
Industrial pollution, 131, 332 Revenue pro ections, 76, 150, 2 6, 2 7 Interest compounded daily, 20 , 210
Inventory, 255 Salaries, 57, 725 Interest compounded monthly, 211,
Inventory build up, 60 Selling price year sold relationship, 212, 224
east-cost input, 75 133 Interest compounded quarterly, 212,
arginal costs, 505, 626, 73 Service call cost, 142, 466 224
 Interest from investments, 154, 1 0 Environmental impacts, 131, 332 Communications, 607, 713
Interest on inactive ban account, 72 Enzyme formation, rate of, 606 Computer viruses, spreading, 175
Investment returns, comparing, 17 Fahrenheit Celsius conversion, 136 Conservation, encouraging, 10
Investments, growth of, 177 Food and nutrition, 57, 143, 156, emographics, 441, 44 , 675
Investments in multiple ventures, 50, 271 Education, enrollment, 74, 134, 504
216 enetics, 3, 3 3, 406 Education, expenses, 106, 256
oan payments, 233 eometry, , 100, 455, 476, 503, Education, income expectations, 633
Net present value, 217 507, 552 Education, promotion committee, 35
Nominal rate of interest, 20 , 211, 224 Health-care recipients, 60 Education, testing, 3 6, 40 , 722
Present value, 220, 223, 224 ife expectancy, 71 Educational scholarships, 235
Reduced payments on debt, 215, 216 ight intensity of clear liquid, 6 3 ambling, 34 , 3 2
Retirement investments, 154, 227 edicine, body elimination of, ames, 424
Savings accounts, 17 , 20 , 210 706 ames, cards, 377, 3 4, 3 6, 404,
Savings rate in US, 62 il slic , size of, 674 407
Stoc portfolio, 26 , 2 3 xygen composition, 200 ames, po er hand, 357, 36 , 377
Trust fund, 215, 220 Predator-pray relation, 202 ames, roll of a die, 350, 36 , 3 1,
Radioactive decay, 1 5 406, 433
Recycling, 1 ames coin tosses, 350, 367, 36 , 370,
Smo ing and sinusitis, 402 376, 402, 435
L S Solutions, mixing, 4 , 157 ender ma eup on committee, 377
Acceleration of ob ects, 563, 632 Survival rates, 403 Hospital discharges, 621
Age dating, 6 7 Temperature, changes in, 53 , 63 essage encoding, 2 0, 2 2
Animal food purchases, 271 Temperature-humidity index, 123 pinion poll, 3 4
Animals, feeding, 57, 156 Throwing ob ects, 14 , 4 , 503 Passenger wait time, 716
Area, finding, Time of murder, 703 Passengers in vehicles, assigning, 364
acteria growth, 176, 1 , 631 Velocity of moving ob ect, 520 Political questionnaire, 353
lood alcohol concentration levels, 3 Weight, absolute value of, 64 Population, growth rate of, 75, 1 1,
ody heat loss, 73 Wind chill factor, 73 1 4, 62 , 6 5, 6
Chemistry, half-life, 1 2, 6 6 east, growth of, 1 4, 675 Population, of fish in habitat, 4 2
ietary supplements, 156, 270, 340 Probability of same day birthday, 3 0
iseases, 411, 551 Rumor spreading, 702
rug concentration in bloodstream, S S Surveys, 3 1
577 Travel routes, determining, 350
rug dosages, 135, 512 Art display in gallery, 364 Vocabulary memorization, 63
Earthqua es, measurement of, 1 , Club membership o cers, 352, 701 Winning a prize, probability of, 3 6,
1 6, 1 7, 202, 536 Committee selections, 357 72
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About the Cover

The Royal Ontario Museum (ROM) in Toronto, Canada, opened in

March 1914. The ROM has undergone several overhauls, the most
dramatic being the addition of Daniel Libeskind’s Lee-Chin Crystal,
finished in June of 2007. The soaring glass and metal structure leads
a visitor from the chaos of the street to the more serene atmosphere
of the museum. Like many modern buildings, the Crystal embodies
application of many areas of mathematics in many ways. Readers of
the linear programming chapter (7) of this book may find it useful to
glance at the cover while contemplating routes, via edges, between
the vertices of similar structures.

www.pearsoncanada.ca 9000

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