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2015

Precalculus
Thomas Tradler
NYC College of Technology

Holly Carley
NYC College of Technology

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 Precalculus

Second Edition (2.7)

Thomas Tradler
Holly Carley

4 y 6 y 5 y

3 5 4

2 4 3

1 3 2

0 x 2 1
-2 -1 0 1 2 3 4 5
-1 1 0 x
-3 -2 -1 0 1 2 3
-2 0 x -1
-5 -4 -3 -2 -1 0 1 2
-3 -1 -2
ii

Copyright c 2012 Thomas Tradler and Holly Carley

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Preface
These are notes for a course in precalculus, as it is taught at New York City
College of Technology - CUNY (where it is offered under the course number
MAT 1375). Our approach is calculator based. For this, we will use the
currently standard TI-84 calculator, and in particular, many of the examples
will be explained and solved with it. However, we want to point out that
there are also many other calculators that are suitable for the purpose of
this course and many of these alternatives have similar functionalities as the
calculator that we have chosen to use. An introduction to the TI-84 calculator
together with the most common applications needed for this course is provided
in appendix A. In the future we may expand on this by providing introductions
to other calculators or computer algebra systems.
This course in precalculus has the overarching theme of “functions.” This
means that many of the often more algebraic topics studied in the previous
courses are revisited under this new function theoretic point of view. However,
in order to keep this text as self contained as possible we always recall all
results that are necessary to follow the core of the course even if we assume
that the student has familiarity with the formula or topic at hand. After a first
introduction to the abstract notion of a function, we study polynomials, ratio-
nal functions, exponential functions, logarithmic functions, and trigonometric
functions with the function viewpoint. Throughout, we will always place par-
ticular importance to the corresponding graph of the discussed function which
will be analyzed with the help of the TI-84 calculator as mentioned above.
These are in fact the topics of the first four (of the five) parts of this precalculus
course.
In the fifth and last part of the book, we deviate from the above theme
and collect more algebraically oriented topics that will be needed in calculus
or other advanced mathematics courses or even other science courses. This
part includes a discussion of the algebra of complex numbers (in particular
complex numbers in polar form), the 2-dimensional real vector space R2 , se-

iii
iv

quences and series with focus on the arithmetic and geometric series (which
are again examples of functions, though this is not emphasized), and finally
the generalized binomial theorem.
In short, here is an outline of the topics in this course and the five parts
into which this course is divided:
Part I: Functions and graphs
|
Part II: Polynomials and rational functions
|
Part III: Exponential and logarithmic functions
|
Part IV: Trigonometric functions
|
Part V: Complex numbers, sequences, and the binomial theorem
The topics in this book are organized in 25 sessions, each session correspond-
ing to one class meeting. Each session ends with a list of exercises that the
student is expected to be able to solve. We cannot overstate the importance
of completing these exercises for a successful completion of this course. These
25 sessions, together with 4 scheduled exams and one review session give a
total of 30 class sessions, which is the number of regularly scheduled class
meetings in one semester. Each of the five parts also ends with a review of
the topics discussed. This may be used as a review for any of the exams
during the semester. Finally, we point out that there is an overview of the
important formulas used in this course at the end of the book.
We would like to thank our colleagues and students for their support dur-
ing the development of this project. In particular, we would like to thank
Henry Africk, Johanna Ellner, Lin Zhou, Satyanand Singh, Jean Camilien, Leo
Chosid, Laurie Caban, Natan Ovshey, Johann Thiel, Wendy Wang, Steven
Karaszewski, Josue Enriquez, and Mohd Nayum Parvez, Akindiji Fadeyi, Is-
abel Martinez, Erik Nowak, Sybil Shaver, Faran Hoosain, Kenia Rodriguez,
Albert Jaradeh, Iftekher Hossain for many useful comments that helped to
improve this text.

Thomas Tradler and Holly Carley

New York City College of Technology - CUNY
September 2015
Contents

Preface iii

Table of contents v

I Functions and graphs 1

1 The absolute value 2
1.1 Background regarding numbers . . . . . . . . . . . . . . . . . . . 2
1.2 The absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Inequalities and intervals . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Absolute value inequalities . . . . . . . . . . . . . . . . . . . . . 6
1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Lines and functions 13

2.1 Lines, slope and intercepts . . . . . . . . . . . . . . . . . . . . . 13
2.2 Introduction to functions . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Functions by formulas and graphs 32

3.1 Functions given by formulas . . . . . . . . . . . . . . . . . . . . 32
3.2 Functions given by graphs . . . . . . . . . . . . . . . . . . . . . 37
3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Introduction to the TI-84 49

4.1 Graphing with the TI-84 . . . . . . . . . . . . . . . . . . . . . . . 49
4.2 Finding zeros, maxima, and minima . . . . . . . . . . . . . . . . 53
4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

v
vi CONTENTS

5 Basic functions and transformations 63

5.1 Graphing basic functions . . . . . . . . . . . . . . . . . . . . . . 63
5.2 Transformation of graphs . . . . . . . . . . . . . . . . . . . . . . 65
5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

6 Operations on functions 76
6.1 Operations on functions given by formulas . . . . . . . . . . . . 76
6.2 Operations on functions given by tables . . . . . . . . . . . . . 81
6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

7 The inverse of a function 86

7.1 One-to-one functions . . . . . . . . . . . . . . . . . . . . . . . . . 86
7.2 Inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
7.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Review of functions and graphs 98

II Polynomials and rational functions 100

8 Dividing polynomials 101

8.1 Long division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
8.2 Dividing by (x − c) . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8.3 Optional section: Synthetic division . . . . . . . . . . . . . . . . 109
8.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

9 Graphing polynomials 113

9.1 Graphs of polynomials . . . . . . . . . . . . . . . . . . . . . . . . 113
9.2 Finding roots of a polynomial with the TI-84 . . . . . . . . . . . 121
9.3 Optional section: Graphing polynomials by hand . . . . . . . . 124
9.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10 Roots of polynomials 130

10.1 Optional section: The rational root theorem . . . . . . . . . . . 130
10.2 The fundamental theorem of algebra . . . . . . . . . . . . . . . . 134
10.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
CONTENTS vii

11 Rational functions 147

11.1 Graphs of rational functions . . . . . . . . . . . . . . . . . . . . . 147
11.2 Optional section: Rational functions by hand . . . . . . . . . . 159
11.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

12 Polynomial and rational inequalities 169

12.1 Polynomial inequalities . . . . . . . . . . . . . . . . . . . . . . . 169
12.2 Rational inequalities and absolute value inequalities . . . . . . 176
12.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

Review of polynomials and rational functions 181

III Exponential and logarithmic functions 183

13 Exponential and logarithmic functions 184

13.1 Exponential functions and their graphs . . . . . . . . . . . . . . 184
13.2 Logarithmic functions and their graphs . . . . . . . . . . . . . . 190
13.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

14 Properties of exp and log 199

14.1 Algebraic properties of exp and log . . . . . . . . . . . . . . . . 199
14.2 Solving exponential and logarithmic equations . . . . . . . . . . 202
14.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

15 Applications of exp and log 208

15.1 Applications of exponential functions . . . . . . . . . . . . . . . 208
15.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

16 Half-life and compound interest 216

16.1 Half-life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
16.2 Compound interest . . . . . . . . . . . . . . . . . . . . . . . . . . 220
16.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

Review of exponential and logarithmic functions 229

viii CONTENTS

IV Trigonometric functions 231

17 Trigonometric functions 232

17.1 Basic trigonometric definitions and facts . . . . . . . . . . . . . 232
17.2 sin, cos, and tan as functions . . . . . . . . . . . . . . . . . . . . 239
17.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

18 Addition of angles and multiple angles 252

18.1 Addition and subtraction of angles . . . . . . . . . . . . . . . . . 252
18.2 Double and half angles . . . . . . . . . . . . . . . . . . . . . . . 256
18.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

19 Inverse trigonometric functions 262

−1 −1 −1
19.1 The functions sin , cos , and tan . . . . . . . . . . . . . . . 262
19.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

20 Trigonometric equations 270

20.1 Basic trigonometric equations . . . . . . . . . . . . . . . . . . . 270
20.2 Equations involving trigonometric functions . . . . . . . . . . . . 278
20.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

Review of trigonometric functions 284

V Complex numbers, sequences, and the binomial theorem

286

21 Complex numbers 287

21.1 Polar form of complex numbers . . . . . . . . . . . . . . . . . . . 287
21.2 Multiplication and division of complex numbers . . . . . . . . . 294
21.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

22 Vectors in the plane 299

22.1 Introduction to vectors . . . . . . . . . . . . . . . . . . . . . . . . 299
22.2 Operations on vectors . . . . . . . . . . . . . . . . . . . . . . . . 303
22.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
CONTENTS ix

23 Sequences and series 311

23.1 Introduction to sequences and series . . . . . . . . . . . . . . . 311
23.2 The arithmetic sequence . . . . . . . . . . . . . . . . . . . . . . . 318
23.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

24 The geometric series 325

24.1 Finite geometric series . . . . . . . . . . . . . . . . . . . . . . . . 325
24.2 Infinite geometric series . . . . . . . . . . . . . . . . . . . . . . . 330
24.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

25 The binomial theorem 337

25.1 The binomial theorem . . . . . . . . . . . . . . . . . . . . . . . . 337
25.2 Binomial expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 342
25.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

Review of complex numbers, sequences, and the binomial theorem 347

A Introduction to the TI-84 349

A.1 Basic calculator functions . . . . . . . . . . . . . . . . . . . . . . 349
A.2 Graphing a function . . . . . . . . . . . . . . . . . . . . . . . . . 353
A.2.1 Rescaling the graphing window . . . . . . . . . . . . . . 354
A.3 Graphing more than one function . . . . . . . . . . . . . . . . . . 357
A.4 Graphing a piecewise defined function . . . . . . . . . . . . . . 359
A.5 Using the table . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
A.6 Solving an equation using the solver . . . . . . . . . . . . . . . 362
A.7 Special functions (absolute value, n-th root, etc.) . . . . . . . . 363
A.8 Programming the calculator . . . . . . . . . . . . . . . . . . . . . 366
A.9 Common errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
A.10 Resetting the calculator to factory settings . . . . . . . . . . . . 373

Answers to exercises 374

Session 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
Session 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
Session 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
Session 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
Session 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
Session 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
Session 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381
x CONTENTS

Review part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

Session 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
Session 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
Session 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
Session 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
Session 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
Review part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
Session 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
Session 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
Session 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Session 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Review part III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Session 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Session 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
Session 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396
Session 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
Review part IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
Session 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
Session 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399
Session 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
Session 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
Session 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
Review part V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

References 402

Important formulas used in precalculus 403

Index 411
 Part I

Functions and graphs

1
Session 1

Absolute value equations and

inequalities

1.1 Background regarding numbers

The natural numbers (denoted by N) are the numbers

1, 2, 3, 4, 5, . . .

The integers (denoted by Z) are the numbers

. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, . . .

The rational numbers (denoted by Q) are the fractions ab of integers a and b

with b 6= 0. Here are some examples of rational numbers:
3 2 3
, − , 17, 0,
5 6 −8
The real numbers (denoted by R) are the numbers on the real number line

√ 2
−3 −2− 2 0 3 1 2 3π 4

Here are some examples of real numbers:

√ 2
3, π, − , 18, 0, 6.789
5
2
1.2. THE ABSOLUTE VALUE 3

A real number that is not a rational number is called an irrational number.

Here are some examples of irrational numbers:
√ 2
π, 2, 5 3 , e

1.2 The absolute value

The absolute value of a real number c, denoted by |c| the non-negative number
which is equal in magnitude (or size) to c, i.e., is the number resulting from
disregarding the sign:

c, if c is positive or zero
|c| =
−c, if c is negative

Example 1.1. | − 4| = 4

Example 1.2. |12| = 12

Example 1.3. | − 3.523| = 3.523

Example 1.4. For which real numbers x do you have |x| = 3?

Solution. Since |3| = 3 and | − 3| = 3, we see that there are two solutions,
x = 3 or x = −3. The solution set is S = {−3, 3}.

Example 1.5. Solve for x: |x| = 5

Solution. x = 5 or x = −5. The solution set is S = {−5, 5}.

Example 1.6. Solve for x: |x| = −7.

Solution. Note that | − 7| = 7 and |7| = 7 so that these cannot give any
solutions. Indeed, there are no solutions, since the absolute value is always
non-negative. The solution set is the empty set S = {}.

Example 1.7. Solve for x: |x| = 0.

Solution. Since −0 = 0, there is only one solution, x = 0. Thus, S = {0}.

Example 1.8. Solve for x: |x + 2| = 6.

4 SESSION 1. THE ABSOLUTE VALUE

Solution. Since the absolute value of x + 2 is 6, we see that x + 2 has to be

either 6 or −6. We evaluate each case,

either x + 2 = 6, or x + 2 = −6,
=⇒ x = 6 − 2, =⇒ x = −6 − 2,
=⇒ x = 4; =⇒ x = −8.

The solution set is S = {−8, 4}.

Example 1.9. Solve for x: |3x − 4| = 5

Solution.
Either 3x − 4 = 5 or 3x − 4 = −5
=⇒ 3x = 9 =⇒ 3x = −1
=⇒ x = 3 =⇒ x = − 13
The solution set is S = {− 31 , 3}.

Example 1.10. Solve for x: −2 · |12 + 3x| = −18

Solution. Dividing both sides by −2 gives |12 + 3x| = 9. With this, we have
the two cases
Either 12 + 3x = 9 or 12 + 3x = −9
=⇒ 3x = −3 =⇒ 3x = −21
=⇒ x = −1 =⇒ x = −7

The solution set is S = {−7, −1}.

1.3 Inequalities and intervals

There is an order relation on the set of real numbers:
4<9 reads as 4 is less than 9,
−3 ≤ 2 reads as −3 is less than or equal to 2,
7
6
>1 reads as 76 is greater than 1,
2 ≥ −3 reads as 2 is greater than or equal to −3.

Example 1.11. We have 2 < 3, but −2 > −3, which can be seen on the
number line above.
1.3. INEQUALITIES AND INTERVALS 5

Example 1.12. We have 5 ≤ 5 and 5 ≥ 5. However the same is not true when
using the symbol <. We write this as 5 ≮ 5.
The set of all real numbers x greater than or equal to some number a
and/or less than or equal to some number b is denoted in different ways by
the following chart:
Inequality notation Number line Interval notation

a≤x≤b a b [a, b]

a<x<b a b (a, b)

a≤x<b a b [a, b)

a<x≤b a b (a, b]

a≤x a [a, ∞)

a<x a (a, ∞)

x≤b b (−∞, b]

x<b b (−∞, b)
Formally, we define the interval [a, b] to be the set of all real numbers x such
that a ≤ x ≤ b:
[a, b] = { x | a ≤ x ≤ b }
There are similar definitions for the other intervals shown in the above table.

Warning 1.13. Be sure to write the smaller number a < b first when writing an interval [a, b]. For example,
the interval [5, 3] = { x | 5 ≤ x ≤ 3 } would be the empty set!

Example 1.14. Graph the the inequality π < x ≤ 5 on the number line and
write it in interval notation.
Solution.

On the number line: π 5

Interval notation: (π, 5]
6 SESSION 1. THE ABSOLUTE VALUE

Example 1.15. Write the following interval as an inequality and in interval

notation: −3

Solution.
Inequality notation: −3 ≤ x
Interval notation: [−3, ∞)

Example 1.16. Write the following interval as an inequality and in interval

notation: −1 0 1 2 3

Solution.
Inequality notation: x<2
Interval notation: (−∞, 2)

Note 1.17. In some texts round and square brackets are also used on the number line to depict an interval.
For example the following displays the interval [2, 5).

2 5

1.4 Absolute value inequalities

Using the notation from the previous section, we now solve inequalities in-
volving the absolute value. These inequalities may be solved in three steps:

• Step 1: Solve the corresponding equality. The solution of the equality

divides the real number line into several subintervals.

• Step 2: Using step 1, check the inequality for a number in each of the
subintervals. This check determines the intervals of the solution set.

• Step 3: Check the endpoints of the intervals.

1.4. ABSOLUTE VALUE INEQUALITIES 7

Here are some examples for the above solution method.

Example 1.18. Solve for x:
a) |x + 7| < 2 b) |3x − 5| ≥ 11 c) |12 − 5x| ≤ 1
Solution. a) We follow the three steps described above. In step 1, we solve
the corresponding equality, |x + 7| = 2.
x+7=2 x + 7 = −2
=⇒ x = −5 =⇒ x = −9
The solutions x = −5 and x = −9 divide the number line into three subin-
tervals:
x < −9 − 9 < x < −5 −5<x

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1

Now, in step 2, we check the inequality for one number in each of these
subintervals.
Check: x = −10 Check: x = −7 Check: x=0
? ? ?
|(−10) + 7| < 2 |(−7) + 7| < 2 |0 + 7| < 2
? ? ?
| − 3| < 2 |0| < 2 |7| < 2
? ? ?
3<2 0<2 7<2
false true false
Since x = −7 in the subinterval given by −9 < x < −5 solves the inequality
|x+7| < 2, it follows that all numbers in the subinterval given by −9 < x < −5
solve the inequality. Similarly, since x = −10 and x = 0 do not solve the
inequality, no number in these subintervals will solve the inequality. For step
3, we note that the numbers x = −9 and x = −5 are not included as solutions
since the inequality is strict (that is we have < instead of ≤).The solution set
is therefore the interval S = (−9, −5). The solution on the number line is:

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1

b) We follow the steps as before. First, in step 1, we solve |3x − 5| = 11.

3x − 5 = 11 3x − 5 = −11
=⇒ 3x = 16 =⇒ 3x = −6
=⇒ x = 16
3
=⇒ x = −2
8 SESSION 1. THE ABSOLUTE VALUE

16
The two solutions x = −2 and x = 3
= 5 13 divide the number line into the
subintervals displayed below.

1 1
x < −2 −2<x<5 5 <x
3 3

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

For step 2, we check a number in each subinterval. This gives:

Check: x = −3 Check: x=1 Check: x=6

? ? ?
|3 · (−3) − 5| ≥ 11 |3 · 1 − 5| ≥ 11 |3 · 6 − 5| ≥ 11
? ? ?
| − 9 − 5| ≥ 11 |3 − 5| ≥ 11 |18 − 5| ≥ 11
? ? ?
| − 14| ≥ 11 | − 2| ≥ 11 |13| ≥ 11
? ? ?
14 ≥ 11 2 ≥ 11 13 ≥ 11
true false true

For step 3, note that we include −2 and 5 31 in the solution set since
the inequality is “greater than or equal to” (that is ≥, as opposed to >).
Furthermore, the numbers −∞ and ∞ are not included, since ±∞ are not
real numbers.
The solution set is therefore the union of the two intervals:
 i h 1 
S= − ∞, −2 ∪ 5 , ∞
3

c) To solve |12 − 5x| ≤ 1, we first solve the equality |12 − 5x| = 1.

12 − 5x = 1 12 − 5x = −1
=⇒ −5x = −11 =⇒ −5x = −13
=⇒ x = −11
−5
= 2.2 =⇒ x = −13
−5
= 2.6

This divides the number line into three subintervals, and we check the original
inequality |12 − 5x| ≤ 1 for a number in each of these subintervals.
1.4. ABSOLUTE VALUE INEQUALITIES 9

Interval: x < 2.2 Interval: 2.2 < x < 2.6 Interval: 2.6 < x
Check: x=1 Check: x = 2.4 Check: x=3
? ? ?
|12 − 5 · 1| ≤ 1 |12 − 5 · 2.4| ≤ 1 |12 − 5 · 3| ≤ 1
? ? ?
|12 − 5| ≤ 1 |12 − 12| ≤ 1 |12 − 15| ≤ 1
? ? ?
|7| ≤ 1 |0| ≤ 1 | − 3| ≤ 1
? ? ?
7≤1 0≤1 3≤1
false true false

The solution set is the interval S = [2.2, 2.6], where we included x = 2.2
and x = 2.6 since the original inequality “less than or equal to” (≤) includes
the equality.
Note: Alternatively, whenever you have an absolute value inequality you
can turn it into two inequalities. Here are a couple of examples.
Example 1.19. Solve for x: |12 − 5x| ≤ 1
Solution. Note that |12 − 5x| ≤ 1 implies that

−1 ≤ 12 − 5x ≤ 1

so that
−13 ≤ −5x ≤ −11
and by dividing by −5 (remembering to switch the direction of the inequalities
when multiplying or dividing by a negative number) we see that
13 11
≥x≥ ,
5 5
or in interval notation, we have the solution set
 
11 13
S= , .
5 5

Example 1.20. If |x + 6| > 2 then either x + 6 > 2 or x + 6 < −2 so

that either x > −4 or x < −8 so that in interval notation the solution is
S = (−∞, −8) ∪ (−4, ∞).
10 SESSION 1. THE ABSOLUTE VALUE

Observation 1.21. There is a geometric interpretation of the absolute value

on the number line as the distance between two numbers:
a b

distance between a and b is |b − a| which is also equal to |a − b|

This interpretation can also be used to solve absolute value equations and
inequalities.
Example 1.22. Solve for x: a) |x − 6| = 4, b)|x − 6| ≤ 4, c) |x − 6| ≥ 4.
Solution. a) Consider the distance between x and 6 to be 4 on a number line:

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

distance 4 distance 4

There are two solutions, x = 2 or x = 10. That is, the distance between 2
and 6 is 4 and the distance between 10 and 6 is 4.
b) Numbers inside the braces above have distance 4 or less. The solution
is given on the number line as:

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

distance ≤ 4 distance ≤ 4

In interval notation, the solution set is the interval S = [2, 10]. One can also
write that the solution set consists of all x such that 2 ≤ x ≤ 10.
c) Numbers outside the braces above have distance 4 or more. The solution
is given on the number line as:

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

distance < 4 distance < 4

In interval notation, the solution set is the interval (−∞, 2] and [10, ∞), or in
short it is the union of the two intervals:
S = (∞, 2] ∪ [10, ∞).
1.5. EXERCISES 11

One can also write that the solution set consists of all x such that x ≤ 2 or
x ≥ 10.

1.5 Exercises
Exercise 1.1. Give examples of numbers that are

a) natural numbers b) integers c) integers but not natural numbers

d) rational numbers e) real numbers f) rational numbers but not integers

Exercise 1.2. Which of the following numbers are natural numbers, integers,
rational numbers, or real numbers? Which of these numbers are irrational?

7 12
√ √
a) 3
b) −5 c) 0 d) 17, 000 e) 4
f) 7 g) 25

Exercise 1.3. Evaluate the following absolute value expressions:

a) | − 8|, b) |10|,√ c) | − 99|, d) −|3|,

e) −| − 2|, f) | − 6|, g) |3 + 4|, h) |2 − 9|,
i) | − 5.4|, j) − 23 , k) −25
, −6
l) − − −3 .

Exercise 1.4. Solve for x:

a) |x| = 8 b) |x| = 0 c) |x| = −3

d) |x + 3| = 10 e) |2x + 5| = 9 f) |2 − 5x| = 22
g) |4x| = −8 h) | − 7x − 3| = 0 i) |4 − 4x| = 44
j) −2 · |2 − 3x| = −12 k) 5 + |2x + 7| = 14 l) −| − 8 − 2x| = −12

Exercise 1.5. Solve for x using the geometric interpretation of the absolute
value:
a) |x| = 8 b) |x| = 0 c) |x| = −3
d) |x − 4| = 2 e) |x + 5| = 9 f) |2 − x| = 5
12 SESSION 1. THE ABSOLUTE VALUE

Exercise 1.6. Complete the table.

Inequality notation Number line Interval notation

2≤x<5
x≤3

12 17

−2
[−2, 6]
(−∞, 0)

√ 4.5
5<x≤ 30
( 13
7
, π)

Exercise 1.7. Solve for x and write the solution in interval notation.
a) |x − 4| ≤ 7 b) |x − 4| ≥ 7 c) |x − 4| > 7
d) |2x + 7| ≤ 13 e) | − 2 − 4x| > 8 f) |4x
√ + 2| √< 17 √
g) |15 − 3x| ≥ 6 h) 5x − 34 > 32 i) 2x − 2 ≤ 8
j) |2x + 3| < −5 k) |5 + 5x| ≥ −2 l) |5 + 5x| > 0
Session 2

Lines and functions

2.1 Lines, slope and intercepts

In this chapter we will introduce the notion of a function. Before we give
the general definition, we recall one special kind of function which is already
familiar to the student. More precisely, we start by recalling functions that
are given by straight lines.
We have seen in the last section that each point on the number line is
represented by a real number x in R. Similarly, each point in the coordinate
plane is represented by a pair of real numbers (x, y). The coordinate plane
is denoted by R2 . Here is a picture of the coordinate plane:
3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

In this section, we will discuss the straight line in the coordinate plane.
We will discuss its slope and intercepts. We will also discuss the line’s

13
14 SESSION 2. LINES AND FUNCTIONS

corresponding algebraic forms: the point-slope form and the slope-intercept

form.
By drawing a line on a coordinate plane we associate every point on the
line with a pair of numbers (a, b), where a is called the x−coordinate and b
is called the y−coordinate. Consider, for example, the picture
4 y
L
3

0 x
-5 -4 -3 -2 -1 0 1 2 3 4
-1

We see that (−2, 0) lies on the line and so does (2, 2). Of course, there are an
infinite number of points on the line. This set of points can also be described
by an equation relating the x− and y− coordinates of points on the line. For
example 2y − x = 2 is the equation of the line above. Notice that (−2, 0)
satisfies the equation since 2(0) − (−2) = 2 and (2, 2) satisfies the equation
since 2(2) − (2) = 2. It is a fact that every line has an equation of the form
px + qy = r, that is, given a line, there are numbers p, q and r so that a point
(a, b) is on the graph of the line if and only if pa + qb = r.
Notice that, in the example, if we put our finger on (−2, 0) and move it up
and over to the point (2, 2) we have moved 2 units up and 4 units to the right.
If from there ((2, 2)) we move another 2 units up and 4 units to the right then
we land again on the line (at (6, 4)). In fact no matter where we start on the
line if from there we move 2 units up and 4 units to the right we always land
on the line again. The ratio rise
run
= 42 is called the slope of this line. The word
‘rise’ indicates vertical movement (down being negative) and ‘run’ indicates
horizontal movement (left being negative). Notice that 24 = 21 = −1 −2
and so
on. So we could also see that if we start at (−2, 0) and rise 1 and run 2 we
land at (0, 1) which is on the line and if we start at (2, 2) and rise −1 (move
down 1) and run −2 (move to the left 2) we end up at (0, 1), which is again
on the line.
Generally, the slope describes how fast the line grows towards the right.
For any two points P1 (x1 , y1 ) and P2 (x2 , y2 ) on the line L, the slope m is
2.1. LINES, SLOPE AND INTERCEPTS 15

rise
given by the following formula (which is run
):

y2
P2
y1 P1

x
x1 x2

y2 − y1
Slope: m= (2.1)
x2 − x1
The slope determines how fast a line grows. When the slope m is negative
the line declines towards the right.
y y y y y

x x x x x

1
m=2 m= 2
m=0 m = − 21 m = −2

From equation (2.1), we see that for a given slope m and a point P1 (x1 , y1 )
y−y1
on the line, any other point (x, y) on the line satisfies m = x−x 1
. Multiplying
(x − x1 ) on both sides gives what is called the point-slope form of the line:

y − y1 = m · (x − x1 ) (2.2)

Other prominent features of a graph of a line (or other graphs) include

where the graph crosses the x-axis and the y−axis. For a given line L, let
us recall what the x- and y-intercepts are. The x-intercept is the point on
the x-axis where the line intersects the x-axis. Similarly, the y-intercept is
the point on the y-axis where the line intersects the y-axis.
16 SESSION 2. LINES AND FUNCTIONS

y
L

y-intercept
x-intercept x

One way to describe a line using the slope and y-intercept is the so-called
slope-intercept form of the line.
Definition 2.1. The slope-intercept form of the line is the equation
y =m·x+b (2.3)

Here, m is the slope and (0, b) is the y-intercept of the line.

Here is an example of a line in slope-intercept form.
Example 2.2. Graph the line y = 2x + 3.
Solution. We calculate y for various values of x. For example, when x is
−2, −1, 0, 1, 2, or 3, we calculate
x −2 −1 0 1 2 3
y −1 1 3 5 7 9
In the above table each y value is calculated by substituting the corresponding
x value into our equation y = 2x + 3:
x = −2 =⇒ y = 2 · (−2) + 3 = −4 + 3 = −1
x = −1 =⇒ y = 2 · (−1) + 3 = −2 + 3 = 1
x=0 =⇒ y = 2 · (0) + 3 = 0 + 3 = 3
x=1 =⇒ y = 2 · (1) + 3 = 2 + 3 = 5
x=2 =⇒ y = 2 · (2) + 3 = 4 + 3 = 7
x=3 =⇒ y = 2 · (3) + 3 = 6 + 3 = 9
In the above calculation, the values for x were arbitrarily chosen. Since a
line is completely determined by knowing two points on it, any two values for
x would have worked for the purpose of graphing the line.
2.1. LINES, SLOPE AND INTERCEPTS 17

Drawing the above points in the coordinate plane and connecting them
gives the graph of the line y = 2x + 3:
10 y

0 x
-4 -3 -2 -1 0 1 2 3 4
-1

-2

Alternatively, note that the y-intercept is (0, 3) (3 is the additive constant in

our initial equation y = 2x + 3) and the slope m = 2 determines the rate at
which the line grows: for each step to the right, we have to move two steps
up.
To plot the graph, we first plot the y-intercept (0, 3). Then from that
point, rise 2 and run 1 so that you find yourself at (1, 5) (which must be on
the graph), and similarly rise 2 and run 1 to get to (2, 7) (which must be on
the graph), etc. Plot these points on the graph and connect the dots to form a
straight line. As noted above, any 2 distinct points on the graph of a straight
line are enough to plot the complete line.
Before giving more examples, we briefly want to justify why, in the ex-
pression y = mx + b, the number (0, b) is the y-intercept and m is the slope.
This proof may be skipped on a first reading.
18 SESSION 2. LINES AND FUNCTIONS

Proof that (0, b) is the y-intercept, and m is the slope.

Given the line y = mx + b, we want to calculate its y-intercept and its
slope. The y-intercept is the value of y where x = 0. Therefore, we have that
the y−coordinate of the y-intercept is

y = m · 0 + b = b.

This shows the first claim.

Next, to see why m is the slope, note that P1 (x1 , y1) lies on the line
exactly when y1 = mx1 + b. Similarly, P2 (x2 , y2) lies on the line exactly when
y2 = mx2 + b. Subtracting y1 = mx1 + b from y2 = mx2 + b, we obtain:

y2 − y1 = (mx2 + b) − (mx1 + b) = mx2 + b − mx1 − b = m · (x2 − x1 ).

Dividing by (x2 − x1 ) gives

y2 − y1 m · (x2 − x1 ) m
= = = m.
(x2 − x1 ) (x2 − x1 ) 1
By Equation (2.1) the fraction xy22 −x
−y1
1
is the slope, establishing that our m is
indeed the slope, as we claimed.
Example 2.3. Find the equation of the line in slope-intercept form.
5 y

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

Solution. The y-intercept can be read off the graph giving us that b = 2. As
for the slope, we use formula (2.1) and the two points on the line P1 (0, 2) and
P2 (4, 0). We obtain
0−2 −2 1
m= = =− .
4−0 4 2
Thus, the line has the slope-intercept form y = − 12 x + 2.
2.1. LINES, SLOPE AND INTERCEPTS 19

Example 2.4. Find the equation of the line in slope-intercept form.

1 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3

-4

-5

-6

-7

Solution. The y-intercept is b = −4. To obtain the slope we can again use
the y-intercept P1 (0, −4). To use (2.1), we need another point P2 on the line.
We may pick any second point on the line, for example, P2 (3, −3). With this,
we obtain
(−3) − (−4) −3 + 4 1
m= = = .
3−0 3 3
1
Thus, the line has the slope-intercept form y = 3 x − 4.
Example 2.5. Find the equation of the line in point-slope form (2.2).
5 y

0 x
-1 0 1 2 3 4 5 6 7 8 9 10 11 12
-1

Solution. We need to identify one point (x1 , y1 ) on the line together with
the slope m of the line so that we can write the line in point-slope form:
y − y1 = m(x − x1 ). By direct inspection, we identify the two points P1 (5, 1)
and P2 (8, 3) on the line, and with this we calculate the slope as
3−1 2
m= = .
8−5 3
20 SESSION 2. LINES AND FUNCTIONS

Using the point (5, 1) we write the line in point-slope form as follows:

2
y − 1 = (x − 5)
3

Note that our answer depends on the chosen point (5, 1) on the line. Indeed,
if we choose a different point on the line, such as (8, 3), we obtain a different
equation, (which nevertheless represents the same line):

2
y − 3 = (x − 8)
3

Note, that we do not need to solve this for y, since we are looking for an
answer in point-slope form.

Example 2.6. Find the slope, find the y-intercept, and graph the line

4x + 2y − 2 = 0.

Solution. We first rewrite the equation in slope-intercept form.

(−4x+2)
4x + 2y − 2 = 0 =⇒ 2y = −4x + 2
(divide 2)
=⇒ y = −2x + 1

We see that the slope is −2 and the y-intercept is (0, 1).

We can then plot the y−intercept (0, 1) and use the slope m = −2
1
to find
another point (1, −1). Plot that point, connect the two plotted points, and
extend to see the graph below.
We can also graph by plotting points. We can calculate the y-values for
some x-values. For example when x = −2, −1, . . . , 3, we obtain:

x −2 −1 0 1 2 3
y 5 3 1 −1 −3 −5
2.1. LINES, SLOPE AND INTERCEPTS 21

This gives the following graph:

5 y

0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1

-2

-3

-4

-5

Example 2.7. Find the slope, y-intercept, and graph the line 5y + 2x = −10.
Solution. Again, we first rewrite the equation in slope-intercept form.
(subtract 2x)
5y + 2x = −10 =⇒ 5y = −2x − 10
(divide 5) −2x − 10
=⇒ y=
5
2
=⇒ y =− x−2
5
Now, the slope is − 52 and the y-intercept is (0, −2).
We can plot the y−intercept and from there move 2 units down and 5 units
to the right to find another point on the line. The graph is given below.
To graph it by plotting points we need to find points on the line. In
fact, any two points will be enough to completely determine the graph of the
line. For some “smart” choices of x or y the calculation of the corresponding
value may be easier than for others. We suggest that you find the x− and
y−intercepts, i.e., point of the form (?, 0) (y = 0 in the equation and find
x) and (0, ?) (set x = 0 in the equation and find y). Plugging x = 0 into
5y + 2x = −10, we obtain
x=0
=⇒ 5y + 2 · 0 = −10 =⇒ 5y = −10 =⇒ y = −2.
22 SESSION 2. LINES AND FUNCTIONS

Similarly, substituting y = 0 into 5y + 2x = −10 gives

y=0
=⇒ 5 · 0 + 2x = −10 =⇒ 2x = −10 =⇒ x = −5.

We obtain the following table:

x 0 −5
y −2 0
This gives the following graph:
1 y

0 x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3
-1

-2

-3

2.2 Introduction to functions

We now formally introduce the notion of a function. A first example was
provided by a straight line such as, for example, y = 5x + 4. Note, that for
each given x we obtain an induced y. (For example, for x = 3, we obtain
y = 5 · 3 + 4 = 19.)
Definition 2.8. A function f consists of two sets, a set D of inputs called
the domain and a set C of possible outputs called the codomain, and an
assignment that assigns to each input x exactly one output y.
A function f with domain D and codomain C is denoted by

f : D → C.

If x is in the domain D (an input), then we denote by f (x) = y the output

that is assigned by f to x.
Sometimes it is of interest to know the set of all elements in the codomain
that actually occur as an output. This set is a subset of the codomain and is
called the range. We have:
2.2. INTRODUCTION TO FUNCTIONS 23

Definition 2.9. The range R of a function f is a subset of the codomain given

by R = {f (x) | x in the domain of f }. That is, the range is the set of all
outputs.

Warning 2.10. Some authors use a slightly different convention by calling the range what we called the
codomain above.

Since we will be dealing with many functions it is convenient to name vari-

ous functions (usually with letters f, g, h, etc). Often we will implicitly assume
that a domain and codomain are given without specifying these explicitly. If
the range can be determined and the codomain is not given explicitly, then we
take the codomain to be the range. If the range can not easily be determined
and the codomain is not explicitly given, then the codomain should be taken
to be a ‘simple’ set which clearly contains the range.
There are several ways to represent a particular function (all of which
may not apply to a specific function): via a table of values (listing the input-
output pairs), via a formula (with the domain and range explicitly or implicitly
given), via a graph (representing input-output pairs on a coordinate plane),
or in words, just to name a few. We have seen examples of the first three of
these in previous sections. Our discussion in this section goes into greater
detail.

Example 2.11. Define the assignment f by the following table

x 2 5 −3 0 7 4
y 6 8 6 4 −1 8

The assignment f assigns to the input 2 the output 6, which is also written
as
f (2) = 6.
Similarly, f assigns to 5 the number 8, in short f (5) = 8, etc:

f (5) = 8, f (−3) = 6, f (0) = 4, f (7) = −1, f (4) = 8.

The domain D is the set of all inputs. The domain is therefore

D = {−3, 0, 2, 4, 5, 7}.
24 SESSION 2. LINES AND FUNCTIONS

The range R is the set of all outputs. The range is therefore

R = {−1, 4, 6, 8}.

The assignment f is indeed a function since each element of the domain gets
assigned exactly one element in the range. Note that for an input number
that is not in the domain, f does not assign an output to it. For example,

f (1) = undefined.

Note also that f (5) = 8 and f (4) = 8, so that f assigns to the inputs 5
and 4 the same output 8. Similarly, f also assigns the same output to the
inputs 2 and −3. Therefore we see that:
• A function may assign the same output to two different inputs!
Example 2.12. Consider the assignment f that is given by the following table.

x 2 5 −3 0 5 4
y 6 8 6 4 −1 8

This assignment does not define a function! What went wrong?

Consider the input value 5. What does f assign to the input 5? The third
column states that f assigns to 5 the output 8, whereas the sixth column
states that f assigns to 5 the output −1,

f (5) = 8, f (5) = −1.

However, by the definition of a function, to each input we have to assign

exactly one output. So, here, to the input 5 we have assigned two outputs 8
and −1. Therefore, f is not a function.
• A function cannot assign two outputs to one input!
We repeat the two bullet points from the last two examples, which are
crucial for the understanding of a function.
Note 2.13.
• A function may assign the same output to two different inputs!

f (x1 ) = y and f (x2 ) = y with x1 6= x2 is allowed!

2.2. INTRODUCTION TO FUNCTIONS 25

• A function cannot assign two outputs to one input!

f (x) = y1 and f (x) = y2 with y1 6= y2 is not allowed!

Example 2.14. A university creates a mentoring program, which matches each

freshman student with a senior student as his or her mentor. Within this pro-
gram it is guaranteed that each freshman gets precisely one mentor, however
two freshmen may receive the same mentor. Does the assignment of freshmen
to mentor, or mentor to freshmen describe a function? If so, what is its domain,
what is its range?

Solution. Since a senior may mentor several freshman, we cannot take a

mentor as an “input,” as he or she would be assigned to several “output”
freshmen students. So freshman is not a function of mentor.
On the other hand, we can assign each freshmen to exactly one mentor,
which therefore describes a function. The domain (the set of all inputs) is
given by the set of all freshmen students. The range (the set of all outputs) is
given by the set of all senior students that are mentors. The function assigns
each “input” freshmen student to his or her unique “output” mentor.

Example 2.15. The rainfall in a city for each of the 12 months is displayed
in the following histogram.
rainfall [in]

✲
J F MA MJ J A S O N D month

a) Is the rainfall a function of the month?

b) Is the month a function of the rainfall?

Solution.

a) Each month has exactly one amount of rainfall associated to it. There-
fore, the assignment that associates to a month its rainfall (in inches)
is a function.
26 SESSION 2. LINES AND FUNCTIONS

b) If we take a certain rainfall amount as our input data, can we associate

a unique month to it? For example, February and March have the same
amount of rainfall. Therefore, to one input amount of rainfall we cannot
assign a unique month. The month is not a function of the rainfall.

Example 2.16. Consider the function f described below.

f

yellow
△ green
♦
blue

Here, the function f maps the input symbol  to the output color blue. Other
assignments of f are as follows:

f () = blue, f (△) = yellow

f (♦) = green, f ( ) = yellow

The domain is the set of symbols D = {, △, ♦, }, and the range is the
set of colors R = {blue, green, yellow}. Notice, in particular, that the inputs
△ and both have the same output yellow, which is certainly allowed for a
function.

Example 2.17. Consider the function y = 5x+4 with domain all real numbers
and range all real numbers. Note that for each input x, we obtain an exactly
one induced output y. For example, for the input x = 3 we get the output
y = 5 · 3 + 4 = 19, etc.

Example 2.18. Consider the function y = x2 with domain all real numbers
and range non-negative numbers. The function takes a real number as an
input and squares it. For example if x = −2 is the input, then y = 4 is the
output.
2.2. INTRODUCTION TO FUNCTIONS 27

Example 2.19. For each real number x, denote by ⌊x⌋ the greatest integer
that is less or equal to x. We call ⌊x⌋ the floor of x. For example, to calculate
⌊4.37⌋, note that all integers 4, 3, 2, . . . are less or equal to 4.37:

. . . , −3, −2, −1, 0, 1, 2, 3, 4 ≤ 4.37

The greatest of these integers is 4, so that ⌊4.37⌋ = 4. We define the floor

function as f (x) = ⌊x⌋. Here are more examples of function values of the
floor function.

⌊7.3⌋ = 7, ⌊π⌋ = 3, ⌊−4.65⌋ = −5,

 
−26
⌊12⌋ = 12, = ⌊−8.667⌋ = −9
3

The domain of the floor function is the set of all real numbers, that is D = R.
The range is the set of all integers, R = Z.

Example 2.20. Let A be the area of an isosceles right triangle with base side
length x. Express A as a function of x.

Solution. Being an isosceles right triangle means that two side lengths are
x, and the angles are 45◦ , 45◦ , and 90◦ (or in radian measure π4 , π4 , and π2 ):

Recall that the area of a triangle is: area = 12 base · height. In this case, we
have base= x, and height= x, so that the area

1 1
A = x · x = x2 .
2 2
1
Therefore, the area A(x) = 2
· x2 .
28 SESSION 2. LINES AND FUNCTIONS

Example 2.21. Consider the equation y = x2 + 3. This equation associates

to each input number a exactly one output number b = a2 + 3. Therefore, the
equation defines a function. For example:

To the input 5 we assign the output 52 + 3 = 25 + 3 = 28.

The domain D is all real numbers, D = R. Since x2 is always ≥ 0, we

see that x2 + 3 ≥ 3, and vice versa every number√ y ≥ 3 can be written as
y = x2 + 3. (To see√this, note that the input x = y − 3 for y ≥ 3 gives the
output x2 − 3 = ( y − 3)2 + 3 = y − 3 + 3 = y.) Therefore, the range is
R = [3, ∞).

Example 2.22. Consider the equation x2 +y 2 = 25. Does this equation define
y as a function of x? That is, does this equation assign to each input x exactly
one output y?
An input number x gets assigned to y with x2 + y 2 = 25. Solving this for
y, we obtain
√
y 2 = 25 − x2 =⇒ y = ± 25 − x2 .

Therefore, there are two possible outputs associated to the input x(6= 5):
√ √
either y = + 25 − x2 or y = − 25 − x2 .

For example, the input x = 0 has two outputs y = 5 and y = −5. However,
a function cannot assign two outputs to one input x! The conclusion is that
x2 + y 2 = 25 does not determine y as a function!

Note 2.23. Note that if y = f (x) then x is called the independent variable
and y is called the dependent variable (since it depends on x). If x = g(y)
then y is the independent variable and x is the dependent variable (since it
depends on y).

2.3 Exercises
Exercise 2.1. Find the slope and y-intercept of the line with the given data.
Using the slope and y-intercept, write the equation of the line in slope-
2.3. EXERCISES 29

intercept form.
5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5
-1 -1

-2 -2

-3 -3

-4 -4

a) -5 b) -5

5 y

3 6 y

2 5

1 4

0 x 3
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1 2

-2 1

-3 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-4 -1

c) -5 d) -2

4 y 6 y

3 5

2 4

1 3

0 x 2
-4 -3 -2 -1 0 1 2 3 4
-1 1

-2 0 x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
-3 -1

e) -4 f) -2

Exercise 2.2. Write the equation of the line in slope-intercept form. Identify
slope and y-intercept of the line.
a) 4x + 2y = 8 b) 9x − 3y + 15 = 0 c) −5x − 10y = 20
d) 3x − 5y = 7 e) −12x + 8y = −60 f) 8x − 9y = 0
30 SESSION 2. LINES AND FUNCTIONS

Exercise 2.3. Find the equation of the line in point-slope form (2.2) using the
indicated point P1 .
5 y 4 y

4 3

3 2

P1 P1
2 1

1 0 x
-1 0 1 2 3 4 5 6 7 8
0 x -1
-1 0 1 2 3 4 5 6 7 8
a) -1 b) -2

3 y 4 y

2 3

1 2

x P1
0 1
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1 0 x
-6 -5 -4 -3 -2 -1 0 1 2 3
P1
-2 -1

-3 -2

c) -4 d) -3

Exercise 2.4. Graph the line by calculating a table (as in Example 2.2). (Solve
for y first, if this is necessary.)
a) y = 2x − 4 b) y = −x + 4 c) y = 12 x + 1
d) y = 3x e) 8x − 4y = 12 f) x + 3y + 6 = 0
Exercise 2.5. Determine if the given table describes a function. If so, deter-
mine its domain and range. Describe which outputs are assigned to which
inputs.
a)
x −5 3 −1 6 0
y 5 2 8 3 7
b)
x 6 17 4 −2 4
y 8 −2 0 3 −1
c)
x 19 7 6 −2 3 −11
y 3 3 3 3 3 3
2.3. EXERCISES 31

d)
x 1 2 3 3 4 5
√
y 5.33 9 13 13 17 19
e)
x 0 1 2 2 3 4
y 0 1 2 3 3 4
Exercise 2.6. We consider children and their (birth) mothers.
a) Does the assignment child to their birth mother constitute a function
(in the sense of Definition 2.8 on page 22)?
b) Does the assignment mother to their children constitute a function?
c) In the case where the assignment is a function, what is the domain?
d) In the case where the assignment is a function, what is the range?
Exercise 2.7. A bank offers wealthy customers a certain amount of interest,
if they keep more than 1 million dollars in their account. The amount is
described in the following table.
dollar amount x in the account interest amount
x ≤ $1, 000, 000 $0
$1, 000, 000 < x ≤ $10, 000, 000 2% of x
$10, 000, 000 < x 1% of x
a) Justify that the assignment cash amount to interest defines a function.
b) Find the interest for an amount of:
i) $50, 000 ii) $5, 000, 000 iii) $1, 000, 000
iv) $30, 000, 000 v) $10, 000, 000 vi) $2, 000, 000
Exercise 2.8. Find a formula for a function describing the given inputs and
outputs.
a) input: the radius of a circle,
output: the circumference of the circle
b) input: the side length in an equilateral triangle,
output: the perimeter of the triangle
c) input: one side length of a rectangle, with other side length being 3,
output: the perimeter of the rectangle
d) input: the side length of a cube,
output: the volume of the cube
Session 3

Functions by formulas and graphs

3.1 Functions given by formulas

Most of the time we will discuss functions that take some real numbers as
inputs, and give real numbers as outputs. These functions are commonly
described with a formula.

Example 3.1. For the given function f , calculate the outputs f (2), f (−3),
and f (−1).
√
a) f (x) = 
3x + 4, b) f (x) = x2 − 3,
5x − 6 , for −1 ≤ x ≤ 1 x+2
c) f (x) = d) f (x) = .
x3 + 2x , for 1 < x ≤ 5 x+3

Solution. We substitute the input values into the function and simplify.

a) f (2) = 3 · 2 + 4 = 6 + 4 = 10,
f (−3) = 3 · (−3) + 4 = −9 + 4 = −5,
f (−1) = 3 · (−1) + 4 = −3 + 4 = 1.

b) Similarly, we calculate
√ √ √
f (2) = 22 − 3 = 4 − 3 = 1 = 1,
p √ √
f (−3) = (−3)2 − 3 = 9 − 3 = 6,
p √ √
f (−1) = (−1)2 − 3 = 1 − 3 = −2 = undefined.

32
3.1. FUNCTIONS GIVEN BY FORMULAS 33

For the last step, we are assuming

√ that we only deal with real numbers. Of
course, as we will see later on, −2 can be defined as a complex number.
But for now, we will only allow
 real solutions outputs.
5x − 6 , for −1 ≤ x ≤ 1
c) The function f (x) = is given as a piece-
x3 + 2x , for 1 < x ≤ 5
wise defined function. We have to substitute the values into the correct
branch:
f (2) = 23 + 2 · 2 = 8 + 4 = 12, since 1 < 2 ≤ 5,
f (−3) = undefined, since −3 is not in any of the two branches,
f (−1) = 5 · (−1) − 7 = −5 − 6 = −11, since − 1 ≤ −1 ≤ 1.
x+2
d) Finally for f (x) = x+3
, we have:
2+2 4
f (2) = = ,
2+3 5
−3 + 2 −1
f (−3) = = = undefined,
−3 + 3 0
−1 + 2 1
f (−1) = = .
−1 + 3 2

Example 3.2. Let f be the function given by f (x) = x2 + 2x − 3. Find the

following function values.
a) f (5)
√ b) f (2)
√ c) f (−2) d) f (0)
e) f ( 5) f) f ( 3 + 1) g) f (a) h) f (a) + 5
i) f (x + h) j) f (x + h) − f (x) k) f (x+h)−f
h
(x)
l) f (x)−f
x−a
(a)

Solution. The first four function values ((a)-(d)) can be calculated directly:
f (5) = 52 + 2 · 5 − 3 = 25 + 10 − 3 = 32,
f (2) = 22 + 2 · 2 − 3 = 4 + 4 − 3 = 5,
f (−2) = (−2)2 + 2 · (−2) − 3 = 4 + −4 − 3 = −3,
f (0) = 02 + 2 · 0 − 3 = 0 + 0 − 3 = −3.

The next two values ((e) and (f)) are similar, but the arithmetic is a bit
more involved.
√ √ 2 √ √ √
f ( 5) = 5 + 2 · 5 − 3 = 5 + 2 · 5 − 3 = 2 + 2 · 5,
34 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS
√ √ √
f ( 3 + 1) = ( 3 + 1)2 + 2 · ( 3 + 1) − 3
√ √ √
= ( 3 + 1) · ( 3 + 1) + 2 · ( 3 + 1) − 3
√ √ √ √
= 3· 3+2· 3+1·1+2· 3+2−3
√ √
= 3+2· 3+1+2· 3+2−3
√
= 3 + 4 · 3.

The last five values ((g)-(l)) are purely algebraic:

f (a) = a2 + 2 · a − 3,
f (a) + 5 = a2 + 2 · a − 3 + 5 = a2 + 2 · a + 2,
f (x + h) = (x + h)2 + 2 · (x + h) − 3
= x2 + 2xh + h2 + 2x + 2h − 3,
f (x + h) − f (x) = (x2 + 2xh + h2 + 2x + 2h − 3) − (x2 + 2x − 3)
= x2 + 2xh + h2 + 2x + 2h − 3 − x2 − 2x + 3
= 2xh + h2 + 2h,
f (x + h) − f (x) 2xh + h2 + 2h
=
h h
h · (2x + h + 2)
= = 2x + h + 2,
h
and
f (x) − f (a) (x2 + 2x − 3) − (a2 + 2a − 3)
=
x−a x−a
2 2
x + 2x − 3 − a − 2a + 3 x2 − a2 + 2x − 2a
= =
x−a x−a
(x + a)(x − a) + 2(x − a) (x − a)(x + a + 2)
= = = x + a + 2.
x−a (x − a)

The quotients f (x+h)−f

h
(x)
and f (x)−f
x−a
(a)
as in the last two examples 3.2(k)
and (l) will become particularly important in calculus. They are called differ-
ence quotients. We now calculate some more examples of difference quotients.
f (x+h)−f (x)
Example 3.3. Calculate the difference quotient h
for
1
a) f (x) = x3 + 2, b) f (x) = .
x
3.1. FUNCTIONS GIVEN BY FORMULAS 35

Solution. We calculate first the difference quotient step by step.

a) f (x + h) = (x + h)3 + 2 = (x + h) · (x + h) · (x + h) + 2
= (x2 + 2xh + h2 ) · (x + h) + 2
= x3 + 2x2 h + xh2 + x2 h + 2xh2 + h3 + 2,
= x3 + 3x2 h + 3xh2 + h3 + 2.

Subtracting f (x) from f (x + h) gives

f (x + h) − f (x) = (x3 + 3x2 h + 3xh2 + h3 + 2) − (x3 + 2)

= x3 + 3x2 h + 3xh2 + h3 + 2 − x3 − 2
= 3x2 h + 3xh2 + h3 .

With this we obtain:

f (x + h) − f (x) 3x2 h + 3xh2 + h3
=
h h
h · (3x2 + 3xh + h2 )
= = 3x2 + 3xh + h2 .
h
We handle part (b) similarly.
1
b) f (x + h) = ,
x+h
so that
1 1
f (x + h) − f (x) = − .
x+h x
We obtain the solution after simplifying the double fraction:
1 x−(x+h)x−x−h −h
f (x + h) − f (x) − x1
x+h (x+h)·x (x+h)·x (x+h)·x
= = = =
h h h h h
−h 1 −1
= · = .
(x + h) · x h (x + h) · x

So far, we have not mentioned the domain and range of the functions
defined above. Indeed, we will often not describe the domain explicitly but
use the following convention:
36 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

Convention 3.4. Unless otherwise stated, a function f has the largest possible
domain, that is the domain is the set of all real numbers x for which f (x) is a
well-defined real number. We refer to this as the standard convention of the
domain. (In particular, under this convention, any polynomial has the domain
R of all real numbers.)
The range is the set of all outputs obtained by f from the inputs (see also
the warning on page 23.)

Example 3.5. Find the domain of each of the following functions.

a) f (x) = 4x3 − 2x + 5, b) f (x) = |x|,

√
√
c) f (x) = x, d) f (x) = x − 3,
1 x−2
e) f (x) = 
x
, f) f (x) = x2 +8x+15 ,
x + 1 , for 2 < x ≤ 4
g) f (x) =
2x − 1 , for 5 ≤ x

Solution.

a) There is no problem taking a real number x to any (positive) power.

Therefore, f is defined for all real numbers x, and the domain is written
as D = R.

b) Again, we can take the absolute value for any real number x. The
domain is all real numbers, D = R.
√
c) The square root x is only defined for x ≥ 0 (remember we are not
using complex numbers yet!). Thus, the domain is D = [0, ∞).

d) Again, the square root is only defined for non-negative numbers. Thus,
the argument in the square root has to be greater or equal to zero:
x − 3 ≥ 0. Solving this for x gives
(add 3)
x−3≥ 0 =⇒ x ≥ 3.

The domain is therefore, D = [3, ∞).

e) A fraction is defined whenever the denominator is not zero, so that in

this case x1 is defined whenever x 6= 0. Therefore the domain is all real
numbers except zero, D = R − {0}.
3.2. FUNCTIONS GIVEN BY GRAPHS 37

f) Again, we need to make sure that the denominator does not become
zero; however we do not care about the numerator. The denominator is
zero exactly when x2 + 8x + 15 = 0. Solving this for x gives:
x2 + 8x + 15 = 0 =⇒ (x + 3) · (x + 5) = 0
=⇒ x + 3 = 0 or x + 5 = 0
=⇒ x = −3 or x = −5.
The domain is all real numbers except for −3 and −5, that is D =
R − {−5, −3}.
g) The function is explicitly defined for all 2 < x ≤ 4 and 5 ≤ x. Therefore,
the domain is D = (2, 4] ∪ [5, ∞).

3.2 Functions given by graphs

In many situations, a function is represented by its graph. The graph of
a function f is the set of all points (on the coordinate plane) of the form
(x, f (x)), where x is in the domain of f . We have already seen an example
of graphing a function when we discussed lines and their graphs. Here is
another example that shows how the formulaic definition relates to the graph
of the function.
Example 3.6. Let y = x2 with domain D = R being the set of all real numbers.
We can graph this after calculating a table as follows:
10 y

x −3 −2 −1 0 1 2 3 5
y 9 4 1 0 1 4 9 y = 44
output P (2, 4)
3

0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1 x = 2 input
38 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

Many function values can be read from this graph. For example, for the input
x = 2, we obtain the output y = 4. This corresponds to the point P (2, 4) on
the graph as depicted above.

In general, an input (placed on the x-axis) gets assigned to an output

(placed on the y-axis) according to where the vertical line at x intersects
with the given graph. This is used in the next example.

Example 3.7. Let f be the function given by the following graph.

5 y

0 x
-1 0 1 2 3 4 5 6 7 8 9 10
-1

Here, the dashed lines show, that the input x = 3 gives an output of y = 2.
Similarly, we can obtain other output values from the graph:

f (2) = 4, f (3) = 2, f (5) = 2, f (7) = 4.

Note, that in the above graph, a closed point means that the point is part of
the graph, whereas an open point means that it is not part of the graph.
The domain is the set of all possible input values on the x-axis. Since
we can take any number 2 ≤ x ≤ 7 as an input, the domain is the interval
D = [2, 7]. The range is the set of all possible output values on the y-axis.
Since any number 1 < y ≤ 4 is obtained as an output, the range is R = (1, 4].
Note in particular, that y = 1 is not an output, since f (5) = 2.
3.2. FUNCTIONS GIVEN BY GRAPHS 39

Example 3.8. Let f be the function given by the following graph.

4 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

Here are some function values that can be read from the graph:
f (−5) = 2, f (−4) = 3, f (−3) and f (−2) are undefined,
f (−1) = 2, f (0) = 1, f (1) = 2, f (2) = −1, f (4) = 0, f (5) = 1.
Note, that the output value f (3) is somewhere between −1 and 0, but we can
only read off an approximation of f (3) from the graph.
To find the domain of the function, we need to determine all possible x-
coordinates (or in other words, we need to project the graph onto the x-axis).
The possible x-coordinates are from the interval [−5, −3) together with the
intervals (−2, 2) and [2, 5]. The last two intervals may be combined. We get
the domain:
D = [−5, −3) ∪ (−2, 5].
For the range, we look at all possible y-values. These are given by the
intervals (1, 3] and (0, 3) and [−1, 1]. Combining these three intervals, we
obtain the range
R = [−1, 3].
Example 3.9. Consider the following graph.
4 y

0 x
0 1 2 3 4 5 6 7
40 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

Consider the input x = 4. There are several outputs that we get for x = 4
from this graph:
f (4) = 1, f (4) = 2, f (4) = 3.

However, in a function, it is not allowed to obtain more than one output for
one input! Therefore, this graph is not the graph of a function!

The reason why the previous example is not a function is due to some
input having more than one output: f (4) = 1, f (4) = 2, f (4) = 3.
4 y

0 x
0 1 2 3 4 5 6 7

In other words, there is a vertical line (x = 4) which intersects the graph in

more than one point. This observation is generalized in the following vertical
line test.

Observation 3.10 (Vertical Line Test). A graph is the graph of a function

precisely when every vertical line intersects the graph at most once.

Example 3.11. Consider the graph of the equation x = y 2:

3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

This does not pass the vertical line test so y is not a function of x. However,
x is a function of y since, if you consider y to be the input, each input has
exactly one output (it passes the ‘horizontal line’ test).
3.2. FUNCTIONS GIVEN BY GRAPHS 41

Example 3.12. Let f be the function given by the following graph.

5 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

a) What is the domain of f ? b) What is the range of f ?

c) For which x is f (x) = 3? d) For which x is f (x) = 2?
e) For which x is f (x) > 2? f) For which x is f (x) ≤ 4?
g) Find f (1) and f (4). h) Find f (1) + f (4).
i) Find f (1) + 4. j) Find f (1 + 4).
Solution. Most of the answers can be read immediately from the graph.
(a) For the domain, we project the graph to the x-axis. The domain consists
of all numbers from −5 to 5 without −3, that is D = [−5, −3) ∪ (−3, 5].
(b) For the range, we project the graph to the y-axis. The domain consists of
all numbers from 1 to 5, that is R = [1, 5].
(c) To find x with f (x) = 3 we look at the horizontal line at y = 3:
5 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

We see that there are two numbers x with f (x) = 3:

f (−2) = 3, f (3) = 3.
Therefore, the answer is x = −2 or x = 3.
42 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

(d) Looking at the horizontal line y = 2, we see that there is only one x
with f (x) = 2; namely f (4) = 2. Note, that x = −3 does not solve the
problem, since f (−3) is undefined. The answer is x = 4.

(e) To find x with f (x) > 2, the graph has to lie above the line y = 2.
5 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

We see that the answer is those numbers x greater than −3 and less than
4. The answer is therefore the interval (−3, 4).

(f) For f (x) ≤ 4, we obtain all numbers x from the domain that are less than
−1 or greater or equal to 1. The answer is therefore,

[−5, −3) ∪ (−3, −1) ∪ [1, 5].

Note that −3 is not part of the answer, since f (−3) is undefined.

(g) f (1) = 4, and f (4) = 2

(h) f (1) + f (4) = 4 + 2 = 6

(i) f (1) + 4 = 4 + 4 = 8

(j) f (1 + 4) = f (5) = 1

Example 3.13. The following graph shows the population size in a small city
3.2. FUNCTIONS GIVEN BY GRAPHS 43

in the years 2001-2011 in thousands of people.

40

30

20

10
year

’01 ’02 ’03 ’04 ’05 ’06 ’07 ’08 ’09 ’10 ’11
a) What was the population size in the years 2004 and 2009?
b) In what years did the city have the most population?
c) In what year did the population grow the fastest?
d) In what year did the population decline the fastest?
Solution. The population size in the year 2004 was approximately 36, 000. In
the year 2009, it was approximately 26, 000. The largest population was in
the year 2006, where the graph has its maximum. The fastest growth in the
population was between the years 2003 and 2004. That is when the graph
has the largest slope. Finally, the fastest decline happened in the years
2006-2007.
Example 3.14. Graph the function described by the following formula:

x + 3 , for −3 ≤ x < −1
f (x) = x2 , for −1 < x < 1

3 , for 2<x≤3
Solution. We really have to graph all three functions y = x + 3, y = x2 , and
y = 3, and then restrict them to their respective domain. Graphing the three
functions, we obtain the following tables and associated graphs, which we
draw in one x-y-plane:
y =x+3 y = x2 y=3
4 y 4 y 4 y

3 3 3
y=3

2 2 2

1 1 y = x2 1

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

-2 -2 -2
44 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS
4 y

3
y=3

1 y = x2

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

-2

However, we need to cut off the functions according to their specific input
domain that is given by the original function.
4 y

0 x
-4 -3 -2 -1 0 1 2 3 4
-1

-2

Note, how the open and closed circles at the endpoints of each branch corre-
spond to the “<” and “≤” rules in the original description of the function.

3.3 Exercises
Exercise 3.1. For each of the following functions,
√
a) f (x) = 3x + 1, b) f (x) = x2 − x, c) f (x) = x2 − 9
d) f (x) = x1 , e) f (x) = x−5
x+2
, f) f (x) = −x3

calculate the function values

√
i) f (3)√ ii) f (5) iii) f (−2) iv) f (0) v) f ( 13)
vi) f ( 2 + 3) vii) f (−x) viii) f (x + 2) ix) f (x) + h x) f (x + h)
3.3. EXERCISES 45

Exercise 3.2. Let f be the piecewise defined function


x − 5 , for −4 < x < 3
f (x) =
x2 , for 3 ≤ x ≤ 6

a) State the domain of the function.

Find the function values

b) f (2), c) f (5), d) f (−3), e) f (3)

Exercise 3.3. Let f be the piecewise defined function


 |x| − x2 , for x<2
f (x) = 7 , for 2 ≤ x < 5
 2
x − 4x + 1 , for 5<x

a) State the domain of the function.

Find the function values
b) f (1), c) f (−2), d) f (3),
e) f (2), f) f (5), g) f (7).
f (x+h)−f (x)
Exercise 3.4. Find the difference quotient h
for the following func-
tions:
a) f (x) = 5x, b) f (x) = 2x − 6, c) f (x) = x2 ,
d) f (x) = x2 + 5x, e) f (x) = x2 + 3x + 4, f) f (x) = 3 − x2 ,
g) f (x) = x2 + 4x − 9, h) f (x) = 3x2 − 2x + 7, i) f (x) = x3
f (x)−f (a)
Exercise 3.5. Find the difference quotient x−a
for the following functions:
1
a) f (x) = 3x, b) f (x) = 4x − 7, c) f (x) = x2 − 3x d) f (x) = x

Exercise 3.6. Find the domains of the following functions.

√
a) f (x) = x2 + 3x + 5, b) f (x) = p
|x − 2|, c) f (x) = x − 2
√ 1
d) f (x) = 8 − 2x, e) f (x) = |x + 3|, f) f (x) = x+6
x−5 x+1 x
g) f (x) = x−7 , h) f (x) = x2 −7x+10 , i) f (x) = |x−2|
 √
|x| for 1 < x < 2 x
√5
j) f (x) = , k) f (x) = x−9
, l) f (x) =
2x for 3 ≤ x x+4
46 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

Exercise 3.7. Below are three graphs for the functions f , g, and h.
4 y = f (x)

x
function f : 0
0 1 2 3 4 5 6 7

4 y = g(x)

function g : 0 x
-1 0 1 2 3 4 5 6 7 8
2 y = h(x)

0 x
-3 -2 -1 0 1 2 3 4
-1

function h : -2

a) Find the domain and range of f .

b) Find the domain and range of g.
c) Find the domain and range of h.

Find the following function values:

d) f (1) e) f (2) f) f (3) g) f (4) h) f (5) i) f (6) j) f (7)

k) g(0) l) g(1) m) g(2) n) g(3) o) g(4) p) g(6) q) g(13.2)√
r) h(−2) s) h(−1) t) h(0) u) h(1) v) h(2) w) h(3) x) h( 2)
3.3. EXERCISES 47

Exercise 3.8. Use the vertical line test to determine which of the following
graphs are the graphs of functions?
4 y 4 y

3 3

2 2

1 1

x x
a) 0 b) 0
0 1 2 3 4 5 0 1 2 3 4 5

4 y 4 y

3 3

2 2

1 1

x x
c) 0 d) 0
0 1 2 3 4 5 0 1 2 3 4 5

Exercise 3.9. Let f be the function given by the following graph.

3 y

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

a) What is the domain of f ? b) What is the range of f ?

c) For which x is f (x) = 0? d) For which x is f (x) = 2?
e) For which x is f (x) ≤ 1? f) For which x is f (x) > 0?
g) Find f (2) and f (5). h) Find f (2) + f (5).
i) Find f (2) + 5. j) Find f (2 + 5).
48 SESSION 3. FUNCTIONS BY FORMULAS AND GRAPHS

Exercise 3.10. The graph below displays the number of students admitted to
a college during the years 1995 to 2007.
6,000

5,000

4,000

3,000
year

’95 ’96 ’97 ’98 ’99 ’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07

a) How many students were admitted in the year 2000?

b) In what years were the most students admitted?
c) In what years did the number of admitted students rise fastest?
d) In what year(s) did the number of admitted students decline?
Exercise 3.11. Consider the function described by the following formula:
 2
x +1 , for −2 < x ≤ 0
f (x) = x−1 , for 0 < x ≤ 2

−x + 4 , for 2 < x ≤ 5

What is the domain of the function f ? Graph the function f .

Session 4

Introduction to the TI-84

4.1 Graphing with the TI-84

We now use the TI-84 to graph functions such as the functions discussed in
the previous chapters. We start by graphing the very well-known function
y = x2 , which is of course
✞ a parabola.
☎ To graph y = x2 , we first have to enter
the function. Press the lpy= lp key to get to the function menu:
✝ ✆

✞ ☎
In the first line, enter the function y = x2 by pressing the buttons lpX,T,θ,nlp
✞ ☎ ✝ ✆
2
for the variable “x,” and then press the lpx lp button. We obtain
✝ ✆

49
50 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎
We now go to the graphing window by pressing the lpgraph lp key. We
✝ ✆
obtain:

Here, the viewing window is displayed in its initial and standard setting
between −10 and 10 on the x-axis, and between✞ −10 and ☎ 10 on the
✞ y-axis.☎
These settings may be changed by pressing the lpwindow lp or the lpzoom lp
✞ ☎ ✝ ✆ ✝ ✆
keys. First, pressing lpwindow lp , we may change the scale by setting Xmin,
✝ ✆ ✞ ☎ ✞ ☎
Xmax, Ymin and Ymax to some new values using lp▽ lp and lpenter lp :
✝ ✆ ✝ ✆

✞ ☎
Note, that negative numbers have to be entered via lp(−) lp and not using
✞ ☎ ✝ ✆
lp− lp .
✝ ✆
✞ ☎ ✞ ☎ ✞ ☎
Note 4.1. The difference between lp(−) lp and lp− lp is that lp(−) lp is
✝ ✆ ✝ ✆✞ ✝☎ ✆
used to denote negative numbers (such as −10), whereas lp− lp is used to
✝ ✆
subtract two numbers (such as 7 − 3).

Note 4.2. A source for a common error occurs when the “Plot1” item in the
function menu is highlighted. When graphing a function as above, always
make sure that none of the “plots” are highlighted.
4.1. GRAPHING WITH THE TI-84 51

After changing the minimum and maximum

✞ x-☎and y-values of our window,
we can see the result after pressing lpgraph lp again:
✝ ✆

We may zoom in✞or out or ☎put the setting back to the standard viewing size
by pressing the lpzoom lp key:
✝ ✆

✞ ☎
We may zoom in by pressing lp2 lp , then choose a center in the graph where
✞ ✝ ☎ ✞✆ ☎✞ ☎✞ ☎
we want to focus (via the lp△ lp , lp▽ lp , lp⊳ lp , lp⊲ lp keys), and confirm
✞ ☎ ✝ ✆✝ ✆✝ ✆✝ ✆
this with lpenter lp :
✝ ✆

✞ ☎
Similarly, we may also zoom out (pressing using lp3 lp in the zoom menu
✝ ✆
and choosing a center). Finally we can
✞ recover
☎ the original setting by using
ZStandard in the zoom menu (press lp6 lp ).
✝ ✆
We can graph more than one function in the same window, which we show
next.
√
Example 4.3. Graph the equations y = 7 − x and y = x3 − 2x2 − 4 in the
same window.
52 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎
Solution. Enter the functions as Y1 and Y2 after pressing lpy= lp . The
✝ ✆
graphs of both functions appear together in the graphing window:

✞ ☎ ✞ ☎
√ 2
Here, the square root symbol “ ” is obtained using lp2nd lp and lpx lp ,
✞ ☎ ✞ ☎ ✝ ✞ ✆☎ ✝ ✆
and the third power via lp∧lp and lp3 lp (followed by lp⊲ lp to continue
✝ ✆ ✝ ✆ ✝ ✆
entering the function on the base line).

Example 4.4. Graph the relation (x − 3)2 + (y − 5)2 = 16.

Solution. Since the above expression is not solved for y, we cannot simply
plug this into the calculator. Instead, we have to solve for y first.

(x − 3)2 + (y − 5)2 = 16 =⇒ (y − 5)2 = 16 − (x − 3)2

p
=⇒ y − 5 = ± 16 − (x − 3)2
p
=⇒ y = 5 ± 16 − (x − 3)2
p
Note, that there
p are two functions that we need to graph: y = 5+ 16 − (x − 3)2
and y = 5 − 16 − (x − 3)2 . Entering these as Y1 and Y2 in the calculator
gives the following function menu and graphing window:

The graph displays a circle of radius 4 with a center at (3, 5). However, due
to the current scaling, the graph looks more like an ellipse instead ✞
of a circle.☎
We can remedy this by using the “zoom square” function; press lpzoom lp
✞ ☎ ✝ ✆
lp5 lp . This adjusts the axis to the same scale in the x- and the y-direction.
✝ ✆
4.2. FINDING ZEROS, MAXIMA, AND MINIMA 53

We obtain the following graph:

Note 4.5. We recall that the equation (x − h)2 + (y − k)2 = r 2 always forms
a circle in the plane. Indeed, this equation describes a circle with center
C(h, k) and radius r.

4.2 Finding zeros, maxima, and minima

In this section, we will show how to locate local maxima and minima of a
function (peaks and valleys of its graph), and the intersection points of two
graphs. Further we will be able to use the calculator to find the x-intercepts of
a graph. The x-intercepts are commonly called zeros or roots of the function
f . In other words, a zero of a function f is a number x for which f (x) = 0.
Example 4.6. Graph the equation y = x3 − 2x2 − 4x + 4.
a) Find points on the graph via the trace function, and by using the table
menu.
b) Approximate the zeros of the function via the calculate menu, that is,
approximate the x−coordinate(s) of the point(s) where the graph crosses
the x−axis.
c) Approximate the (local) maximum and minimum via the calculate menu.
(A (local) maximum or minimum is also called an (local) extremum.)
Solution.
a) Graphing the function in the standard window gives the following graph:
54 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎
Pressing lptrace lp , we can move a cursor to the right and left via
✞ ☎ ✝ ✞ ☎✆
lp⊲ lp and lp⊳ lp . We can see several function values as displayed
✝ ✆ ✝ ✆
below:

For the graph on the left we have displayed the point (x, y) ≈
(−.638, 5.478), and for the graph on the right, we have displayed the
point (x, y) ≈ (2.128, −3.933).
Another way
✞ to find
☎✞ function ☎values is to look at the table menu by
pressing lp2nd lp lpgraph lp :
✝ ✆✝ ✆

We see the y-values for specific ✞ x-values

☎ are
✞ displayed.
☎ More values
can be displayed by pressing lp△ lp and lp▽ lp . Furthermore, the
✝ ✆ ✝ ✆
settings
✞ for
☎✞ the table can
☎ be changed in the table-set menu by pressing
lp2nd lp lpwindow lp . In particular, by changing the the “Indpnt”
✝ ✆✝ ✆
setting in the table-set menu from “Auto” to “Ask”, we can have spe-
cific values calculated in the table menu. For example, below, we have
calculated the value of y for the independent variable x set to x = 4.

b) Note, that the function f has three x-intercepts (the places on a graph
intersects with the x-axis). They are called the zeros of the function
4.2. FINDING ZEROS, MAXIMA, AND MINIMA 55

and can be approximated with the TI-84. We have to find each ✞

zero, one☎
at a time. To this end, we need to switch the calculate menu lp2nd lp
✞ ☎ ✝ ✆
lptrace lp :
✝ ✆

✞ ☎
Press lp2 lp to find a zero. We need to specify a left bound (using the
✞✝ ☎✆✞ ☎ ✞ ☎
keys lp⊳ lp , lp⊲ lp and lpenter lp ), that is a point on the graph that
✝ ✆✝ ✆ ✝ ✆
is a bit left of the zero that we seek. Next, we also need to specify a
right bound for our zero, and also a “guess” that is close to the zero:

The zero displayed is approximately at x ≈ −1.709 (rounded to the

nearest thousandth).
Similarly,
✞ we can
☎✞ also approximate
☎ the other zeros using the calculate
menu lp2nd lp lptrace lp . After choosing lower bounds, upper bounds,
✝ ✆✝ ✆
and guesses, we obtain:

The zeros are approximately at x ≈ 0.806 and x ≈ 2.903.

c) The TI-84 can also approximate the maximum and minimum of the func-
tion (the places on a graph where there is a ’peak’ or a ’valley’). This
56 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎✞ ☎
is again done in the calculate menu lp2nd lp lptrace lp :
✝ ✆✝ ✆

✞ ☎
Now, press lp3 lp for the minimum. After specifying a left bound, a right
✝ ✆
bound, and a guess, we obtain the following answer for our minimum:

The minimum displayed is at (1.9999985, −4). The actual answer on

your calculator may be different, depending on the bounds and the
guess chosen. However, it is important to note that the exact mini-
mum is at (x, y) = (2, −4), whereas the calculator only provides an
approximation of the minimum!
Similarly,
✞ ☎✞we can also
☎✞approximate
☎ the maximum in the calculate menu
lp2nd lp lptrace lp lp4 lp . After choosing lower and upper bound,
✝ ✆✝ ✆✝ ✆
we obtain:

Therefore, the maximum is approximately at (x, y) ≈ (−0.667, 5.481),

after rounding to the nearest thousandth.

This concludes example 4.6.

Note
✞ 4.7.☎✞We can always
☎ return to the home screen by pressing quit, that is
lp2nd lp lpmodelp .
✝ ✆✝ ✆
4.2. FINDING ZEROS, MAXIMA, AND MINIMA 57

Example 4.8. Graph the equation y = x4 − x3 − 4x2 + 4x.

a) Approximate the zeros of the function via the calculate menu.

b) Approximate the (local) maxima and the minima via the calculate menu.

c) Calculate the zeros of the function via the solve function.

Solution.
✞ ☎
a) First, press lpy= lp and enter the function y = x4 − x3 − 4x2 + 4x. We
✝ ✆
find the zeros using item 2 from the calculate menu. Here are two of
the four zeros:

Note, in particular, the left zero has a y value of “1.6E−12,” that is

y = 1.6 · 10−12 = 0.0000000000016. The value x = −2 is an actual
zero as can be seen by direct computation, whereas the calculator only
obtains an approximation!
✞ ☎✞ ☎
b) From the calculate menu, lp2nd lp lptrace lp , items 3 and 4 can be
✝ ✆✝ ✆
used to approximate the maximum and the two minima, (choosing a lower
bound, upper bound, and a guess each time):

c) There is also a non-graphical procedure to calculate the zeros of the

function y = x4 −x3 −4x2 + 4x. Algebraically, we want to find a number
x so that y = 0. In other words:

0 = x4 − x3 − 4x2 + 4x.
58 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎
Press the lpmathlp button, scroll down to “Solver...” and press the
✞ ✝
☎ ✆
lpenter lp key. You will end up in one of the following two screens:
✝ ✆

✞ ☎
If you obtained the screen on the right, press lp△ lp to get to the screen
✞ ✝ ✆☎
on the left. Enter the equation and press lpenter lp .
✝ ✆

We now need to specify a number X which is our “guess” for the zero.
In other words, the calculator will try to identify a zero that is close
to a specified number X. For example, we✞may enter☎✞ x = 3. Then,
☎ the
solve command is executed by pressing lpalpha lp lpenter lp ; (the
✞ ☎ ✝ ✆✝ ✆
lpalpha lp -key gives access to the commands written in green). We
✝ ✆
obtain the following zero:

✞ ✞
☎ ☎
lpalpha lp lpenter lp
✝ ✆
✝ ✆

Choosing other values for x (for example x = ✞1.2, x = 0.2☎✞and x = −3)

☎
produces the other three zeros after pressing lpalpha lp lpenter lp .
✝ ✆✝ ✆
4.2. FINDING ZEROS, MAXIMA, AND MINIMA 59

Note, that the exact zeros are x = 2, x = 1, x = 0, and x = −2, whereas

again the calculator only supplied an approximation!

The equation solver used in the last example (part (c)) is a slightly faster
method for finding zeros than using the graph and the calculate menu. How-
ever, the disadvantage is that we first need to have some knowledge about
the zeros (such as how many zeros there are!) before we can effectively use
this tool. Generally, the graph and the calculate menu will give us a much
better idea of where the zeros are located. For this reason, we recommend to
use the calculate menu as the main method of finding zeros.

Example 4.9. Solve the equation. Approximate your answer to the nearest
hundredth.

a) x4 + 3x2 − 5x − 7 = 0, b) x3 − 4 = 7x − 3x .

Solution. a) We need to find all numbers x so that x4 +3x2 −5x−7 = 0. Note,

that these are precisely the zeros of the function f (x) = x4 + 3x2 − 5x − 7,
since the zeros are the values x for which f (x) = 0. Graphing the function
f (x) = x4 + 3x2 − 5x − 7 shows that there are two zeros.

Using the calculate menu, we can identify both zeros.

The answer is therefore, x ≈ −0.85 and x ≈ 1.64 (rounded to the nearest

hundredth).
 60 SESSION 4. INTRODUCTION TO THE TI-84

b) We use the same method as in part (a). To this end we rewrite the
equation so that one side becomes zero:

x3 − 4 − 7x + 3x = 0

We can now graph the function f (x) = x3 − 4 − 7x + 3x and find its zeros.

We see that there are three solutions, x ≈ −2.30, x ≈ −0.51, and x ≈

2.06.

By a method similar to the above, we can also find the intersection points
of two given graphs.

Example 4.10. Graph the equations y = x2 − 3x + 2 and y = x3 + 2x2 − 1.

a) Approximate the intersection point of the two graphs via the calculate menu.
b) Calculate the intersection point of the two graphs via the solve function.

☎ First, enter the two equations for Y1 and Y2 after pressing the
Solution.
✞
lpy= lp key. Both graphs are displayed in the graphing window.
✝ ✆

a) The procedure for finding the intersection of the graphs for Y1 and Y2
is
✞ similar☎✞
to finding ☎minima, maxima, or zeros. In the calculate menu,
lp2nd lp lptrace lp , choose intersect (item 5). Choose the first curve
✝ ✆✞
✝ ✆
☎ ✞ ☎ ✞ ☎
Y1 (with lp△ lp , lp▽ lp , and lpenter lp ), and the second curve
✝ ✆ ✝ ✆ ✝ ✆ ✞ ☎
Y2. Finally choose a guess of the intersection point (with lp⊳ lp ,
✝ ✆
4.3. EXERCISES 61
✞ ☎ ✞ ☎
lp⊲ lp , and lpenter lp ). The intersection is approximated as (x, y) ≈
✝ ✆ ✝ ✆
(.71134574, .37197554):

b) The intersection is the point where the two y-values y = x2 − 3x + 2

and y = x3 + 2x2 − 1 coincide. Therefore, we want to find an x with
x2 − 3x + 2 = x3 + 2x2 − 1
The TI-84 equation solver requires an equation with zero on the left.
Therefore, we subtract the left-hand side, and obtain:
(subtract (x2 − 3x + 2)) =⇒ 0 = (x3 + 2x2 − 1) − (x2 − 3x + 2)
=⇒ 0 = x3 + 2x2 − 1 − x2 + 3x − 2
=⇒ 0 = x3 + x2 + 3x − 3
✞ ☎✞ ☎
Entering 0 = x3 + x2 + 3x−3 into the equation solver, lpmathlp lp△ lp
✞ ☎ ✝ ✆✝ ✆
lpenter lp , and choosing an approximation x = 1 for X, the calculator
✝ ✆ ✞ ☎✞ ☎
solves this equation (using lpalpha lp lpenter lp ) as follows:
✝ ✆✝ ✆

The approximation provided by the calculator is x ≈ .71134573927 . . . .

4.3 Exercises
Exercise 4.1. Graph the equation in the standard window.
a) y = 3x
√ −5 b) y = x2 − 3x − 2 c) y = x4 − 3x3 + 2x − 1
1
d) y = x2 − 4 e) y = x+2 f) y = |x + 3|
62 SESSION 4. INTRODUCTION TO THE TI-84
✞ ☎
For the last exercise, the absolute value is obtained by pressing lpmathlp
✞ ☎✞ ☎ ✝ ✆
lp⊲ lp lpenter lp .
✝ ✆✝ ✆
Exercise 4.2. Solve the equation for y and graph all branches in the same
window.
a) x2 + y 2 = 4 b) (x + 5)2 + y 2 = 15 c) (x − 1)2 + (y − 2)2 = 9
d) y + x − 8x − 14 = 0 e) y 2 = x2 + 3
2 2
f) y 2 = −x2 + 77
Exercise 4.3. Find all zeros of the given function. Round your answer to the
nearest hundredth.
a) f (x) = x2 + 3x + 1 b) f (x) = x4 − 3x2 + 2
c) f (x) = x3 + 2x − 6 x5 − 11x4 + 20x3 + 88x2 − 224x + 1
d) f (x) = √
e) f (x) = x3 − 5x2 + 2x + 3 f) f (x) = 2x − 3 − 2x + 3

g) 0 = 0.04x3 − 0.02x2 − 0.5174x + 0.81

h) 0 = 0.04x3 − 0.02x2 − 0.5175x + 0.81
i) 0 = 0.04x3 − 0.02x2 − 0.5176x + 0.81
Exercise 4.4. Find all solutions of the equation. Round your answer to the
nearest thousandth.
a) x3 + 3 = x5 + 7 b) 4x3 + 6x2 − 3x − 2 = 0
2x 2 +2
c) x−3 = xx+1 d) 53x+1 = x5 + 6
e) x3 + x2 = x4 − x2 + x f) 3x2 = x3 − x2 + 3x
Exercise 4.5. Graph the equation. Determine how many maxima and minima
the graph has. To this end, resize the graphing window (via the zoom-in,
zoom-out, and zoom-box functions of the calculator) to zoom into the maxima
or minima of the graph.
a) y = x2 − 4x + 13 b) y = −x2 + x − 20
c) y = 2x − 5x + 3x d) y = x4 − 5x3 + 8x2 − 5x + 1
3 2

Exercise 4.6. Approximate the (local) maxima and minima of the graph. Round
your answer to the nearest tenth.
a) y = x3 + 2x2 − x + 1 b) y = x3 − 5x2 + 8x − 3
c) y = −x4 + 3x3 + x2 + 2 d) y = x4 − x3 − 4x2 + 6x + 2
e) y = x − x − 4x + 8x + 2 f) y = x4 − x3 − 4x2 + 7x + 2
4 3 2
Session 5

Basic functions and transformations

5.1 Graphing basic functions

It will be useful to study the shape of graphs of some basic functions, which
may then be taken as building blocks for more advanced and complicated
functions. In this section, we consider the following basic functions:
√ 1
y = |x|, y = x2 , y = x3 , y= x, y= .
x
We can either graph these functions by hand by calculating a table, or by
using the TI-84, via the table and graph buttons.

• We begin with the absolute value function y = |x|. Recall that the ab-
solute
✞ value
☎✞ is obtained
☎✞ on the
☎ calculator in the math menu by pressing
lpmathlp lp⊲ lp lpenter lp . The domain of y = |x| is all real numbers,
✝ ✆✝ ✆✝ ✆
D = R.
3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

We have drawn the graph a second time in the x-y-plane on the right.

63
64 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

• Similarly, we obtain the graph for y = x2 , which is a parabola. The

domain is D = R. 3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

• Here is the graph for y = x3 . The domain is D = R.

3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

√
• Next we graph y = x. The domain is D = [0, ∞).
3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3

• Finally, here is the graph for y = x1 . The domain is D = R − {0}.

3 y

0 x
-3 -2 -1 0 1 2 3
-1

-2

-3
5.2. TRANSFORMATION OF GRAPHS 65

These graphs together with the line y = mx + b studied in Section 2.1 are
our basic building blocks for more complicated graphs in the next sections.
Note in particular, that the graph of y = x is the diagonal line.
y y
y = mx + c y=x

c x
x

5.2 Transformation of graphs

For a given function, we now study how the graph of the function changes when
performing elementary operations, such as adding, subtracting, or multiplying
a constant number to the input or output. We will study the behavior in five
specific examples.
1. Consider the following graphs:
y = x2 y = x2 + 2 y = x2 − 2
4 y 4 y 4 y

3 3 3

2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

-2 -2 -2

-3 -3 -3

We see that the function y = x2 is shifted up by 2 units, respectively

down by 2 units. In general, we have:
Observation 5.1. Consider the graph of a function y = f (x). Then, the
graph of y = f (x) + c is that of y = f (x) shifted up or down by c. If c
is positive, the graph is shifted up, if c is negative, the graph is shifted
down.
66 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

2. Next, we consider the transformation of y = x2 given by adding or

subtracting a constant to the input x.
y = x2 y = (x + 1)2 y = (x − 1)2
3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

Now, we see that the function is shifted to the left or right. Note, that
y = (x + 1)2 shifts the function to the left, which can be seen to be
correct, since the input x = −1 gives the output y = ((−1) + 1)2 = 02 =
0.
Observation 5.2. Consider the graph of a function y = f (x). Then, the
graph of y = f (x + c) is that of y = f (x) shifted to the left or right by
c. If c is positive, the graph is shifted to the left, if c is negative, the
graph is shifted to the right.

3. Another transformation is given by multiplying the function by a fixed

positive factor.
1
y = x3 + 1 y = 2 · (x3 + 1) y= 2
· (x3 + 1)
3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

-2 -2 -2

-3 -3 -3

This time, the function is stretched away or compressed towards the

x-axis.
Observation 5.3. Consider the graph of a function y = f (x) and let
c > 0. Then, the graph of y = c · f (x) is that of y = f (x) stretched
away or compressed towards the x-axis by a factor c. If c > 1, the graph
is stretched away from the x-axis, if 0 < c < 1, the graph is compressed
towards the x-axis.
5.2. TRANSFORMATION OF GRAPHS 67

4. Similarly, we can multiply the input by a positive factor.

y = x3 + 1 y = (2 · x)3 + 1 y = ( 21 · x)3 + 1
3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

-2 -2 -2

-3 -3 -3

This time, the function is stretched away or compressed towards the

y-axis.
Observation 5.4. Consider the graph of a function y = f (x) and let
c > 0. Then, the graph of y = f (c · x) is that of y = f (x) stretched
away or compressed towards the y-axis by a factor c. If c > 1, the graph
is compressed towards the y-axis, if 0 < c < 1, the graph is stretched
away from the y-axis.

5. The last transformation is given by multiplying (−1) to the input or

output, as displayed in the following chart.

y = x3 + 1 y = −(x3 + 1) y = (−x)3 + 1
3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

-2 -2 -2

-3 -3 -3

Here, the function is reflected either about the x-axis or the y-axis.
Observation 5.5. Consider the graph of a function y = f (x). Then,
the graph of y = −f (x) is that of y = f (x) reflected about the x-axis.
Furthermore, the graph of y = f (−x) is that of y = f (x) reflected about
the y-axis.
68 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

Example 5.6. Guess the formula for the function, based on the basic graphs
in Section 5.1 and the transformations described above.
4 y 3 y

3 2

2 1

1 0 x
-4 -3 -2 -1 0 1 2 3 4
0 x -1
-4 -3 -2 -1 0 1 2 3 4
-1 -2

a) -2 b) -3
4 y 4 y

3 3

2 2

1 1

0 x 0 x
-4 -3 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 4 5
c) -1 d) -1

Solution. a) This is the square-root function shifted

√ to the left by 2. Thus,
by Observation 5.1, this is the function f (x) = x + 2.
b) This is the graph of y = x1 reflected about the x-axis (or also y = 1
x
reflected about the y-axis). In either case, we obtain the rule y = − x1 .
c) This is a parabola reflected about the x-axis and then shifted up by 3.
Thus, we get:
y = x2
reflecting about the x-axis gives y = −x2
shifting the graph up by 3 gives y = −x2 + 3

d) Starting from the graph of the cubic equation y = x3 , we need to reflect

about the x-axis (or also y-axis), then shift up by 2 and to the right by 3.
These transformations affect the formula as follows:
y = x3
reflecting about the x-axis gives y = −x3
shifting up by 2 gives y = −x3 + 2
shifting the the right by 3 gives y = −(x − 3)3 + 2
All these answers can be checked by graphing the function with the TI-
84.
5.2. TRANSFORMATION OF GRAPHS 69

Example 5.7. Sketch the graph of the function, based on the basic graphs in
Section 5.1 and the transformations described above.
a) y = x2 +
√3 b) y = (x + 2)2
c) y = |x − 3| − 2
d) y = 2 · x + 1 e) y = − x1 + 2 f) y = (−x + 1)3

Solution. a) This is the parabola y = x2 shifted up by 3. The graph is shown

below.

b) y = (x + 2)2 is the parabola y = x2 shifted 2 units to the left.

5 y 4 y

4 3

3 2

2 1

1 0 x
-5 -4 -3 -2 -1 0 1 2 3
a) -3 -2 -1
0
0 1 2
x
3
b) -1

c) The graph of the function f (x) = |x − 3| − 2 is the absolute value shifted

to the right by 3 and down by 2. (Alternatively, we can first shift down by
2 and then to the right by 3.)
√ √
d) Similarly, to get from the graph of y = x to the √ graph of y = x + 1, we
shift the graph to the left, and then for y = 2 · x + 1, we need to stretch
the graph by a factor 2 away from the x-axis. (Alternatively, we could first
stretch the the graph away from the x-axis, then shift the graph by 1 to
the left.)
3 y 5 y

2 4

1 3

0 x 2
-2 -1 0 1 2 3 4 5
-1 1

-2 0 x
-2 -1 0 1 2 3 4 5
c) -3 d) -1

e) For y = −( x1 + 2), we start with y = x1 and add 2, giving y = x1 + 2, which

shifts the graph up by 2. Then, taking the negative gives y = −( x1 + 2),
which corresponds to reflecting the graph about the x-axis.
70 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

Note, that in this case, we cannot perform these transformations in the

opposite order, since the negative of y = x1 gives y = − x1 , and adding 2
gives y = − x1 + 2 which is not equal to −( x1 + 2).
f) We start with y = x3 . Adding 1 in the argument, y = (x + 1)3 , shifts its
graph to the left by 1. Then, taking the negative in the argument gives
y = (−x + 1)3 , which reflects the graph about the y-axis.
Here, the order in which we perform these transformations is again im-
portant. In fact, if we first take the negative in the argument, we obtain
y = (−x)3 . Then, adding one in the argument would give y = (−(x+1))3 =
(−x − 1)3 which is different than our given function y = (−x + 1)3 .
4 y 4 y

3 3

2 2

1 1

0 x 0 x
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 -1

-2 -2

-3 -3

e) -4 f) -4

All these solutions may also easily be checked by using the graphing
function of the calculator.
Example 5.8. a) The graph of f (x) = |x3 − 5| is stretched away from the
y-axis by a factor of 3. What is the formula for the new function?
√
b) The graph of f (x) = 6x2 + 3 is shifted up 5 units, and then reflected
about the x-axis. What is the formula for the new function?
c) How are the graphs of y = 2x3 + 5x − 9 and y = 2(x − 2)3 + 5(x − 2) − 9
related?
d) How are the graphs of y = (x − 2)2 and y = (−x + 3)2 related?
Solution. a) By Observation 5.4 on page 67, we have to multiply the argu-
ment by 31 . The new function is therefore:
1   1 3 1
f ·x = ·x −5 = · x3 − 5
3 3 27
5.2. TRANSFORMATION OF GRAPHS 71
√
b) After the shift, we have the graph of a new function y = 6x2 + 3 + 5.
Then, √a reflection about the x-axis gives the graph of the function
y = −( 6x2 + 3 + 5).
c) By Observation 5.2 on page 66, we see that we need to shift the graph of
y = 2x3 + 5x − 9 by 2 units to the right.
d) The formulas can be transformed into each other as follows:
We begin with y = (x − 2)2 .
Replacing x by x + 5 gives y = ((x + 5) − 2)2 = (x + 3)2 .
Replacing x by −x gives y = ((−x) + 3)2 = (−x + 3)2 .
Therefore, we have performed a shift to the left by 5, and then a reflection
about the y-axis.
We want to point out that there is a second solution for this problem:
We begin with y = (x − 2)2 .
Replacing x by −x gives
y = ((−x) − 2)2 = (−x − 2)2 .
Replacing x by x − 5 gives
y = (−(x − 5) − 2)2 = (−x + 5 − 2)2 = (−x + 3)2 .
Therefore, we could also first perform a reflection about the y-axis, and
then shift the graph to the right by 5.

Some of the above functions have special symmetries, which we investigate

now.
Definition 5.9. A function f is called even if f (−x) = f (x) for all x.
Similarly, a function f is called odd if f (−x) = −f (x) for all x.
Example 5.10. Determine, if the following functions are even, odd, or neither.
f (x) = x2 , g(x) = x3 , h(x) = x4 , k(x) = x5 ,
l(x) = 4x5 + 7x3 − 2x, m(x) = x2 + 5x.

Solution. The function f (x) = x2 is even, since f (−x) = (−x)2 = x2 . Simi-

larly, g(x) = x3 is odd, h(x) = x4 is even, and k(x) = x5 is odd, since

g(−x) = (−x)3 = −x3 = −g(x)

72 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

h(−x) = (−x)4 = x4 = h(x)

k(−x) = (−x)5 = −x5 = −k(x)

Indeed, we see that a function y = xn is even, precisely when n is even, and

y = xn is odd, precisely when n is odd. (These examples are in fact the
motivation behind defining even and odd functions as in definition 5.9 above.)
Next, in order to determine if the function l is even or odd, we calculate
l(−x) and compare it with l(x).

l(−x) = 4(−x)5 + 7(−x)3 − 2(−x) = −4x5 − 7x3 + 2x

= −(4x5 + 7x3 − 2x) = −l(x)

Therefore, l is an odd function.

Finally, for m(x) = x2 + 5x, we calculate m(−x) as follows:

m(−x) = (−x)2 + 5(−x) = x2 − 5x

Note, that m is not an even function, since x2 − 5x 6= x2 + 5x. Furthermore,

m is also not an odd function, since x2 − 5x 6= −(x2 + 5x). Therefore, m is a
function that is neither even nor odd.

Observation 5.11. An even function f is symmetric with respect to the y−axis

(if you reflect the graph of f about the y−axis you get the same graph back),
since even functions satisfy f (−x) = f (x):
y
f (−x) = f (x)

Example y = x2 :

x
−x x

An odd function f is symmetric with respect to the origin (if you reflect
the graph of f about the y−axis and then about the x−axis you get the same
5.3. EXERCISES 73

graph back), since odd functions satisfy f (−x) = −f (x):

y

f (x)
3
Example y = x :
−x x
x

−f (x)

5.3 Exercises
Exercise 5.1. Find a possible formula of the graph displayed below.
4 y 3 y

3 2

2 1

1 0 x
-3 -2 -1 0 1 2 3 4
0 x -1
-4 -3 -2 -1 0 1 2 3 4
-1 -2

a) -2 b) -3

4 y 6 y

5
3
4

2 3

2
1
1

0 x 0 x
-3 -2 -1 0 1 2 3 4 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4
-1
-1
-2

c) -2 d) -3

3 y 4 y

2 3

1 2

0 x 1
-4 -3 -2 -1 0 1 2 3
-1 0 x
-2 -1 0 1 2 3 4 5 6
-2 -1

e) -3 f) -2
74 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS

Exercise 5.2. Sketch the graph of the function, based on the basic graphs in
Section 5.1 and the transformations described above. Confirm your answer
by graphing the function with the calculator.
1
a) f (x) = |x| − 3 b) f (x) = x+2
2
c) f (x) = −x
√ d) f (x) = (x − 1)3
e) f (x) = √ −x f) f (x) = 4 · |x − 3|
g) f (x) = − x + 1 h) f (x) = ( 12 · x)2 + 3

Exercise 5.3. Consider the graph of f (x) = x2 − 7x + 1. Find the formula of

the function that is given by performing the following transformations on the
graph.

a) Shift the graph of f down by 4.

b) Shift the graph of f to the left by 3 units.
c) Reflect the graph of f about the x-axis.
d) Reflect the graph of f about the y-axis.
e) Stretch the graph of f away from the y-axis by a factor 3.
f) Compress the graph of f towards the y-axis by a factor 2.

Exercise 5.4. How are the graphs of f and g related?

√ √
a) f (x) = x, g(x) = x − 5
b) f (x) = |x|, g(x) = 2 · |x|
c) f (x) = (x + 1)3 , g(x) = (x − 3)3
d) f (x) = x2 + 3x + 5, g(x) = (2x)2 + 3(2x)2 + 5
1
e) f (x) = x+3 , g(x) = − x1
f) f (x) = 2 · |x|, g(x) = |x + 1| + 1
Exercise 5.5. Determine, if the function is even, odd, or neither.
a) y = 2x3 , b) y = 5x2 , c) y = 3x4 − 4x2 + 5,
d) y = 2x3 +3 5x
y
2
, e) y = |x|, 3 y
f) y = x1 , 3 y
2 2 2

1 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1

-2 -2 -2

g) -3 h) -3 i) -3
5.3. EXERCISES 75

Exercise 5.6. The graph of the function y = f (x) is displayed below.

4 y

0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1

-2

-3

Sketch the graph of the following functions.

a) y = f (x) + 1, b) y = f (x − 3), c) y = −f (x),

d) y = 2f (x), e) y = f (2x), f) y = f ( 21 x)
Session 6

Operations on functions

6.1 Operations on functions given by formulas

We can combine functions in many different ways. First, note that we can
add, subtract, multiply, and divide functions.

Example 6.1. Let f (x) = x2 + 5x and g(x) = 7x − 3. Find the following

functions, and state their domain.
 
f
(f + g)(x), (f − g)(x), (f · g)(x), and (x).
g

Solution. The functions are calculated by adding the functions (or subtract-
ing, multiplying, dividing).

(f + g)(x) = (x2 + 5x) + (7x − 3) = x2 + 12x − 3,

(f − g)(x) = (x2 + 5x) − (7x − 3)
= x2 + 5x − 7x + 3 = x2 − 2x + 3,
(f · g)(x) = (x2 + 5x) · (7x − 3)
= 7x3 − 3x2 + 35x2 − 15x = 7x3 + 32x2 − 15x,
 
f x2 + 5x
(x) = .
g 7x − 3

The calculation of these functions was straightforward. To state their domain

is also straightforward, except for the domain of the quotient fg . Note, that
f + g, f − g, and f · g are all polynomials. By the standard convention

76
6.1. OPERATIONS ON FUNCTIONS GIVEN BY FORMULAS 77

(Convention 3.4 on page 36) all these functions have the domain R, that is
their domain is all real numbers.
Now, for the domain of fg , we have to be a bit more careful, since the
x2 +5x
denominator of a fraction cannot be zero. The denominator of fg (x) = 7x−3
is zero, exactly when
3
7x − 3 = 0 =⇒ 7x = 3 =⇒ x= .
7
3 f
We have to exclude 7
from the domain. The domain of the quotient g
is
therefore R − { 73 }.
We can formally state the observation we made in the previous example.
Observation 6.2. Let f be a function with domain Df , and g be a function
with domain Dg . A value x can be used as an input of f + g, f − g, and f · g
exactly when x is an input of both f and g. Therefore, the domains of the
combined functions are the intersection of the domains Df and Dg :
Df +g = Df ∩ Dg = {x | x ∈ Df and x ∈ Dg },
Df −g = Df ∩ Dg ,
Df ·g = Df ∩ Dg .
For the quotient fg , we also have to make sure that the denominator g(x) is
not zero.
D f = {x | x ∈ Df , x ∈ Dg , and g(x) 6= 0}.
g
√
Example 6.3. Let f (x) = x + 2, and let g(x) = x2 − 5x + 4. Find the
functions fg and fg and state their domains.
Solution. First, the domain of f consists of those numbers x for which the
square root is defined. In other words, we need x + 2 ≥ 0, that is x ≥ −2, so
that the domain of f is Df = [−2, ∞). On the other hand, the domain of g is
all real numbers, Dg = R. Now, we have the quotients
  √  
f x+2 g x2 − 5x + 4
(x) = 2 , and (x) = √ .
g x − 5x + 4 f x+2
For the domain of fg , we need to exclude those numbers x for which x2 −
5x + 4 = 0. Thus,
x2 − 5x + 4 = 0 =⇒ (x − 1)(x − 4) = 0
78 SESSION 6. OPERATIONS ON FUNCTIONS

=⇒ x = 1, or x = 4.

We obtain the domain for fg as the combined domain for f and g, and exclude
1 and 4. Therefore, D f = [−2, ∞) − {1, 4}.
g
g x2√
−5x+4
Now, for f
(x) = x+2
, the denominator becomes zero exactly when

x + 2 = 0, =⇒ x = −2.

Therefore, we need to exclude −2 from the domain, that is

D fg = [−2, ∞) − {−2} = (−2, ∞).

Example 6.4. To form the quotient fg (x) where f (x) = x2 −1 and g(x) = x+1
2
we write fg (x) = xx+1
−1
= (x+1)(x−1)
x+1
= x − 1. One might be tempted to say
that the domain is all real numbers. But it is not! The domain is all real
numbers except −1, and the last step of the simplification performed above is
only valid for x 6= −1.
Another operation we can perform is the composition of two functions.
Definition 6.5. Let f and g be functions, and assume that g(x) is in the
domain of f . Then define the composition of f and g at x to be

(f ◦ g)(x) := f (g(x)).

We can take any x as an input of f ◦ g which is an input of g and for which

g(x) is an input of f . Therefore, if Df is the domain of f and Dg is the domain
of g, the domain of Df ◦g is

Df ◦g = { x | x ∈ Dg , g(x) ∈ Df }.

Example 6.6. Let f (x) = 2x2 + 5x and g(x) = 2 − x. Find the following
compositions

a) f (g(3)), b) g(f (3)), c) f (f (1)), d) f (2 · g(5)), e) g(g(4) + 5).

Solution. We evaluate the expressions, one at a time, as follows:

a) f (g(3)) = f (2 − 3) = f (−1) = 2 · (−1)2 + 5 · (−1)

6.1. OPERATIONS ON FUNCTIONS GIVEN BY FORMULAS 79

= 2 − 5 = −3,
b) g(f (3)) = g(2 · 32 + 5 · 3) = g(18 + 15) = g(33)
= 2 − 33 = −31,
c) f (f (1)) = f (2 · 12 + 5 · 1) = f (2 + 5) = f (7)
= 2 · 72 + 5 · 7 = 98 + 35 = 133,
d) f (2 · g(5)) = f (2 · (2 − 5)) = f (2 · (−3)) = f (−6)
= 2 · (−6)2 + 5 · (−6) = 72 − 30 = 42,
e) g(g(4) + 5) = g((2 − 4) + 5) = g((−2) + 5) = g(3) = 2 − 3 = −1.

We can also calculate composite functions for arbitrary x in the domain.

Example 6.7. Let f (x) = x2 + 1 and g(x) = x + 3. Find the following
compositions
a) (f ◦ g)(x), b) (g ◦ f )(x), c) (f ◦ f )(x), d) (g ◦ g)(x).
Solution. a) There are essentially two ways to evaluate (f ◦ g)(x) = f (g(x)).
We can either first use the explicit formula for f (x) and then the one for g(x),
or vice versa. We will evaluate f (g(x)) by substituting g(x) into the formula
for f (x):
(f ◦ g)(x) = f (g(x)) = (g(x))2 + 1 = (x + 3)2 + 1
= x2 + 6x + 9 + 1 = x2 + 6x + 10.
Similarly, we evaluate the other expressions (b)-(d):
b) (g ◦ f )(x) = g(f (x)) = f (x) + 3 = x2 + 1 + 3 = x2 + 4,
c) (f ◦ f )(x) = f (f (x)) = (f (x))2 + 1 = (x2 + 1)2 + 1
= x4 + 2x2 + 1 + 1 = x4 + 2x2 + 2,
d) (g ◦ g)(x) = g(g(x)) = g(x) + 3 = x + 3 + 3 = x + 6.

Example 6.8. Find (f ◦ g)(x) and (g ◦ f )(x) for the following functions, and
state their domains.
3
a) f (x) = x+2 , and g(x) = x2 − 3x,
b) f (x) = q
|3x − 2| − 6x + 4, and g(x) = 5x + 1,
1
c) f (x) = 2
· (x − 4), and g(x) = 2x2 + 4.
80 SESSION 6. OPERATIONS ON FUNCTIONS

Solution.
a) Composing f ◦ g, we obtain
3 3
(f ◦ g)(x) = f (g(x)) = = 2 .
g(x) + 2 x − 3x + 2
The domain is the set of numbers x for which the denominator is non-zero.

x2 − 3x + 2 = 0 =⇒ (x − 2)(x − 1) = 0
=⇒ x = 2 or x = 1
=⇒ Df ◦g = R − {1, 2}.

Similarly,
 2
2 3 3
(g ◦ f )(x) = g(f (x)) = f (x) − 3f (x) = −3
x+2 x+2
9 9 9 − 9(x + 2)
= 2
− =
(x + 2) x+2 (x + 2)2
9 − 9x − 18 −9x − 9 −9 · (x + 1)
= 2
= 2
=
(x + 2) (x + 2) (x + 2)2
The domain is all real numbers except x = −2, that is Dg◦f = R − {−2}.

b) We calculate the compositions as follows.

(f ◦ g)(x) = f (g(x)) = |3g(x) − 2| − 6g(x) + 4

= |3(5x + 1) − 2| − 6(5x + 1) + 4 = |15x + 1| − 30x − 2,
(g ◦ f )(x) = g(f (x)) = 5f (x) + 1 = 5 · (|3x − 2| − 6x + 4) + 1
= 5 · |3x − 2| − 30x + 20 + 1 = 5 · |3x − 2| − 30x + 21.

Since the domains of f and g are all real numbers, so are also the domains
for both f ◦ g and g ◦ f .

c) Again we calculate the compositions.

r r
1 1
(f ◦ g)(x) = f (g(x)) = · (g(x) − 4) = · (2x2 + 4 − 4)
2 2
r
1 √
= · 2x2 = x2 = |x|.
2
6.2. OPERATIONS ON FUNCTIONS GIVEN BY TABLES 81

The domain of g is all real numbers, and the outputs g(x) = 2x2 + 4 are
all ≥ 4, (since 2x2 ≥ 0). Therefore, g(x) is in the domain of f , and we
have a combined domain of f ◦ g of Df ◦g = R. On the other hand,
r 2
2 1
(g ◦ f )(x) = g(f (x)) = 2(f (x)) + 4 = 2 · · (x − 4) +4
2
 
1
= 2· · (x − 4) + 4 = (x − 4) + 4 = x.
2
The domain of g ◦ f consists of all numbers x which are in the domain of f
and for which f (x) is in the domain of g. Now, the domain of f consists of
all real numbers x that give a non-negative argument in the square-root,
that is: 21 (x − 4) ≥ 0. Therefore we must have x − 4 ≥ 0, so that x ≥ 4,
and we obtain the domain Df = [4, ∞). Since the domain Dg = R, the
composition g ◦ f has the same domain as f :

Dg◦f = Df = [4, ∞).

We remark that at a first glance, we might have expected that (g ◦ f ) = x

has a domain of all real numbers. However, the composition g(f (x)) can
only have those inputs that are also allowed inputs of f . We see that the
domain of a composition is sometimes smaller than the domain that we use
via the standard convention (Convention 3.4).

6.2 Operations on functions given by tables

We can also combine functions that are defined using tables.
Example 6.9. Let f and g be the functions defined by the following table.

x 1 2 3 4 5 6 7
f (x) 6 3 1 4 0 7 6
g(x) 4 0 2 5 −2 3 1

Describe the following functions via a table:

a) 2 · f (x) + 3, b) f (x) − g(x), c) f (x + 2), d) g(−x).

82 SESSION 6. OPERATIONS ON FUNCTIONS

Solution. For (a) and (b), we obtain by immediate calculation

x 1 2 3 4 5 6 7
2 · f (x) + 3 15 9 5 11 3 17 15
f (x) − g(x) 2 3 −1 −1 2 4 5

For example, for x = 3, we obtain 2 · f (x) + 3 = 2 · f (3) + 3 = 2 · 1 + 3 = 5

and f (x) − g(x) = f (3) − g(3) = 1 − 2 = −1.
For part (c), we have a similar calculation of f (x + 2). For example, for
x = 1, we get f (1 + 2) = f (1 + 2) = f (3) = 1.

x 1 2 3 4 5 6 7 −1 0
f (x + 2) 1 4 0 7 6 undef. undef. 6 3

Note that for the last two inputs x = 6 and x = 7 the expression f (x + 2)
is undefined, since, for example for x = 6, it is f (x + 2) = f (6 + 2) = f (8)
which is undefined. However, for x = −1, we obtain f (x + 2) = f (−1 + 2) =
f (1) = 6. If we define h(x) = f (x + 2), then the domain of h is therefore
Dh = {−1, 0, 1, 2, 3, 4, 5}.
Finally, for part (d), we need to take x as inputs, for which g(−x) is
defined via the table for g. We obtain the following answer.

x −1 −2 −3 −4 −5 −6 −7
g(−x) 4 0 2 5 −2 3 1

Example 6.10. Let f and g be the functions defined by the following table.

x 1 3 5 7 9 11
f (x) 3 5 11 4 9 7
g(x) 7 −6 9 11 9 5

Describe the following functions via a table:

a) f ◦ g, b) g ◦ f, c) f ◦ f, d) g ◦ g.

Solution. The compositions are calculated by repeated evaluation. For ex-

ample,
(f ◦ g)(1) = f (g(1)) = f (7) = 4.
6.3. EXERCISES 83

The complete answer is displayed below.

x 1 3 5 7 9 11
(f ◦ g)(x) 4 undef. 9 7 9 11
(g ◦ f )(x) −6 9 5 undef. 9 11
(f ◦ f )(x) 5 11 7 undef. 9 4
(g ◦ g)(x) 11 undef. 9 5 9 9

6.3 Exercises
Exercise 6.1. Find f + g, f − g, f · g for the functions below. State their
domain.
a) f (x) = x2 + 6x, and g(x) = 3x − 5,
b) f (x) = x3 + 5,√ and g(x) = 5x2 + 7,√
c) f (x) = 3x + 7 x, and g(x) = 2x2 + 5 x,
1 5x
d) f (x) = √x+2
, and g(x) = x+2
√,
e) f (x) = x − 3, and g(x) = 2 x − 3,
f) f (x) = x2 + 2x + 5, and g(x) = 3x − 6,
g) f (x) = x2 + 3x, and g(x) = 2x2 + 3x + 4.

Exercise 6.2. Find fg , and g

f
for the functions below. State their domain.

a) f (x) = 3x + 6, and g(x) = 2x − 8,

b) f (x) = x + 2, and g(x) = x2 − 5x + 4,
1
c) f (x) = x−5
√ , and g(x) = x−2
x+3
,
d) f (x) = x + 6, and g(x) = √
2x + 5,
e) f (x) = x2 + 8x − 33, and g(x) = x.

Exercise 6.3. Let f (x) = 2x − 3 and g(x) = 3x2 + 4x. Find the following
compositions

a) f (g(2)), b) g(f (2)), c) f (f (5)), d) f (5g(−3)),

e) g(f (2) − 2), f) f (f (3) + g(3)), g) g(f (2 + x)), h) f (f (−x)),
i) f (f (−3) − 3g(2)), j) f (f (f (2))), k) f (x + h), l) g(x + h).
84 SESSION 6. OPERATIONS ON FUNCTIONS

Exercise 6.4. Find the composition (f ◦ g)(x) for the functions:

a) f (x) = 3x − 5, and g(x) = 2x + 3,
b) f (x) = x2 + 2, and g(x) = x + 3,
c) f (x) = x2 − 3x
√ + 2, and g(x) = 2x + 1,
d) f (x) = x2 + x + 3, and g(x) = x2 + 2x,
2
e) f (x) = x+4 , and g(x) = x + h,
2
f) f (x) = x + 4x + 3, and g(x) = x + h.
Exercise 6.5. Find the compositions
(f ◦ g)(x), (g ◦ f )(x), (f ◦ f )(x), (g ◦ g)(x)
for the following functions:
a) f (x) = 2x + 4, and g(x) = x − 5,
b) f (x) = x + 3, and x2 − 2x,
g(x) = √
c) f (x) = 2x2 − x − 6, and g(x) = 3x + 2,
1 1
d) f (x) = x+3 , and g(x) = √
x
− 3,
x+7
e) f (x) = (2x − 7)2 , and g(x) = 2 .
Exercise 6.6. Let f and g be the functions defined by the following table.
Complete the table given below.
x 1 2 3 4 5 6 7
f (x) 4 5 7 0 −2 6 4
g(x) 6 −8 5 2 9 11 2
f (x) + 3
4g(x) + 5
g(x) − 2f (x)
f (x + 3)
Exercise 6.7. Let f and g be the functions defined by the following table.
Complete the table by composing the given functions.
x 1 2 3 4 5 6
f (x) 3 1 2 5 6 3
g(x) 5 2 6 1 2 4
(g ◦ f )(x)
(f ◦ g)(x)
(f ◦ f )(x)
(g ◦ g)(x)
6.3. EXERCISES 85

Exercise 6.8. Let f and g be the functions defined by the following table.
Complete the table by composing the given functions.

x 0 2 4 6 8 10 12
f (x) 4 8 5 6 12 −1 10
g(x) 10 2 0 −6 7 2 8
(g ◦ f )(x)
(f ◦ g)(x)
(f ◦ f )(x)
(g ◦ g)(x)
Session 7

The inverse of a function

7.1 One-to-one functions

We have seen that some functions f may have the same outputs for different
inputs. For example for f (x) = x2 , the inputs x = 2 and x = −2 have the
same output f (2) = 4 and f (−2) = 4. A function is one-to-one, precisely
when this is not the case.
Definition 7.1. A function f is called one-to-one (or injective), if two different
inputs x1 6= x2 always have different outputs f (x1 ) 6= f (x2 ).
Example 7.2. As was noted above, the function f (x) = x2 is not one-to-one,
because, for example, for inputs 2 and −2, we have the same output
f (−2) = (−2)2 = 4, f (2) = 22 = 4.
On the other hand, g(x) = x3 is one-to-one, since, for example, for inputs −2
and 2, we have different outputs:
g(−2) = (−2)3 = −8, g(2) = 23 = 8.
The difference between the functions f and g can be seen from their graphs.
y = x2 y = x3

y0
y0

x −x0 x
−x0 x0 x0

−y0

86
7.1. ONE-TO-ONE FUNCTIONS 87

The graph of f (x) = x2 on the left has for different inputs (x0 and −x0 ) the
same output (y0 = (x0 )2 = (−x0 )2 ). This is shown in the graph since the
horizontal line at y0 intersects the graph at two different points. In general,
two inputs that have the same output y0 give two points on the graph which
also lie on the horizontal line at y0 .
Now, the graph of g(x) = x3 on the right intersects with a horizontal line
at some y0 only once. This shows that for two different inputs, we can never
have the same output y0 , so that the function g is one-to-one.

We can summarize the observation of the last example in the following

statement.

Observation 7.3 (Horizontal Line Test). A function is one-to-one exactly when

every horizontal line intersects the graph of the function at most once.
y = f (x)

y0

x
x0
Example 7.4. Which of the following are or represent one-to-one functions?
5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-3 -2 -1 0 1 2 3 4 -1 0 1 2 3 4 5
a) -1 b) -1

c) f (x) = −x3 + 6x2 − 13x + 12 d) f (x) = x3 − 2x2 + 3

Solution. We use the horizontal line test to see which functions are one-to-
one. For (a) and (b), we see that the functions are not one-to-one since there
88 SESSION 7. THE INVERSE OF A FUNCTION

is a horizontal line that intersects with the graph more than once:
5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-3 -2 -1 0 1 2 3 4 -1 0 1 2 3 4 5
a) -1 b) -1

For (c), using the calculator to graph the function f (x) = −x3 +6x2 −13x+12,
we see that all horizontal lines intersect the graph exactly once. Therefore
the function in part (c) is one-to-one. The function in part (d) however has
a graph that intersects some horizontal line in several points. Therefore
f (x) = x3 − 2x2 + 3 is not one-to-one:
5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-1 0 1 2 3 4 5 -2 -1 0 1 2 3 4
c) -1 d) -1

7.2 Inverse function

A function is one-to-one, when each output is determined by exactly one
input. Therefore we can construct a new function, called the inverse function,
where we reverse the roles of inputs and outputs. For example, when y = x3 ,
7.2. INVERSE FUNCTION 89

each y0 comes from exactly one x0 as shown in the picture below:

y

y0

x
x0

The inverse function assigns to the input y0 the output x0 .

Definition 7.5. Let f be a function with domain Df and range Rf , and assume
that f is one-to-one. The inverse of f is a function f −1 so that
f (x) = y means precisely that f −1 (y) = x.

input f output
x y
output f −1 input

Here the outputs of f are the inputs of f −1 , and the inputs of f are the outputs
of f −1 . Therefore, the inverse function f −1 has a domain equal to the range
of f , Df −1 = Rf , and f −1 has a range equal to the domain of f , Rf −1 = Df .
In short, when f is a function f : Df → Rf , then the inverse function f −1 is
a function f −1 : Rf → Df .
The inverse function reverses the roles of inputs and outputs.
Example 7.6. Find the inverse of the following function via algebra.
√
a) f (x) = 2x − 7 b) g(x) = x + 2
1
c) h(x) = x+4 d) j(x) = x+1
x+2
e) k(x) = (x − 2)2 + 3 for x ≥ 2
Solution. a) First, reverse the role of input and output in y = 2x − 7 by
exchanging the variables x and y. That is, we write x = 2y − 7. We need to
solve this for y:
(add 7) x+7
=⇒ x + 7 = 2y =⇒ y= .
2
90 SESSION 7. THE INVERSE OF A FUNCTION

Therefore, we obtain that the inverse of f is f −1 (x) = x+7

2
.
For the other
√ parts, we always exchange x and y and solve for y:
b) Write y = x + 2 and exchange x and y:
p
x = y + 2 =⇒ x2 = y + 2 =⇒ y = x2 − 2
=⇒ g −1 (x) = x2 − 2
1
c) Write y = x+4
and exchange x and y:
1 1 1
x= =⇒ y+4= =⇒ y = −4
y+4 x x
1
=⇒ h−1 (x) = −4
x
x+1
d) Write y = x+2
and exchange x and y:
y+1 ×(y+2)
x= =⇒ x(y + 2) = y + 1 =⇒ xy + 2x = y + 1
y+2
=⇒ xy − y = 1 − 2x =⇒ y(x − 1) = 1 − 2x
1 − 2x 1 − 2x
=⇒ y= =⇒ j −1 (x) = ,
x−1 x−1
e) Write y = (x − 2)2 + 3 and exchange x and y:
√
x = (y − 2)2 + 3 =⇒ x − 3 = (y − 2)2 =⇒ x − 3 = y − 2
√ √
=⇒ y =2+ x−3 =⇒ k −1 (x) = 2 + x − 3

The function in the last example is not one-to-one when allowing x to be

any real number. This is why we had to restrict the example to the inputs
x ≥ 2. We exemplify the situation in the following example.
Example 7.7. Note, that the function y = x2 can be restricted to a one-to-one
function by choosing the domain to be all non-negative numbers [0, ∞) or by
choosing the domain to be all non-positive numbers (−∞, 0].
y

x
7.2. INVERSE FUNCTION 91

Let f : [0, ∞) → [0, ∞) be the function f (x) = x2 , so that f has a domain

√
of all non-negative numbers. Then, the inverse is the function f −1 (x) = x.
On the other hand, we can take g(x) = x2 whose domain consists of all
non-positive numbers (−∞, 0], that is g : (−∞, 0] → [0, ∞). Then, the inverse
function g −1 must reverse domain and range, that is g −1 : [0, ∞) → (−∞, 0].
The inverse is obtained by exchanging x and y in y = x2 as follows:
√ √
x = y2 =⇒ y = ± x =⇒ g −1 (x) = − x.

Example 7.8. Restrict the function to a one-to-one function. Find the inverse
function, if possible.
1
a) f (x) = (x + 3)2 + 1, b) g(x) = , c) h(x) = x3 − 3x2 .
(x − 2)2

Solution. The graphs of f and g are displayed below.

4 y 4 y

3 3

2 2

1 1

0 x 0 x
-6 -5 -4 -3 -2 -1 0 1 2 -2 -1 0 1 2 3 4 5
a) -1 b) -1

a) The graph shows that f is one-to-one when restricted to all numbers

x ≥ −3, which is the choice we make to find an inverse function. Next, we
replace x and y in y = (x + 3)2 + 1 to give x = (y + 3)2 + 1. When solving
this for y, we must now remember that our choice of x ≥ −3 becomes y ≥ −3
after replacing x with y. We now solve for y.
√
x = (y + 3)2 + 1 =⇒ x − 1 = (y + 3)2 =⇒ y + 3 = ± x − 1
√
=⇒ y = −3 ± x − 1

Since we have chosen the restriction

√ of y ≥ −3, we use the expression with
the positive
√ sign, y = −3 + x − 1, so that the inverse function is f −1 (x) =
−3 + x − 1.
b) For the function g, the graph shows that we can restrict g to x > 2 to
obtain a one-to-one function. The inverse for this choice is given as follows.
92 SESSION 7. THE INVERSE OF A FUNCTION

1 1
Replacing x and y in y = (x−2)2
gives x = (y−2)2
, which we solve this for y
under the condition y > 2.
1 1 1
x= =⇒ (y − 2)2 = =⇒ y − 2 = ± √
(y − 2)2 x x
1 1
=⇒ y =2± √ =⇒ g −1(x) = 2 + √
x x
c) Finally, h(x) = x3 − 3x2 can be graphed with the calculator as follows:

The picture on the right shows that the approximation of the local minimum is
(approximately) at x = 2. Therefore, if we restrict h to all x ≥ 2, we obtain a
one-to-one function. We replace x and y in y = x3 − 3x2 , so that the inverse
function is obtained by solving the equation x = y 3 − 3y 2 for y. However,
this equation is quite complicated and solving it is beyond our capabilities
at this time. Therefore we simply say that h−1 (x) is that y ≥ 2 for which
y 3 − 3y 2 = x, and leave the example with this.
Let f be a one-to-one function. If f maps x0 to y0 = f (x0 ), then f −1
maps y0 to x0 . In other words, the inverse function is precisely the function
for which
f −1 (f (x0 )) = f −1 (y0 ) = x0 and f (f −1(y0 )) = f (x0 ) = y0 .

f
x0 y0
f −1

Observation 7.9. Let f and g be two one-to-one functions. Then f and g are
inverses of each other exactly when

f (g(x)) = x and g(f (x)) = x for all x.

In this case we write that g = f −1 and f = g −1.

7.2. INVERSE FUNCTION 93

Example 7.10. Are the following functions inverse to each other?

a) f (x) = 5x + 7, g(x) = x−7

5
,
3 3
b) f (x) = x−6 , g(x) = x + 6,
√
c) f (x) = x − 3, g(x) = x2 + 3.

Solution. We calculate the compositions f (g(x)) and g(f (x)).

x−7 x−7
a) f (g(x)) = f ( )=5· + 7 = (x − 7) + 7 = x,
5 5
(5x + 7) − 7 5x
g(f (x)) = g(5x + 7) = = = x,
5 5
3 3 3 x
b) f (g(x)) = f ( + 6) = 3 = 3 = 3 · = x,
x ( x + 6) − 6 x
3
3 3 x−6
g(f (x)) = g( )= 3 +6=3· + 6 = (x − 6) + 6 = x.
x−6 x−6
3

Using the Observation 7.9, we see that in both part (a) and (b), the functions
are inverse to each other. For part (c), we calculate for a general x in the
domain of g:
√
f (g(x)) = x2 + 3 − 3 6= x.
It is enough to show that for one composition (f ◦ g)(x) does not equal x to
conclude that f and g are not inverses. (It is not necessary to also calculate
the other composition g(f (x)).)

Warning 7.11. It is true that when we have one relation f (g(x)) = x, then we also have the other relation
g(f (x)) = x. Nevertheless, we recommend to always check both equations. The reason is that it is easy
to mistake one of the relations, when not being careful about the domain and range.
2
√ √ 2
√ = x and g(x) = − x. Then, naively, we would calculate f (g(x)) = (− x) = x
For example, let f (x)
and g(f (x)) = − x2 = −|x|, so that the first equation would say f and g are inverses, whereas the
second equation says they are not inverses.

We can resolve the apparent contradiction by restricting f to a one-to-one function. For example,
√ if we
take the domain of f to be all positive numbers or zero, Df = [0, ∞), then f (g(x)) = f (− x) which
√
is undefined, since f only takes non-negative inputs. Also, it is g(f (x)) = − x2 = −x. Therefore, the
functions f and g are not inverse to each other!

On the other hand, if√we restrict the function f (x) = x2 to all negative numbers and
√ zero, Df = (−∞, 0],
then f (g(x)) = (− x)2 = x, since √ now f is defined for the negative input − x. Also, for a negative
number x < 0, it is g(f (x)) = − x2 = −|x| = x. So, in this case, f and g are inverse to each other!
94 SESSION 7. THE INVERSE OF A FUNCTION

Our last observation concerns the graph of inverse functions. If f (x0 ) = y0

then f −1 (y0 ) = x0 , and the point P (x0 , y0 ) is on the graph of f , whereas the
point Q(y0 , x0 ) is on the graph of f −1 .
y
P
y0

x0 Q

x
x0 y0

We see that Q is the reflection of P along the diagonal y = x. Since this is

true for any point on the graph of f and f −1 , we have the following general
observation.
Observation 7.12. The graph of f −1 is the graph of f reflected along the
diagonal. y f

f −1

x
l
na
o
ag
di

Example 7.13. Find the graph of the inverse function of the function given
below.
5 y

0 x
-3 -2 -1 0 1 2 3 4 5 6 7
-1

a) -2 b) f (x) = (x + 1)3
7.3. EXERCISES 95

Solution. Carefully reflecting the graphs given in part (a) and (b) gives the
following solution. The function f (x) = (x + 1)3 in part (b) can be graphed
with a calculator first.
7 y

6
3 y
5

4 2

3
1
2
0 x
1
-3 -2 -1 0 1 2 3
0 x
-1
-3 -2 -1 0 1 2 3 4 5 6 7
-1
-2
-2

a) -3 b) -3

7.3 Exercises
Exercise 7.1. Use the horizontal line test to determine if the function is one-
to-one.
5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-2 -1 0 1 2 3 4 5 -1 0 1 2 3 4 5 6 7 8
a) -1 b) -1

c) f (x) = x2 + 2x + 5 d) f (x) = x2 − 14x + 29

x2
e) f (x) = x3 − 5x2 f) f (x) = xp
2 −3
√
g) f (x) = x + 2 h) f (x) = |x + 2|
Exercise 7.2. Find the inverse of the function f and check your solution.
a) f (x) = √
4x + 9 b) f (x) = −8x
√ −3
c) f (x) = x√ +8 d) f (x) = 3x + 7
e) f (x) = 6 · −x − 2 f) f (x) = x3
g) f (x) = (2x + 5)3 h) f (x) = 2 · x3 + 5
96 SESSION 7. THE INVERSE OF A FUNCTION

1 1 √1
i) f (x) = x
j) f (x) =x−1
k) f (x) = x−2
−5 x 3x
l) f (x) = 4−x m) f (x) = x+2 n) f (x) = x−6
x+2
o) f (x) = x+3 p) f (x) = 7−x
x−5
q) f given by the table below:

x 2 4 6 8 10 12
f (x) 3 7 1 8 5 2

Exercise 7.3. Restrict the domain of the function f in such a way that f
becomes a one-to-one function. Find the inverse of f with the restricted
domain.
a) f (x) = x2 b) f (x) = (x + 5)2 + 1
c) f (x) = |x| d) f (x) = |x − 4| − 2
1 −3
e) f (x) = x2
f) f (x) = (x+7)2
(x−3)4
g) f (x) = x4 h) f (x) = 10

Exercise 7.4. Determine whether the following functions f and g are inverse
to each other.

a) f (x) = x + 3, and g(x) = x − 3,

b) f (x) = −x − 4, and g(x) = 4 − x,
c) f (x) = 2x + 3, and g(x) = x − 23 ,
x+1
d) f (x) = 6x − 1, and g(x) = 6
,
√
e) f (x) = x3 − 5, and g(x) = 5 + 3
x,
1 1
f) f (x) = x−2
, and g(x) = x
+ 2.

Exercise 7.5. Draw the graph of the inverse of the function given below.
5 y

4 5 y

3 4

2 3

1 2

0 x 1
-2 -1 0 1 2 3 4 5 6
-1 0 x
-2 -1 0 1 2 3 4 5
a) -2 b) -1
7.3. EXERCISES 97
5 y

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6
c) -1

√
d) f (x) = x e) f (x) = x3 − 4
f) f (x) = 2x − 4 g) f (x) = 2x
1 1
h) f (x) = x−2 for x > 2 i) f (x) = x−2 for x < 2.
Review of functions and graphs

Exercise I.1. Find all solutions of the equation: 3x − 9 = 6

Exercise I.2. Find the equation of the line displayed below.

4 y

0 x
-4 -3 -2 -1 0 1 2 3 4
-1

-2

Exercise I.3. Find the solution of the equation: x3 − 3x2 + 2x − 2 = 0

Approximate your answer to the nearest hundredth.

Exercise I.4. Let f (x) = x2 − 2x + 5. Simplify the difference quotient

f (x+h)−f (x)
h
as much as possible.

Exercise I.5. Consider the following graph of a function f .

5 y

0 x
-1 0 1 2 3 4 5 6 7 8 9 10

Find: domain of f , range of f , f (3), f (5), f (7), f (9).

98
Exercise I.6. Find the formula of the graph displayed below.
3 y

0 x
-4 -3 -2 -1 0 1 2 3 4
-1

Exercise

I.7. Let f (x) = 5x + 4 and g(x) = x2 + 8x + 7. Find the quotient
f
g
(x) and state its domain.
√
Exercise I.8. Let f (x) = x2 + x − 3 and g(x) = 2x−3. Find the composition
(f ◦ g)(x) and state its domain.

Exercise I.9. Consider the assignments for f and g given by the table below.

x 2 3 4 5 6
f (x) 5 0 2 4 2
g(x) 6 2 3 4 1

Is f a function? Is g a function? Write the composed assignment for (f ◦ g)(x)

as a table.
1
Exercise I.10. Find the inverse of the function f (x) = 2x+5
.

99
 Part II

Polynomials and rational functions

100
Session 8

Dividing polynomials

We now start our discussion of specific classes of examples of functions. The

first class of functions that we discuss are polynomials and rational functions.
Let us first recall the definition of polynomials and rational functions.
Definition 8.1. A monomial is a number, a variable, or a product of numbers
and variables. A polynomial is a sum (or difference) of monomials.
Example 8.2. The following are examples of monomials:
√
5, x, 7x2 y, −12x3 y 2z 4 , 2 · a3 n2 xy
The following are examples of polynomials:
x2 + 3x − 7, 4x2 y 3 + 2x + z 3 + 4mn2 , −5x3 − x2 − 4x − 9, 5x2 y 4
In particular, every monomial is also a polynomial.
We are mainly interested in polynomials in one variable x, and consider
these as functions. For example, f (x) = x2 + 3x − 7 is such a function.
Definition 8.3. A polynomial is a function f of the form
f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 ,
for some real (or complex) numbers a0 , a1 , . . . , an . The domain of a polynomial
f is all real numbers (see our standard convention 3.4).
The numbers a0 , a1 , . . . , an are called coefficients. For each k, the number
ak is the coefficient of xk . The number an is called the leading coefficient
and n is the degree of the polynomial.
The zeros of a polynomial are usually referred to as roots. Therefore x is
a root of a polynomial f precisely when f (x) = 0.

101
102 SESSION 8. DIVIDING POLYNOMIALS

Definition 8.4. A rational function is a fraction of two polynomials f (x) =

g(x)
h(x)
, where g(x) and h(x) are both polynomials. The domain of f is all real
numbers for which the denominator h(x) is not zero:
Df = { x | h(x) 6= 0 }.
Example 8.5. The following are examples of rational functions:
−3x2 + 7x − 5 1
f (x) = 3 , f (x) = , f (x) = −x2 + 3x + 5.
2x + 4x2 + 3x + 1 x

8.1 Long division

We now show how to divide two polynomials. The method is similar to the
long division of natural numbers. Our first example shows the procedure in
full detail.
Example 8.6. Divide the following fractions via long division:
3571 x3 + 5x2 + 4x + 2
a) b)
11 x+3
Solution. a) Recall the procedure for long division of natural numbers:
324
11 3571
−33
271
−22
51
−44
7 = remainder
The steps above are performed as follows. First, we find the largest multiple
of 11 less or equal to 35. The answer 3 is written as the first digit on the
top line. Multiply 3 times 11, and subtract the answer 33 from the first two
digits 35 of the dividend. The remaining digits 71 are copied below to give
271. Now we repeat the procedure, until we arrive at the remainder 7. In
short, what we have shown is that:
3571 7
3571 = 324 · 11 + 7 or alternatively, = 324 + .
11 11
8.1. LONG DIVISION 103

b) We repeat the steps from part (a) as follows. First, write the dividend
and divisor in the above format:
x + 3 x3 +5x2 +4x +2
Next, consider the highest term x3 of the dividend and the highest term x of
3
the divisor. Since xx = x2 , we start with the first term x2 of the quotient:
x2
Step 1:
x + 3 x3 +5x2 +4x +2
Multiply x2 by the divisor x + 3 and write it below the dividend:
x2
Step 2: x + 3 x3 +5x2 +4x +2
x3 +3x2
Since we need to subtract x3 + 3x2 , so we equivalently add its negative (don’t
forget to distribute the negative):
x2
x+3 x3 +5x2 +4x +2
Step 3:
−(x3 +3x2 )
2x2
Now, carry down the remaining terms of the dividend:
x2
x+3 x3 +5x2 +4x +2
Step 4:
−(x3 +3x2 )
2x2 +4x +2
Now, repeat steps 1-4 for the remaining polynomial 2x2 +4x+2. The outcome
after going through steps 1-4 is the following:
2x2
x2 +2x (add x
= 2x)
3
x+3 x +5x2 +4x +2
−(x3 +3x2 )
2x2 +4x +2
−(2x2 +6x) (multiply 2x by (x + 3))
−2x +2 (subtract from the above)
104 SESSION 8. DIVIDING POLYNOMIALS

Since x can be divided into −2x, we can proceed with the above steps 1-4
one more time. The outcome is this:
x2 +2x −2
x+3 x3 +5x2 +4x +2
−(x3 +3x2 )
2x2 +4x +2
−(2x2 +6x)
−2x +2
−(−2x −6)
8 = remainder

Note that now x cannot be divided into 8 so we stop here. The final term 8 is
called the remainder. The term x2 + 2x − 2 is called the quotient. In analogy
with our result in part (a), we can write our conclusion as:

x3 + 5x2 + 4x + 2 = (x2 + 2x − 2) · (x + 3) + 8.

Alternatively, we could also divide this by (x + 3) and write it as:

x3 + 5x2 + 4x + 2 8
= x2 + 2x − 2 + .
x+3 x+3

Note 8.7. Just as with a division operation involving numbers, when dividing
f (x)
g(x)
, f (x) is called the dividend and g(x) is called the divisor. As a result
of dividing f (x) by g(x) via long division with quotient q(x) and remainder
r(x), we can write
f (x) r(x)
= q(x) + . (8.1)
g(x) g(x)
If we multiply this equation by g(x), we obtain the following alternative ver-
sion:
f (x) = q(x) · g(x) + r(x) (8.2)
Example 8.8. Divide the following fractions via long division.
x2 +4x+5 x4 +3x3 −5x+1
a) x−4
, b) x+1
,
4x3 +2x2 +6x+18 x3 +x2 +2x+1
c) 2x+3
, d) x2 +3x+1
.
8.1. LONG DIVISION 105

Solution. For part (a), we calculate:

x +8
2
x−4 x +4x +5
−(x2 −4x)
8x +5
−(8x −32)
37
Therefore, x2 + 4x + 5 = (x + 8) · (x − 4) + 37.
Now, for part (b), there is no x2 term in the dividend. This can be resolved
by adding +0 x2 to the dividend:
x3 +2x2 −2x −3
4
x+1 x +3x3 +0x2 −5x +1
−(x4 +x3 )
2x3 +0x2 −5x +1
−(2x3 +2x2 )
−2x2 −5x +1
−(−2x2 −2x)
−3x +1
−(−3x −3)
4
Therefore, we showed:
x4 + 3x3 − 5x + 1 4
= x3 + 2x2 − 2x − 3 + .
x+1 x+1
For (c), calculate:
2x2 −2x +6
3
2x + 3 4x +2x2 +6x +18
−(4x3 +6x2 )
−4x2 +6x +18
−(−4x2 −6x)
12x +18
−(12x +18)
0
106 SESSION 8. DIVIDING POLYNOMIALS

Since the remainder is zero, we succeeded in factoring 4x3 + 2x2 + 6x + 18:

4x3 + 2x2 + 6x + 18 = (2x2 − 2x + 6) · (2x + 3)

d) The last example has a divisor that is a polynomial of degree 2. There-

fore, the remainder is not a number, but a polynomial of degree 1.

x −2
2 3 2
x + 3x + 1 x +x +2x +1
−(x3 +3x2 +x)
−2x2 +x +1
−(−2x2 −6x −2)
7x +3

Here, the remainder is r(x) = 7x + 3.

x3 + x2 + 2x + 1 7x + 3
2
= x−2+ 2
x + 3x + 1 x + 3x + 1

Note 8.9. The divisor g(x) is a factor of f (x) exactly when the remainder
r(x) is zero, that is:

f (x) = q(x) · g(x) ⇐⇒ r(x) = 0.

For example, in the above Example 8.8, only part (c) results in a factorization
of the dividend, since this is the only part with remainder zero.

8.2 Dividing by (x − c)
We now restrict our attention to the case where the divisor is g(x) = x − c
for some real number c. In this case, the remainder r of the division f (x) by
g(x) is a real number. We make the following observations.

Observation 8.10. Assume that g(x) = x − c, and the long division of f (x)
by g(x) has remainder r, that is,

Assumption: f (x) = q(x) · (x − c) + r.

8.2. DIVIDING BY (X − C) 107

When we evaluate both sides in the above equation at x = c we see that

f (c) = q(c) · (c − c) + r = q(c) · 0 + r = r. In short:

The remainder when dividing f (x) by (x − c) is r = f (c). (8.3)

In particular:

f (c) = 0 ⇐⇒ g(x) = x − c is a factor of f (x). (8.4)

The above statement (8.3) is called the remainder theorem, and (8.4) is called
the factor theorem.
Example 8.11. Find the remainder of dividing f (x) = x2 + 3x + 2 by

a) x − 3, b) x + 4, c) x + 1, d) x − 21 .

Solution. By Observation 8.10, we know that the remainder r of the division

by x − c is f (c). Thus, the remainder for part (a), when dividing by x − 3 is

r = f (3) = 32 + 3 · 3 + 2 = 9 + 9 + 2 = 20.

For (b), note that g(x) = x + 4 = x − (−4), so that we take c = −4 for

our input giving that r = f (−4) = (−4)2 + 3 · (−4) + 2 = 16 − 12 + 2 = 6.
Similarly, the other remainders are:

c) r = f (−1) = (−1)2 + 3 · (−1) + 2 = 1 − 3 + 2 = 0,

 1   1 2 1 1 3 1+6+8 15
d) r= f = + ·3+2= + +2= = .
2 2 2 4 2 4 4
Note that in part (c), we found a remainder 0, so that (x + 1) is a factor of
f (x).
Example 8.12. Determine whether g(x) is a factor of f (x).

a) f (x) = x3 + 2x2 + 5x + 1, g(x) = x − 2,

b) f (x) = x4 + 4x3 + x2 + 18, g(x) = x + 3,
c) f (x) = x5 + 3x2 + 7, g(x) = x + 1.

Solution. a) We need to determine whether 2 is a root of f (x) = x3 + 2x2 +

5x + 1, that is, whether f (2) is zero.

f (2) = 23 + 2 · 22 + 5 · 2 + 1 = 8 + 8 + 10 + 1 = 27.
108 SESSION 8. DIVIDING POLYNOMIALS

Since f (2) = 27 6= 0, we see that g(x) = x − 2 is not a factor of f (x).

b) Now, g(x) = x + 3 = x − (−3), so that we calculate:

f (−3) = (−3)4 + 4 · (−3)3 + (−3)2 − 18 = 81 − 108 + 9 + 18 = 0.

Since the remainder is zero, we see that x+3 is a factor of x4 +4·x3 +x2 +18.
Therefore, if we wanted to find the other factor, we could use long division to
obtain the quotient.
c) Finally, we have:

f (−1) = (−1)5 + 3 · (−1)2 + 7 = −1 + 3 + 7 = 9.

g(x) = x + 1 is not a factor of f (x) = x5 + 3x2 + 7.

Example 8.13.
a) Show that −2 is a root of f (x) = x5 − 3 · x3 + 5x2 − 12, and use this to
factor f .

b) Show that 5 is a root of f (x) = x3 − 19x − 30, and use this to factor f
completely.
Solution. a) First, we calculate that −2 is a root.

f (−2) = (−2)5 − 3 · (−2)3 + 5 · (−2)2 + 12 = −32 + 24 + 20 − 12 = 0.

So we can divide f (x) by g(x) = x − (−2) = x + 2:

x4 −2x3 +x2 +3x −6

5
x+2 x +0x4 −3x3 +5x2 +0x −12
−(x5 +2x4 )
−2x4 −3x3 +5x2 +0x −12
−(−2x4 −4x3 )
x3 +5x2 +0x −12
−(x3 +2x2 )
3x2 +0x −12
−(3x2 +6x)
−6x −12
−(−6x −12)
0
8.3. OPTIONAL SECTION: SYNTHETIC DIVISION 109

So we factored f (x) as f (x) = (x4 − 2x3 + x2 + 3x − 6) · (x + 2).

b) Again we start by calculating f (5):
f (5) = 53 − 19 · 5 − 30 = 125 − 95 − 30 = 0.
Long division by g(x) = x − 5 gives:
x2 +5x +6
3
x−5 x +0x2 −19x −30
−(x3 −5x2 )
5x2 −19x −30
−(5x2 −25x)
6x −30
−(6x −30)
0
Thus, x3 − 19x − 30 = (x2 + 5x + 6) · (x − 5). To factor f completely, we also
factor x2 + 5x + 6.
f (x) = (x2 + 5x + 6) · (x − 5) = (x + 2) · (x + 3) · (x − 5).

8.3 Optional section: Synthetic division

When dividing a polynomial f (x) by g(x) = x−c, the actual calculation of the
long division has a lot of unnecessary repetitions, and we may want to reduce
this redundancy as much as possible. In fact, we can extract the essential
part of the long division, the result of which is called synthetic division.
5x3 +7x2 +x+4
Example 8.14. Our first example is the long division of x+2
.
5x2 −3x +7
x+2 5x3 +7x 2
+x +4
−(5x3 2
+10x )
−3x2 +x +4
2
−(−3x −6x)
7x +4
−(7x +14)
−10
110 SESSION 8. DIVIDING POLYNOMIALS

Here, the first term 5x2 of the quotient is just copied from the first term of
the dividend. We record this together with the coefficients of the dividend
5x3 + 7x2 + x + 4 and of the divisor x + 2 = x − (−2) as follows:

5 7 1 4 (dividend (5x3 + 7x2 + x + 4))

−2 (divisor (x − (−2)))
5 (quotient)

The first actual calculation is performed when multiplying the 5x2 term with
2, and subtracting it from 7x2 . We record this as follows.

5 7 1 4
−2 −10 (−10 is the product of −2 · 5)
5 −3 (−3 is the sum 7 + (−10))

Similarly, we obtain the next step by multiplying the 2x by (−3) and sub-
tracting it from 1x. Therefore, we get

5 7 1 4
−2 −10 6 (6 is the product of −2 ·( −3))
5 −3 7 (7 is the sum 1 + (6))

The last step multiplies 7 times 2 and subtracts this from 4. In short, we write:

5 7 1 4
−2 −10 6 −14 (−14 is the product of −2 ·( 7) )
5 −3 7 −10 (−10 is the sum 4 + (−14))

The answer can be determined from these coefficients. The quotient is 5x2 −
3x + 7, and the remainder is −10.

Example 8.15. Find the following quotients via synthetic division.

4x3 − 7x2 + 4x − 8 x4 − x2 + 5
a) , b) .
x−4 x+3
Solution. a) We need to perform the synthetic division.

4 −7 4 −8
4 16 36 160
4 9 40 152
8.4. EXERCISES 111

Therefore we have
4x3 − 7x2 + 4x − 8 152
= 4x2 + 9x + 40 + .
x−4 x−4

Similarly, we calculate part (b). Note that some of the coefficients are
now zero.
1 0 −1 0 5
−3 −3 9 −24 72
1 −3 8 −24 77
We obtain the following result.

x4 − x2 + 5 77
= x3 − 3x2 + 8x − 24 +
x+3 x+5

Note 8.16. Synthetic division only works when dividing by a polynomial of

the form x − c. Do not attempt to use this method to divide by other forms
like x2 + 2.

8.4 Exercises
Exercise 8.1. Divide by long division.
x3 −4x2 +2x+1 x3 +6x2 +7x−2 x2 +7x−4
a) x−2
, b) x+3
, c) x+1
,
x3 +3x2 +2x+5 2x3 +x2 +3x+5 2x4 +7x3 +x+3
d) x+2
, e) x−1
, f) x+5
,
2x4 −31x2 −13 x3 +27 3x4 +7x3 +5x2 +7x+4
g) x−4
, h) x+3
, i) 3x+1
,
8x3 +18x2 +21x+18 x3 +3x2 −4x−5 x5 +3x4 −20
j) 2x+3
, k) x2 +2x+1
, l) x2 +3
.

Exercise 8.2. Find the remainder when dividing f (x) by g(x).

a) f (x) = x3 + 2x2 + x − 3, g(x) = x − 2,

b) f (x) = x3 − 5x + 8, g(x) = x − 3,
c) f (x) = x5 − 1, g(x) = x + 1,
d) f (x) = x5 + 5x2 − 7x + 10, g(x) = x + 2.
112 SESSION 8. DIVIDING POLYNOMIALS

Exercise 8.3. Determine whether the given g(x) is a factor of f (x). If so,
name the corresponding root of f (x).

a) f (x) = x2 + 5x + 6, g(x) = x + 3,
b) f (x) = x3 − x2 − 3x + 8, g(x) = x − 4,
c) f (x) = x4 + 7x3 + 3x2 + 29x + 56, g(x) = x + 7,
d) f (x) = x999 + 1, g(x) = x + 1.

Exercise 8.4. Check that the given numbers for x are roots of f (x) (see
Observation 8.10). If the numbers x are indeed roots, then use this information
to factor f (x) as much as possible.

a) f (x) = x3 − 2x2 − x + 2, x = 1,
b) f (x) = x3 − 6x2 + 11x − 6, x = 1, x = 2, x = 3,
c) f (x) = x3 − 3x2 + x − 3, x = 3,
d) f (x) = x3 + 6x2 + 12x + 8, x = −2,
e) f (x) = x3 + 13x2 + 50x + 56, x = −3, x = −4,
f) f (x) = x3 + 3x2 − 16x − 48, x = 2, x = −4,
g) f (x) = x5 + 5x4 − 5x3 − 25x2 + 4x + 20, x = 1, x = −1,
x = 2, x = −2.

Exercise 8.5. Divide by using synthetic division.

2x3 +3x2 −5x+7 4x3 +3x2 −15x+18 x3 +4x2 −3x+1
a) x−2
, b) x+3
, c) x+2
,
x4 +x3 +1 x5 +32 x3 +5x2 −3x−10
d) x−1
, e) x+2
, f) x+5
.
Session 9

Graphing polynomials

9.1 Graphs of polynomials

We now discuss the shape of the graphs of polynomial functions. Recall that
a polynomial function f of degree n is a function of the form

f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 .

Observation 9.1. We already know from Section 2.1, that the graphs of poly-
nomials f (x) = ax + b of degree 1 are straight lines.

y = 2x + 3 y = −2x + 3
y y

x x

Polynomials of degree 1 have only one root.

It is also easy to sketch the graphs of functions f (x) = xn .

113
114 SESSION 9. GRAPHING POLYNOMIALS

Observation 9.2. Graphing y = x2 , y = x3 , y = x4 , y = x5 , we obtain:

y = x2 y = x3 y = x4 y = x5

From this, we see that the shape of the graph of f (x) = xn depends on n
being even or odd.
3 y 3 y

2 2

1 1

0 x 0 x
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 -1

-2 -2

-3 -3

n n
y = x , for n even y = x , for n odd
If x approaches ±∞, If x approaches ±∞,
=⇒ y approaches +∞. =⇒ y approaches ±∞.

We now make some observations about graphs of general polynomials of

degrees 2, 3, 4, 5, and, more generally, of any degree n. In particular, we will
be interested in the number of roots and the number of extrema (that is the
number of maxima or a minima) of the graph of a polynomial f .

Observation 9.3. Let f (x) = ax2 + bx + c be a polynomial of degree 2. The

graph of f is a parabola.

• f has at most 2 roots. f has one extremum (that is one maximum or

minimum).
3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-2 -1 0 1 2 3 4 5 -2 -1 0 1 2 3 4 5 -2 -1 0 1 2 3 4 5
-1 -1 -1

-2 -2 -2

-3 -3 -3
9.1. GRAPHS OF POLYNOMIALS 115

• If a > 0 then f opens upwards, if a < 0 then f opens downwards.

y = x2 − 4x + 2 y = −x2 + 4x − 2
3 y 3 y

2 2

1 1

0 x 0 x
-2 -1 0 1 2 3 4 5 -2 -1 0 1 2 3 4 5
-1 -1

-2 -2

-3 -3

Observation 9.4. Let f (x) = ax3 + bx2 + cx + d be a polynomial of degree 3.

The graph may change its direction at most twice.
• f has at most 3 roots. f has at most 2 extrema.
4 y 4 y 4 y

3 3 3

2 2 2

1 1 1

0 x 0 x 0 x
-1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6
-1 -1 -1

-2 -2 -2

-3 -3 -3

-4 -4 -4

3 roots, 2 extrema 2 roots, 2 extrema 1 root, 0 extrema

• If a > 0 then f (x) approaches +∞ when x approaches +∞ (that is,

f (x) gets large when x gets large), and f (x) approaches −∞ when x
approaches −∞. If a < 0 then f (x) approaches −∞ when x approaches
+∞, and f (x) approaches +∞ when x approaches −∞.

y = x3 − 6x2 + 11x − 4 y = −x3 + 6x2 − 11x + 7

5 y 5 y

4 4

3 3

2 2

1 1

0 x 0 x
-2 -1 0 1 2 3 4 5 6 -2 -1 0 1 2 3 4 5 6
-1 -1

-2 -2
116 SESSION 9. GRAPHING POLYNOMIALS

Above, we have a first instance of a polynomial of degree n which “changes

its direction” one more time than a polynomial of one lesser degree n − 1.
Below, we give examples of this phenomenon for higher degrees as well.

Observation 9.5. Let f (x) = ax4 + bx3 + cx2 + dx + e be a polynomial of

degree 4.

• f has at most 4 roots. f has at most 3 extrema. If a > 0 then f opens

upwards, if a < 0 then f opens downwards.

y = x4 − 6.5x3 + 12.5x2 − 7x y = x4 − 4x3 + 5x2 − 1 y = −x4 + 11x3 − 44x2 + 76x − 50

3 y 5 y 2 y

2 4 1

1 3 0 x
-1 0 1 2 3 4 5 6
0 x 2 -1
-2 -1 0 1 2 3 4 5
-1 1 -2

-2 0 x -3
-3 -2 -1 0 1 2 3 4
-3 -1 -4

-4 -2 -5

4 roots, 3 extrema 2 roots, 1 extremum 0 roots, 3 extrema

Observation 9.6. Let f be a polynomial of degree 5.

• f has at most 5 roots. f has at most 4 extrema. If a > 0 then f (x)

approaches +∞ when x approaches +∞, and f (x) approaches −∞
when x approaches −∞. If a < 0 then f (x) approaches −∞ when x
approaches +∞, and f (x) approaches +∞ when x approaches −∞.

y = −x5
4 y 6 y 4 y

3 5 3

2 4 2

1 3 1

0 x 2 0 x
-1 0 1 2 3 4 5 6 -3 -2 -1 0 1 2 3
-1 1 -1

-2 0 x -2
-1 0 1 2 3 4 5
-3 -1 -3

-4 -2 -4

5 roots, 4 extrema 2 roots, 4 extrema 1 root, 0 extrema

9.1. GRAPHS OF POLYNOMIALS 117

Observation 9.7.

• Let f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 be a polynomial of

degree n. Then f has at most n roots, and at most n − 1 extrema.

degree 2 degree 3 degree 4

3 y 3 y 3 y

2 2 2

1 1 1

0 x 0 x 0 x
-1 0 1 2 3 4 5 -1 0 1 2 3 4 5 -1 0 1 2 3 4 5
-1 -1 -1

-2 -2 -2

-3 -3 -3

• Assume the degree of f is even n = 2, 4, 6, . . . . If an > 0, then the poly-

nomial opens upwards. If an < 0 then the polynomial opens downwards.

f (x) = x4 f (x) = −x4

3 y 3 y

2 2

1 1

0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1

-2 -2

-3 -3

• Assume the degree of f is odd, n = 1, 3, 5, . . . . If an > 0, then f (x)

approaches +∞ when x approaches +∞, and f (x) approaches −∞
as x approaches −∞. If an < 0, then f (x) approaches −∞ when x
approaches +∞, and f (x) approaches +∞ as x approaches −∞.

f (x) = x5 f (x) = −x5

3 y 3 y

2 2

1 1

0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1

-2 -2

-3 -3
118 SESSION 9. GRAPHING POLYNOMIALS

• The domain of a polynomial f is all real numbers, and f is continuous

for all real numbers (there are no jumps in the graph). The graph of
f has no horizontal or vertical asymptotes, no discontinuities (jumps in
the graph), and no corners. Furthermore, f (x) approaches ±∞ when x
approaches ±∞. Therefore, the following graphs cannot be graphs of
polynomials.

5 y 5 y 5 y 5 y

4 4 4 4

3 3 3 3

2 2 2 2

1 1 1 1

0 x 0 x 0 x 0 x
-1 0 1 2 3 4 5 -1 0 1 2 3 4 5 -1 0 1 2 3 4 5 -1 0 1 2 3 4 5
-1 -1 -1 -1

vertical discontinuity corner horizontal

asymptote / pole asymptote

Example 9.8. Which of the following graphs could be the graphs of a poly-
nomial? If the graph could indeed be a graph of a polynomial then determine
a possible degree of the polynomial.

4 y 6 y 5 y

3 5 4

2 4 3

1 3 2

0 x 2 1
-3 -2 -1 0 1 2 3 4
-1 1 0 x
0 1 2 3 4 5 6
-2 0 x -1
-1 0 1 2 3 4 5
a) -3 b) -1 c)-2

5 y 4 y

4 3

3 2

2 1

1 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
d) -4 -3 -2 -1
0
0 1 2 3 4 5
x
6
e) -1
9.1. GRAPHS OF POLYNOMIALS 119

Solution.
a) Yes, this could be a polynomial. The degree could be, for example, 4.
b) No, since the graph has a pole.
c) Yes, this could be a polynomial. A possible degree would be degree 3.
d) No, since the graph has a corner.
e) No, since f (x) does not approach ∞ or −∞ as x approaches ∞. (In fact
f (x) approaches 0 as x approaches ±∞ and we say that the function (or
graph) has a horizontal asymptote y = 0.)

Example 9.9. Identify the graphs of the polynomials in (a), (b) and (c) with
the functions (i), (ii), and (iii).
4 y 4 y 4 y

3 3 3

2 2 2

1 1 1

0 x 0 x 0 x
0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6
-1 -1 -1

a)-2 b)-2 c)-2

i) f (x) = −x3 + 9x2 − 27x + 29

ii) f (x) = −x2 + 6x − 7
iii) f (x) = x4 − 12x3 + 54x2 − 108x + 83
Solution. The only odd degree function is (i) and it must therefore correspond
to graph (c). For (ii), since the leading coefficient is negative, we know that
the function opens downwards, so that it corresponds to graph (a). Finally
(iii) opens upwards, since the leading coefficient is positive, so that its graph
is (b).

When graphing a function, we need to make sure to draw the function

in a window that expresses all the interesting features of the graph. More
precisely, we need to make sure that the graph includes all essential parts of
the graph, such as all intercepts (both x-intercepts and y-intercept), all roots,
all asymptotes (this will be discussed in the next part), and the long range
behavior of the function (that is how the function behaves when x approaches
±∞). We also want to include all extrema (that is all maxima and minima)
of the function.
120 SESSION 9. GRAPHING POLYNOMIALS

Example 9.10. Graph of the given function with the TI-84. Include all extrema
and intercepts of the graph in your viewing window.

a) f (x) = −x3 + 26x2 − 129x + 175

b) f (x) = .1x4 − 2.4x2 + 6.4x − 35
c) f (x) = −5x3 + 75x2 − 374.998x + 630
d) f (x) = −.01x4 + .4x3 + .0025x2 − 160x + 1600

Solution. a) The graph in the standard window looks as follows:

However, since the function is of degree 3, this cannot be the full graph, as
f (x) has to approach −∞ when x approaches ∞. Zooming out, and rescaling
appropriately for the following setting the following graph.

b) The graph in the standard window is drawn on the left below. After
rescaling to −10 ≤ x ≤ 10 and −100 ≤ y ≤ 30 we obtain the graph on the
right.

c) The graph in the standard window is drawn on the left. Zooming to the
interesting part of the graph, we obtain the graph on the right. (Below, we
9.2. FINDING ROOTS OF A POLYNOMIAL WITH THE TI-84 121

have chosen a window of 4.94 ≤ x ≤ 5.06 and 5.00995 ≤ y ≤ 5.01005.)

d) The standard window (see left graph) shows an empty coordinate system
without any part of the graph. However, zooming out to −30 ≤ x ≤ 40 and
−1000 ≤ y ≤ 4000, we obtain the middle graph. There is another interesting
part of the graph displayed on the right, coming from zooming to the plateau
on the right (here shown at 19.2 ≤ x ≤ 20.8 and 0.9 ≤ y ≤ 1.1).

9.2 Finding roots of a polynomial with the TI-84

We have seen that the roots are important features of a polynomial. Recall
that the roots of the polynomial f are those x0 for which f (x0 ) = 0. These
are, of course, precisely the x-intercepts of the graph. Therefore we may use
the graph of a polynomial for finding its roots as we did in section 4.
Example 9.11. Find the roots of the polynomial from its graph.
a) f (x) = x3 − 7x2 + 14x − 8, b) f (x) = −x3 + 8x2 − 21x + 18,
c) f (x) = x4 + 3x3 − x + 6, d) f (x) = x5 − 3x4 − 8x3 + 24x2 + 15x − 45.
Solution. a) We start by graphing the polynomial f (x) = x3 − 7x2 + 14x − 8.
122 SESSION 9. GRAPHING POLYNOMIALS

The graph suggests that the roots are at x = 1, x = 2, and x = 4. This may
easily be checked by looking at the function table.

Since the polynomial is of degree 3, there cannot be any other roots.

b) Graphing f (x) = −x3 + 8x2 − 21x + 18 with the calculator shows the
following display.

Zooming into the graph reveals that there are in fact two roots, x = 2 and
x = 3, which can be confirmed from the table.

Note, that the root x = 3 only “touches” the x-axis. This is due to the fact
that x = 3 appears as a multiple root. Indeed, since 3 is a root, we can divide
f (x) by x − 3 without remainder and factor the resulting quotient to see that
that
f (x) = (x − 2)(x − 3)(x − 3) = (x − 2)(x − 3)2 .
We say that x = 3 is a root of multiplicity 2.
c) The graph of f (x) = x4 +3x3 −x+6 in the standard window is displayed
as follows.
9.2. FINDING ROOTS OF A POLYNOMIAL WITH THE TI-84 123

Since this is a polynomial of degree 4, all of the essential features are already
displayed in the above graph. The roots can be seen by zooming into the
graph.

From the table and the graph we see that there is a root at x = −2 and
another root at between −3 and −2. Finding the exact value of this second
root can be quite difficult, and we will say more about this in section 10.2
below. At this point, we can only approximate the root with the “zero” function
from the “calc” menu:

d) The graph of f (x) = x5 − 3x4 − 8x3 + 24x2 + 15x − 45 in the standard

window is displayed below.

Zooming into the x-axis, and checking the table shows that the only obvious
root is x = 3.

The other four roots are more difficult to find. However they can be approxi-
mated using the “zero” function from the “calc” menu.
124 SESSION 9. GRAPHING POLYNOMIALS

It will often be necessary to find the roots of a quadratic polynomial. For

this reason, we recall the well-known quadratic formula.

Theorem 9.12 (The quadratic formula).

The solutions of the equation ax2 + bx + c = 0 for some real
numbers a, b, and c are given by
√ √
−b+ b2 −4ac −b− b2 −4ac
x1 = 2a
, and x2 = 2a
.

We may combine the two solutions x1 and x2 and simply write this as:
√
−b ± b2 − 4ac
x1/2 =
2a

Note 9.13. We may always use the roots x1 and x2 of a quadratic polynomial
f (x) = ax2 + bx + c from the quadratic formula and rewrite the polynomial as
√ √
2
 −b + b2 − 4ac  −b − b2 − 4ac 
f (x) = ax + bx + c = a · x − x− .
2a 2a

9.3 Optional section: Graphing polynomials by hand

In this section we will show how to sketch the graph of a factored polynomial
without the use of a calculator.

Example 9.14. Sketch the graph of the following polynomial without using
the calculator:
p(x) = −2(x + 10)3 (x + 9)x2 (x − 8)

Solution. Note, that on the calculator it is impossible to get a window which

will give all of the features of the graph (while focusing on getting the max-
imum in view the other features become invisible). We will sketch the graph
by hand so that some of the main features are visible. This will only be a
sketch and not the actual graph up to scale. Again, the graph can not be
drawn to scale while being able to see the features.
We first start by putting the x-intercepts on the graph in the right order,
but not to scale necessarily. Then note that

p(x) = −2x7 + . . . (lower terms) ≈ −2x7 for large |x|.

9.3. OPTIONAL SECTION: GRAPHING POLYNOMIALS BY HAND 125

This is the leading term of the polynomial (if you expand p it is the term with
the largest power) and therefore dominates the polynomial for large |x|. So
the graph of our polynomial should look something like the graph of y = −2x7
on the extreme left and right side.

−10 −9 8

Now we look at what is going on at the roots. Near each root the factor
corresponding to that root dominates. So we have
for x ≈ p(x) ≈
−10 C1 (x + 10)3 cubic
−9 C2 (x + 9) line
0 C3 x2 parabola
8 C4 (x − 8) line
where C1 , C2 , C3 , and C4 are constants which can, but need not, be calculated.
For example, whether or not the parabola near 0 opens up or down will depend
on whether the constant C3 = −2·(0+10)3(0+9)(0−8) is negative or positive.
In this case C3 is positive so it opens upward but we will not use this fact to
graph. We will see this independently which is a good check on our work.
Starting from the left of our graph where we had determined the behavior
for large negative x, we move toward the left most zero, −10. Near −10 the
graph looks cubic so we imitate a cubic curve as we pass through (−10, 0).

−10 −9 8
126 SESSION 9. GRAPHING POLYNOMIALS

Now we turn and head toward the next zero, −9. Here the graph looks like
a line, so we pass through the point (−9, 0) as a line would.

−10 −9 8

Now we turn and head toward the root 0. Here the graph should look like a
parabola. So we form a parabola there. (Note that as we had said before the
parabola should be opening upward here–and we see that it is).

−10 −9 8

Now we turn toward the final zero 8. We pass through the point (8, 0) like a
line and we join (perhaps with the use of an eraser) to the large x part of the
graph. If this doesn’t join nicely (if the graph is going in the wrong direction)
then there has been a mistake. This is a check on your work. Here is the
final sketch.

−10 −9 8
9.4. EXERCISES 127

What can be understood from this sketch? Questions like ’when is p(x) >
0?’ can be answered by looking at the sketch. Further, the general shape of
the curve is correct so other properties can be concluded. For example, p has
a local minimum between x = −10 and x = −9 and a local maximum between
x = −9 and x = 0, and between x = 0 and x = 8. The exact point where the
function reaches its maximum or minimum can not be decided by looking at
this sketch. But it will help to decide on an appropriate window so that the
minimum or maximum finder on the calculator can be used.

9.4 Exercises
Exercise 9.1. Which of the graphs below could be the graphs of a polynomial?
5 y 5 y 6 y

4 4 5

3 3 4

2 2 3

1 1 2

0 x 0 x 1
-1 0 1 2 3 4 5 -1 0 1 2 3 4 5
a) -1 b) -1 c) -4 -3 -2 -1
0
0 1 2 3
x
4
5 y 5 y 3 y

4 4 2

3 3 1

2 2 0 x
0 1 2 3 4 5 6 7 8
1 1 -1

0 x 0 x -2
-1 0 1 2 3 4 5 -1 0 1 2 3 4 5
d) -1 e) -1 f)-3

Exercise 9.2. Identify each of the graphs (a)-(e) with its corresponding as-
signment from (i)-(vi) below.
5 y 5 y 4 y

4 4 3

3 3 2

2 2 1

1 1 0 x
-3 -2 -1 0 1 2 3
0 x 0 x -1
-1 0 1 2 3 4 5 -1 0 1 2 3 4 5
a) -1 b) -1 c) -2
128 SESSION 9. GRAPHING POLYNOMIALS
3 y 4 y

2 3

1 2

0 x 1
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1 0 x
-4 -3 -2 -1 0 1 2 3 4 5
-2 -1

d) -3 e) -2

i) f (x) = −x2
ii) f (x) = −0.2x2 + 1.8
iii) f (x) = −0.6x + 3.8
iv) f (x) = −0.2x3 + 0.4x2 + x − 0.6
v) f (x) = x3 − 6x2 + 11x − 4
vi) f (x) = x4

Exercise 9.3. Identify the graph with its assignment below.

3 y 5 y 4 y

2 4 3

1 3 2

0 x 2 1
-1 0 1 2 3 4 5
-1 1 0 x
-1 0 1 2 3 4 5 6
-2 0 x -1
-1 0 1 2 3 4 5
a) -3 b) -1 c) -2

i) f (x) = x6 − 14x5 + 78.76x4 − 227.5x3 + 355.25x2 − 283.5x + 93

ii) f (x) = −2x5 + 30x4 − 176x3 + 504x2 − 704x + 386
iii) f (x) = x5 − 13x4 + 65x3 − 155x2 + 174x − 72

Exercise 9.4. Sketch the graph of the function with the TI-84, which includes
all extrema and intercepts of the graph.

a) f (x) = 0.002x3 + 0.2x2 − 0.05x − 5

b) f (x) = x3 + 4x + 50
c) f (x) = 0.01x4 − 0.101x3 − 3x2 + 50.3x
d) f (x) = x3 − .007x
e) f (x) = x3 + .007x
f) f (x) = 0.025x4 + 0.0975x3 − 1.215x2 + 2.89x − 22
9.4. EXERCISES 129

Exercise 9.5. Find the exact value of at least one root of the given polynomial.

a) f (x) = x3 − 10x2 + 31x − 30,

b) f (x) = −x3 − x2 + 8x + 8,
c) f (x) = x3 − 11x2 − 3x + 33,
d) f (x) = x4 + 9x3 − 6x2 − 136x − 192,
e) f (x) = x2 + 6x + 3,
f) f (x) = x4 − 6x3 + 3x2 + 5x.

Exercise 9.6. Graph the following polynomials without using the calculator.

a) f (x) = (x + 4)2 (x − 5),

b) f (x) = −3(x + 2)3 x2 (x − 4)5 ,
c) f (x) = 2(x − 3)2 (x − 5)3 (x − 7),
d) f (x) = −(x + 4)(x + 3)(x + 2)2 (x + 1)(x − 2)2 .
Session 10

Roots of polynomials

10.1 Optional section: The rational root theorem

Example 10.1. Consider the equation 10x3 − 6x2 + 5x − 3 = 0. Let x be a
rational solution of this equation, that is x = pq is a rational number such that
 p 3  p 2 p
10 · −6· +5· − 3 = 0.
q q q

We assume that x = pq is completely reduced, that is, p and q have no

common factors that can be used to cancel the numerator and denominator of
the fraction pq . Now, simplifying the above equation, and combining terms, we
obtain:
p3 p2 p
10 · − 6 · + 5 · −3 =0
q3 q2 q
(multiply by q 3 ) =⇒ 10p3 − 6p2 q + 5pq 2 − 3q 3 = 0
(add 3q 3 ) =⇒ 10p3 − 6p2 q + 5pq 2 = 3q 3
(factor p on the left) =⇒ p · (10p2 − 6pq + 5q 2 ) = 3q 3 .

Therefore, p is a factor of 3q 3 (with the other factor being (10p2 − 6pq + 5q 2 )).
Since p and q have no common factors, p must be a factor of 3. That is, p is
one of the following integers: p = +1, +3, −1, −3.
Similarly starting from 10p3 − 6p2 q + 5pq 2 − 3q 3 = 0, we can write

(add +6p2 q − 5pq 2 + 3q 3 ) =⇒ 10p3 = 6p2 q − 5pq 2 + 3q 3

130
10.1. OPTIONAL SECTION: THE RATIONAL ROOT THEOREM 131

(factor q on the right) =⇒ 10p3 = (6p2 − 5pq + 3q 2 ) · q.

Now, q must be a factor of 10p3 . Since q and p have no common factors, q
must be a factor of 10. In other words, q is one of the following numbers: q =
±1, ±2, ±5, ±10. Putting this together with the possibilities for p = ±1, ±3,
we see that all possible rational roots are the following:
1 1 1 1 3 3 3 3
± , ± , ± , ± , ± , ± , ± , ± .
1 2 5 10 1 2 5 10
The observation in the previous example holds for a general polynomial
equation with integer coefficients.
Observation 10.2 (Rational root theorem). Consider the equation
an xn + an−1 xn−1 + · · · + a1 x + a0 = 0, (10.1)
where every coefficient an , an−1 , . . . , a0 is an integer and a0 6= 0, an 6= 0.
Assume that x = pq is a solution of (10.1) and the fraction x = pq is completely
reduced. Then a0 is an integer multiple of p, and an is an integer multiple of
q.
Therefore, all possible rational solutions of (10.1) are fractions x = pq
where p is a factor of a0 and q is a factor of an .
We can use this observation to find good candidates for the roots of a
given polynomial.
Example 10.3.
a) Find all rational roots of f (x) = 7x3 + x2 + 7x + 1.
b) Find all real roots of f (x) = 2x3 + 11x2 − 2x − 2.
c) Find all real roots of f (x) = 4x4 − 23x3 − 2x2 − 23x − 6.
Solution. a) If x = pq is a rational root, then p is a factor of 1, that is p = ±1,
and q is a factor of 7, that is q = ±1, ±7. The candidates for rational roots
are therefore x = ± 11 , ± 71 . To see which of these candidates are indeed roots
of f we plug these numbers into f via the calculator. We obtain the following:
132 SESSION 10. ROOTS OF POLYNOMIALS

Note that we entered the x-value as a fraction “(−)1/7” on the right. The
only root among ±1, ± 17 is x = − 71 .
b) We need to identify all real roots of f (x) = 2x3 + 11x2 − 2x − 2. In
general, it is a quite difficult task to find a root of a polynomial of degree 3,
so that it will be helpful if we can find the rational roots first. If x = pq is a
rational root then p is a factor of −2, that is p = ±1, ±2, and q is a factor of
2, that is q = ±1, ±2. The possible rational roots x = pq of f are:

1
±1, ±2, ±
2
Using the calculator, we see that the only rational root is x = 21 .

Therefore, by the factor theorem (Observation 8.10), we see that (x − 21 ) is

a factor of f , that is f (x) = q(x) · (x − 12 ). To avoid fractions in the long
division, we rewrite this as
1 2x − 1 q(x)
f (x) = q(x) · (x − ) = q(x) · = · (2x − 1),
2 2 2
so that we may divide f (x) by (2x − 1) instead of (x − 12 ) (note that this can
not be done with synthetic division). We obtain the following quotient.
x2 +6x +2
2x − 1 2x3 +11x2 −2x −2
−(2x3 −x2 )
12x2 −2x −2
−(12x2 −6x)
4x −2
−(4x −2)
0

Therefore, f (x) = (x2 + 6x + 2)(2x − 1), and any root of f is either a root
of x2 + 6x + 2 or of 2x − 1. We know that the root of 2x − 1 is x = 12 , and
10.1. OPTIONAL SECTION: THE RATIONAL ROOT THEOREM 133

that x2 + 6x + 2 has no other rational roots. Nevertheless, we can identify all

other real roots of x2 + 6x + 2 via the quadratic formula, (see Theorem 9.12
below).
√
2 −6 ± 62 − 4 · 1 · 2
x + 6x + 2 = 0 =⇒ x1/2 =
√ 2 √
−6 ± 36 − 8 −6 ± 28
= =
2
√ 2√
−6 ± 4 · 7 −6 ± 2 7
= =
2√ 2
= −3 ± 7

Therefore, the roots of f are precisely the following

√ √ 1
x1 = −3 + 7, x2 = −3 − 7, x3 = .
2

c) First we find the rational roots x = pq of f (x) = 4x4 −23x3 −2x2 −23x−6.
Since p is a factor of −6 it must be p = ±1, ±2, ±3, ±6, and since q is a factor
of 4 it must be q = ±1, ±2, ±4. All candidates for rational roots x = pq are
the following (where we excluded repeated ways of writing x):

1 3 1 3
±1, ±2, ±3, ±6, ± , ± , ± , ±
2 2 4 4

Checking all these candidates with the calculator produces exactly two ra-
tional roots: x = 6 and x = − 41 . Therefore, we may divide f (x) by both
(x − 6) and by (x + 14 ) without remainder. To avoid fractions, we use the
term 4 · (x + 41 ) = (4x + 1) instead of (x + 14 ) for our factor of f . Therefore,
f (x) = q(x) · (x − 6) · (4x + 1). The quotient q(x) is determined by performing
a long division by (x − 6) and then another long division by (4x + 1), or
alternatively by only one long division by

(x − 6) · (4x + 1) = 4x2 + x − 24x − 6 = 4x2 − 23x − 6.

Dividing f (x) = 4x4 − 23x3 − 2x2 − 23x − 6 by 4x2 − 23x − 6 produces the
134 SESSION 10. ROOTS OF POLYNOMIALS

quotient q(x):

x2 +1
4x2 − 23x − 6 4x4 −23x2 2
−2x −23x −6
−(4x4 −23x3 −6x2 )
4x2 −23x −6
−(4x2 −23x −6)
0

We obtain the factored expression for f (x) as f (x) = (x2 + 1)(4x + 1)(x − 6).
The only remaining real roots we need to find are those of x2 + 1. However,

x2 + 1 = 0 =⇒ x2 = −1

has no real solution. In other words there are only complex solutions of
x2 = −1, which are x = i and x = −i (we will discuss complex solutions in
more detail in the next section). Since the problem requires us to find the real
roots of f , our answer is that the only real roots are x1 = 6 and x2 = − 41 .

In the above example, we used the quadratic formula 9.12 from page 124.

10.2 The fundamental theorem of algebra

We have seen in Observation 8.10 on page 106 that every root c of a poly-
nomial f (x) gives a factor (x − c) of f (x). There is a theorem which says
something about the existence of roots and factors but we will need to discuss
complex numbers briefly before stating that theorem.
As stated above, we know that there is no real number that satisfies
2
x = −1. So we define i to be a solution to this equation. This i is not a real
number
√ but a new kind of number called a complex number. We can think of
i as −1. We can now consider numbers of the form a + ib where a and b are
real numbers. Numbers of this form constitutes the set of complex numbers,
denoted by C. a is called the real part and b is called the imaginary part of
the complex number a + ib. We can add two complex numbers by adding their
real and imaginary parts to form the real and imaginary parts of the sum. We
can multiply two complex numbers by ordinary distribution (FOIL) then use
the property that i2 = −1.
10.2. THE FUNDAMENTAL THEOREM OF ALGEBRA 135

Example 10.4. We have

(2 − 3i) − (4 + 3i) = (2 − 4) + (−3 − 3)i = −2 − 6i
and
(2 − 3i)(4 + 3i) = 8 + 6i − 12i − 9i2 = 8 − 6i − 9(−1) = 17 − 6i.
√ √
We can see by writing, for example, −6 = i 6, that these numbers
arise naturally as roots of quadratic equations. But there is more! The
following fundamental theorem of algebra guarantees the existence of roots
of any polynomial.
Theorem 10.5 (Fundamental theorem of algebra).
Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a non-constant
polynomial. Then there exists a complex number c which is a root
of f .
Let us make two remarks about the fundamental theorem of algebra to
clarify the statement of the theorem.
Note 10.6.
• In the above Theorem 10.5 we did not specify what kind of coefficients
a0 , . . . an are allowed for the theorem to hold. In fact, to be precise, the
fundamental theorem of algebra states that for any complex numbers
a0 , . . . an , the polynomial f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 has
a root.
• In general there may not exist a real root c of a given polynomial,
but the root c may only be a complex number. For example, consider
f (x) = x2 + 1, and consider the roots c of f , that is c2 + 1 = 0. Then,
for any real number c, we have always c2 ≥ 0, so that f (c) = c2 + 1 ≥ 1,
so that there cannot be a real root c of f . However, if we multiply
(x − i)(x + i) = x2 + xi − xi − i2 = x2 + 1,
we see that f has complex roots c = i and c = −i.
Now, while the fundamental theorem of algebra guarantees a root c of a
polynomial f , we can use the remainder theorem from Observation 8.10 to
check possible candidates c for the roots, and the factor theorem to factor
f (x) = q(x) · (x − c).
136 SESSION 10. ROOTS OF POLYNOMIALS

Example 10.7. Find roots of the given polynomial and use this information
to factor the polynomial completely.

a) f (x) = 2x3 − 8x2 − 6x + 36, b) f (x) = x4 − 3x3 − 36x2 + 68x + 240,

c) f (x) = x3 + 1, d) f (x) = x4 − 16.

Solution. a) In order to find a root, we use the graph to make a guess for one
of the roots.

The graph suggests that the roots may be at x = −2 and x = 3. This is also
supported by looking at the table for the function.

We check that these are roots by plugging the numbers into the function.

f (−2) = 2 · (−2)3 − 8 · (−2)2 − 6 · (−2) + 36 = −16 − 32 + 12 + 36 = 0,

f (3) = 2 · 33 − 8 · 32 − 6 · 3 + 36 = 54 − 72 − 18 + 36 = 0.

By the factor theorem we can divide f (x) = 2x3 − 8x2 − 6x + 36, for example,
by (x − 3).
2x2 −2x −12
3 2
x−3 2x −8x −6x +36
3 2
−(2x −6x )
−2x2 −6x +36
2
−(−2x +6x)
−12x +36
−(−12x +36)
0
10.2. THE FUNDAMENTAL THEOREM OF ALGEBRA 137

Therefore, f (x) = (2x2 − 2x − 12)(x − 3). To factor f completely, we still

need to factor the first polynomial 2x2 − 2x − 12:

f (x) = (2x2 − 2x − 12)(x − 3) = 2 · (x2 − x − 6) · (x − 3)

= 2 · (x − 3) · (x + 2) · (x − 3) = 2 · (x + 2) · (x − 3)2 .

b) The graph of f (x) = x4 −3x3 −36x2 +68x+240 in the standard window

is the following:

We suspect that x = −5, x = −2, x = 4, and x = 6 are the roots of the

function. This is confirmed by checking the table on the calculator, and also
by direct computation.

f (−5) = (−5)4 − 3(−5)3 − 36(−5)2 + 68(−5) + 240 = 0,

f (−2) = (−2)4 − 3(−2)3 − 36(−2)2 + 68(−2) + 240 = 0,
f (4) = 44 − 3 · 43 − 36 · 42 + 68 · 4 + 240 = 0,
f (6) = 64 − 3 · 63 − 36 · 62 + 68 · 6 + 240 = 0.

Therefore, each (x+5), (x+2), (x−4), and (x−6) is a factor of f (x), and then
their product is also a factor of f (x). The product (x+5)(x+2)(x−4)(x−6) is
calculated as follows. Since (x+ 5)(x+ 2) = x2 + 7x+ 10 and (x−4)(x−6) =
x2 − 10x + 24, we obtain

(x + 5)(x + 2)(x − 4)(x − 6) = (x2 + 7x + 10)(x2 − 10x + 24)

= x4 − 10x3 + 24x2 +7x3 − 70x2 + 168x+10x2 − 100x + 240
= x4 − 3x3 − 36x2 + 68x + 240 = f (x).
138 SESSION 10. ROOTS OF POLYNOMIALS

We see that this is precisely f (x), so that no more polynomial division is

necessary. The answer is f (x) = (x + 5)(x + 2)(x − 4)(x − 6).
c) We can graph the function, or just make a guess for a root of f (x) =
3
x + 1. An immediate guess would be to try x = −1. This is indeed a root,
since f (−1) = (−1)3 + 1 = −1 + 1 = 0, which is also supported by the graph.

Therefore, we can divide x3 + 1 by x + 1.

x2 −x +1
x+1 x3 2
+0x +0x +1
−(x3 +x2 )
−x2 +0x +1
−(−x2 −x)
x +1
−(x +1)
0
We obtain f (x) = x3 + 1 = (x2 − x + 1)(x + 1). Now x + 1 cannot be factored
any further, however, we may continue to factor x2 − x + 1 = (x − x1 )(x − x2 ).
Using the quadratic formula 9.12, we obtain the solutions of x2 − x + 1 = 0
as p √ √
−(−1) ± (−1)2 − 4 · 1 · 1 1 ± −3 1 3
x1/2 = = = ±i .
2·1 2 2 2
√ √
Therefore, x3 + 1 = (x − ( 21 + i 23 )) · (x − ( 12 − i 23 )) · (x + 1).
d) For f (x) = x4 − 16 there are the two obvious candidates x = 2 and
x = −2 that we can identify as roots, either by guessing (since 24 = 16) or
by looking at the graph.
10.2. THE FUNDAMENTAL THEOREM OF ALGEBRA 139

Indeed the algebra confirms these guesses.

f (2) = 24 − 16 = 16 − 16 = 0,
f (−2) = (−2)4 − 16 = 16 − 16 = 0.

Therefore, f (x) = q(x) · (x − 2) · (x + 2) = q(x) · (x2 − 4). To find q(x) we

perform the long division (x4 − 16)/(x2 − 4).

x2 +4
x2 − 4 x4 +0x3 2
+0x +0x −16
−(x4 −4x2 )
4x2 +0x −16
−(4x2 −16)
0

We obtain f (x) = (x2 + 4)(x + 2)(x − 2). The roots of x2 + 4 are given by
the quadratic formula as
√ √
−0 ± 02 − 4 · 1 · 4 −16 4i
x1/2 = =± = ± = ±2i,
2·1 2 2
or alternatively, x2 + 4 = 0 =⇒ x2 = −4 =⇒ x = ±2i. We obtain the
factored polynomial f (x) = (x + 2i)(x − 2i)(x + 2)(x − 2).

As we have seen in the last example, we can use the roots to factor a
polynomial completely so that all factors are polynomials of degree 1 only.
Furthermore, whenever a complex root, a + ib, appeared, its conjugate, a − ib,
was also a root. These remarks hold more generally, as we state now.

Observation 10.8.

(1) Every polynomial f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 of degree n

can be factored as

f (x) = m · (x − c1 ) · (x − c2 ) · · · · · (x − cn ).

(2) In particular, every polynomial of degree n has at most n roots. (However,

these roots may be real or complex.)
140 SESSION 10. ROOTS OF POLYNOMIALS

(3) The factor (x − c) for a root c could appear multiple times in the above
product, that is, we may have (x − c)k as a factor of f . The multiplicity
of a root c is the number of times k that a root appears in the factored
expression for f as in (1).

(4) If f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 has only real coefficients

a0 , . . . , an , and c = a+bi is a complex root of f , then the complex conjugate
c̄ = a − bi is also a root of f .
Proof. If x is any root, then an xn + an−1 xn−1 + · · · + a1 x + a0 = 0. Applying the complex
conjugate to this and using that u · v = ū · v̄ gives an x̄n + an−1 x̄n−1 + · · · + a1 x̄ + a0 = 0. Since
the coefficients aj are real, we have that aj = aj , so that an x̄n + an−1 x̄n−1 + · · · + a1 x̄ + a0 = 0.
This shows that the complex conjugate x̄ is a root of f as well.

Example 10.9. Find the roots of the polynomial and sketch its graph including
all roots.

a) f (x) = x3 + 2x2 − 14x − 3 b) f (x) = x4 − 7x3 + 11x2 − 7x + 10

Solution. a) To find a root, we first graph the function f (x) = x3 +2x2 −14x−3
with the calculator. The graph and the table suggest that we have a root at
x = 3.

Therefore we divide f (x) by (x − 3). We obtain:

x2 +5x +1
3
x−3 x +2x2 −14x −3
−(x3 −3x2 )
5x2 −14x −3
−(5x2 −15x)
x −3
−(x −3)
0
10.2. THE FUNDAMENTAL THEOREM OF ALGEBRA 141

This shows that f (x) = (x − 3)(x2 + 5x + 1). To find the roots of f , we also
have to find the roots of the second factor x2 + 5x + 1, i.e., the solutions to
x2 + 5x + 1 = 0. The quadratic formula gives:
√ √
−5 ± 52 − 4 · 1 · 1 −5 ± 21
x= =
2·1 2
The roots are therefore:
√ √
−5 + 21 −5 − 21
x = 3, x= ≈ −0.2, x= ≈ −4.8
2 2
We put these roots together with the graph in an appropriate window.

The graph including the roots is displayed below.

y

x
√ √
−5− 21 −5+ 21
2 2
3

b) The graph of f (x) = x4 −7x3 + 11x2 −7x+ 10 in an appropriate window

suggests the roots x = 2 and x = 5.
142 SESSION 10. ROOTS OF POLYNOMIALS

Dividing f (x) by (x − 2)(x − 5) = x2 − 7x + 10 gives:

x2 +1
2 4 3
x − 7x + 10 x −7x +11x2 −7x +10
−(x4 −7x3 +10x2 )
x2 −7x +10
−(x2 −7x +10)
0

Since the roots of x2 + 1 are all complex, we see that the only roots of f are
x = 2 and x = 5. The graph including its roots is displayed below.
y

2 5

Example 10.10. Find a polynomial f with the following properties.

a) f has degree 3, the roots of f are precisely 4, 5, 6, and the leading

coefficient of f is 7
b) f has degree 3 with real coefficients, f has roots 3i, −5 (and possibly
other roots as well), and f (0) = 90
c) f has degree 4 with complex coefficients, f has roots i + 1, 2i, 3
d) f has degree 5 with real coefficients, the leading coefficient is 1, and
the roots are determined by its graph:
4 y

0 x
-1 0 1 2 3 4 5 6
-1

-2
10.2. THE FUNDAMENTAL THEOREM OF ALGEBRA 143

Solution. a) In general a polynomial f of degree 3 is of the form f (x) =

m · (x − c1 ) · (x − c2 ) · (x − c3 ). Identifying the roots and the leading coefficient,
we obtain the polynomial

f (x) = 7 · (x − 4) · (x − 5) · (x − 6).

b) A polynomial f of degree 3 is of the form f (x) = m · (x − c1 ) · (x − c2 ) ·

(x − c3 ). Roots of f are 3i and −5, and since the coefficients of f are real
it follows from Observation 10.8(4), that the complex conjugate −3i is also a
root of f . Therefore, f (x) = m · (x + 5) · (x − 3i) · (x + 3i). To identify m, we
use the last condition f (0) = 90.

90 = m · (0 + 5) · (0 − 3i) · (0 + 3i) = m · 5 · (−9)i2 = m · 5 · 9 = 45m

Dividing by 45, we obtain m = 2, so that

f (x) = 2 · (x + 5) · (x − 3i) · (x + 3i) = 2 · (x + 5) · (x2 + 9),

which clearly has real coefficients.

c) Since f is of degree 4 it can be written as f (x) = m · (x − c1 ) · (x −
c2 ) · (x − c3 ) · (x − c4 ). Three of the roots are identified as i + 1, 2i, and 3:

f (x) = m · (x − (1 + i)) · (x − 2i) · (x − 3) · (x − c4 )

However, we have no further information on the fourth root c4 or the leading

coefficient m. (Note that Observation 10.8(4) cannot be used here, since we
are not assuming that the polynomial has real coefficients. Indeed it cannot
have real coefficients since that would mean we would have also the complex
conjugates of 1 + i and 2i in addition to the three that we are given, giving us
a total of 5 roots which cannot be accommodated by a polynomial of degree
4.) We can therefore choose any number for these remaining variables. For
example, a possible solution of the problem is given by choosing m = 3 and
c4 = 2, for which we obtain:

f (x) = 3 · (x − (1 + i)) · (x − 2i) · (x − 3) · (x − 2)

d) f is of degree 5, and we know that the leading coefficient is 1. The

graph is zero at x = 1, 2, 3, and 4, so that the roots are 1, 2, 3, and 4.
Moreover, since the graph just touches the root x = 4, this must be a multiple
root, that is, it must occur more than once (see Section 9.3 for a discussion
144 SESSION 10. ROOTS OF POLYNOMIALS

of multiple roots and their graphical consequences). We obtain the following

solution:
f (x) = (x − 1)(x − 2)(x − 3)(x − 4)2 .
Note that the root x = 4 is a root of multiplicity 2.

Note that to see that a polynomial has real coefficients, it may be neces-
sary to multiply factors like (x − (2 + 3i))(x − (2 − 3i)). We suggest a way
of doing this for which we use the fact that a2 − b2 = (a + b)(a − b). We have

(x − (2 + 3i))(x − (2 − 3i)) = ((x − 2) − 3i)((x − 2) + 3i) = (x − 2)2 + 9,

which clearly has real coefficients.

10.3 Exercises
Exercise 10.1.

a) Find all rational roots of f (x) = 2x3 − 3x2 − 3x + 2.

b) Find all rational roots of f (x) = 3x3 − x2 + 15x − 5.
c) Find all rational roots of f (x) = 6x3 + 7x2 − 11x − 12.
d) Find all real roots of f (x) = 6x4 + 25x3 + 8x2 − 7x − 2.
e) Find all real roots of f (x) = 4x3 + 9x2 + 26x + 6.

Exercise 10.2. Find a root of the polynomial by guessing possible candidates

of the root.

a) f (x) = x5 − 1 b) f (x) = x4 − 1 c) f (x) = x3 − 27

d) f (x) = x3 + 1000 e) f (x) = x4 − 81 f) f (x) = x3 − 125
g) f (x) = x5 + 32 h) f (x) = x777 − 1 i) f (x) = x2 + 64

Exercise 10.3. Find the roots of the polynomial and use it to factor the
polynomial completely.

a) f (x) = x3 − 7x + 6, b) f (x) = x3 − x2 − 16x − 20,

c) f (x) = x4 − 5x2 + 4, d) f (x) = x3 + x2 − 5x − 2,
e) f (x) = 2x3 + x2 − 7x − 6, f) f (x) = 12x3 + 49x2 − 2x − 24,
g) f (x) = x4 − 1, h) f (x) = x5 − 6x4 + 8x3 + 6x2 − 9x,
i) f (x) = x3 − 27, j) f (x) = x4 + 2x2 − 15.
10.3. EXERCISES 145

Exercise 10.4. Find the exact roots of the polynomial; write the roots in
simplest radical form, if necessary. Sketch a graph of the polynomial with all
roots clearly marked.
a) f (x) = x3 − 2x2 − 5x + 6, b) f (x) = x3 + 5x2 + 3x − 4,
c) f (x) = −x3 + 5x2 + 7x − 35, d) f (x) = x3 + 7x2 + 13x + 7,
e) f (x) = 2x3 − 8x2 − 18x − 36, f) f (x) = x4 − 4x2 + 3,
g) f (x) = −x4 + x3 + 24x2 − 4x − 80, h) f (x) = 7x3 − 11x2 − 10x + 8,
i) f (x) = −15x3 + 41x2 + 15x − 9, j) f (x) = x4 − 6x3 + 6x2 + 4x.
Exercise 10.5. Find a polynomial f that fits the given data.
a) f has degree 3. The roots of f are precisely 2, 3, 4.
The leading coefficient of f is 2.
b) f has degree 4. The roots of f are precisely −1, 2, 0, −3.
The leading coefficient of f is −1.
c) f has degree 3. f has roots −2, −1, 2, and f (0) = 10.
d) f has degree 4. f has roots 0, 2, −1, −4, and f (1) = 20.
e) f has degree 3. The coefficients of f are all real.
The roots of f are precisely 2 + 5i, 2 − 5i, 7.
The leading coefficient of f is 3.
f) f has degree 3. The coefficients of f are all real.
f has roots i, 3, and f (0) = 6.
g) f has degree 4. The coefficients of f are all real.
f has roots 5 + i and 5 − i of multiplicity 1, the root 3 of multipli-
city 2, and f (5) = 7.
h) f has degree 4. The coefficients of f are all real.
f has roots i and 3 + 2i.
i) f has degree 6. f has complex coefficients.
f has roots 1 + i, 2 + i, 4 − 3i of multiplicity 1 and the root −2
of multiplicity 3.
j) f has degree 5. f has complex coefficients.
f has roots i, 3, −7 (and possibly other roots).
k) f has degree 3. The roots of f are determined by its graph:
3 y

0 x
-1 0 1 2 3 4 5 6
-1

-2

-3
146 SESSION 10. ROOTS OF POLYNOMIALS

l) f has degree 4. The coefficients of f are all real. The leading

coefficient of f is 1. The roots of f are determined by its graph:
4 y

0 x
-2 -1 0 1 2 3 4 5
-1

(see Section 9.3).

m) f has degree 4. The coefficients of f are all real. f has the following
graph:
4 y

0 x
-2 -1 0 1 2 3 4 5 6
-1

-2

-3
Session 11

Rational functions

11.1 Graphs of rational functions

Recall from the beginning of this chapter that a rational function is a fraction
of polynomials:

an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0

In this section, we will study some characteristics of graphs of rational func-

tions. Asymptotes are important features of graphs of rational functions. An
asymptote is a line that is approached by the graph of a function: x = a is a
vertical asymptote if f (x) approaches ±∞ as x approaches a from either the
left of from the right, and y = b is a horizontal asymptote if f (x) approaches
b as x approaches ∞ or −∞.
To obtain a better idea of some of these features including asymptotes, we
start by graphing some rational functions. In particular it is useful to know
the graphs of the basic functions y = x1n .

Observation 11.1. Graphing y = x1 , y = 1

x2
, y= 1
x3
, y= 1
x4
, we obtain:

1 1 1 1
y= x
y= x2
y= x3
y= x4

147
148 SESSION 11. RATIONAL FUNCTIONS

In general we see that x = 0 is a vertical asymptote and y = 0 is a horizontal

asymptote. The shape of y = x1n depends on n being even or odd. We have:
3 y 3 y

2 2

1 1

0 x 0 x
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 -1

-2 -2

-3 -3

1 1
y= xn
, n odd y= xn
, n even
We now study more general rational functions.
1
Example 11.2. a) Our first graph is f (x) = x−3
.
4 y

0 x
-2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

-4

Here, the domain is all numbers where the denominator is not zero, that is
D = R − {3}. There is a vertical asymptote, x = 3. Furthermore, the graph
approaches 0 as x approaches ±∞. Therefore, f has a horizontal asymptote,
y = 0. Indeed, whenever the denominator has a higher degree than the
numerator, the line y = 0 will be the horizontal asymptote.
8x2 −8
b) Next, we graph f (x) = 4x 2 −16 .
6 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3
11.1. GRAPHS OF RATIONAL FUNCTIONS 149

Here, the domain is all x for which 4x2 − 16 6= 0. To see where this happens,
calculate

4x2 − 16 = 0 =⇒ 4x2 = 16 =⇒ x2 = 4 =⇒ x = ±2.

Therefore, the domain is D = R − {−2, 2}. As before, we see from the graph,
that the domain reveals the vertical asymptotes x = 2 and x = −2 (the vertical
dashed lines). To find the horizontal asymptote (the horizontal dashed line),
we note that when x becomes very large, the highest terms of both numerator
and denominator dominate the function value, so that

8x2 − 8 8x2
for |x| very large =⇒ f (x) = ≈ =2
4x2 − 16 4x2

Therefore, when x approaches ±∞, the function value f (x) approaches 2, and
therefore the horizontal asymptote is at y = 2 (the horizontal dashed line).
2
c) Our next graph is f (x) = x −8x+15
x−3
.
2 y

0 x
-2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

-4

-5

-6

We see that there does not appear to be any vertical asymptote, despite the
fact that 3 is not in the domain. The reason for this is that we can “remove
the singularity” by cancelling the troubling term x − 3 as follows:

x2 − 8x + 15 (x − 3)(x − 5) x−5
f (x) = = = = x − 5, x 6= 3
x−3 (x − 3) 1

Therefore, the function f reduces to x − 5 for all values where it is defined.

2
However, note that f (x) = x −8x+15
x−3
is not defined at x = 3. We denote this
in the graph by an open circle at x = 3, and call this a removable singularity
(or a hole).
150 SESSION 11. RATIONAL FUNCTIONS

d) Our fourth and last graph before stating the rules in full generality is
2x3 −8
f (x) = 3x 2 −16 .

The graph indicates that there is no horizontal asymptote, as the graph ap-
pears to increase towards ∞ and decrease towards −∞. To make this obser-
vation precise, we calculate the behavior when x approaches ±∞ by ignoring
the lower terms in the numerator and denominator

2x3 − 8 2x3 2x
for |x| very large =⇒ f (x) = ≈ =
3x2 − 16 3x2 3

Therefore, when x becomes very large, f (x) behaves like 32 x, which ap-
proaches ∞ when x approaches ∞, and approaches −∞ when x approaches
2x3 −8 2 r(x)
−∞. (In fact, after performing a long division we obtain 3x 2 −16 = 3 ·x+ 3x2 −16 ,

which would give rise to what is called a slant asymptote y = 23 · x; see also
remark 11.4 below.) Indeed, whenever the degree of the numerator is greater
than the degree of the denominator, we find that there is no horizontal asymp-
tote, but the graph blows up to ±∞. (Compare this also with example (c)
above).

Combining what we have learned from the above examples, we state our
observations as follows.

Observation 11.3. Let f (x) = p(x)

q(x)
be a rational function with polynomials
p(x) and q(x) in the numerator and denominator, respectively.

• The domain of f is all real numbers x for which the denominator is not
zero,
D = { x ∈ R | q(x) 6= 0 }

• Assume that q(x0 ) = 0, so that f is not defined at x0 . If x0 is not a root

of p(x), or if x0 is a root of p(x) but of a lesser multiplicity than the
11.1. GRAPHS OF RATIONAL FUNCTIONS 151

root in q(x), then f has a vertical asymptote x = x0 .

x−2 1 (x−3)(x−4)
f (x) = x−3
f (x) = (x−3)2
f (x) = (x−3)2
4 y 4 y 4 y

3 3 3

2 2 2

1 1 1

0 x 0 x 0 x
-1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6
-1 -1 -1

-2 -2 -2

• If p(x0 ) = 0 and q(x0 ) = 0, and the multiplicity of the root x0 in p(x)

is at least the multiplicity of the root in q(x), then these roots can
be canceled, and it is said that there is a removable discontinuity (or
sometimes called a hole) at x = x0 .
(x−1)(x−3) (x−2)(x−3)
f (x) = (x−3)
f (x) = (x−3)(x−4)
4 y 4 y

3 3

2 2

1 1

0 x 0 x
-1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6 7 8
-1 -1

-2 -2

• To find the horizontal asymptotes, we need to distinguish the cases

where the degree of p(x) is less than, equal to, or greater than q(x).

deg(p) >deg(q) deg(p) =deg(q) deg(p) <deg(q)

x3 −1 6x2 −5 5x−1
f (x) = 3x2 −1
f (x) = 3x2 +1
f (x) = x3 +4x
3 y 3 y 4 y

2 2 3

1 1 2

0 x 0 x 1
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 -1 0 x
-4 -3 -2 -1 0 1 2 3 4
-2 -2 -1

-3 -3 -2

no horizontal asymptote: asymptote:

highest coeff. of p
asymptote y= highest coeff. of q
y=0
152 SESSION 11. RATIONAL FUNCTIONS

In addition, it is sometimes useful to determine the x- and y-intercepts.

• If 0 is in the domain of f , then the y-intercept is (0, f (0)).
x+2
f (x) = x+1
4 y

0+2
2 f (0) = 0+1
=2

0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1

-2

• If p(x0 ) = 0 but q(x0 ) 6= 0, then f (x0 ) = p(x 0)

q(x0 )
= q(x00 ) = 0, so that (x0 , 0)
is an x-intercept, that is the graph intersects with the x-axis at x0 .
x−2 (x−2)(x−4)(x−5)
f (x) = x−3
f (x) = (x−3)3
4 y 4 y

3 3

2 2

1 1

0 x 0 x
-1 0 1 2 3 4 5 6 -1 0 1 2 3 4 5 6 7 8
-1 -1

-2 -2

Remark 11.4. We can calculate the asymptotic behavior of a rational function

f (x) = p(x)
q(x)
in the case where the degree of p is greater than the degree of q
by performing a long division. If the the quotient is m(x), and the remainder
is r(x), then
p(x) r(x)
f (x) = = m(x) + .
q(x) q(x)
r(x)
Now, since deg(r) <deg(q), the fraction q(x)
approaches zero as x approaches
±∞, so that f (x) ≈ m(x) for large x.
Example 11.5. Find the domain, all horizontal asymptotes, vertical asymp-
totes, removable singularities, and x- and y-intercepts. Use this information
together with the graph of the calculator to sketch the graph of f .
−x2 5x x3 −9x2 +26x−24
a) f (x) = x2 −3x−4
b) f (x) = x2 −2x
c) f (x) = x2 −x−2
x−4 3x2 −12
d) f (x) = (x−2)2
e) f (x) = 2x2 +1
11.1. GRAPHS OF RATIONAL FUNCTIONS 153

Solution. a) We combine our knowledge of rational functions and its algebra

with the particular graph of the function. The calculator gives the following
graph.

To find the domain of f we only need to exclude from the real numbers those
x that make the denominator zero. Since x2 − 3x − 4 = 0 exactly when
(x + 1)(x − 4) = 0, which gives x = −1 or x = 4, we have the domain:

domain D = R − {−1, 4}

The numerator has a root exactly when −x2 = 0, that is x = 0. Therefore,

x = −1 and x = 4 are vertical asymptotes, and since we cannot cancel terms
in the fraction, there is no removable singularity. Furthermore, since f (x) = 0
exactly when the numerator is zero, the only x-intercept is (0, 0).
To find the horizontal asymptote, we consider f (x) for large values of x
by ignoring the lower order terms in numerator and denominator,

−x2
|x| large =⇒ f (x) ≈ = −1
x2

We see that the horizontal asymptote is y = −1. Finally, for the y-intercept,
we calculate f (0):

−02 0
f (0) = 2
= = 0.
0 −3·0−4 −4
154 SESSION 11. RATIONAL FUNCTIONS

Therefore, the y-intercept is (0, 0). The function is then graphed as follows.
4 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
-1

-2

-3

-4

-5

5x
b) The graph of f (x) = x2 −2x
as drawn with the TI-84 is the following.

For the domain, we find the roots of the denominator,

x2 − 2x = 0 =⇒ x(x − 2) = 0 =⇒ x = 0 or x = 2.

The domain is D = R − {0, 2}. For the vertical asymptotes and removable
singularities, we calculate the roots of the numerator,

5x = 0 =⇒ x = 0.

Therefore, x = 2 is a vertical asymptote, and x = 0 is a removable singularity.

Furthermore, the denominator has a higher degree than the numerator, so that
y = 0 is the horizontal asymptote. For the y-intercept, we calculate f (0) by
evaluating the fraction f (x) at 0
5·0 0
= ,
02 −2·0 0
which is undefined. Therefore, there is no y-intercept (we, of course, already
noted that there is a removable singularity when x = 0). Finally for the
11.1. GRAPHS OF RATIONAL FUNCTIONS 155

x-intercept, we need to analyze where f (x) = 0, that is where 5x = 0. The

only candidate is x = 0 for which f is undefined. Again, we see that there is
no x-intercept. The function is then graphed as follows. (Notice in particular
the removable singularity at x = 0.)
5 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1

-2

-3

-4

-5

3 2
c) We start again by graphing the function f (x) = x −9x +26x−24
x2 −x−2
with the
calculator. After zooming to an appropriate window, we get:

To find the domain of f , we find the zeros of the denominator

x2 − x − 2 = 0 =⇒ (x + 1)(x − 2) = 0 =⇒ x = −1 or x = 2.

The domain is D = R − {−1, 2}. The graph suggests that there is a vertical
asymptote x = −1. However the x = 2 appears not to be a vertical asymptote.
This would happen when x = 2 is a removable singularity, that is, x = 2 is
a root of both numerator and denominator of f (x) = p(x)
q(x)
. To confirm this, we
calculate the numerator p(x) at x = 2:

p(2) = 23 − 9 · 22 + 26 · 2 − 24 = 8 − 36 + 52 − 24 = 0
156 SESSION 11. RATIONAL FUNCTIONS

Therefore, x = 2 is indeed a removable singularity. To analyze f further, we

also factor the numerator. Using the factor theorem, we know that x − 2 is a
factor of the numerator. Its quotient is calculated via long division.

x2 −7x +12
3
x−2 x −9x2 +26x −24
−(x3 −2x2 )
−7x2 +26x −24
−(−7x2 +14x)
12x −24
−(12x −24)
0

With this, we obtain

(x − 2)(x2 − 7x + 12) (x − 2)(x − 3)(x − 4)

f (x) = 2
= .
x −x−2 (x + 1)(x − 2)

Therefore, we conclude that x = −1 is a vertical asymptote and x = 2 is a

removable singularity. We also see that the x-intercepts are (3, 0) and (4, 0)
(that is x− values where the numerator is zero).
Now, the long range behavior is determined by ignoring the lower terms
in the fraction,

x3
|x| large =⇒ f (x) ≈ = x =⇒ no horizontal asymptote
x2

Finally, the y-coordinate of the y-intercept is given by

03 − 9 · 02 + 26 · 0 − 24 −24
y = f (0) = 2
= = 12.
0 −0−2 −2
11.1. GRAPHS OF RATIONAL FUNCTIONS 157

We draw the graph as follows:

y

12

x
−1 3 4

x−4
d) We first graph f (x) = (x−2) 2 . (Don’t forget to set the viewing window

back to the standard settings!)

The domain is all real numbers except where the denominator becomes zero,
that is, D = R − {2}. The graph has a vertical asymptote x = 2 and no hole.
The horizontal asymptote is at y = 0, since the denominator has a higher
0−4 −4
degree than the numerator. The y-intercept is at y0 = f (0) = (0−2)2 = 4 =

−1. The x-intercept is where the numerator is zero, x − 4 = 0, that is at

x = 4. Since the above graph did not show the x-intercept, we can confirm
this by changing the window size as follows:
158 SESSION 11. RATIONAL FUNCTIONS

Note in particular that the graph intersects the x-axis at x = 4 and then
changes its direction to approach the x-axis from above. A graph of the
function f which includes all these features is displayed below.
2 y

0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1

-2

-3

-4

3x2 −12
e) We graph f (x) = 2x2 +1
.

For the domain, we determine the zeros of the denominator.

1
2x2 + 1 = 0 =⇒ 2x2 = −1 =⇒ x2 = − .
2
The only solutions of this equation are given by complex numbers, but not by
any real numbers. In particular, for any real number x, the denominator of
f (x) is not zero. The domain of f is all real numbers, D = R. This implies
in turn that there are no vertical asymptotes, and no removable singularities.
The x-intercepts are determined by f (x) = 0, that is where the numerator
is zero,
3x2 − 12 = 0 =⇒ 3x2 = 12 =⇒ x2 = 4 =⇒ x = ±2.
3x2
The horizontal asymptote is given by f (x) ≈ 2x2
= 23 , that is, it is at y = 3
2
=
1.5. The y-intercept is at
3 · 02 − 12 −12
y = f (0) = 2
= = −12.
2·0 +1 1
11.2. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 159

We sketch the graph as follows:

y 3
2
= 1.5

x
−2 2
−2

−12

Since the graph is symmetric with respect to the y-axis, we can make one
more observation, namely that the function f is even (see observation 5.11 on
page 72):
3(−x)2 − 12 3x2 − 12
f (−x) = = = f (x)
2(−x)2 + 1 2x2 + 1

11.2 Optional section: Graphing rational functions

by hand
In this section we will show how to sketch the graph of a factored rational
function without the use of a calculator. It will be helpful to the reader to have
read section 9.3 on graphing a polynomial by hand before continuing in this
section. In addition to having the same difficulties as polynomials, calculators
often have difficulty graphing rational functions near an asymptote.

−3x2 (x−2)3 (x+2)

Example 11.6. Graph the function p(x) = (x−1)(x+1)2 (x−3)3
.

Solution. We can see that p has zeros at x = 0, 2, and −2 and vertical

asymptotes x = 1, x = −1 and x = 3. Also note that for large |x|, p(x) ≈ −3.
So there is a horizontal asymptote y = −3. We indicate each of these facts
160 SESSION 11. RATIONAL FUNCTIONS

on the graph:

−2 0 2

−3

n(x)
We can in fact get a more precise statement by performing a long division and writing p(x) = d(x)
=
r(x)
−3 + d(x)
. If you drop all but the leading order terms in the numerator and the denominator of the second
12
term, we see that p(x) ≈ −3 − x
whose graph for large |x| looks like

−3

This sort of reasoning can make the graph a little more accurate but is not necessary for a sketch.

We also have the following table:

for a near a, p(x) ≈ type sign change at a

−2 C1 (x + 2) linear changes
−1 C2 /(x + 1)2 asymptote does not change
0 C3 x2 parabola does not change
1 C4 /(x − 1) asymptote changes
2 C5 (x − 2)3 cubic changes
3 C6 /(x − 3)3 asymptote changes
11.2. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 161

Note that if the power appearing in the second column is even then the
function does not change from one side of a to the other. If the power is odd,
the sign changes (either from positive to negative or from negative to positive).
Now we move from large negative x values toward the right, taking into
account the above table. For large negative x, we start our sketch as follows:

−2 0 2

−3

And noting that near x = −2 the function p(x) is approximately linear, we

have

−2 0 2

−3

Then noting that we have an asymptote (noting that we can not cross the
162 SESSION 11. RATIONAL FUNCTIONS

x-axis without creating an x-intercept) we have

−2 0 2

−3

Now, from the table we see that there is no sign change at −1 so we have

−2 0 2

−3

and from the table we see that near x = 0 the function p(x) is approximately
quadratic and therefore the graph looks like a parabola. This together with
11.2. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 163

the fact that there is an asymptote at x = 1 gives

−2 0 2

−3

Now, from the table we see that the function changes sign at the asymptote,
so while the graph “hugs” the top of the asymptote on the left hand side, it
“hugs” the bottom on the right hand side giving

−2 0 2

−3

Now, from the table we see that near x = 2, p(x) is approximately cubic.
164 SESSION 11. RATIONAL FUNCTIONS

Also, there is an asymptote x = 3 so we get

−2 0 2

−3

Finally, we see from the table that p(x) changes sign at the asymptote x = 3
and has a horizontal asymptote y = −3, so we complete the sketch:

−2 0 2

−3

Note that if we had made a mistake somewhere there is a good chance

that we would have not been able to get to the horizontal asymptote on the
right side without creating an additional x-intercept.
What can we conclude from this sketch? This sketch exhibits only the
general shape which can help decide on an appropriate window if we want to
11.2. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 165

investigate details using technology. Furthermore, we can infer where p(x)

is positive and where p(x) is negative. However, it is important to notice,
that there may be wiggles in the graph that we have not included in our
sketch.

We now give one more example of graphing a rational function where the
horizontal asymptote is y = 0.

Example 11.7. Sketch the graph of

2x2 (x − 1)3 (x + 2)
r(x) = .
(x + 1)4 (x − 2)3

Solution. Here we see that there are x-intercepts at (0, 0), (0, 1), and (0, −2).
There are two vertical asymptotes: x = −1 and x = 2. In addition, there is a
horizontal asymptote at y = 0. (Why?) Putting this information on the graph
gives

−2 0 1

In this case, it is easy to get more information for large |x| that will be helpful
in sketching the function. Indeed, when |x| is large, we can approximate r(x)
by dropping all but the highest order term in the numerator and denominator
6
which gives r(x) ≈ 2x x7
= x2 . So for large |x|, the graph of r looks like

−2 0 1
166 SESSION 11. RATIONAL FUNCTIONS

The function gives the following table:

for a near a, p(x) ≈ type sign change at a

−2 C1 (x + 2) linear changes
−1 C2 /(x + 1)4 asymptote does not change
0 C3 x2 parabola does not change
1 C4 (x − 1)3 cubic changes
2 C5 /(x − 2)3 asymptote changes

Looking at the table for this function, we see that the graph should look like
a line near the zero (0, −2) and since it has an asymptote x = −1, the graph
looks something like:

−2 0 1

Then, looking at the table we see that r(x) does not change its sign near
x = −1, so that we obtain:

−2 0 1

Now, the function is approximately quadratic near x = 0 so the graph looks

11.2. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 167

like:

−2 0 1

Turning to head toward the root at 1 and noting that the function is approxi-
mately cubic there, and that there is an asymptote x = 2, we have:

−2 0 1

Finally, we see that the function changes sign at x = 2 (see the table). So
since the graph “hugs” the asymptote near the bottom of the graph on the
left side of the asymptote, it will “hug” the asymptote near the top on the
right side. So this together with the fact that y = 0 is an asymptote gives the
sketch (perhaps using an eraser to match the part of the graph on the right
that uses the large x):

−2 0 1

Note that if the graph couldn’t be matched at the end without creating an
extra x-intercept, then a mistake has been made.
168 SESSION 11. RATIONAL FUNCTIONS

11.3 Exercises
Exercise 11.1. Find the domain, the vertical asymptotes and removable dis-
continuities of the functions.
2 x2 +2
a) f (x) = x−2
b) f (x) = x2 −6x+8
3x+6 (x−2)(x+3)(x+4)
c) f (x) = x3 −4x
d) f (x) = (x−2)2 (x+3)(x−5)
x−1 2
e) f (x) = x3 −1
f) f (x) = x3 −2x2 −x+2

Exercise 11.2. Find the horizontal asymptotes of the functions.

8x2 +2x+1 1
a) f (x) = 2x2 +3x−2
b) f (x) = (x−3)2
x2 +3x+2 12x3 −4x+2
c) f (x) = x−1
d) f (x) = −3x3 +2x2 +1

Exercise 11.3. Find the x- and y-intercepts of the functions.

x−3 x3 −4x
a) f (x) = x−1
b) f (x) = x2 −8x+15
(x−3)(x−1)(x+4) x2 +5x+6
c) f (x) = (x−2)(x−5)
d) f (x) = x2 +2x

Exercise 11.4. Sketch the graph of the function f by using the domain of
f , the horizontal and vertical asymptotes, the removable singularities, the x-
and y-intercepts of the function, together with a sketch of the graph obtained
from the calculator.
6x−2 x−3
a) f (x) = 2x+4
b) f (x) = x3 −3x2 −6x+8
x4 −10x2 +9 x3 −3x2 −x+3
c) f (x) = x2 −3x+2
d) f (x) = x3 −2x2

Exercise 11.5. Find a rational function f that satisfies all the given proper-
ties.
a) vertical asymptote at x = 4 and horizontal asymptote y = 0
b) vertical asymptotes at x = 2 and x = 3 and horizontal asymptote y = 5
c) removable singularity at x = 1 and no horizontal asymptote
Session 12

Polynomial and rational

inequalities

12.1 Polynomial inequalities

We now consider inequalities. Solving inequalities is quite similar to solving
equalities. There is one extra consideration, that multiplying or dividing by a
negative number on both sides of an inequality changes the direction of the
inequality sign.
−2x ≤ −6 =⇒ x≥3
but 2x ≤ 6 =⇒ x ≤ 3
Example 12.1. Solve for x.

a) −3x + 7 > 19, b) 2x + 5 ≥ 4x − 11

c) 3 < −6x − 4 ≤ 13, d) −2x − 1 ≤ 3x + 4 < 4x − 20

Solution. The first three calculations are straightforward.

(−7) (÷(−3))
a) −3x + 7 > 19 =⇒ −3x > 12 =⇒ x < −4
(−4x−5) (÷(−2))
b) 2x + 5 ≥ 4x − 11 =⇒ −2x ≥ −16 =⇒ x≤8
(+4) (÷(−6)) 7 17
c) 3 < −6x − 4 ≤ 13 =⇒ 7 < −6x ≤ 17 =⇒ −6
>x≥ −6
=⇒ − 17
6
≤ x < − 76 ,

The last implication was obtained by switching the right and left terms of the
inequality. The solution set is the interval [− 17
6
, − 67 ).

169
170 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

For part (d), it is best to consider both inequalities separately.

(−3x+1) (÷(−5))
−2x − 1 ≤ 3x + 4 =⇒ −5x ≤ 5 =⇒ x ≥ −1,
(−4x−4) (·(−1))
3x + 4 < 4x − 20 =⇒ −x < −24 =⇒ x > 24.

The solution has to satisfy both inequalities x ≥ −1 and x > 24. Both
inequalities are true for x > 24 (since then also x ≥ −1), so that this is in
fact the solution: x > 24.

When dealing with polynomial inequalities, we use the same three-step

strategy that we used in section 1.4. More precisely, the first step is to solve
the corresponding equality, and the second step is to determine the solution
by investigating the subintervals induced from step 1. For both of these steps
we may now also use the graph of the function and its display on the graphing
calculator. The third step is to check the endpoints of each interval.

Example 12.2. Solve for x.

a) x2 − 3x − 4 ≥ 0 b) x3 − 9x2 + 23x − 15 ≤ 0
c) x4 − x2 > 5(x3 − x) d) x3 + 15x > 7x2 + 9
e) x5 − 6x4 − 26x3 + 144x2 − 47x − 210 ≤ 0

Solution. a) We can find the roots of the polynomial on the left by factoring.

x2 − 3x − 4 = 0 =⇒ (x − 4)(x + 1) = 0 =⇒ x = 4 or x = −1

To see where f (x) = x2 − 3x − 4 is ≥ 0, we graph it with the calculator.

3 y

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6 7
-1

-2

-3

-4

-5

-6

-7
12.1. POLYNOMIAL INEQUALITIES 171

We see that f (x) ≥ 0 when x ≤ −1 and when x ≥ 4 (the parts of the

graph above the x-axis). The solution set is therefore
{x|x ≤ −1, or x ≥ 4} = (−∞, −1] ∪ [4, ∞).
b) Here is the graph of the function f (x) = x3 − 9x2 + 23x − 15 with the
TI-84 in the standard window.
4 y

0 x
-1 0 1 2 3 4 5 6 7
-1

-2

-3

-4

This graph shows that there are two intervals where f (x) ≤ 0 (the parts of
the graph below the x-axis). To determine the exact intervals, we calculate
where f (x) = x3 − 9x2 + 23x − 15 = 0. The graph suggests that the roots of
f (x) are at x = 1, x = 3, and x = 5. This can be confirmed by a calculation:
f (1) = 13 − 9 · 12 + 23 · 1 − 15 = 1 − 9 + 23 − 15 = 0,
f (3) = 33 − 9 · 32 + 23 · 3 − 15 = 27 − 81 + 69 − 15 = 0,
f (5) = 53 − 9 · 52 + 23 · 5 − 15 = 125 − 225 + 115 − 15 = 0.
Since f is a polynomial of degree 3, the roots x = 1, 3, 5 are all of the roots
of f . (Alternatively, we could have divided f (x), for example, by x − 1 and
used this to completely factor f and with this obtain all the roots of f .) With
this, we can determine the solution set to be the set:
solution set = {x ∈ R|x ≤ 1, or 3 ≤ x ≤ 5}
= (−∞, 1] ∪ [3, 5].
Note that we include the roots 1, 3, and 5 in the solution set since the
original inequality was “≤” (and not “<”), which includes the solutions of the
corresponding equality.
c) In order to use the graphing calculator, we rewrite the inequality to
obtain zero on one side of the inequality.
x4 − x2 > 5(x3 − x) (distribute 5) =⇒ x4 − x2 > 5x3 − 5x
172 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

(subtract 5x3 , add 5x) =⇒ x4 − 5x3 − x2 + 5x > 0.

We graph f (x) = x4 − 5x3 − x2 + 5x with the TI-84.

The graph suggests the roots x = −1, 0, 1, and 5. This can be confirmed by
a straightforward calculation.

f (−1) = (−1)4 − 5 · (−1)3 − (−1)2 + 5 · (−1) = 1 + 5 − 1 − 5 = 0,

f (0) = 04 − 5 · 03 − 02 − 5 · 0 = 0,
f (1) = 14 − 5 · 13 − 12 + 5 · 1 = 1 − 5 − 1 + 5 = 0,
f (5) = 54 − 5 · 53 − 52 + 5 · 5 = 125 − 125 − 25 + 25 = 0.

The roots x = −1, 0, 1, and 5 are the only roots since f is of degree 4. The
intervals of the solution for f (x) > 0 may be read off from the graph:

solution set = (−∞, −1) ∪ (0, 1) ∪ (5, ∞)

(Notice that the roots −1, 0, 1, and 5 are not included in the solution set
since our inequality reads f (x) > 0 and not f (x) ≥ 0.)
d) Again, we bring all terms to one side:

x3 + 15x > 7x2 + 9 =⇒ x3 − 7x2 + 15x − 9 > 0

(Here it does not matter whether we bring the terms to the right or the left
side of the inequality sign! The resulting inequality is different, but the
solution to the problem is the same.) With this, we now use the TI-84 to find
the graph of the function f (x) = x3 − 7x2 + 15x − 9.
12.1. POLYNOMIAL INEQUALITIES 173

The graph suggests at least one root (the left most intersection point), but
possibly one or two more roots. To gain a better understanding of whether
the graph intersects the x-axis on the right, we rescale the window size of
the previous graph.

This viewing window suggests that there are two roots x = 1 and x = 3. We
confirm that these are the only roots with an algebraic computation. First,
we check that x = 1 and x = 3 are indeed roots:

f (1) = 13 − 7 · 12 + 15 · 1 − 9 = 1 − 7 + 15 − 9 = 0,
f (3) = 33 − 7 · 32 + 15 · 3 − 9 = 27 − 63 + 45 − 9 = 0.

To confirm that these are the only roots (and we have not just missed one of
the roots which might possibly become visible after sufficiently zooming into
the graph), we factor f (x) completely. We divide f (x) by x − 1:

x2 −6x +9
3
x−1 x −7x2 +15x −9
−(x3 −x2 )
−6x2 +15x −9
−(−6x2 +6x)
9x −9
−(9x −9)
0

and use this to factor f :

f (x) = x3 − 7x2 + 15x − 9 = (x − 1)(x2 − 6x + 9)

= (x − 1)(x − 3)(x − 3)

This shows, that 3 is a root of multiplicity 2, and so f has no other roots than
x = 1 and x = 3. The solution set consists of those numbers x for which
f (x) > 0. From the graph we see that this is the case when 1 < x < 3 and
174 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

when x > 3 (the roots x = 1 and x = 3 are not included as solutions). We

can write the solution set in several different ways:

solution set = {x|1 < x < 3 or x > 3} = {x|1 < x} − {3},

or in interval notation:

solution set = (1, 3) ∪ (3, ∞) = (1, ∞) − {3}.

e) If we set f (x) = x5 − 6x4 − 26x3 + 144x2 − 47x − 210 then we need to

find those numbers x with f (x) ≤ 0. We first graph f in the standard window.

To get a better view of the graph, we rescale to an appropriate window:

There appear to be three intervals, where f (x) ≤ 0. To determine the exact

numbers, we guess some of the roots from the graph. These would be x =
−5, −1, 2, 3, 7. To confirm these roots, we calculate the following function
values.

f (−5) = (−5)5 − 6 · (−5)4 − 26 · (−5)3 + 144 · (−5)2 − 47 · (−5) − 210

= −3125 − 3750 + 3250 + 3600 + 235 − 210 = 0,
f (−1) = (−1)5 − 6 · (−1)4 − 26 · (−1)3 + 144 · (−1)2 − 47 · (−1) − 210
= −1 − 6 + 26 + 576 + 47 − 210 = 0,
f (2) = 25 − 6 · 24 − 26 · 23 + 144 · 22 − 47 · 2 − 210
= 32 − 96 − 208 + 576 − 94 − 210 = 0,
f (3) = 35 − 6 · 34 − 26 · 33 + 144 · 32 − 47 · 3 − 210
12.1. POLYNOMIAL INEQUALITIES 175

= 243 − 486 − 702 + 1296 − 141 − 210 = 0,

f (7) = 75 − 6 · 74 − 26 · 73 + 144 · 72 − 47 · 7 − 210
= 16807 − 14406 − 8918 + 7056 − 329 − 210 = 0.
Since f (x) is of degree 5, we know that these are all of the roots of f (x).
The solution set for f (x) ≤ 0 can be read from the graphs above:
solution set = (−∞, −5] ∪ [−1, 2] ∪ [3, 7]
(Note again, that the roots are all included in the solution set.)
Polynomial inequalities come up, for example, when finding the domain of
functions involving a square root, as we will show in the next example.
Example 12.3. Find the domain of the given functions.
√ √
a) f (x) = x2 − 4 b) g(x) = x3 − 5x2 + 6x
√
Solution. a) The domain of f (x) = x2 − 4 is given by all x for which the
square root is non-negative. In other words the domain is given by numbers
x with x2 − 4 ≥ 0. Graphing the function y = x2 − 4 = (x + 2)(x − 2), we see
that this is precisely the case, when x ≤ −2 or x ≥ 2.

Therefore, the domain is Df = (−∞,

√ −2] ∪ [2, ∞).
b) For the domain of g(x) = x3 − 5x2 + 6x, we need find those x with
x3 − 5x2 + 6x ≥ 0. To this end, we graph y = x3 − 5x2 + 6x and check for its
roots.

From the graph and table above, we calculate the roots of y = x3 − 5x2 + 6x
at x = 0, x = 2, and x = 3. Furthermore, the graph and table show that
x3 − 5x2 + 6x ≥ 0 precisely when 0 ≤ x ≤ 2 or 3 ≤ x. The domain is
therefore, Dg = [0, 2] ∪ [3, ∞).
176 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

12.2 Rational inequalities and absolute value in-

equalities
Rational inequalities are solved with the same three-step process that was
used to solve the polynomial and absolute value inequalities before (see page
6). That is, in step 1, we find the solution of the corresponding equality, and
then, in step 2, we use sample points of the graph to determine the intervals
of the solution. Finally, in step 3, we check the endpoints of each interval.

Example 12.4. Solve for x.

x2 −5x+6 5
a) x2 −5x
≥ 0 b) x−2
≤3
4 3
c) x+5
< x−3 d) |2x − 3| > 7

x2 −5x+6
Solution. a) Here is the graph of x2 −5x
in the standard window.

Factoring numerator and denominator, we can determine vertical asymptotes,

holes, and x-intercepts.

x2 − 5x + 6 (x − 2)(x − 3)
2
=
x − 5x x(x − 5)

The vertical asymptotes are at x = 0 and x = 5, the x-intercepts are at x = 2

2
and x = 3. Since for large x, the fraction reduces to xx2 = 1, we see that the
2
horizontal asymptote is at y = 1. Thus, x x−5x+6
2 −5x ≥ 0 for x < 0 and x > 5. To
see where the graph is ≥ 0 between 0 and 5, we zoom into the graph:
12.2. RATIONAL INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES177

Combining all of the above information, we obtain the solution set:

solution set = (−∞, 0) ∪ [2, 3] ∪ (5, ∞)
Notice that the x−coordinate of the x−intercepts are x = 2 and x = 3 are
included in the solution set, whereas the values x = 0 and x = 5 associated
with the vertical asymptotes are not included since the fraction is not defined
for x = 0 and x = 5.
5
b) To find the numbers x where x−2 ≤ 3, we can graph the two functions
on the right and left of the inequality.

However, this can sometimes be confusing, and we recommend rewriting the

inequality so that one side becomes zero. Then, we graph the function on the
other side of the new inequality.
5 5 5 − 3(x − 2)
≤3 =⇒ − 3 ≤ 0 =⇒ ≤0
x−2 x−2 x−2
5 − 3x + 6 11 − 3x
=⇒ ≤ 0 =⇒ ≤0
x−2 x−2
11−3x
Therefore, we graph the function f (x) = x−2
.

The vertical asymptote is x = 2, and the x-intercept found thus

11
11 − 3x = 0 =⇒ 11 = 3x =⇒ x= .
3
This together with the graph and the fact that f is undefined at 2 and f ( 11
2
)=
0 gives the following solution set:
  h 11 
solution set = − ∞, 2 ∪ ,∞
3
178 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

4 3
c) We want to find those numbers x for which x+5 < x−3 . One way to do
4 3
this is given by graphing both functions f1 (x) = x+5 and f2 (x) = x−3 , and
by trying to determine where f1 (x) < f2 (x). However this can sometimes be
quite confusing, as the two graphs for f1 and f2 show below.

As before, we recommend rewriting the inequality so that one side of the

inequality becomes zero:
4 3 4 3 4(x − 3) − 3(x + 5)
< =⇒ − < 0 =⇒ <0
x+5 x−3 x+5 x−3 (x + 5)(x − 3)
4x − 12 − 3x − 15 x − 27
=⇒ < 0 =⇒ <0
(x + 5)(x − 3) (x + 5)(x − 3)
x−27
We therefore graph the function f (x) = (x+5)(x−3)
.

x−27
The vertical asymptotes of f (x) = (x+5)(x−3) are x = −5 and x = 3. The
x-intercept is (27, 0). We see from the graph that f (x) < 0 for x < −5. To
see the graph at x > 3, we zoom to the x-intercept at x = 27.

Therefore, the solution set is

solution set = {x|x < −5, or 3 < x < 27} = (−∞, −5) ∪ (3, 27).
12.3. EXERCISES 179

(The x-intercept x = 27 is not included in the solution set since the original
equation had a “<” and not “≤” sign.)
d) To analyze |2x − 3| > 7, we graph the function f (x) = |2x − 3| − 7.

To see where f (x) > 0, we find the zeros of f (x).

|2x − 3| − 7 = 0 =⇒ |2x − 3| = 7 =⇒ 2x − 3 = ±7

=⇒ 2x − 3 = 7 =⇒ 2x − 3 = −7
(add 3) =⇒ 2x = 10 (add 3) =⇒ 2x = −4
(divide by 2) =⇒ x = 5 (divide by 2) =⇒ x = −2
Thus, we read off the solution set for f (x) > 0 from the graph.

solution set = (−∞, −2) ∪ (5, ∞)

12.3 Exercises
Exercise 12.1. Solve for x.
a) 5x + 6 ≤ 21, b) 3 + 4x > 10x
c) 2x + 8 ≥ 6x + 24, d) 9 − 3x < 2x − 13
e) −5 ≤ 2x + 5 ≤ 19, f) 15 > 7 − 2x ≥ 1
g) 3x + 4 ≤ 6x − 2 ≤ 8x + 5, h) 5x + 2 < 4x − 18 ≤ 7x + 11
Exercise 12.2. Solve for x.
a) x2 − 5x − 14 > 0, b) x2 − 2x ≥ 35
c) x2 − 4 ≤ 0, d) x2 + 3x − 3 < 0
e) 2x2 + 2x ≤ 12, f) 3x2 < 2x + 1
g) x2 − 4x + 4 > 0, h) x3 − 2x2 − 5x + 6 ≥ 0
i) x3 + 4x2 + 3x + 12 < 0, j) −x3 − 4x < −4x2
k) x4 − 10x2 + 9 ≤ 0, l) x4 − 5x3 + 5x2 + 5x < 6
m) x4 − 5x3 + 6x2 > 0, n) x5 − 6x4 + x3 + 24x2 − 20x ≤ 0
180 SESSION 12. POLYNOMIAL AND RATIONAL INEQUALITIES

o) x5 − 15x4 + 85x3 − 225x2 + 274x − 120 ≥ 0,

p) x11 − x10 + x − 1 ≤ 0
Exercise 12.3. Find the domain of the functions below.
√ √
a) f (x) = x2 − 8x + 15 b) f (x) = 9x − x3
p p
c) f (x) = (x − 1)(4 − x) d) f (x) = (x − 2)(x − 5)(x − 6)
e) f (x) = √ 5 f) f (x) = √ 1
6−2x x2 −6x−7

Exercise 12.4. Solve for x.

x−5 4x−4 x−2 x2 −9
a) 2−x
>0 b) x2 −4
≥0 c) x2 −4x−5
<0 d) x2 −4
≥0
x−3 1 2 5 x2
e) x+3
≤4 f) x+10
>5 g) x−2
≤ x+1
h) x+4
≤x

Exercise 12.5. Solve for x.

a) |2x + 7| > 9 b) |6x + 2| < 3
c) |5 − 3x| ≥ 4 d) | − x − 7| ≤ 5
x
e) |1 − 8x| ≥ 3 f) 1 > 2 + 5
Review of polynomials and rational
functions

2x3 +x2 −9x−8

Exercise II.1. Divide the polynomials: 2x+3

Exercise II.2. Find the remainder when dividing x3 + 3x2 − 5x + 7 by x + 2.

Exercise II.3. Which of the following is a factor of x400 − 2x99 + 1:

x − 1, x + 1, x−0

Exercise II.4. Identify the polynomial with its graph.

4 y 4 y

3 3

2 2

1 1

0 x 0 x
-2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4
-1 -1

a) -2 b) -2

4 y 4 y

3 3

2 2

1 1

0 x 0 x
-2 -1 0 1 2 3 4 -2 -1 0 1 2 3 4
-1 -1

c) -2 d) -2

181
 i) f (x) = −x2 + 2x + 1, graph:
ii) f (x) = −x3 + 3x2 − 3x + 2, graph:
iii) f (x) = x3 − 3x2 + 3x + 1, graph:
iv) f (x) = x4 − 4x3 + 6x2 − 4x + 2, graph:

Exercise II.5. Sketch the graph of the function:

f (x) = x4 − 10x3 − 0.01x2 + 0.1x

• What is your viewing window?

• Find all roots, all maxima and all minima of the graph with the calculator.

Exercise II.6. Find all roots of f (x) = x3 + 6x2 + 5x − 12.

Use this information to factor f (x) completely.

Exercise II.7. Find a polynomial of degree 3 whose roots are 0, 1, and 3, and
so that f (2) = 10.

Exercise II.8. Find a polynomial of degree 4 with real coefficients, whose

roots include −2, 5, and 3 − 2i.
2
Exercise II.9. Let f (x) = x3x −12
2 −2x−3 . Sketch the graph of f . Include all vertical

and horizontal asymptotes, all holes, and all x- and y-intercepts.

Exercise II.10. Solve for x:

x+1
a) x4 + 2x < 2x3 + x2 , b) x2 + 3x ≥ 7, c) x+4
≤2

182
 Part III

Exponential and logarithmic

functions

183
Session 13

Exponential and logarithmic

functions

13.1 Exponential functions and their graphs

We now consider functions that differ a great deal from polynomials and ratio-
nal fractions in their complexity. More precisely, we will explore exponential
and logarithmic functions from a function theoretic point of view. We start by
recalling the definition of exponential functions and by studying their graphs.
Definition 13.1. A function f is called an exponential function if it is of the
form
f (x) = c · bx
for some real number c and positive real number b. The constant b is called
the base.
Example 13.2. Graph the functions
 x  x
x x x 1 1
f (x) = 2 , g(x) = 3 , h(x) = 10 , k(x) = , l(x) = .
2 10
First, we will graph the function f (x) = 2x by calculating the function
values in a table and then plotting the points in the x-y-plane. We can
calculate the values by hand, or simply use the table function of the calculator
to find the function values.

f (0) = 20 = 1

184
13.1. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 185

f (1) = 21 = 2
f (2) = 22 = 4
f (3) = 23 = 8
f (−1) = 2−1 = 0.5
f (−2) = 2−2 = 0.25

Similarly, we can calculate the table for the other functions g, h, k and l by
entering the functions in the spots at Y2, Y3, Y4, and Y5. The values in the
table✞ for these
☎ functions can be seen by moving the cursor to the right with
the lp⊲ lp key.
✝ ✆

✞ ☎
We can see the graphs by pressing the lpgraph lp key.
✝ ✆

In order to see these graphs more clearly, we adjust the graphing window to
a more appropriate size.
186 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Since all these functions are graphed in the same window, it is difficult to as-
sociate the graphs with their corresponding functions. In order to distinguish
between the graphs, we may use ✞ the TI-84
☎ to draw the graphs in different
line styles. In the function menu ( lpy= lp ), move the cursor to the left with
✞ ☎ ✞ ☎ ✝ ✆
lp⊳ lp , and press lpenter lp until the line becomes a thick line, or a dotted
✝ ✆ ✝ ✆ ✞ ☎ ✞ ☎
line, as desired. By moving the cursor down lp▽ lp , and pressing lpenter lp
✝ ✆ ✝ ✆
, we can adjust the line styles of the other graphs as well.

Note that the function k can also be written as

 x
1
k(x) = = (2−1 )x = 2−x ,
2


1 x
and similarly, l(x) = 10 = 10−x .

This example shows that the exponential function has the following prop-
erties.

Observation 13.3. The graph of the exponential function f (x) = bx with b > 0
and b 6= 1 has a horizontal asymptote at y = 0.
6 y
1 x
y = ( 10 ) y = 10x
5
y = ( 31 )x y = 3x
4
y = ( 21 )x y = 2x
3
1 x
y = ( 1.2 ) y = 1.2x
2

1 y = 1x

0 x
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1
13.1. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 187

• If b > 1, then f (x) approaches +∞ when x approaches +∞, and f (x)

approaches 0 when x approaches −∞.

• If 0 < b < 1, then f (x) approaches 0 when x approaches +∞, and f (x)
approaches +∞ when x approaches −∞.

An important base that we will frequently need to consider is the base of

e, where e is the Euler number.

Definition 13.4. The Euler number e is an irrational number that is approx-

imately
e = 2.718281828459045235 . . .

To be precise,

x we can define e as the number which is the horizontal asymptote of the function f (x) =
1 + x1 when x approaches +∞.

4 y

3
e

0 x

-1 0 1 2 3 4 5 6 7 8 9 10 11 12
-1

One can show that f has, indeed, a horizontal asymptote, and this limit is defined as e.

 x
1
e := lim 1+
x→∞ x

Furthermore, one can show that the exponential function with base e has a similar limit expression.

 r x
er = lim 1+ (13.1)
x→∞ x

Alternatively, the Euler number and the exponential function with base e may also be defined using an
r2 r3 r4
infinite series, namely, er = 1 + r + 1·2 + 1·2·3 + 1·2·3·4 + . . . . These ideas will be explored further in
a course in calculus.

Example 13.5. Graph the functions.

2 ex +e−x
a) y = ex , b) y = e−x , c) y = e−x , d) y = 2
188 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Solution. Using the calculator, we obtain

✞ the ☎
desired
✞ ☎graphs. The exponential
x
function y = e may be entered via lp2nd lp lplnlp .
✝ ✆✝ ✆

y = ex y = e−x

Note that the minus sign is entered

✞ in the
☎ last expression (and also in the
following two functions) via the lp(−) lp key.
✝ ✆
2 ex +e−x
y = e−x y= 2

ex +e−x
The last function y = 2
is called the hyperbolic cosine, and is denoted
x −x
by cosh(x) = e +e2
.

We now study how different multiplicative factors c affect the shape of an

exponential function.

Example 13.6. Graph the functions.

a) y = 2x , b) y = 3 · 2x , c) y = (−3) · 2x
d) y = 0.2 · 2x , e) y = (−0.2) · 2x

Solution. We graph the functions in one viewing window.

13.1. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 189

Here are the graphs of functions f (x) = c · 2x for various choices of c.

6 y

5
y = 3 · 2x
4
y = 2x
3
y = 0.2 · 2x
2

0 x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1

-2

y = (−0.2) · 2x
-3

y = (−1) · 2x
-4

y = (−3) · 2x
-5

-6

Note that for f (x) = c · 2x , the y-intercept is given at f (0) = c.

Finally, we can combine our knowledge of graph transformations to study

exponential functions that are shifted and stretched.

Example 13.7. Graph the functions.

1
a) y = 3x − 5, b) y = ex+4 , c) y = 4
· ex−3 + 2

Solution. The graphs are displayed below.

1
y = 3x − 5 y = ex+4 y= 4
· ex−3 + 2
190 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

The first graph y = 3x − 5 is the graph of y = 3x shifted down by 5.

3 y

0 x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5
-1

-2

-3

-4

-5

-6

The graph of y = ex+4 is the graph of y = ex shifted to the left by 4.

4 y

0 x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4
-1

Finally, y = 14 ex−3 + 2 is the graph of y = ex shifted to the right by 3 (see

the graph of y = ex−3 ), then compressed by a factor 4 towards the x-axis (see
the graph of y = 41 ex−3 ), and then shifted up by 2.
5 y

1
y = ex y = ex−3 y = 14 ex−4 x
0
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

13.2 Logarithmic functions and their graphs

The logarithmic function is closely related to the exponential function. Specif-
ically, the logarithm is the inverse function of the exponential function.
13.2. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 191

Definition 13.8. Let 0 < b 6= 1 be a positive real number that is not equal to
1. For x > 0, the logarithm of x with base b is defined by the equivalence

y = logb (x) ⇔ by = x (13.2)

For the particular base b = 10 we use the short form

log(x) := log10 (x)

For the particular base b = e, where e ≈ 2.71828 is the Euler number, we call
the logarithm with base e the natural logarithm, and write

ln(x) := loge (x)

The logarithmic function is the function y = logb (x) with domain D =

{x ∈ R|x > 0} of all positive real numbers, and range R = R of all real
numbers. It is the inverse of the exponential function y = bx with base b.

Example 13.9. Rewrite the equation as a logarithmic equation.

a) 34 = 81, b) 103 = 1000,

c) ex = 17, d) 27·a = 53.

Solution. We can immediately apply equation (13.2). For part (a), we have
b = 3, y = 4, and x = 81. Therefore we have:

34 = 81 ⇔ log3 (81) = 4

Similarly, we obtain the solutions for (b), (c), and (d).

b) 103 = 1000 ⇔ log(1000) = 3

c) ex = 17 ⇔ ln(17) = x
d) 27a = 53 ⇔ log2 (53) = 7a

Example 13.10. Evaluate the expression by rewriting it as an exponential

expression.

a) log2 (16), b) log5 (125), c) log13 (1), d) log4 (4)

e) log(10, 000), f) log(0.001), g) ln(e7 ), h) logb (bx )
192 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Solution. a) If we set y = log2 (16), then this is equivalent to 2y = 16. Since,

clearly, 24 = 16, we see that y = 4. Therefore, we have log2 (16) = 4.

b) log5 (125) = y ⇔ 5y = 125

(since 53 =125)
=⇒ 3 = y = log5 (125)
c) log13 (1) = y ⇔ 13y = 1
(since 130 =1)
=⇒ 0 = y = log13 (1)
d) log4 (4) = y ⇔ 4y = 4
(since 41 =4)
=⇒ 1 = y = log4 (4)
e) log(100, 000) = y ⇔ 10y = 100, 000
(since 105 =100,000)
=⇒ 5 = y = log(100, 000)
f) log(0.001) = y ⇔ 10y = 0.001
(since 10−3 =0.001)
=⇒ −3 = y = log(0.001)
g) ln(e7 ) = y ⇔ ey = e7 =⇒ 7 = y = ln(e7 )
h) logb (bx ) = y ⇔ by = bx =⇒ x = y = logb (bx )

In fact, the last example, in which we obtained logb (bx ) = x, combines all of
the previous examples.

In the previous example (in parts (c), (d), and (h)), we were able to find
certain elementary logarithms. We record these in the next observation.

Observation 13.11. We have the elementary logarithms:

logb (bx ) = x logb (b) = 1 logb (1) = 0 (13.3)

In general, when the argument is not a power of the base, we can use the
calculator to approximate the values of a logarithm via the formulas:

log(x) ln(x)
logb (x) = or logb (x) = (13.4)
log(b) ln(b)

The last two formulas will be proved in proposition 14.1 below. For now,
we want to show how they can be used to calculate any logarithmic expression
with the calculator.
13.2. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 193

Example 13.12. Calculate: a) log3 (13), b) log2.34 (98.765)

Solution. We calculate log3 (13) by using the first formula in (13.4).
log(13)
log3 (13) = ≈ 2.335
log(3)
Alternatively, we can also calculate this with the second formula in (13.4).
ln(13)
log3 (13) = ≈ 2.335
ln(3)
log(98.765)
For part (b) we have log2.34 (98.765) = log(2.34)
≈ 5.402.
We also want to gain an understanding of the graph of a logarithmic
function. Consider the graph of y = 2x from the previous section. Recall that
the graph of the inverse of a function is the reflection of the graph of the
function about the diagonal line y = x. So in this case we have:
6 y

5
y=x
4

2
y = log2 (x)
x 1
y=2
0 x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1

-2

-3

-4

The graph can also be obtained using a graphing calculator.

Example 13.13.
a) Graph the functions f (x) = ln(x), g(x) = log(x), h(x) = log2 (x), and
k(x) = log0.5 (x). What are the domains of f , g, h, and k? How do these
functions differ?
b) Graph the function p(x) = −3 · ln(x) + 4. What is the domain of p?
c) Graph the function q(x) = ln(5 − x). What is the domain of q?
d) Graph the function r(x) = log7 (2x + 8). What is the domain of r?
194 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Solution. a) We know from the definition that the domain of f , g, and h is all
real positive numbers, Df = Dg = Dh = Dk = {x|x > 0}. The functions f
and g can immediately be entered into the calculator. The standard window
gives the following graphs.

f (x) = ln(x) g(x) = log(x)

Note that we can rewrite g(x), h(x), and k(x) as a constant times f (x):

ln(x) 1
g(x) = log(x) = log10 (x) = = · f (x)
ln(10) ln(10)
ln(x) 1
h(x) = log2 (x) = = · f (x)
ln(2) ln(2)
ln(x) 1
k(x) = log0.5 (x) = = · f (x)
ln(0.5) ln(0.5)
1
Since ln(10) ≈ 0.434 < 1, we see that the graph of g is that of f compressed
1 1
towards the x-axis by a factor ln(10) . Similarly, ln(2) ≈ 1.443 > 1, so that
1
the graph of h is that of f stretched away from the x-axis by a factor ln(2) .
1 1 1 1
Finally, ln(0.5) ≈ −1.443, or more precisely, ln(0.5) = ln(2−1 ) = − ln(2) , so that
the graph of k is that of h reflected about the x-axis.

f (x) = ln(x) h(x) = log2 (x) k(x) = log0.5 (x)

Note that all these graphs have a common x-intercept at x = 1:

f (1) = g(1) = h(1) = k(1) = 0

13.2. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 195

To visualize the differences between the graphs, we graph them together in

one coordinate system.
4 y

3
h(x) = log2 (x)
f (x) = ln(x)
2

1
g(x) = log(x)

0 x
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
-1

-2

-3
k(x) = log0.5 (x)
-4

b) Using our knowledge of transformations of graphs, we expect that p(x) =

−3 · ln(x) + 4 is that of y = ln(x) reflected and stretched away from the x-axis
(by a factor 3), and then shifted up by 4. The stretched and reflected graph
is in the middle below, whereas the graph of the shifted function p is on the
right below.
y = ln(x) y = −3 · ln(x) p(x) = −3 · ln(x) + 4

The domain consist of numbers x for which the ln(x) is defined, that is,
Dp = {x|x > 0}.
c) To determine the domain of q(x) = ln(5 − x), we have to see for which
x the logarithm has a positive argument. More precisely, we need 5 − x > 0,
that is, 5 > x, so that the domain is Dq = {x|x < 5}.
The calculator displays the following graph.
196 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Note that the graph, as displayed by the calculator, appears to end at a

point that is approximately at (5, −2.5). However, the actual graph of the
logarithm does not stop at any point, since it has a vertical asymptote at x =
5, that is, the graph approaches −∞ as x approaches 5. The calculator only
displays an approximation, which may be misleading since this approximation
is determined by the window size and the size of each pixel. The graph would
therefore look as follows.
3 y

0 x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

-4

-5

d) The domain of r(x) = log7 (2x+8) consists of those numbers x for which
the argument of the logarithm is positive.
(subtract 8) (divide by 2)
2x + 8 > 0 =⇒ 2x > −8 =⇒ x > −4

Therefore, the domain is Dr = {x|x > −4}. To graph the function r(x) =
log7 (2x + 8), we can enter r(x) = ln(2x+8)
ln(7)
into the calculator. The graph looks
as follows.
3 y

0 x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

The previous example analyzes the graph of the logarithm function. Here
is the summary.
13.3. EXERCISES 197

Observation 13.14. The graph of a logarithmic function y = logb (x) with base
b is that of the natural logarithm y = ln(x) stretched away from the x-axis,
or compressed towards the x-axis when b > 1. When 0 < b < 1, the graph is
furthermore reflected about the x-axis.
4 y

3
y = log2 (x)
y = ln(x)
2

1
y = log10 (x)

0 x
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
-1
y = log0.1 (x)
-2

-3
y = log0.5 (x)
-4

The graph y = logb (x) has domain D = {x|x > 0}, and a vertical asymptote
at x = 0. There is no horizontal asymptote, as f (x) approaches +∞ when
x approaches +∞ for b > 1, and f (x) approaches −∞ when x approaches
+∞ for 0 < b < 1. Finally, there is an x-intercept at x = 1.

13.3 Exercises
Exercise 13.1. Graph the following functions with the calculator.

a) y = 5x b) y = 1.01x c) y = ( 31 )x d) y = 0.97x
2
e) y = 3−x f) y = ( 31 )−x g) y = ex h) y = 0.01x
ex −e−x ex −e−x
i) y = 1x j) y = ex + 1 k) y = 2
l) y = ex +e−x

x −x
The last two functions are known as the hyperbolic sine, sinh(x) = e −e
2
,
ex −e−x
and the hyperbolic tangent, tanh(x) = ex +e−x . Recall that the hyperbolic
x −x
cosine cosh(x) = e +e
2
was already graphed in example 13.5.

Exercise 13.2. Graph the given function. Describe how the graph is obtained
by a transformation from the graph of an exponential function y = bx (for
198 SESSION 13. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

appropriate base b).

a) y = 0.1 · 4x b) y = 3 · 2x c) y = (−1) · 2x
d) y = 0.006 · 2x e) y = e−x f) y = e−x + 1
g) y = ( 21 )x + 3 h) y = 2x−4 i) y = 2x+1 − 6

Exercise 13.3. Use the definition of the logarithm to write the given equation
as an equivalent logarithmic equation.

a) 42 = 16 b) 28 = 256 c) ex = 7 d) 10−1 = 0.1

x
e) 3x = 12 f) 57·x = 12 g) 32a+1 = 44 h) 12 h = 30

Exercise 13.4. Evaluate the following expressions without using a calculator.

a) log7 (49) b) log3 (81) c) log2 (64) d) log50 (2, 500)

e) log2 (0.25) f) log(1, 000) g) ln(e4 ) h) log13 (13)
1
i) log(0.1) j) log6 ( 36 ) k) ln(1) l) log 1 (8)
2

Exercise 13.5. Using a calculator, approximate the following expressions to

the nearest thousandth.

a) log3 (50) b) log3 (12) c) log17 (0.44) d) log0.34 (200)

Exercise 13.6. State the domain of the function f and sketch its graph.

a) f (x) = log(x) b) f (x) = log(x + 7)

c) f (x) = ln(x + 5) − 1 d) f (x) = ln(3x − 6)
e) f (x) = 2 · log(x + 4) f) f (x) = −4 · log(x + 2)
g) f (x) = log3 (x − 5) h) f (x) = − ln(x − 1) + 3
i) f (x) = log0.4 (x) j) f (x) = log3 (−5x) − 2
k) f (x) = log |x| l) f (x) = log |x + 2|
Session 14

Properties of exponential and

logarithmic functions

14.1 Algebraic properties of exponential and log-

arithmic functions
We recall some of the main properties of the exponential function:

bx+y = bx · by
bx
bx−y = by
(14.1)
(bx )n = b nx

Writing the above properties in terms of f (x) = bx we have f (x + y) =

f (x)f (y), f (x − y) = f (x)/f (y), and f (nx) = f (x)n .
Due to the fact that the logarithm is the inverse function of the exponential,
we have corresponding properties whose proof follows:
Proposition 14.1. The logarithm behaves well with respect to products, quo-
tients, and exponentiation. Indeed, for all positive real numbers 0 < b 6= 1,
x > 0, y > 0, and real numbers n, we have:

logb (x · y) = logb (x) + logb (y)

logb ( xy ) = logb (x) − logb (y) (14.2)
logb (xn ) = n · logb (x)

199
200 SESSION 14. PROPERTIES OF EXP AND LOG

In terms of the logarithmic function g(x) = logb (x), the properties in the
table above can be written: g(xy) = g(x) + g(y), g(x/y) = g(x) − g(y), and
g(xn ) = n · g(x).
Furthermore, for another positive real number 0 < a 6= 1, we have the
change of base formula:

loga (x)
logb (x) = (14.3)
loga (b)

In particular, we have the formulas from equation (13.4) on page 192 when
taking the base a = 10 and a = e:

log(x) ln(x)
logb (x) = and logb (x) =
log(b) ln(b)
Proof. We start with the first formula logb (x · y) = logb (x) + logb (y). If we call u = logb (x) and
v = logb (y), then the equivalent exponential formulas are bu = x and bv = y. With this, we have

x · y = bu · bv = bu+v .

Rewriting this in logarithmic form, we obtain

logb (x · y) = u + v = logb (x) + logb (y).

This is what we needed to show.

Next, we prove the formula logb ( xy ) = logb (x) − logb (y). We abbreviate u = logb (x) and v = logb (y)
as before, and their exponential forms are bu = x and bv = y. Therefore, we have

x bu
= v = bu−v .
y b

Rewriting this again in logarithmic form, we obtain the desired result.

x
logb = u − v = logb (x) − logb (y)
y

For the third formula, logb (xn ) = n · logb (x), we write u = logb (x), that is in exponential form bu = x.
Then:
xn = (bu )n = bn·u =⇒ logb (xn ) = n · u = n · logb (x)

For the last formula (14.3), we write u = logb (x), that is, bu = x. Applying the logarithm with base a
to bu = x gives loga (bu ) = loga (x). As we have just shown before, loga (bu ) = u · loga (b). Combining
these identities with the initial definition u = logb (x), we obtain

loga (x) = loga (bu ) = u · loga (b) = logb (x) · loga (b)

loga (x)
Dividing both sides by loga (b) gives the result loga (b)
= logb (x).
14.1. ALGEBRAIC PROPERTIES OF EXP AND LOG 201

Example 14.2. Combine the terms using the properties of logarithms so as

to write as one logarithm.
1
a) 2
ln(x) + ln(y) b) 23 (log(x2 y) − log(xy 2 ))
c) 2 ln(x) − 13 ln(y) − 75 ln(z) d) 5 + log2 (a2 − b2 ) − log2 (a + b)
Solution. Recall that a fractional exponent can also be rewritten with an nth
root.
1 √ 1 √ p 1 √
x2 = x and xn = n x =⇒ x q = (xp ) q = q xp
We apply the rules from proposition 14.1.
1 1 1 √
a) 2
ln(x) + ln(y) = ln(x 2 ) + ln(y) = ln(x 2 y) = ln( x · y)
  2    
2 2 x y 2
b) 3
(log(x2
y) − log(xy 2
)) = 3
log xy 2
= 3
log xy
  2  q 
x 3 3 x2
= log y
= log y2

√ √  2 
2 ln(x) − 31 ln(y) − 75 ln(z) = ln(x2 ) − ln( 3 y) − ln( z 7 ) = ln √3 x√5 7
5
c)
y· z

d) 5 + log2 (a2 − b2 ) − log2 (a + b) = log2 (25 ) + log2 (a2 − b2 ) − log2 (a + b)

 5 2 2   
= log2 2 ·(aa+b−b ) = log2 32·(a+b)(a−b)
a+b
= log2 (32 · (a − b))

Example 14.3. Write the expressions in terms of elementary logarithms u =

logb (x), v = logb (y), and, in part (c), also w = logb (z). Assume that x, y, z > 0.
q  s !
√ √ x2
a) ln( x5 · y 2 ), b) log x · y3 c) log2 3 √ .
y z
Solution. In a first step, we rewrite the expression with fractional exponents,
and then apply the rules from proposition 14.1.
√ 5 5
a) ln( x5 · y 2 ) = ln(x 2 · y 2 ) = ln(x 2 ) + ln(y 2 )
5
= ln(x) + 2 ln(y) = 52 u + 2v
2
p√    21   1 
1
3 1
b) log 3
x · y = log x y
2 = 2 log x 2 y 3
 1


= 21 log(x 2 ) + log(y 3 ) = 12 12 log(x) + 3 log(y)
1
= 4
log(x) + 32 log(y) = 41 u + 32 v
202 SESSION 14. PROPERTIES OF EXP AND LOG

q    31 !  
x√2 x2 1 x2
c) log2 3
y z
= log2 1 = log2
3 1
y·z 2 y·z 2
 1

1
= 3
log2 (x2 ) − log2 (y) − log2 (z 2 )
 
= 31 2 log2 (x) − log2 (y) − 21 log2 (z)
2
= 3
log2 (x) − 13 log2 (y) − 16 log2 (z)
2
= 3
u − 31 v − 16 w

14.2 Solving exponential and logarithmic equations

We can solve exponential and logarithmic equations by applying logarithms
and exponentials. Since the exponential and logarithmic functions are invert-
ible (they are inverses of each other), applying these operations can also be
reversed. In short we have observed that
Observation 14.4.
x=y =⇒ bx = by , and bx = by =⇒ x=y
x=y =⇒ logb (x) = logb (y), and logb (x) = logb (y) =⇒ x=y

We can use the above implications whenever we have a common basis, or

when we can convert the terms to a common basis.
Example 14.5. Solve for x.

a) 2x+7 = 32 b) 102x−8 = 0.01

c) 72x−3 = 75x+4 d) 53x+1 = 254x−7
e) ln(3x − 5) = ln(x − 1) f) log2 (x + 5) = log2 (x + 3) + 4
g) log6 (x) + log6 (x + 4) = log6 (5) h) log3 (x − 2) + log3 (x + 6) = 2

Solution. In these examples, we can always write both sides of the equation
as an exponential expression with the same base.

a) 2x+7 = 32 =⇒ 2x+7 = 25 =⇒ x + 7 = 5 =⇒ x = −2
b) 102x−8 = 0.01 =⇒ 10 2x−8 −2
= 10 =⇒ 2x − 8 = −2
=⇒ 2x = 6 =⇒ x=3
14.2. SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS 203

Here it is useful to recall the powers of 10, which were also used to solve the
equation above.

103 = 1, 000
102
 n
= 100  10 = 1 |00 ·{z
· · 00}

101 = 10 
 n zeros
100 = 1 In general (n ≥ 1) :

10−1 = 0.1 −n
10 = |0.0 ·{z· · 00} 1


10−2 = 0.01 n zeros
10−3 = 0.001

(−5x+3)
c) 72x−3 = 75x+4 =⇒ 2x − 3 = 5x + 4 =⇒ −3x = 7
7
=⇒ x=−
3
d) 53x+1 = 254x−7 =⇒ 53x+1 = 52·(4x−7) =⇒ 3x + 1 = 2 · (4x − 7)
(−8x−1)
=⇒ 3x + 1 = 8x − 14 =⇒ −5x = −15
=⇒ x=3

By a similar reasoning we can solve equations involving logarithms whenever

the bases coincide.
(−x+5)
e) ln(3x − 5) = ln(x − 1) =⇒ 3x − 5 = x − 1 =⇒ 2x = 4
=⇒ x=2

For part (f), we have to solve log2 (x+5) = log2 (x+3)+4. To combine the right-
hand side, recall that 4 can be written as a logarithm, 4 = log2 (24 ) = log2 16.
With this remark we can now solve the equation for x.

log2 (x + 5) = log2 (x + 3) + 4 =⇒ log2 (x + 5) = log2 (x + 3) + log2 (16)

=⇒ log2 (x + 5) = log2 (16 · (x + 3)) =⇒ x + 5 = 16(x + 3)
(−16x−5) 43
=⇒ x + 5 = 16x + 48 =⇒ −15x = 43 =⇒ x=−
15
Next, in part (g), we start by combining the logarithms.

log6 (x) + log6 (x + 4) = log6 (5) =⇒ log6 (x(x + 4)) = log6 (5)
204 SESSION 14. PROPERTIES OF EXP AND LOG

remove log6
=⇒ x(x + 4) = 5
=⇒ x2 + 4x − 5 = 0
=⇒ (x + 5)(x − 1) = 0
=⇒ x = −5 or x = 1
Since the equation became a quadratic equation, we ended up with two pos-
sible solutions x = −5 and x = 1. However, since x = −5 would give a neg-
ative value inside a logarithm in our original equation log6 (x) + log6 (x + 4) =
log6 (5), we need to exclude this solution. The only solution is x = 1.
Similarly we can solve the next part, using that 2 = log3 (32 ):
h) log3 (x − 2) + log3 (x + 6) = 2 =⇒ log3 ((x − 2)(x + 6)) = log3 (32 )
=⇒ (x − 2)(x + 6) = 32
=⇒ x2 + 4x − 12 = 9
=⇒ x2 + 4x − 21 = 0
=⇒ (x + 7)(x − 3) = 0
=⇒ x = −7 or x = 3
We exclude x = −7, since we would obtain a negative value inside a loga-
rithm, so that the solution is x = 3.
When the two sides of an equation are not exponentials with a common
base, we can solve the equation by first applying a logarithm and then solving
for x. Indeed, recall from Observation 7.9 in section 7.2 on inverse functions,
that since f (x) = logb (x) and g(x) = bx are inverse functions, we have
logb (bx ) = x and blogb (x) = x whenever the compositions of the left sides
make sense. That is, the action of the logarithm cancels out the action of the
exponential function, and vice versa. So we can think of applying a logarithm
(an exponentiation) on both sides of an equation to cancel an exponentiation
(a logarithm) much like squaring both sides of an equation to cancel a square
root.
Example 14.6. Solve for x.
a) 3x+5 = 8 b) 132x−4 = 6 c) 5x−7 = 2x
d) 5.1x = 2.72x+6 e) 17x−2 = 3x+4 f) 72x+3 = 113x−6
Solution. We solve these equations by applying a logarithm (both log or ln
will work for solving the equation), and then we use the identity log(ax ) =
14.2. SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS 205

x · log(a).

a) 3x+5 = 8 =⇒ ln 3x+5 = ln 8 =⇒ (x + 5) · ln 3 = ln 8
ln 8 ln 8
=⇒ x+5= =⇒ x= − 5 ≈ −3.11
ln 3 ln 3
b) 132x−4 = 6 =⇒ ln 132x−4 = ln 6 =⇒ (2x − 4) · ln 13 = ln 6
ln 6 ln 6
=⇒ 2x − 4 = =⇒ 2x = +4
ln 13 ln 13
ln 6
+4 ln 6
=⇒ x = ln 13 = + 2 ≈ 2.35
2 2 · ln 13
c) 5x−7 = 2x =⇒ ln 5x−7 = ln 2x =⇒ (x − 7) · ln 5 = x · ln 2

At this point, the calculation will proceed differently than the calculations in
parts (a) and (b). Since x appears on both sides of (x − 7) · ln 5 = x · ln 2, we
need to separate terms involving x from terms without x. That is, we need to
distribute ln 5 on the left and separate the terms. We have

(x − 7) · ln 5 = x · ln 2 =⇒ x · ln 5 − 7 · ln 5 = x · ln 2
(add + 7 · ln 5 − x · ln 2) =⇒ x · ln 5 − x · ln 2 = 7 · ln 5
=⇒ x · (ln 5 − ln 2) = 7 · ln 5
7 · ln 5
=⇒ x= ≈ 12.30
ln 5 − ln 2
We need to apply the same solution strategy for the remaining parts (d)-(f)
as we did in (c).

d) 5.1x = 2.72x+6 =⇒ ln 5.1x = ln 2.72x+6

=⇒ x · ln 5.1 = (2x + 6) · ln 2.7
=⇒ x · ln 5.1 = 2x · ln 2.7 + 6 · ln 2.7
=⇒ x · ln 5.1 − 2x · ln 2.7 = 6 · ln 2.7
=⇒ x · (ln 5.1 − 2 · ln 2.7) = 6 · ln 2.7
6 · ln 2.7
=⇒ x= ≈ −16.68
ln 5.1 − 2 · ln 2.7

e) 17x−2 = 3x+4 =⇒ ln 17x−2 = ln 3x+4

=⇒ (x − 2) · ln 17 = (x + 4) · ln 3
206 SESSION 14. PROPERTIES OF EXP AND LOG

=⇒ x · ln 17 − 2 · ln 17 = x · ln 3 + 4 · ln 3
=⇒ x · ln 17 − x · ln 3 = 2 · ln 17 + 4 · ln 3
=⇒ x · (ln 17 − ln 3) = 2 · ln 17 + 4 · ln 3
2 · ln 17 + 4 · ln 3
=⇒ x= ≈ 5.80
ln 17 − ln 3

f) 72x+3 = 113x−6 =⇒ ln 72x+3 = ln 113x−6

=⇒ (2x + 3) · ln 7 = (3x − 6) · ln 11
=⇒ 2x · ln 7 + 3 · ln 7 = 3x · ln 11 − 6 · ln 11
=⇒ 2x · ln 7 − 3x · ln 11 = −3 · ln 7 − 6 · ln 11
=⇒ x · (2 · ln 7 − 3 · ln 11) = −3 · ln 7 − 6 · ln 11
−3 · ln 7 − 6 · ln 11
=⇒ x= ≈ 6.13
2 · ln 7 − 3 · ln 11

Note that in the problems above we could have also changed the base as
we did earlier in the section. For example, in part f) above, we could have
(3x−6) )
begun by writing the right hand side as 113x−6 = 7log7 (11 . We chose to
simply apply a log to both sides instead, because the notation is somewhat
simpler.

14.3 Exercises
Exercise 14.1. Combine the terms and write your answer as one logarithm.

a) 3 ln(x) + ln(y) b) log(x) − 32 log(y)

1
c) 3
log(x) − log(y) + 4 log(z) d) log(xy 2 z 3 ) − log(x4 y 3 z 2 )
1
e) 4
ln(x) − 12 ln(y) + 23 ln(z) f) − ln(x2 − 1) + ln(x − 1)
g) 5 ln(x) + 2 ln(x4 ) − 3 ln(x) h) log5 (a2 + 10a + 9) − log5 (a + 9) + 2

Exercise 14.2. Write the expressions in terms of elementary logarithms u =

logb (x), v = logb (y), and w = logb (z) (whichever are applicable). Assume that
14.3. EXERCISES 207

x, y, z > 0.
√3
p p √

a) log(x3 · y) b) log( x2 · 4 y 7 ) c) log x· 3 y

 3  2  q 3

x·y
d) ln xy4 e) ln √xy·z 2 f) log3 √
z
√   √  q √ 
4 3 y·z 4
x ·z 100 5 z 3
g) log2 y3
h) log y2
i) ln e2

Exercise 14.3. Solve for x without using a calculator.

a) 6x−2 = 36 b) 23x−8 = 16
1
c) 105−x = 0.0001 d) 55x+7 = 125
e) 2x = 64x+1 f) 4 x+3
= 32x
g) 134+2x = 1 h) 3x+2 = 27x−3
i) 257x−4 = 52−3x j) 95+3x = 278−2x

Exercise 14.4. Solve for x without using a calculator.

a) ln(2x + 4) = ln(5x − 5) b) ln(x + 6) = ln(x − 2) + ln(3)

c) log2 (x + 5) = log2 (x) + 5 d) log(x) + 1 = log(5x + 380)
e) log(x + 5) + log(x) = log(6) f) log2 (x) + log2 (x − 6) = 4
g) log6 (x) + log6 (x − 16) = 2 h) log5 (x − 24) + log5 (x) = 2
i) log4 (x) + log4 (x + 6) = 2 j) log2 (x + 3) + log2 (x + 5) = 3

Exercise 14.5. Solve for x. First find the exact answer as an expression
involving logarithms. Then approximate the answer to the nearest hundredth
using the calculator.

a) 4x = 57 b) 9x−2 = 7 c) 2x+1 = 31
d) 3.82x+7 = 63 e) 5x+5 = 8x f) 3x+2 = 0.4x
g) 1.022x−9 = 4.35x h) 4x+1 = 5x+2 i) 93−x = 4x−6
j) 2.47−2x = 3.83x+4 k) 49x−2 = 92x−4 l) 1.95−3x−4 = 1.24−7x
Session 15

Applications of the exponential and

the logarithm

15.1 Applications of exponential functions

Before giving specific applications, we note that the exponential function y =
c · bx is uniquely determined by providing any two of its function values.
Example 15.1. Let f (x) = c · bx . Determine the constant c and base b under
the given conditions.
a) f (0) = 5, f (1) = 20 b) f (0) = 3, f (4) = 48
c) f (2) = 160, f (7) = 5 d) f (−2) = 55, f (1) = 7
Solution. a) Applying f (0) = 5 to f (x) = c · bx , we get
5 = f (0) = c · b0 = c · 1 = c

Indeed, in general, we always have f (0) = c for any exponential function. The
base b is then determined by substituting the second equation f (1) = 20.
(÷5)
20 = f (1) = c · b1 = 5 · b =⇒ b=4
Therefore, f (x) = 5 · 4x . Note that in the last implication, we used that the
base must be positive.
b) As before, we get 3 = f (0) = c · b0 = c, and
1
(÷3) (exponentiate by ) 1
48 = f (4) = c · b4 = 3 · b4 =⇒ 16 = b4 =⇒ 4
b = 16 4 = 2

208
15.1. APPLICATIONS OF EXPONENTIAL FUNCTIONS 209

Therefore, f (x) = 3 · 2x .
c) When f (0) is not given, it is easiest to solve for b first. We can see this
as follows. Since 160 = f (2) = c · b2 and 5 = f (7) = c · b7 , the quotient of
these equations eliminates c.

160 c · b2 1
= 7
= 5 =⇒ 32 = b−5
5 c·b b
(exponentiate by (− 15 )) 1 1 1
=⇒ b = 32− 5 = 1 =
32 5 2

Then c is determined by any of the original equations.

 1 2 1
2
160 = f (2) = c · b = c · =c· =⇒ c = 4 · 160 = 640
2 4
 x
Therefore, f (x) = 640 · 21 .
d) This solution is similar to part (c).

55 f (−2) c · b−2 1 7
= = 1
= 3 =⇒ b3 =
7 f (1) c·b b 55
 7  13
=⇒ b= ≈ 0.503
55

 7  13 −2  7  −2
3
55 = f (−2) = c · b−2 = c · =c·
55 55
55  7 32  1 2 √ √
3 3
=⇒ c =   −2 = 55 · 2 = 55 3 · 7 3 = 55 · 72 = 2695 ≈ 13.916
3
7 55 3
55

√
3
q
7
x
therefore, f (x) = 2695 · 3
55
.

Many situations are modeled by exponential functions.

Example 15.2. The mass of a bacteria sample is 2 · 1.02t grams after t hours.

a) What is the mass of the bacteria sample after 4 hours?

b) When will the mass reach 10 grams?
210 SESSION 15. APPLICATIONS OF EXP AND LOG

Solution. a) The formula for the mass y in grams after t hours is y(t) = 2·1.02t.
Therefore, after 4 hours, the mass in grams is

y(4) = 2 · 1.024 ≈ 2.16.

b) We are seeking the number of hours t for which y = 10 grams. Therefore,

we have to solve:
(÷2)
10 = 2 · 1.02t =⇒ 5 = 1.02t

We need to solve for the variable in the exponent. In general, to solve for a
variable in the exponent requires an application of a logarithm on both sides
of the equation.
(apply log)
5 = 1.02t =⇒ log(5) = log(1.02t )

Recall one of the main properties that we need to solve for t:

log(xt ) = t · log(x) (15.1)

Using this fact, we can now solve for t:

log(5) = log(1.02t ) =⇒ log(5) = t · log(1.02)

(divide by log(1.02)) log(5)
=⇒ =t
log(1.02)
✞ ☎
Using the lploglp key on the calculator, we may approximate this as
✝ ✆

log(5)
t= log(1.02)
≈ 81.3

After approximately 81.3 hours, the mass will be 10 grams.

Note that the formula 2 · 1.02t comes from the following calculations. Sup-
pose that after each hour there is an increase of 2 percent of the mass and
that the initial mass of the bacteria is 2 grams. Then after the first hour we
see that the mass in grams is

2 + 2 · (.02) = 2(1 + .02) = 2(1.02)

15.1. APPLICATIONS OF EXPONENTIAL FUNCTIONS 211

and that after the second hour the mass is

2(1.02) + 2(1.02)(.02) = 2(1.02)(1 + .02) = 2(1.02)2
and that after three hours the mass is
2(1.02)2 + 2(1.02)2(.02) = 2(1.02)2 (1 + .02) = 2(1.02)3 .
Continuing in this manner we can see that, at least for whole numbers of
hours t, the weight is given by
2 · 1.02t .
The same sort of idea can be used in all of the following applications.
Example 15.3. The population size of a country was 12.7 million in the year
2000, and 14.3 million in the year 2010.
a) Assuming an exponential growth for the population size, find the formula
for the population depending on the year t.
b) What will the population size be in the year 2015, assuming the formula
holds until then?
c) When will the population reach 18 million?
Solution. a) The growth is assumed to be exponential, so that y(t) = c · bt
describes the population size depending on the year t, where we set t = 0
corresponding to the year 2000. Then the example describes y(0) = c as
c = 12.7, which we assume in units of millions. To find the base b, we
substitute the data of t = 10 and y(t) = 14.3 into y(t) = c · bt .
14.3  14.3  101 1
14.3 = 12.7 · b10 =⇒ = b10 =⇒ = (b10 ) 10 = b
12.7 12.7
 14.3  101
=⇒ b = ≈ 1.012
12.7
The formula for the population size is y(t) ≈ 12.7 · 1.012t .
b) We calculate the population size in the year 2015 by setting t = 15:
y(15) = 12.7 · 1.01215 ≈ 15.2
c) We seek t so that y(t) = 18. We solve for t using the logarithm.
18  18 
18 = 12.7 · 1.012t =⇒ = 1.012t =⇒ log = log(1.012t )
12.7 12.7
212 SESSION 15. APPLICATIONS OF EXP AND LOG
 18 
=⇒ log = t · log(1.012)
12.7
 
18
log 12.7
=⇒ t= ≈ 29.2
log(1.012)

The population will reach 18 million in the year 2029.

In many instances the exponential function f (x) = c · bx is given via a rate

of growth r.

Definition 15.4. An exponential function with a rate of growth r is a function

f (x) = c · bx with base
b=1+r .

For an example of why this is a reasonable definition see the note following
Example 15.2.

Note 15.5. We want to point out that some authors use a different convention than the one given in
definition 15.4. Indeed, sometimes a function with rate of growth r is defined as an exponential function
2
with base b = er , whereas for us it has base b = 1+r. Since er can be expanded as er = 1+r + r2 +. . . ,
we see that the two versions only differ by an error of order 2.

Example 15.6. The number of PCs that are sold in the U.S. in the year 2011
is approximately 350 million with a rate of growth of 3.6% per year. Assuming
the rate stays constant over the next years, how many PCs will be sold in the
year 2015?

Solution. Since the rate of growth is r = 3.6% = 0.036, we obtain a base of

b = 1 + r = 1.036, giving the number of PCs to be modeled by c(1.036)t . If
we set t = 0 for the year 2011, we find that c = 350, so the number of sales
is given by y(t) = 350 · 1.036t . Since the year 2015 corresponds to t = 4, we
can calculate the number of sales in the year 2015 as

y(4) = 350 · 1.0364 ≈ 403.

Approximately 403 million PCs will be sold in the year 2015.

Example 15.7. The size of an ant colony is decreasing at a rate of 1% per

month. How long will it take until the colony has reached 80% of its original
size?
15.1. APPLICATIONS OF EXPONENTIAL FUNCTIONS 213

Solution. Since r = −1% = −0.01, we obtain the base b = 1+r = 1−0.01 =

0.99. We have a colony size of y(t) = c · 0.99t after t months, where c is the
original size. We need to find t so that the size is at 80% of its original size
c, that is, y(t) = 80% · c = 0.8 · c.
(÷c)
0.8 · c = c · 0.99t =⇒ 0.8 = 0.99t =⇒ log(0.8) = log(0.99t )
=⇒ log(0.8) = t · log(0.99)
log(0.8)
=⇒ t= ≈ 22.2
log(0.99)
After approximately 22.2 months, the ant colony has decreased to 80% of its
original size.
Example 15.8. a) The population size of a country is increasing at a rate
of 4% per year. How long does it take until the country has doubled
its population size?
b) The number of new flu cases is decreasing at a rate of 5% per week.
After how much time will the number of new flu cases reach a quarter
of its current level?
Solution. a) The rate of change is r = 4% = 0.04, so that population size is an
exponential function with base b = 1 + 0.04 = 1.04. Therefore, f (x) = c · 1.04x
denotes the population size, with c being the initial population in the year
corresponding to x = 0. In order for the population size to double, f (x) has
to reach twice its initial size, that is:
(÷c)
f (x) = 2c =⇒ c · 1.04x = 2c =⇒ 1.04x = 2
=⇒ log(1.04x ) = log(2) =⇒ x log(1.04) = log(2)
log(2)
=⇒ x= ≈ 17.7
log(1.04)
It will take about 17.7 years until the population size has doubled.
b) Since the number of new flu cases is decreasing, the rate of growth
is negative, r = −5% = −0.05 per week, so that we have an exponential
function with base b = 1 + r = 1 + (−0.05) = 0.95. To reach a quarter of its
initial number of flu cases, we set f (x) = c · 0.95x equal to 14 c.
1 (÷c) 1 1
c · 0.95x = c =⇒ 0.95x = =⇒ x log(0.95) = log( )
4 4 4
214 SESSION 15. APPLICATIONS OF EXP AND LOG

log( 41 )
=⇒ x= ≈ 27.0
log(0.95)

It will therefore take about 27 weeks until the number of new flu cases has
decreased to a quarter of its current level.

15.2 Exercises
Exercise 15.1. Assuming that f (x) = c · bx is an exponential function, find
the constants c and b from the given conditions.

a) f (0) = 4, f (1) = 12 b) f (0) = 5, f (3) = 40

c) f (0) = 3, 200, f (6) = 0.0032 d) f (3) = 12, f (5) = 48
e) f (−1) = 4, f (2) = 500 f) f (2) = 3, f (4) = 15

Exercise 15.2. The number of downloads of a certain software application

was 8.4 million in the year 2005 and 13.6 million in the year 2010.

a) Assuming an exponential growth for the number of downloads, find the

formula for the downloads depending on the year t.
b) Assuming the number of downloads will follow the formula from part (a),
what will the number of downloads be in the year 2015?
c) In which year will the number of downloaded applications reach the 20
million barrier?

Exercise 15.3. The population size of a city was 79, 000 in the year 1990
and 136, 000 in the year 2005. Assume that the population size follows an
exponential function.

a) Find the formula for the population size.

b) What is the population size in the year 2010?
c) What is the population size in the year 2015?
d) When will the city reach its expected maximum capacity of 1, 000, 000
residents?

Exercise 15.4. The population of a city decreases at a rate of 2.3% per year.
After how many years will the population be at 90% of its current size? Round
your answer to the nearest tenth.
15.2. EXERCISES 215

Exercise 15.5. A big company plans to expand its franchise and, with this,
its number of employees. For tax reasons it is most beneficial to expand the
number of employees at a rate of 5% per year. If the company currently has
4, 730 employees, how many years will it take until the company has 6, 000
employees? Round your answer to the nearest hundredth.

Exercise 15.6. An ant colony has a population size of 4, 000 ants and is
increasing at a rate of 3% per week. How long will it take until the ant
population has doubled? Round your answer to the nearest tenth.

Exercise 15.7. The size of a beehive is decreasing at a rate of 15% per month.
How long will it take for the beehive to be at half of its current size? Round
your answer to the nearest hundredth.

Exercise 15.8. If the population size of the world is increasing at a rate of

0.5% per year, how long does it take until the world population doubles in
size? Round your answer to the nearest tenth.
Session 16

More applications: Half-life and

compound interest

We have already encountered applications of the logarithm in solving equa-

tions in section 15.1. In this chapter we solve more equations related to
applications by using the logarithm.

16.1 Half-life
Recall from definition 15.4 on page 212 that a function with rate of growth r
is an exponential function f (x) = c · bx with base b = 1 + r. There is also
another important way of determining the base of an exponential function,
which is given by the notion of half-life. We start with a motivating example.

x
Example 16.1. Consider the function f (x) = 200 · 12 7 . We calculate the
function values f (x), for x = 0, 7, 14, 21, and 28.
  70
1
f (0) = 200 · = 200 · 1 = 200
2
  77
1 1
f (7) = 200 · = 200 · = 100
2 2
  147
1 1
f (14) = 200 · = 200 · = 50
2 4
  217
1 1
f (21) = 200 · = 200 · = 25
2 8

216
16.1. HALF-LIFE 217

  287
1 1
f (28) = 200 · = 200 · = 12.5
2 16

From this calculation, we can see how the function values of f behave: starting
from f (0) = 200, the function takes half of its value whenever x is increased by
7. For this reason, we say that f has a half-life of 7. (The general definition
will be given below.) The graph of the function is displayed below.
300 y

250

200

150

100

50

0 x
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-50

We collect the ideas that are displayed in the above example in the defi-
nition and observation below.
Definition 16.2. Let f be an exponential function f (x) = c · bx with a domain
of all real numbers, D = R. Then we say that f has a half-life of h, if the
base is given by
  h1
1
b= (16.1)
2
Note that we can also write h in terms of b. Converting (16.1) into a loga-
log b log 12

rithmic equation gives h1 = log 1 (b) = log 1 , so that h = log b = logb
1
2
.
2 2

Observation 16.3. Let f be the exponential function given for some real con-
stants c > 0 and half-life h > 0, that is
  h1 !x   hx
1 1
f (x) = c · =c· .
2 2
Then we can calculate f (x + h) as follows:
  x+h   hx + hh   hx +1
1 h 1 1
f (x + h) = c · =c· =c·
2 2 2
218 SESSION 16. HALF-LIFE AND COMPOUND INTEREST
  hx  1
1 1 1
= c· · = · f (x)
2 2 2

To summarize, f has the following property:

1
f (x + h) = f (x) for all x ∈ R. (16.2)
2

The above equation shows that, whenever we add an amount of h to an input

x, the effect on f is that the function value decreases by half its previous
value. This is also displayed in the graph below.
y

h 2h 3h 4h

We will sometimes use a different letter for the input variable. In particular,
x t
the function f (x) = c · ( 12 ) h is the same as the function f (t) = c · ( 21 ) h .

Many radioactive isotopes decay with well-known half-lives.

Example 16.4.

a) Chromium-51 has a half-life of 27.7 days1 . How much of 3 grams of

chromium-51 will remain after 90 days?

b) An isotope decays within 20 hours from 5 grams to 2.17 grams. Find the
half-life of the isotope.

t
Solution. a) We use the above formula y = c · 12 h , where c = 3 grams is the
initial amount of chromium-51, h = 27.7 days is the half-life of chromium-51,
1
Half-lives are taken from: http://en.wikipedia.org/wiki/List of isotopes by half-life
16.1. HALF-LIFE 219

and t = 90 days is time that the isotope decayed. Substituting these numbers
into the formula for y, we obtain:
90
  27.7
1
y = 3· ≈ 0.316
2
Therefore, after 90 days, 0.316 grams of the chromium-51 is remaining.
b) We have an initial amount of c = 5 grams and a remaining amount of
y = 2.17 grams after t = 20 hours. The half-life can be obtained as follows.
  20h   20h
1 (÷5) 1
2.17 = 5 · =⇒ 0.434 =
2 2
(apply ln)
 20 
=⇒ ln(0.434) = ln 0.5 h
20
=⇒ ln(0.434) = · ln (0.5)
h
h
(× ln(0.434) ) 20 · ln(0.5)
=⇒ h=
ln(0.434)
=⇒ t ≈ 16.6

Therefore, the half-life of the isotope is approximately 16.6 hours.

Note 16.5. An important isotope is the radioisotope carbon-14. It decays
with a half-life of 5730 years with an accuracy of ±40 years. For definiteness
we will take 5730 years as the half-life of carbon-14.

The half-life of carbon-14 is 5730 years.

One can use the knowledge of the half-life of carbon-14 in dating organic
materials via the so called carbon dating method. Carbon-14 is produced by
a plant during the process of photosynthesis at a fixed level until the plant
dies. Therefore by measuring the remaining amount of carbon-14 in a dead
plant one can determine the date when the plant died. Furthermore, since
humans and animals consume plants, the same argument can be applied to
determine their (approximate) dates of death.

Example 16.6.
a) A dead tree trunk has 86% of its original carbon-14. (Approximately) how
many years ago did the tree die?
220 SESSION 16. HALF-LIFE AND COMPOUND INTEREST

b) A dead animal at an archeological site has lost 41.3% of its carbon-14.

When did the animal die?

t
Solution. a) Using the function y = c · 21 h , where c is the amount of carbon-
14 that was produced by the tree until it died, y is the remaining amount
to date, t is the time that has passed since the tree has died, and h is the
half-life of carbon-14. Since 86% of the carbon-14 is left, we have y = 86%·c.
Substituting the half-life h = 5730 of carbon-14, we can solve for t.
t
  5730 t
  5730
1 (÷c) 1
0.86 · c = c · =⇒ 0.86 =
2 2
(apply ln)
 t

=⇒ ln(0.86) = ln 0.5 5730

t
=⇒ ln(0.86) = · ln (0.5)
5730
5730
(× ln(0.5) ) 5730
=⇒ · ln(0.86) = t
ln(0.5)
=⇒ t ≈ 1247
Therefore, the tree died approximately 1247 years ago.
b) Since 41.3% of the carbon-14 is gone, 100% − 41.3% = 58.7% is

t
remaining. Using y = c · 21 h with y = 58.7% · c and h = 5730, we obtain
t
  5730 t
  5730
1 (÷c) 1
0.587 · c = c · =⇒ 0.587 =
2 2
(apply ln)
 t

=⇒ ln(0.587) = ln 0.5 5730
t
=⇒ ln(0.587) = · ln (0.5)
5730
5730
(× ln(0.5) ) 5730
=⇒ · ln(0.587) = t
ln(0.5)
=⇒ t ≈ 4404
The animal died 4404 years ago.

16.2 Compound interest

There is another interesting and important application of the exponential
function given by calculating the interest on an investment. We start with the
16.2. COMPOUND INTEREST 221

following motivating example.

Example 16.7.
• We invest an initial amount of P = $500 for 1 year at a rate of r = 6%.
The initial amount P is also called the principal.
After 1 year, we receive the principal P together with the interest r ·
P generated from the principal. The final amount A after 1 year is
therefore
A = $500 + 6% · $500 = $500 · (1 + 0.06) = $530.

• We change the setup of the previous example by taking a quarterly

compounding. This means, that instead of receiving interest on the
principal once at the end of the year, we receive the interest 4 times
within the year (after each quarter). However, we now receive only 14 of
the interest rate of 6%. We break down the amount received after each
quarter.
 0.06 
after first quarter: 500 · 1 + = 500 · 1.015
4
 0.06 
after second quarter: (500 · 1.015) · 1 + = 500 · 1.0152
4
 0.06 
2
after third quarter: (500 · 1.015 ) · 1 + = 500 · 1.0153
4
 0.06 
3
after fourth quarter: A = (500 · 1.015 ) · 1 + = 500 · 1.0154
4
=⇒ A ≈ 530.68
Note that in the second quarter, we receive interest on the amount we
had after the first quarter, and so on. So, in fact, we keep receiving inter-
est on the interest of the interest, etc. For this reason, the final amount
received after 1 year A = $530.68 is slightly higher when compounded
quarterly than when compounded annually (where A = $530.00).
• We make an even further variation from the above. Instead of investing
money for 1 year, we invest the principal for 10 years at a quarterly
compounding. We then receive interest every quarter for a total of
4 · 10 = 40 quarters.
 0.06 
after first quarter: 500 · 1 + = 500 · 1.015
4
222 SESSION 16. HALF-LIFE AND COMPOUND INTEREST
 0.06 
after second quarter: (500 · 1.015) · 1 + = 500 · 1.0152
4
..
.
 0.06 
after fortieth quarter: A = (500 · 1.01539 ) · 1 + = 500 · 1.01540
4
=⇒ A ≈ 907.01

We state our observations from the previous example in the following

general observation.

Observation 16.8. A principal (=initial amount) P is invested for t years at

a rate r and compounded n times per year. The final amount A is given by


 P = principal (=initial) amount

 r n·t  A = final amount

A = P · 1+ where r = annual interest rate
n 


 n = number of compoundings per year

t = number of years

We can consider performing the compounding in smaller and smaller incre-

ments. Instead of quarterly compounding, we may take monthly compounding,
or daily, hourly, secondly compounding or compounding in even smaller time
intervals. Note, that in this case the number of compoundings n in the above
formula tends to infinity. In the limit when the time intervals go to zero, we
obtain what is called continuous compounding.

Observation 16.9. A principal amount P is invested for t years at a rate r

and with continuous compounding. The final amount A is given by


 P = principal amount

A = final amount
A = P · er·t where

 r = annual interest rate

t = number of years

Note 16.10. The reason the exponential function appears in the above formula is that the exponential is
the limit of the previous formula in observation 16.8, when n approaches infinity; compare this with equation
(13.1) on page 187.  r n
er = lim 1 +
n→∞ n
A more detailed discussion of limits (including its definition) will be provided in a calculus course.
16.2. COMPOUND INTEREST 223

Example 16.11. Determine the final amount received on an investment under

the given conditions.

a) $700, compounded monthly, at 4%, for 3 years

b) $2500, compounded semi-annually, at 5.5%, for 6 years
c) $1200, compounded continuously, at 3%, for 2 years

Solution. We can immediately use the formula by substituting the given val-
ues. For part (a), we have P = 700, n = 12, r = 4% = 0.04, and t = 3.
Therefore, we calculate
 12·3  36
0.04 0.04
A = 700 · 1 + = 700 · 1 + ≈ 789.09
12 12

b) We have P = 2500, n = 2, r = 5.5% = 0.055, and t = 6.

 2·6
0.055
A = 2500 · 1 + ≈ 3461.96
2

c) We have P = 1200, r = 3% = 0.03, t = 2, and we use the formula for

continuous compounding.

A = 1200 · e0.03·2 = 1200 · e0.06 ≈ 1274.20

✞ ☎
(Note that the Euler number is entered in the TI-84 via the keys lp2nd lp
✞ ☎ ✝ ✆
and lplnlp .)
✝ ✆
224 SESSION 16. HALF-LIFE AND COMPOUND INTEREST

Instead of asking to find the final amount, we may also ask about any of
the other variables in the above formulas for investments.
Example 16.12.

a) Find the amount P that needs to be invested at 4.275% compounded

annually for 5 years to give a final amount of $3000. (This amount P is
also called the present value of the future amount of $3000 in 5 years.)
b) At what rate do we have to invest $800 for 6 years compounded quarterly
to obtain a final amount of $1200?
c) For how long do we have to invest $1000 at a rate of 2.5% compounded
continuously to obtain a final amount of $1100?
d) For how long do we have to invest at a rate of 3.2% compounded monthly
until the investment doubles its value?

Solution. a) We have the following data: r = 4.275% = 0.04275, n = 1,

t = 5, and A = 3000. We want to find the present value P . Substituting the
given numbers into the appropriate formula, we can solve this for P .
 1·5
0.04275
3000 = P · 1 + =⇒ 3000 = P · (1.04275)5
1
(divide by 1.042755 ) 3000
=⇒ P = ≈ 2433.44
1.042755
Therefore, if we invest $2433.44 today under the given conditions, then this
will be worth $3000 in 5 years.
b) Substituting the given numbers (P = 800, t = 6, n = 4, A = 1200) into
the formula gives:
 r 4·6 (divide by 800) 1200  r 24
1200 = 800 · 1 + =⇒ = 1+
4 800 4
 r 24 3
=⇒ 1+ =
4 2
Next, we have to get the exponent 24 to the right side. This is done by taking
q   241
1
a power of 24 , or in other words, by taking the 24th root, 24 32 = 32 .

 1   241   241
r 24 24 3  r 24· 241 3
1+ = =⇒ 1+ =
4 2 4 2
16.2. COMPOUND INTEREST 225

  241   241   241 !

r 3 r 3 3
=⇒ 1+ = =⇒ = −1 =⇒ r = 4· −1
4 2 4 2 2

We
✞ plug this
☎ into
✞ the☎calculator. Note that the nth root is given by pressing
lpmathlp and lp5 lp .
✝ ✆ ✝ ✆

More precisely, the 24th root of 3/2 (highlighted above) can be entered by
pressing the following keys:
✞ ☎
✞ ☎ ✞ ☎
✞ ☎ ✞ ☎
✞ ☎
✞ ☎ ✞ ☎
lp2 lp lp4 lp lpmathlp lp5 lp lp3 lp lp÷ lp lp2 lp lpenter lp
✝ ✆
✝ ✆ ✝ ✆
✝ ✆ ✝ ✆
✝ ✆
✝ ✆ ✝ ✆

Therefore our answer is r ≈ 0.06815 = 6.815%.

c) Again we substitute the given values, P = 1000, r = 2.5% = 0.025,
A = 1100, but now we use the formula for continuous compounding.
1100
1100 = 1000 · e0.025·t =⇒ = e0.025·t =⇒ e0.025·t = 1.1
1000
To solve for the variable t in the exponent, we need to apply the logarithm.
Here, it is most convenient to apply the natural logarithm, because ln(x) is
inverse to ex (which appears in the equation)
✞ ☎and this logarithm is directly
implemented on the calculator via the lplnlp key. By applying ln to both
✝ ✆
sides, since ln(x) is inverse to ex , we see that .025t = ln(1.1).
Alternatively,

ln(e0.025·t ) = ln(1.1) =⇒ 0.025 · t · ln(e) = ln(1.1)

Here we have used that logb (xn ) = n · logb (x) for any number n as we have
seen in proposition 14.1. Observing that ln(e) = 1, which is the special case
of the second equation in (13.3) on page 192 for the base b = e, the above
becomes
ln(1.1)
0.025 · t = ln(1.1) =⇒ t= ≈ 3.81
0.025
Therefore, we have to wait 4 years until the investment is worth (more than)
$1100.
226 SESSION 16. HALF-LIFE AND COMPOUND INTEREST

d) We are given that r = 3.2% = 0.032 and n = 12, but no initial amount P
is provided. We are seeking to find the time t when the investment doubles.
This means that the final amount A is twice the initial amount P , or as a
formula: A = 2 · P . Substituting this into the investment formula and solving
gives the wanted answer.
 12·t  12·t
0.032 (divide by P ) 0.032
2P = P · 1 + =⇒ 2= 1+
12 12
 12·t !  
(apply ln) 0.032 0.032
=⇒ ln(2) = ln 1+ =⇒ ln(2) = 12·t·ln 1 +
12 12
0.032
ln(2)


(divide by 12 · ln 1 + 12
)
=⇒ t= 0.032

≈ 21.69
12 · ln 1 + 12

So, after approximately 21.69 years, the investment will have doubled in value.

16.3 Exercises
Exercise 16.1. An unstable element decays at a rate of 5.9% per minute. If
40mg of this element has been produced, then how long will it take until 2mg
of the element are left? Round your answer to the nearest thousandth.

Exercise 16.2. A substance decays radioactively with a half-life of 232.5 days.

How much of 6.8 grams of this substance is left after 1 year?

Exercise 16.3. Fermium-252 decays in 10 minutes to 76.1% of its original

mass. Find the half-life of fermium-252.

Exercise 16.4. How long do you have to wait until 15mg of beryllium-7 have
decayed to 4mg, if the half-life of beryllium-7 is 53.12 days?

Exercise 16.5. If Pharaoh Ramses II died in the year 1213 BC, then what
percent of the carbon-14 was left in the mummy of Ramses II in the year
2000?

Exercise 16.6. In order to determine the age of a piece of wood, the amount
of carbon-14 was measured. It was determined that the wood had lost 33.1%
of its carbon-14. How old is this piece of wood?
16.3. EXERCISES 227

Exercise 16.7. Archaeologists uncovered a bone at an ancient resting ground.

It was determined that 62% of the carbon-14 was left in the bone. How old
is the bone?

Exercise 16.8. An investment of $5, 000 was locked in for 30 years. According
to the agreed conditions, the investment will be worth $5, 000 · 1.08t after t
years.

a) How much is the investment worth after 5 years?

b) After how many years will the investment be worth $20, 000?

Exercise 16.9. Determine the final amount in a savings account under the
given conditions.

a) $700, compounded quarterly, at 3%, for 7 years

b) $1400, compounded annually, at 2.25%, for 5 years
c) $1400, compounded continuously, at 2.25%, for 5 years
d) $500, compounded monthly, at 3.99%, for 2 years
e) $5000, compounded continuously, at 7.4%, for 3 years
f) $1600, compounded daily, at 3.333%, for 1 year
g) $750, compounded semi-annually, at 4.9%, for 4 years

Exercise 16.10.

a) Find the amount P that needs to be invested at a rate of 5% compounded

quarterly for 6 years to give a final amount of $2000.
b) Find the present value P of a future amount of A = $3500 invested at
6% compounded annually for 3 years.
c) Find the present value P of a future amount of $1000 invested at a rate
of 4.9% compounded continuously for 7 years.
d) At what rate do we have to invest $1900 for 4 years compounded monthly
to obtain a final amount of $2250?
e) At what rate do we have to invest $1300 for 10 years compounded con-
tinuously to obtain a final amount of $2000?
f) For how long do we have to invest $3400 at a rate of 5.125% compounded
annually to obtain a final amount of $3700?
g) For how long do we have to invest $1000 at a rate of 2.5% compounded
continuously to obtain a final amount of $1100?
228 SESSION 16. HALF-LIFE AND COMPOUND INTEREST

h) How long do you have to invest a principal at a rate of 6.75% com-

pounded daily until the investment doubles its value?
i) An certain amount of money has tripled its value while being in a savings
account that has an interest rate of 8% compounded continuously. For
how long was the money in the savings account?
Review of exponential and
logarithmic functions

Exercise III.1. The population of a country grows exponentially at a rate of

1% per year. If the population was 35.7 million in the year 2010, then what
is the population size of this country in the year 2015?

Exercise III.2. A radioactive substance decays exponentially at a rate of 7%

per hour. How long do you have to wait until the substance has decayed to
1
4
of its original size?

Exercise III.3. Combine to an expression with only one logarithm.

2 1 3
a) ln(x) + 4 ln(y), b) log2 (x) − log2 (y) + 3 log2 (z)
3 2 4
Exercise III.4. Assuming that x, y > 0, write the following expressions in
terms of u = log(x) and v = log(y):
√3
!
x4  p  p 
a) log 2
, 5
b) log x y , c) log 5 xy 4
y

Exercise III.5. Solve without using the calculator: log3 (x) + log3 (x−8) = 2

Exercise III.6.
a) Find the exact solution of the equation: 6x+2 = 7x
b) Use the calculator to approximate your solution from part (a).

Exercise III.7. 45mg of fluorine-18 decay in 3 hours to 14.4mg. Find the

half-life of fluorine-18.

Exercise III.8. A bone has lost 35% of its carbon-14. How old is the bone?

229
Exercise III.9. How much do you have to invest today at 3% compounded
quarterly to obtain $2, 000 in return in 3 years?

Exercise III.10. $500 is invested with continuous compounding. If $692.01 is

returned after 5 years, what is the interest rate?

230
 Part IV

Trigonometric functions

231
Session 17

Trigonometric functions

17.1 Review of basic trigonometric definitions and

facts
In this part, we will investigate trigonometric functions, such as y = sin(x),
y = cos(x), and y = tan(x) in terms of their function theoretic aspects. Before
we graph these functions, we recall the basic definitions and the main facts,
which we will consider as known background material.

Definition 17.1. An angle can be expressed in degree or in radian measure.

The relationship between degrees and radians is given by

π = 180◦ (17.1)

π
= 90◦ π
2π
3
= 120◦ 2
3
= 60◦
π
3π
4
= 135◦ 4
= 45◦
π
5π
6
= 150◦ 6
= 30◦

π = 180◦ 0 = 0◦

7π 11π
6
= 210◦ 6
= 330◦
5π 7π
4
= 225◦ 4
= 315◦
4π 5π
= 240◦ 3π = 300◦
3
2
= 270◦ 3

232
17.1. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS 233

Definition 17.2. Let x be an angle. Consider the terminal side of the angle
x, and assume that the point P (a, b) is a point on the terminal side of x. If
r is the distance from P to the origin (0, 0), then we define the sine, cosine,
tangent, cosecant, secant, and cotangent as follows:

terminal side of x a2 + b2 = r 2
√
P (a, b) =⇒ r = a2 + b2
b

r
b r
sin(x) = r
csc(x) = b
x
a r
cos(x) = r
sec(x) = a
a
b a
tan(x) = a
cot(x) = b

There are some immediate consequences from the above definition.

1 1
csc(x) = sec(x) =
sin(x) cos(x)
sin(x) cos(x) 1
tan(x) = cot(x) = =
cos(x) sin(x) tan(x)

If we take r = 1 in the above we have the following ‘unit circle’ definition

of sin and cos:
(cos(x), sin(x))

sin(x)

r=1

cos(x)
234 SESSION 17. TRIGONOMETRIC FUNCTIONS

That is to say, that the point where the terminal side of the angle x
intersects the unit circle is (cos(x), sin(x)) (which can serve as a definition
cos and sin as functions of x).
For the following, we use the short hand notation:

Definition 17.3. Define

sin2 α := (sin α)2 , cos2 α := (cos α)2 , and tan2 α := (tan α)2 .

From these definitions, we obtain that

sin2 (x) + cos2 (x) = 1 sec2 (x) = 1 + tan2 (x) (17.2)

2
2 2 2 2
since sin2 (x) + cos2 (x) = rb + ar = b r+a 2 = rr2 = 1, and, similarly,

2 2 2 2
2
1 + tan2 (x) = 1 + ab = a a+b 2 = ar 2 = ar = sec2 (x). Furthermore, we can
see from the definition that

sin(x + 2π) = sin(x) sin(−x) = − sin(x) sin(π − x) = sin(x)

cos(x + 2π) = cos(x) cos(−x) = cos(x) cos(π − x) = − cos(x)

(17.3)
For a point P (a, b) the x-coordinate a is positive when P is in quadrant I
and IV, whereas the y-coordinate b is positive when P is in quadrant I and II.
The length of the hypothenuse r is always positive. Thus the trigonometric
functions are positive/negative according to the chart:

Quadrant II Quadrant I

sin(x) is positive sin(x) is positive

cos(x) is negative cos(x) is positive

tan(x) is negative tan(x) is positive

Quadrant III Quadrant IV

sin(x) is negative sin(x) is negative

cos(x) is negative cos(x) is positive

tan(x) is positive tan(x) is negative

17.1. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS 235

There are two basic triangles that are used to calculate exact values of
the trigonometric functions. These two triangles are the 45◦ − 45◦ − 90◦ and
30◦ − 60◦ − 90◦ triangles, which can be described with specific side lengths.

45◦ − 45◦ − 90◦ 30◦ − 60◦ − 90◦

√
2 45◦ 2 60◦
1 1

45◦ 90◦ 90◦

◦
30
√
1 3

These basic triangles may be used to calculate the trigonometric functions of

certain angles.

Example 17.4. Find sin(x), cos(x), and tan(x) for the angles

x = 30◦ , x = 45◦ , x = 60◦ , x = 90◦ , x = 0◦ .

Solution. We draw the angle x = 30◦ in the plane, and use the special
triangle to find a point on the terminal side of the angle. From this, we then
read off the trigonometric functions. Here are the function values for 30◦ , 45◦ ,
and 60◦ .

b 1
sin(30◦ ) = r
= 2
1 √
2 cos(30◦ ) = a
= 3
30◦ r 2
√ √
b √1 3
3 tan(30◦ ) = a
= 3
= 3

√ √
2 b √1 2
sin(45◦ ) = r
= 2
= 2
1 √
a √1 2
cos(45◦ ) = r
= 2
= 2
45◦

b 1
1 tan(45◦ ) = a
= 1
=1
236 SESSION 17. TRIGONOMETRIC FUNCTIONS
√ √
3 sin(60◦ ) = b
= 3
r 2
2
a 1
60◦ cos(60◦ ) = r
= 2

b
√
3
√
1 tan(60◦ ) = a
= 1
= 3

Here are the trigonometric functions for 90◦ and 0◦ .

b 1
sin(90◦ ) = r
= 1
=1
1
a 0
90◦
cos(90◦ ) = r
= 1
=0
0 b 1
tan(90◦ ) = a
= 0
is undefined

b 0
sin(0◦ ) = r
= 1
=0
a 1
◦
cos(0◦ ) = r
= 1
=1
0

0 b 0
1 tan(0◦ ) = a
= 1
=0

We collect the results of the previous example in a table:

π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √
1 2 3
sin(x) 0 2 2 2
1
√ √
3 2 1
cos(x) 1 2 2 2
0
√
3
√
tan(x) 0 3
1 3 undef.

Note 17.5. There is an√easy

√way
√ to √
remember
√ the function value given in the above table by writing the
numerator of sin(x) as 0, 1, 2, 3, 4 in increasing order, and for cos(x) in decreasing order:

π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √ √ √ √
0 1 2 3 4
sin(x) 2 2 2 2 2
√ √ √ √ √
4 3 2 1 0
cos(x) 2 2 2 2 2
17.1. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS 237

sin(x)
The values of tan(x) are then determined from this by tan(x) = cos(x)
:

π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √
0
1 √ 2 3 √
tan(x) =0 2
√ = √1 = 3 √2
=1 2
1 = 3 “ 10 ” is undef.
1 3 3 3 2
2 2 2

We can use the special 45◦ − 45◦ − 90◦ and 30◦ − 60◦ − 90◦ triangles to
also find other values of the trigonometric functions.

Example 17.6. Find sin(x), cos(x), and tan(x) for the following angles.
11π −9π
a) x = 240◦, b) x = 495◦ , c) x = 6
, d) x = 4

Solution. a) We graph the angle x = 240◦, and identify a special 30◦ − 60◦ −
90◦ triangle using the terminal side of x.

−1 240◦

2
√
P − 3

We identify
√ a point P on the terminal side of x. This point P has coordinates
P (−1, − 3), which can be seen by considering one of two 30◦ − 60◦ − 90◦
triangles in the plane (shaded above). The length
√ of the line segment from P
to the origin (0, 0) is 2. Thus, a = −1, b = − 3, r = 2. We get
√ √
◦ − 3 ◦ −1 ◦ − 3 √
sin(240 ) = , cos(240 ) = , tan(240 ) = = 3
2 2 −1
b) The angle x = 495◦ is greater than 360◦ . However, the trigonometric
functions are invariant under addition or subtraction of 360◦ :

sin(x ± 360◦ ) = sin(x) cos(x ± 360◦ ) = cos(x) tan(x ± 360◦) = tan(x)

Since 495◦ − 360◦ = 135◦ , we have sin(495◦) = sin(135◦ ), and similarly

for cos and tan. Graphing the terminal side of x, and identifying a special
238 SESSION 17. TRIGONOMETRIC FUNCTIONS

45◦ − 45◦ − 90◦ triangle, we find the coordinates of a point P on the terminal
side with coordinates P (−1, 1).

P
√ 1
2 135◦

−1

√
Therefore, a = −1, b = 1, r = 2. We obtain
√ √
1 2 −1 − 2 1
sin(495◦ ) = √ = , cos(495◦ ) = √ = , tan(495◦) = = −1
2 2 2 2 −1
11π
c) Converting 6
into degrees, we have

11π 11π 180◦

x= = · = 11 · 30◦ = 330◦ .
6 6 π
Drawing the terminal side, we obtain the point P .

√
330◦ 3

2
−1
P

√ √
The point P has coordinates P ( 3, −1). Therefore, a = 3, b = −1, r = 2,
and
 11π  −1  11π  √3  11π  −1 √
− 3
sin = , cos = , tan =√ =
6 2 6 2 6 3 3
◦
d) Converting the angle x = −9π 4
= −9π
4
· 180
π
= −9 · 45◦ = −405◦ , we see
that the trigonometric functions of x are the same as of −405◦ + 360◦ = −45◦ ,
17.2. SIN, COS, AND TAN AS FUNCTIONS 239

and also of −45◦ + 360◦ = 315◦. We can draw either −45◦ or 315◦ to find a
point P on the terminal side of x.

315◦ 1
−45◦
√
2
−1
P
√
The point P (1, −1) determines a = 1, b = −1, r = 2, and so
 −9π  −1 √  −9π  √
− 2 1 2
sin =√ = , cos =√ = ,
4 2 2 4 2 2
 −9π  −1
tan = = −1
4 1

17.2 sin, cos, and tan as functions

We now turn to function theoretic aspects of the trigonometric functions de-
fined in the last section. In particular, we will be interested in understanding
the graphs of the functions y = sin(x), y = cos(x), and y = tan(x). With an
eye toward calculus, we will take the angles x in radian measure.
Example 17.7. We graph the functions y = sin(x), y = cos(x), and y =
tan(x).
One way to proceed is to calculate and collect various function values in a
table and then graph them. However, this is quite elaborate, as the following
table shows.
x 0 π6 π
4
π
3
π
2
2π
3
3π
4
5π
6
π ...
√ √ √ √
1 2 3 3 2 1
sin(x) 0 2 2 2
1 2 2 2
0 ...
√ √ √ √
3 2 1 −1 − 2 − 3
cos(x) 1 2 2 2
0 2 2 2
−1 ...
√
3
√ √ √
− 3
tan(x) 0 3
1 3 undef. − 3 −1 3
0 ...
240 SESSION 17. TRIGONOMETRIC FUNCTIONS

An easier approach is to use the TI-84 to graph each function. Before entering
the function, we✞need to make
☎ sure that the calculator is in radian mode. For
this, press the lpmodelp key, and in case the third item is set on degree,
✝ ✆ ✞ ☎
switch to radian, and press lpenter lp .
✝ ✆

We may now enter the function y = sin(x) and study its graph.

From this, we can make some immediate observations (which can also be
seen using definition 17.2). First, the graph is bounded between −1 and +1,
because by definition 17.2, we have sin(x) = rb with −r ≤ b ≤ r, so that

−1 ≤ sin(x) ≤ 1, for all x.

We therefore change the window of the graph to display y’s between −2 and
2, to get a closer view.

y = sin(x)

Second, we see that y = sin(x) is a periodic function with period 2π, since
the function doesn’t change its value when adding 360◦ = 2π to its argument
(and this is the smallest non-zero number with that property):

sin(x + 2π) = sin(x)

17.2. SIN, COS, AND TAN AS FUNCTIONS 241

The graph of y = sin(x) has the following specific values:

y = sin(x)

1
−5π −π 3π
2 2 2 2π x
−3π π 5π
−3π −2π 2 −π 2 π 2 3π
−1

| {z }
period 2π

The graph of y = sin(x) has a period of 2π, and an amplitude of 1.

Similarly, we can graph the function y = cos(x). Since −1 ≤ cos(x) ≤ 1
for all x, we graph it also with the zoomed window setting.

y = cos(x)

We see that y = cos(x) is also periodic with period 2π, that is

cos(x + 2π) = cos(x),

and y = cos(x) also has an amplitude of 1, since −1 ≤ cos(x) ≤ 1.
y = cos(x)

1
−3π −π π 3π x
−5π −2π −3π −π π 3π 2π 5π
2 2 2 2 2 2
−1

| {z }
period 2π

Indeed, the graph of y = cos(x) is that of y = sin(x) shifted to the left by π2 .

The reason for this is that
 π
cos(x) = sin x + (17.4)
2
242 SESSION 17. TRIGONOMETRIC FUNCTIONS

Many other properties of sin and cos can be observed from the graph (or
from the unit circle definition). For example, recalling Observation 5.11 from
page 72, we see that the sine function is an odd function, since the graph
of y = sin(x) is symmetric with respect to the origin. Similarly, the cosine
function is an even function, since the graph of y = cos(x) is symmetric with
respect to the y-axis. Algebraically, Definition 5.9 from page 71 therefore
shows that these functions satisfy the following relations:

sin(−x) = − sin(x) and cos(−x) = cos(x) (17.5)

Finally, we come to the graph of y = tan(x). Graphing this function in

the standard window, we obtain:

y = tan(x)

Zooming into this graph, we see that y = tan(x) has vertical asymptotes
x = π2 ≈ 1.6 and x = −π
2
≈ −1.6.

y = tan(x)

This is also supported by the fact that tan( π2 ) and tan(− π2 ) are undefined.
The graph of y = tan(x) with some more specific function values is shown
17.2. SIN, COS, AND TAN AS FUNCTIONS 243

below.
y = tan(x)

1
−π
4 x
−5π −3π −π π π 3π 5π
2 −2π 2 −π 2 4 2 π 2 2π 2

−1

| {z }
period π

We see that the tangent is periodic with a period of π:

tan(x + π) = tan(x) (17.6)

There are vertical asymptotes: x = π2 , −π

2
, 3π
2
, −3π
2
, 5π
2
, −5π
2
, . . . , or, in short
π
asymptotes of y = tan(x) : x=n· , where n = ±1, ±3, ±5, . . .
2
Furthermore, the tangent is an odd function, since it is symmetric with respect
to the origin (see Observation 5.11):

tan(−x) = − tan(x) (17.7)

Recall from section 5.2 how changing the formula of a function affects the
graph of the function, such as:

• the graph of c · f (x) (for c > 0) is the graph of f (x) stretched away from
the x-axis by a factor c (or compressed when 0 < c < 1)

• the graph of f (c · x) (for c > 0) is the graph of f (x) compressed towards

the y-axis by a factor c (or stretched away the y-axis when 0 < c < 1)

• graph of f (x) + c is the graph of f (x) shifted up by c (or down when

c < 0)
244 SESSION 17. TRIGONOMETRIC FUNCTIONS

• graph of f (x + c) is the graph of f (x) shifted to the left by c (or to the

right when c < 0)
With this we can graph some variations of the basic trigonometric functions.
Example 17.8. Graph the functions:
f (x) = sin(x) + 3, g(x) = 4 · sin(x), h(x) = sin(x + 2), i(x) = sin(3x),
j(x) = 2 · cos(x) + 3, k(x) = cos(2x − π), l(x) = tan(x + 2) + 3
Solution. The functions f , g, h, and i have graphs that are variations of the
basic y = sin(x) graph. The graph of f (x) = sin(x) + 3, shifts the graph of
y = sin(x) up by 3, whereas the graph of g(x) = 4·sin(x) stretches y = sin(x)
away from the x-axis.
y

4
f (x) = sin(x) + 3
3

1
−5π −π 3π
2 2 2 2π x
−3π π 5π
−3π −2π 2
−π 2 π 2 3π
−1

−2

−3
g(x) = 4 · sin(x)
−4

The graph of h(x) = sin(x + 2) shifts the graph of y = sin(x) to the left by 2,
and i(x) = sin(3x) compresses y = sin(x) towards the y-axis.
y
h(x) = sin(x + 2)

x
−2π −π π 2π
π−2

y
i(x) = sin(3 · x)

x
−2π −π π 2π
π 2π
3 3
17.2. SIN, COS, AND TAN AS FUNCTIONS 245

Next, j(x) = 2 · cos(x) + 3 has a graph of y = cos(x) stretched by a factor

2 away from the x-axis, and shifted up by 3.
y
j(x) = 2 · cos(x) + 3
5

1
x
−5π −3π −π π 3π 5π
−3π 2 −2π 2 −π 2 2 π 2 2π 2 3π
−1

−2

For the graph of k(x) = cos(2x−π) = cos(2·(x− π2 )), we need to compress

the graph of y = cos(x) by a factor 2 (we obtain the graph of the function
y = cos(2x)) and then shift this by π2 to the right.
y
k(x) = cos(2 · x − π)

x
−2π −π π 2π

We will explore this case below in more generality. In fact, whenever y =

cos(b · x + c) = cos(b · (x − −c
b
)), the graph of y = cos(x) is shifted to the right
−c
by b , and compressed by a factor b, so that it has a period of 2π b
.
Finally, l(x) = tan(x + 2) + 3 shifts the graph of y = tan(x) up by 3 and
to the left by 2.
y

x
−2π −π π 2π
246 SESSION 17. TRIGONOMETRIC FUNCTIONS

We collect some of the observations that were made in the above examples
in the following definition.
Definition 17.9. Let f be one of the functions:

f (x) = a · sin(b · x + c) or f (x) = a · cos(b · x + c)

2π
The number |a| is called the amplitude, the number b
is the period, and
the number −c
b
is called the phase shift.
2π
In physical applications, the period is sometimes denoted by T = b
and the frequency is the reciprocal

1
f = .
T

Using the amplitude, period and phase-shift, we can draw the graph of,
for example, a function f (x) = a · sin(b · x + c) over one period by shifting the
graph of sin(x) by the phase-shift −c b
to the right, then marking one full period
of length 2πb
, and then drawing the basic sin(x) graph with an amplitude of
a (or the reflected graph of − sin(x) when a is negative).
y = a · sin(b · x + c)

|a|

−c −c 2π
b b
+ b x

−|a|

| {z }
period | 2πb |

The zero(s), maxima, and minima of the graph within the drawn period can
be found by calculating the midpoint between ( −c b
, 0) and ( −c
b
+ 2π
b
, 0) and
midpoints between these and the resulting midpoint.
A similar graph can be obtained for f (x) = a · cos(b · x + c) by replacing
the sin(x) graph with the cos(x) graph over one period.
17.2. SIN, COS, AND TAN AS FUNCTIONS 247

Example 17.10. Find the amplitude, period, and phase-shift, and sketch the
graph over one full period.

a) f (x) = 3 sin(2x) b) f (x) = −2 cos(π · x)

c) f (x) = cos(x − 2)
d) f (x) = cos(2x + π)
e) f (x) = −4 · sin x2 + 3 f) f (x) = 3 sin(−2x)

Solution. a) The amplitude is |3| = 3, and since f (x) = 3 · sin(2 · x + 0), it

is b = 2 and c = 0, so that the period is 2π 2
= π and the phase-shift is
−0
2
= 0. This is also supported by graphing the function with the calculator;
in particular the graph repeats after one full period, which is at π ≈ 3.1.
y = 3 · sin(2 · x)

3π
4 x
π π
4 2 π

−3

b) For f (x) = −2 cos(π · x), the amplitude is | − 2| = 2, the period is

2π
π
= 2, and the phase-shift is 0. Note, that the first coefficient is negative,
so that the cos(x) has to be reflected about the x-axis. One full period
is graphed below with a solid line (and the continuation of the graph is
dashed). This graph is again supported by the graph and table as given by
the calculator.
y = −2 · cos(π · x)

1 2

−2

c) The amplitude of f (x) = cos(x − 2) is 1, the period is 2π, and the

248 SESSION 17. TRIGONOMETRIC FUNCTIONS

−(−2)
phase-shift is 1
= 2. The graph over one period is displayed below.
y = cos(x − 2)

1
2+π x

2 π 3π 2+2·π
2+ 2+
−1 2 2

The minimum of the function f in this period is given by taking the value in
the middle of the interval [2, 2 + 2π], that is at 2 + π. With this, the zeros
are given by taking the value in the middle of the intervals [2, 2 + π] and
[2 + π, 2 + 2π], which are at 2 + π2 and 2 + 3π
2
.
d) The function f (x) = cos(2x + π) has amplitude 1, period 2π 2
= π, and
phase-shift 2 . We may thus draw the graph over one period from −π
−π
2
to
−π π
2
+ π = 2 ; see graph on the left below. However, we may also graph the
function over another full period, such as [0, π] (see the graph on the right).
y = cos(2x + π) y = cos(2x + π)

1 1
x π x
−π −π π π π π 3π
2 4 4 2 4 2 4
−1 −1

e) The graph of f (x) = −4 · sin x2 + 3 has an amplitude of | − 4| = 4, a
period of 2π −3
1 = 4π, and a phase shift of 1 = −6. Using this information, and
2 2
confirming this with the calculator, we can draw one full period of the graph
of f .
y = −4 sin( x2 + 3)

−6 −6 + π x

−6 + 2π −6 + 3π −6 + 4π

−4
17.3. EXERCISES 249

2π
f) For f (x) = 3 sin(−2x), we have an amplitude of 3, a period of −2 =π
−0
and a phase-shift of −2 = 0. Comparing this with example (a) above, we
see that we have the same data as for the function y = 3 sin(2x) from part
(a). However, in this case, we have to draw the sine function over one period
moving to the left, which means that the graph has to start the period as a
decreasing function. This can also be seen by observing that the sine function
is an odd function, that is equation (17.5), so that

f (x) = 3 sin(−2x) = −3 sin(2x).

The graph is therefore drawn as follows.

y = 3 · sin(−2 · x)

π
4 x
π 3π
2 4
π

−3

17.3 Exercises
Exercise 17.1. Find sin(x), cos(x), and tan(x) for the following angles.

a) x = 120◦ , b) x = 390◦ , c) x = −150◦ , d) x = −45◦

5π 5π
e) x = 1050◦ , f) x = −810◦ , g) x = 4
, h) x = 6
10π 15π −π −54π
i) x = 3
, j) x = 2
, k) x = 6
, l) x = 8

Exercise 17.2. Graph the function, and describe how the graph can be ob-
tained from one of the basic graphs y = sin(x), y = cos(x), or y = tan(x).

a) f (x) = sin(x) + 2 b) f (x) = cos(x − π) c) f (x) = tan(x) − 4

d) f (x) = 5 · sin(x) e) f (x) = cos(2 · x) f) f (x) = sin(x − 2) − 5
250 SESSION 17. TRIGONOMETRIC FUNCTIONS

Exercise 17.3. Identify the formulas with the graphs.

f (x) = sin(x) + 2, g(x) = tan(x − 1), h(x) = 3 sin(x),

i(x) = 3 cos(x), j(x) = cos(x − π), k(x) = tan(x) − 1,
y y

x x

a) y
b) y

x x

c) y
d) y

x x

e) f)

Exercise 17.4. Find the formula of a function whose graph is the one dis-
played below.

a) b) c)

d) e) f)
17.3. EXERCISES 251

Exercise 17.5. Find the amplitude, period, and phase-shift of the function.

a) f (x) = 5 sin(2x + 3) b) f (x) = sin(πx − 5) c) f (x) = 6 sin(4x)

d) f (x) = −2 cos(x + π4 ) e) f (x) = 8 cos(2x − 6) f) f (x) = 3 sin( x4 )
g) f (x) = − cos(x + 2) h) f (x) = 7 sin( 2π
5
x − 6π
5
) i) f (x) = cos(−2x)

Exercise 17.6. Find the amplitude, period, and phase-shift of the function.
Use this information to graph the function over a full period. Label all maxima,
minima, and zeros of the function.

a) y = 5 cos(2x) b) y = 4 sin(πx) c) y = 2 sin( 2π

3
x)
d) y = cos(2x − π) e) y = cos(πx − π) f) y = −6 cos(− x4 )


g) y = − cos(4x + π) h) y = 7 sin x + π4 i) y = 5 cos(x + 3π 2
)


j) y = 4 sin(5x − π) k) y = −3 cos(2πx − 4) l) y = 7 sin 14 x + π4



m) y = cos(3x − 4π) n) y = 2 sin 15 x − 10
π
o) y = 31 cos 14 5
x − 6π
5
Session 18

Addition of angles and multiple

angle formulas

18.1 Addition and subtraction of angles

In the previous section we found exact values of the trigonometric functions for
specific angles of 0, π3 , π4 , π6 plus possibly any multiple of π2 . Using these val-
ues, we can find many other values of trigonometric functions via the following
addition and subtraction of angles formulas, which we state now.

Proposition 18.1. For any angles α and β, we have

sin(α + β) =
sin α cos β + cos α sin β
sin(α − β) =
sin α cos β − cos α sin β
cos(α + β) =
cos α cos β − sin α sin β
cos(α − β) =
cos α cos β + sin α sin β
tan α + tan β
tan(α + β) =
1 − tan α tan β
tan α − tan β
tan(α − β) =
1 + tan α tan β

Proof. We start with the proof of the formulas for sin(α + β) and cos(α + β) when α and β are angles
between 0 and π2 = 90◦ . We prove the addition formulas (for α, β ∈ (0, π2 )) in a quite elementary way,
and then show that the addition formulas also hold for arbitrary angles α and β.

252
18.1. ADDITION AND SUBTRACTION OF ANGLES 253

To find sin(α + β), consider the following setup.

y

α b
f
d

c e a
β γ

α x

Note, that there are vertically opposite angles, labelled by γ, which are therefore equal. These angles are
angles in two right triangles, with the third angle being α. We therefore see that the angle α appears
again as the angle among the sides b and f . With this, we can now calculate sin(α + β).

opposite e+f e f a f a c f b
sin(α + β) = = = + = + = · + ·
hypotenuse d d d d d c d b d
= sin(α) cos(β) + cos(α) sin(β)

The above figure displays the situation when α + β ≤ π2 . There is a similar figure for π2 < α + β < π.
(We recommend as an exercise to draw the corresponding figure for the case of π2 < α + β < π.)
Next, we prove the addition formula for cos(α + β). The following figure depicts the relevant objects.
y

α b

d
k

c
β
α x

g h

We calculate cos(α + β) as follows.

adjacent g g+h h g+h k g+h c k b

cos(α + β) = = = − = − = · − ·
hypotenuse d d d d d c d b d
= cos(α) cos(β) − sin(α) sin(β)
π
Again, there is a corresponding figure when the angle α + β is greater than 2
. (We encourage the student
to check the addition formula for this situation as well.)
We therefore have proved the addition formulas for sin(α + β) and cos(α + β) when α and β are angles
between 0 and π2 , which we will now extend to all angles α and β. First, note that the addition formulas are
trivially true when α or β are 0. (Check this!) Now, by observing that sin(x) and cos(x) can be converted
to each other via shifts of π2 , (or, alternatively, by using the identities (17.3) and (17.4)), we obtain, that

π π
sin(x + )= cos x, cos(x + ) = − sin(x),
2 2
254 SESSION 18. ADDITION OF ANGLES AND MULTIPLE ANGLES

π π
sin(x − ) = − cos x, cos(x − )= sin(x).
2 2

With this, we extend the addition identities for α by ± π2 as follows:

 π  π
sin (α + ) + β = sin(α + β + ) = cos(α + β) = cos(α) cos(β) − sin(α) sin(β)
2 2
π π
= sin(α + ) cos(β) + cos(α + ) sin(β),
2 2
 π  π
sin (α − ) + β = sin(α + β − ) = − cos(α + β) = − cos(α) cos(β) + sin(α) sin(β)
2 2
π π
= sin(α − ) cos(β) + cos(α − ) sin(β),
2 2
 π  π
cos (α + ) + β = cos(α + β + ) = − sin(α + β) = − sin(α) cos(β) − cos(α) sin(β)
2 2
π π
= cos(α + ) cos(β) − sin(α + ) sin(β),
2 2
 π  π
cos (α − )+β = cos(α + β − ) = sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
2 2
π π
= cos(α − ) cos(β) − sin(α − ) sin(β).
2 2

There are similar proofs to extend the identities for β. An induction argument shows the validity of the
addition formulas for arbitrary angles α and β.
The remaining formulas now follow via the use of trigonometric identities.

sin α cos β+cos α sin β sin α sin β

sin(α + β) sin α cos β + cos α sin β cos α cos β cos α
+ cos β
tan(α + β) = = = cos α cos β−sin α sin β
= sin α sin β
cos(α + β) cos α cos β − sin α sin β 1 − cos
cos α cos β α cos β

tan α+tan β
This shows that tan(α + β) = 1−tan α tan β
. For the relations with α − β, we use the fact that sin and
tan are odd functions, whereas cos is an even function, see identities (17.5) and (17.7).

sin(α − β) = sin(α + (−β)) = sin(α) cos(−β) + cos(α) sin(−β) = sin α cos β − cos α sin β,
cos(α − β) = cos(α + (−β)) = cos(α) cos(−β) − sin(α) sin(−β) = cos α cos β + sin α sin β,
tan(α) + tan(−β) tan α − tan β
tan(α − β) = tan(α + (−β)) = = .
1 − tan(α) tan(−β) 1 + tan α tan β

This completes the proof of the proposition.

Before giving examples of the above proposition, we recall the elementary

function values of the sin, cos, and tan from the previous section:
π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦ π = 180◦
√ √
1 2 3
sin(x) 0 2 2 2
1 0
√ √
3 2 1
cos(x) 1 2 2 2
0 −1
√
3
√
tan(x) 0 3
1 3 undef. 0
18.1. ADDITION AND SUBTRACTION OF ANGLES 255

Example 18.2. Find the exact values of the trigonometric functions.

π




a) cos 12 b) tan 5π12
c) cos 11π
12

π
Solution. a) The key is to realize the angle 12 as a sum or difference of angles
with known trigonometric function values. Note, that π3 − π4 = 4π−3π12
π
= 12 , so
that
π π π  π π π π
cos = cos − = cos cos + sin sin
12 √3 4√ √ 3 √ 4 √ 3 √ 4 √
1 2 3 2 2 6 2+ 6
= · + · = + =
2 2 2 2 4 4 4
We remark that the last expression is in the simplest form and cannot be
simplified any further.
b) Again we can write the angle 5π
12
as a sum involving only special angles
5π
given in the table above: 12 = 12 + 12 = π6 + π4 . Therefore,
2π 3π

  √
5π π π tan π6 + tan π4 3
3
+1
tan = tan + = π π = √
12 6 4 1 − tan 6 tan 4 1 − 33 · 1
√
3+3
√ √
3√ 3+3 3 3+3
= = · √ = √
3− 3 3 3− 3 3− 3
3

Here, we may simplify the last expression√further by rationalizing the denom-

inator. This is done by multiplying (3 + 3) to numerator and denominator.
  √ √ √ √ 2 √
5π ( 3 + 3) · (3 + 3) 3 3+ 3 +9+3 3
tan = √ √ = √ 2
12 (3 − 3) · (3 + 3) 32 − 3
√ √
3+9+6 3 12 + 6 3 √
= = =2+ 3
9−3 6
c) Again we need to write the angle 11π12
as a sum or difference of the above
angles. In fact, we can do so in at least two different ways: 11π
12
= 6π
12
+ 5π
12
and
11π 12π π π
also 12 = 12 − 12 = π − 12 . In either case, we first need to calculate some
other trigonometric functions, namely those for either 5π 12
π
or 12 . We choose
the second solution, and using part (a), we have
π √ √
2+ 6
cos = , and,
12 4
256 SESSION 18. ADDITION OF ANGLES AND MULTIPLE ANGLES
π π π π π π π
sin = sin − = sin cos − cos sin
12 √ 3√ 4 √ 3 √ 4 √ 3√ 4 √
3 2 1 2 6 2 6− 2
= · − · = − =
2 2 2 2 4 4 4
With this, we have
  
11π π π π
cos = cos π − = cos π cos + sin π sin
12 12 12 12
√ √ √ √ √ √
2+ 6 6− 2 −( 2 + 6)
= (−1) · +0· =
4 4 4

π
Generalizing the previous example, we can obtain any multiple of 12 as
◦ ◦ ◦
a sum or difference of known angles coming from the 30 − 60 − 90 and
45◦ − 45◦ − 90◦ triangles:

Note 18.3. Starting from the angles π6 = 2π , π =

12 4
3π π
,
12 3
= 4π
12
, and π
2
= 6π
12
, we obtain multiples of
π
the angle 12 by addition and subtraction, such as:

π 4π 3π 5π 2π 3π 7π 4π 3π
12
= 12
− 12
, 12
= 12
+ 12
, 12
= 12
+ 12
,

8π 4π 4π 9π 6π 3π 10π 6π 4π
12
= 12
+ 12
, 12
= 12
+ 12
, 12
= 12
+ 12
,

11π 6π 5π 12π π
and 12
= 12
+ 12
= 12
− 12
.

Here, the trigonometric function values of the last fraction is obtained from the previously obtained trigono-
π
metric values, as in part (c) of example 18.2. Higher multiples of 12 can be obtained from the above list by
adding multiples of π to it. Note also that in many instances there are several ways of writing an angle
as a sum or difference. For example: 8π 12
= 4π12
+ 4π
12
= 6π
12
+ 2π
12
= 12π
12
− 4π
12

Using proposition 18.1 we can prove other related trigonometric identities

with the addition and subtraction formulas.
Example 18.4. We rewrite cos(x + π2 ) by using the subtraction formula.
 π π π
cos x + = cos x · cos − sin x · sin = cos x · 0 + sin x · 1 = sin(x)
2 2 2

18.2 Double and half angles

We can write formulas for the trigonometric functions of twice an angle and
half an angle.
18.2. DOUBLE AND HALF ANGLES 257

Proposition 18.5. Let α be an angle. Then we have the half-angle formulas:

r
α 1 − cos α
sin = ±
2 2
r
α 1 + cos α
cos = ±
2 2
r
α 1 − cos α sin α 1 − cos α
tan = = = ±
2 sin α 1 + cos α 1 + cos α
α
Here, the signs “±” are determined by the quadrant in which the angle 2
lies. (For more on the signs, see also page 234.)
Furthermore, we have the double angle formulas:

sin(2α) = 2 sin α cos α

cos(2α) = cos2 α − sin2 α = 1 − 2 sin2 α = 2 cos2 α − 1
2 tan α
tan(2α) =
1 − tan2 α

Proof. We start with the double angle formulas, which we prove using Proposition 18.1.

sin(2α) = sin(α + α) = sin α cos α + cos α sin α = 2 sin α cos α

cos(2α) = cos(α + α) = cos α cos α − sin α sin α = cos2 α − sin2 α
tan α + tan α 2 tan α
tan(2α) = tan(α + α) = =
1 − tan α tan α 1 − tan2 α

Notice that cos(2α) = cos2 α − sin2 α can be rewritten using sin2 α + cos2 α = 1 as follows:

cos2 α − sin2 α = (1 − sin2 α) − sin2 α = 1 − 2 sin2 α

2 2
and cos α − sin α = cos2 α − (1 − cos2 α) = 2 cos2 α − 1

This shows the double angle formulas. These formulas can now be used to prove the half-angle formulas.

1 − cos(2α)
cos(2α) = 1 − 2 sin2 α=⇒ 2 sin2 α = 1 − cos(2α) =⇒ sin2 α =
2
r r
1 − cos(2α) replace α by α2 α 1 − cos α
=⇒ sin α = ± =⇒ sin = ±
2 2 2
1 + cos(2α)
cos(2α) = 2 cos2 α − 1 =⇒ 2 cos2 α = 1 + cos(2α) =⇒ 2
cos α =
2
r r
1 + cos(2α) replace α by α2 α 1 + cos α
=⇒ cos α = ± =⇒ cos = ±
2 2 2
q
α sin( α ) ± 1−cos2
α r
1 − cos α
2
in particular: tan = α = q =±
2 cos( 2 ) ± 1+cos α 1 + cos α
2
258 SESSION 18. ADDITION OF ANGLES AND MULTIPLE ANGLES

α
For the first two formulas for tan 2
we simplify sin(2α) · tan(α) and (1 + cos(2α)) · tan(α) as follows.

sin α
sin(2α) · tan(α) = 2 sin α cos α · = 2 sin2 α = 1 − cos(2α)
cos α
1 − cos(2α) replace α by α2 α 1 − cos(α)
=⇒ tan(α) = =⇒ tan( ) =
sin(2α) 2 sin(α)
2 sin α
(1 + cos(2α)) · tan(α) = 2 cos α · = 2 sin α cos α = sin(2α)
cos α
sin(2α) replace α by α α sin(α)
=⇒ tan(α) = =⇒ 2 tan( ) =
1 + cos(2α) 2 1 + cos(α)

This completes the proof of the proposition.

Here is an example involving the half-angle identities.

Example 18.6. Find the trigonometric functions using the half-angle formulas.




a) sin π8 b) cos 9π8
c) tan π
24

π
π
Solution. a) Since 8
= 4
2
, we use the half-angle formula with α = π4 .

r s √
s √
π π 2 2− 2
π  1 − cos 1−
sin = sin 4 = ± =± 4 2
=± 2
8 2 2 2 2
s
√ p √
2− 2 2− 2
= ± =± .
4 2
π ◦
Since = 180 = 22.5◦ is in the first quadrant, the sine is positive, so that
8 √ 8√
2− 2
sin( π8 ) = 2
.
9π ◦
b) Note that 9π
8
= 24 . So we use α = 9π 4
. Now, 9π
8
= 9·1808
= 202.5◦ is in
the third quadrant, so that the cosine is negative. We have:
s
   9π 
9π 1 + cos 9π
cos = cos 4 = − 4
.
8 2 2
√
Now, cos( 9π
4
) = cos( 8π+π
4
) = cos(2π + π4 ) = cos( π4 ) = 2
2
, so that
s s s
  √
2
√
2+ 2
√ p √
9π 1+ 2 2 2+ 2 2+ 2
cos =− =− =− =− .
8 2 2 4 2
18.2. DOUBLE AND HALF ANGLES 259
π
π
c) Note that 24 = 122 , and we already calculated the trigonometric function
π
values of α = 12 in example 18.2(c). So that we obtain:
  √ √ √ √
π π
π 1 − cos 12
12
1 − 2+4 6 4− 2− 6
tan = tan = π = √ √ = √ 4√
24 2 sin 12 6− 2 6− 2
4 4
√ √ √ √
4− 2− 6 4 4− 2− 6
= ·√ √ = √ √
4 6− 2 6− 2

We can √rationalize
√ the denominator by multiplying numerator and denomina-
tor by ( 6 + 2).

π √ √ √ √
4− 2− 6 6+ 2
tan = √ √ ·√ √
24 6− 2 6+ 2
√ √ √ √ √ √ √ √ √
4 6 + 4 2 − 12 − 4 − 36 − 12 4 6 + 4 2 − 2 12 − 2 − 6
= =
6−2 4
√ √ √
4 6+4 2−4 3−8 √ √ √
= = 6 + 2 − 3 − 2.
4
Although we used the first formula for tan α2 from the proposition, we could
as well have used the other two formulas.

Example 18.7. Find the trigonometric functions of 2α when α has the prop-
erties below.
a) sin(α) = 35 , and α is in quadrant II
b) tan(α) = 125
, and α is in quadrant III

Solution. a) From sin2 (α) + cos2 (α) = 1, we find that cos2 (α) = 1 − sin2 (α),
and since α is in the second quadrant, cos(α) is negative, so that
q r  3 2 r
2 9
cos(α) = − 1 − sin (α) = − 1 − =− 1−
5 25
r r
25 − 9 16 4
= − =− =− ,
25 25 5
and
3
sin α 5 3 5 3
tan(α) = = −4 = · =−
cos α 5
5 −4 4
260 SESSION 18. ADDITION OF ANGLES AND MULTIPLE ANGLES

From this we can calculate the solution by plugging these values into the
double angle formulas.
3 (−4) −24
sin(2α) = 2 sin α cos α = 2 · · =
5 5 25
 −4 2  3 2 16 9 7
cos(2α) = cos2 (α) − sin2 (α) = − = − =
−3

5 −3
5 25 25
−3
25
2 tan α 2· 4 2 2 −3 16 −24
tan(2α) = 2
=
2 = 9 = 16−9 = · =
1 − tan α 1− −3 1 − 16 16
2 7 7
4

b) Similar to the calculation in part (a), we first calculate sin(α) and

cos(α), which are both negative in the third quadrant. Recall from equation
(17.2) on page 234 that sec2 α = 1 + tan2 α, where sec α = cos1 α . Therefore,
 12 2 144 25 + 144 169 13
sec2 α = 1 + =1+ = = =⇒ sec α = ±
5 25 25 25 5
Since cos(α) is negative (in quadrant III), so is sec(α), so that we get,
1 1 5
cos α = = 13 = −
sec α −5 13
Furthermore, sin2 α = 1 − cos2 α, and sin α is negative (in quadrant III), we
have
r r
√  5 2 25
sin α = − 1 − cos2 α = − 1 − − =− 1−
13 169
r r
169 − 25 144 12
= − =− =−
169 169 13
Thus, we obtain the solution as follows:
(−12) (−5) 120
sin(2α) = 2 sin α cos α = 2 · · =
13 13 169
 −5 2  −12 2 25 144 −119
2 2
cos(2α) = cos (α) − sin (α) = − = − =
13 13 169 169 169
2 tan α 2 · 12
5
24
5
24
5
tan(2α) = =
= 144 = 25−144
1 − tan2 α 12 2 1 − 25
1− 5 25
24 25 120
= · =
5 −119 −119
18.3. EXERCISES 261

18.3 Exercises
Exercise 18.1. Find the trigonometric function values.





a) sin 5π
12
b) cos 5π
12
c) tan 12π
d) sin 7π
12





e) cos 11π
12
f) sin 2π
3
g) sin 5π
6
h) cos 3π
4
13π

π

11π

29π


i) tan 12 j) cos − 12 k) sin 12 l) sin 12

Exercise 18.2. Simplify the function f using the addition and subtraction
formulas.



a) f (x) = sin x + π2 b) f (x) = cos x − π4 c) f (x) = tan (π − x)




d) f (x) = sin π6 − x e) f (x) = cos x + 11π12
f) f (x) = cos 2π
3
− x

Exercise 18.3. Find the exact values of the trigonometric functions applied
to the given angles by using the half-angle formulas.
π





a) cos 12 b) tan π8 π
c) sin 24 π
d) cos 24





e) sin 9π
8
f) tan 3π
8
g) sin −π
8
h) sin 13π
24

α
Exercise 18.4. Find the exact values of the trigonometric functions of 2
and
of 2α by using the half-angle and double angle formulas.

a) sin(α) = 54 , and α in quadrant I

7
b) cos(α) = 13 , and α in quadrant IV
−3
c) sin(α) = 5 , and α in quadrant III
d) tan(α) = 34 , and α in quadrant III
e) tan(α) = −512
, and α in quadrant II
−2
f) cos(α) = 3 , and α in quadrant II
Session 19

Inverse trigonometric functions

19.1 The functions sin−1, cos−1 , and tan−1

The inverse trigonometric functions are the inverse functions of the y = sin x,
y = cos x, and y = tan x functions restricted to appropriate domains. In this
section we give a precise definition of these functions.

The function y = tan−1 (x)

We start with the inverse to the tangent function y = tan(x). Recall that the
graph of y = tan(x) is the following:
y = tan(x)

x
−5π −3π −π π 3π 5π
2 −2π 2 −π 2 2 π 2 2π 2

It has vertical asymptotes at x = ± π2 , ± 3π

2
, ± 5π
2
, . . . . Note, that y = tan(x)
is not a one-to-one function in the sense of defintion 7.1 on page 86. (For

262
19.1. THE FUNCTIONS SIN−1 , COS−1 , AND TAN−1 263

example, the horizontal line y = 1 intersects the graph at x = π4 , x = π4 ± π,

x = π4 ± 2π, etc.) However, when we restrict the function to the domain
D = ( −π , π ) the restricted function is one-to-one, and, for this restricted
2 2
function, we may take its inverse function.
Definition 19.1. The inverse of the function y = tan(x) with restricted domain
D = ( −π , π ) and range R = R is called the inverse tangent or arctangent
2 2
function. It is denoted by
 π π
y = tan−1 (x) or y = arctan(x) ⇐⇒ tan(y) = x, y ∈ − ,
2 2
The arctangent reverses the input and output of the tangent function, so that
the arctangent has domain D = R and range R = ( −π , π ). The graph is
2 2
displayed below.
3 y = tan−1 (x) = arctan(x)

π
1 2

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
−π
2
-2

-3

Warning 19.2. The notation of tan−1 (x) and tan2 (x) is slightly inconsistent, since the exponentiation
symbol is used above in two different ways. In fact, tan−1 (x) = arctan(x) refers to the inverse function
of the tan(x) function. However, when we write tan2 (x), we mean

tan2 (x) = (tan(x))2 = tan(x) · tan(x)

Therefore, tan−1 (x) is the inverse function of tan(x) with respect to the composition operation, whereas
tan2 (x) is the square with respect to the usual product in R. Note also that the inverse function of the
1
tangent with respect to the product in R is y = tan(x) = cot(x), which is the cotangent.

Observation 19.3. The inverse tangent function is an odd function:

tan−1 (−x) = − tan−1 (x) (19.1)

264 SESSION 19. INVERSE TRIGONOMETRIC FUNCTIONS

This can be seen by observing that the tangent y = tan(x) is an odd function
(that is tan(−x) = − tan(x)), or directly from the symmetry of the graph with
respect to the origin (0, 0).
The next example calculates function values of the inverse tangent function.
Example 19.4. Recall the exact values of the tangent function from section
17.1:
x 0 = 0◦ π6 = 30◦ π4 = 45◦ π3 = 60◦ π2 = 90◦
√
3
√
tan(x) 0 3
1 3 undef.
From this, we can deduce function values by reversing inputs and outputs,
such as:
 π  √3  √3  π
−1
tan = =⇒ tan =
6  3 3 6
π −1

π
tan = 1 =⇒ tan 1 =
4 4
−1 −1
Also, since tan (−x) = − tan (x), we obtain the inverse tangent of negative
numbers.
√ √ π
tan−1 (− 3) = − tan−1 ( 3) = −
3
−1 −1 π
tan (−1) = − tan (1) = −
4
We may ✞ calculate
☎ the✞inverse☎tangent of specific values with the calculator
using the lp2nd lp and lptanlp keys. For example, tan−1 (4.3) ≈ 1.34.
✝ ✆ ✝ ✆

Note, that the answer differs, when changing the mode from radians to degree,
since tan−1 (4.3) ≈ 76.9◦ ≈ 1.34.
19.1. THE FUNCTIONS SIN−1 , COS−1 , AND TAN−1 265

The function y = sin−1 (x)

Next, we define the inverse sine function. For this, we again first recall the
graph of the y = sin(x) function, and note that it is also not one-to-one.
y = sin(x)

1
−π 3π
2 2 2π x
−3π π
−2π 2 −π 2 π
−1

 
However, when restricting the sine to the domain −π , π , the restricted func-
2 2
tion
 −π is  one-to-one. Note furthermore, that when restricting the domain to
π
,
2 2
, the range is [−1, 1], and therefore we cannot extend this to a larger
domain in a way suchthat the function remains a one-to-one function. We
use the domain −π , π to define the inverse sine function.
2 2
Definition
 −π 19.5.
 The inverse of the function y = sin(x) with restricted domain
π
D = 2 , 2 and range R = [−1, 1] is called the inverse sine or arcsine
function. It is denoted by
 
−1 −π π
y = sin (x) or y = arcsin(x) ⇐⇒ sin(y) = x, y ∈ ,
2 2
The arcsine reverses the input and output of the sine function,
 so that the
arcsine has domain D = [−1, 1] and range R = −π ,
2 2
π
. The graph of the
arcsine is drawn below. 3 y = sin−1 (x)

= arcsin(x)
2
π
2
1

0 x
-3 -2 -1 0 1 2 3
-1
−π
-2 2

-3
266 SESSION 19. INVERSE TRIGONOMETRIC FUNCTIONS

Observation 19.6. The inverse sine function is odd:

sin−1 (−x) = − sin−1 (x) (19.2)
This can again be seen by observing that the sine y = sin(x) is an odd function
(that is sin(−x) = − sin(x)), or alternatively directly from the symmetry of
the graph with respect to the origin (0, 0).
We now calculate specific function values of the inverse sine.
Example 19.7. We first recall the known values of the sine.
π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √
1 2 3
sin(x) 0 2 2 2
1
These values together with the fact that the inverse sine is odd, that is
sin−1 (−x) = − sin−1 (x), provides us with examples of its function values.
 √2  π π
−1
sin = , sin−1 (1) = ,
2 4 2
 −1  1 π
sin−1 (0) = 0, sin−1 = − sin−1 =− .
2 2 6
−1
Note, that the domain of y = sin (x) is D = [−1, 1], so that input numbers
that are not in this interval give undefined outputs of the inverse sine:
sin−1 (3) is undefined
Input values✞that are☎✞
not in the
☎ above table may be found with the cal-
culator via the lp2nd lp lpsinlp keys. We point out, that the output values
✝ ✆✝ ✆
depend on wether the calculator is ✞set to radian
☎ or degree mode. (Recall that
the mode may be changed via the lpmodelp key).
✝ ✆
19.1. THE FUNCTIONS SIN−1 , COS−1 , AND TAN−1 267

The function y = cos−1 (x)

Finally, we define the inverse cosine. Recall the graph of y = cos(x), and
note again that the function is not one-to-one.
y = cos(x)

1
−π π x
−2π −3π −π π 3π 2π
2 2 2 2
−1

In this case, the way to restrict the cosine to a one-to-one function is not
as clear as in the previous cases for the sine and tangent. By convention,
the cosine is restricted to the domain [0, π]. This provides a function that is
one-to-one, which is used to define the inverse cosine.
Definition 19.8. The inverse of the function y = cos(x) with restricted domain
D = [0, π] and range R = [−1, 1] is called the inverse cosine or arccosine
function. It is denoted by

y = cos−1 (x) or y = arccos(x) ⇐⇒ cos(y) = x, y ∈ [0, π]

The arccosine reverses the input and output of the cosine function, so that
the arccosine has domain D = [−1, 1] and range R = [0, π]. The graph of the
arccosine is drawn below.
5 y = cos−1 (x)

= arccos(x)
4

3
π
2
π
2
1

0 x
-3 -2 -1 0 1 2 3
-1
268 SESSION 19. INVERSE TRIGONOMETRIC FUNCTIONS

Observation 19.9. The inverse cosine function is neither even nor odd. That
is, the function cos−1 (−x) cannot be computed by simply taking ± cos−1 (x).
But it does have some symmetry given algebraically by the more complicated
relation
cos−1 (−x) = π − cos−1 (x) (19.3)
π
Proof. We can see that if we shift the graph down by 2
the resulting function is odd. That is to say the
function with the rule cos−1 (x) − π2 is odd:

π π
cos−1 (−x) − = −(cos−1 (x) − ),
2 2
π
which yields 19.3 upon distributing and adding 2
.
Another, more formal approach is as follows. The bottom right relation of (17.3) on page 234 states, that
we have the relation cos(π − y) = − cos(y) for all y. Let −1 ≤ x ≤ 1, and denote by y = cos−1 (x).
That is y is the number 0 ≤ y ≤ π with cos(y) = x. Then we have

−x = − cos(y) = cos(π − y) (by equation (17.3))

Applying cos−1 to both sides gives:

cos−1 (−x) = cos−1 (cos(π − y)) = π − y

The last equality follows, since cos and cos−1 are inverse to each other, and 0 ≤ y ≤ π, so that
0 ≤ π − y ≤ π are also in the range of the cos−1 . Rewriting y = cos−1 (x) gives the wanted result:

cos−1 (−x) = π − cos−1 (x)

This is the equation (19.3) which we wanted to prove.

We now calculate specific function values of the inverse cosine.

Example 19.10. We first recall the known values of the cosine.
π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √
3 2 1
cos(x) 1 2 2 2
0

Here are some examples for function values of the inverse cosine.
 √3  π π
cos−1 = , cos−1 (1) = 0, cos−1 (0) = .
2 6 2
Negative inputs to the arccosine can be calculated with equation (19.3), that
is cos−1 (−x) = π − cos−1 (x), or by going back to the unit circle definition.
 1 1 π 3π − π 2π
cos−1 − = π − cos−1 =π− = = ,
2 2 3 3 3
19.2. EXERCISES 269

cos−1 (−1) = π − cos−1 (1) = π − 0 = π.

Furthermore, the domain of y = cos−1 (x) is D = [−1, 1], so that input numbers
not in this interval give undefined outputs of the inverse cosine.
cos−1 (17) is undefined
✞ Other☎✞
input values
☎ can be obtained with the calculator by pressing the
lp2nd lp lpcoslp keys. For example, we obtain the following function values
✝ ✆✝ ✆
(here using radian measure).

19.2 Exercises
Exercise 19.1. Graph the function with the calculator. Use both radian and
degree mode to display your graph. Zoom to an appropriate window for each
mode to display a graph which includes the main features of the graph.
a) y = sin−1 (x) b) y = cos−1 (x) c) y = tan−1 (x)
Exercise 19.2. Find the exact value of the inverse trigonometric function.
√
a) tan−1 (√ 3) b) sin−1 ( 12 )√ c) cos−1 ( 21 ) d) tan−1 (0)
2
√
e) cos−1 ( 2√ ) f) cos−1 (−√22 ) g) sin−1 (−1) √
h) tan−1 (− 3)
i) cos (− 23 )
−1
j) sin−1 (− 22 ) k) sin−1 (− 23 ) l) tan−1 (− √13 )
Exercise 19.3. Find the inverse trigonometric function value using the calcu-
lator. Approximate your answer to the nearest hundredth.
• For parts (a)-(f), write your answer in radian mode.
a) cos−1 (0.2) b) sin−1 (−0.75) c) cos−1 ( 31 )
d) tan−1 (100, 000) e) tan−1 (−2) f) cos−1 (−2)
• For parts (g)-(l), write your answer in degree mode.
√ √
g) cos−1 (0.68) h) tan−1 (−1)
√ i) sin−1 ( 2+ 6
4
)
√
2− 2 √ √ √
j) tan−1 (100, 000) k) cos−1 ( 2
) l) tan−1 (2 + 3− 6− 2)
Session 20

Trigonometric equations

20.1 Basic trigonometric equations

In this section we solve equations such as:

sin(x) = 0.5

We know that x = π6 solves this equation. However, there are also other
solutions such as x = 5π6
or x = 13π
6
that can easily be checked with the
calculator. Below, we will study how to find all solutions of equations of the
form
sin(x) = c, cos(x) = c, and tan(x) = c.
We start with equations involving the tangent.

The equation tan(x) = c

√
Example 20.1. Solve for x: tan(x) = 3
y = tan(x)

√
3

x
π π π π π
3
− 2π 3
−π 3 3
+π 3
+ 2π

270
20.1. BASIC TRIGONOMETRIC EQUATIONS 271
√
Solution. There is an obvious solution given by x = tan−1 ( 3) = π3 , as
we studied√ in the last section. However, we can look for all solutions of
tan(x) = 3 by studying the graph of the tangent function, that is, by finding
all points√where the graph of the y = tan(x) intersects with the horizontal
line y = 3. Since the function y = √ tan(x) is periodic with period π, we see
that the other solutions of tan(x) = 3 besides x = π3 are
π π π π π π
+ π, + 2π, + 3π, . . . , and − π, − 2π, − 3π, . . .
3 3 3 3 3 3
In general, we write the solution as
π
x = + n · π, where n = 0, ±1, ±2, ±3, . . .
3
√
The graph also shows that these are indeed all solutions of tan(x) = 3.
By the same argument we also get the general solution of tan(x) = c.
Observation 20.2. To solve tan(x) = c, we first determine one solution x =
tan−1 (c). Then the general solution is given by
x = tan−1 (c) + n · π where n = 0, ±1, ±2, ±3, . . . (20.1)
Example 20.3. Solve for x:
a) tan(x) = 1, b) tan(x) = −1, c) tan(x) = 5.1, d) tan(x) = −3.7
Solution. a) First, we find tan−1 (1) = π4 . The general solution is thus:
π
x = + n · π where n = 0, ±1, ±2, ±3, . . .
4
b) First, we need to find tan−1 (−1). Recall from equation (19.1) that
tan−1 (−c) = − tan−1 (c), and recall further that tan−1 (1) = π4 . With this we
have
π
tan−1 (−1) = − tan−1 (1) = −
4
The general solution of tan(x) = −1 is therefore,
π
x = − + n · π, where n = 0, ±1, ±2, . . .
4
For parts (c) and (d), we do not have an exact solution, so that the solution
can only be approximated with the calculator.
c) x = tan−1 (5.1) + nπ ≈ 1.377 + nπ, where n = 0, ±1, ±2, . . .
d) x = tan−1 (−3.7) + nπ ≈ −1.307 + nπ, where n = 0, ±1, ±2, . . .
272 SESSION 20. TRIGONOMETRIC EQUATIONS

The equation cos(x) = c

We start again with an example.
1
Example 20.4. Solve for x: cos(x) = 2
y = cos(x)

x
−π − π3 π π
3

| {z }
one full period

Solution. We have the obvious solution to the equation x = cos−1 ( 21 ) = π3 .

However, since cos(−x) = cos(x), there is another solution given by taking
x = − π3 :
 π π  1
cos − = cos =
3 3 2
Moreover, the y = cos(x) function is periodic with period 2π, that is, we have
cos(x + 2π) = cos(x). Thus, all of the following numbers are solutions of the
equation cos(x) = 12 :
π π π π π
..., 3
− 4π, 3
− 2π, 3
, 3
+ 2π, 3
+ 4π, ...,
and: . . . , − π3 − 4π, π
−3 π
− 2π, − 3 , π
−3 + 2π, − π3 + 4π, ....

From the graph we see that there are only two solutions of cos(x) = 21 within
one period. Thus, the above list constitutes all solutions of the equation.
With this observation, we may write the general solution as:
π π
x= + 2n · π, or x = − + 2n · π, where n = 0, ±1, ±2, ±3, . . .
3 3
In short, we write this as: x = ± π3 + 2n · π with n = 0, ±1, ±2, ±3, . . . .
We generalize this example as follows.
Observation 20.5. To solve cos(x) = c, we first determine one solution x =
cos−1 (c). Then the general solution is given by

x = cos−1 (c)+2n·π, or x = − cos−1 (c)+2n·π, where n = 0, ±1, ±2, ±3, . . .

20.1. BASIC TRIGONOMETRIC EQUATIONS 273

In short, we write

x = ± cos−1 (c) + 2n · π where n = 0, ±1, ±2, ±3, . . . (20.2)

Example 20.6. Solve for x.

√
2
a) cos(x) = − 2
, b) cos(x) = 0.6, c) cos(x) = −3, d) cos(x) = −1
√
Solution. a) First, we need to find cos−1 (− 22 ). From equation (19.3) we
know that cos−1 (−c) = π − cos−1 (c), so that:
 √  √ 
2 2 π 4π − π 3π
cos−1 − = π − cos−1 = π− = =
2 2 4 4 4
The solution is therefore,
3π
x=± + 2nπ, where n = 0, ±1, ±2, ±3, . . .
4
b) We calculate cos−1 (0.6) ≈ 0.927 with the calculator. The general solu-
tion is therefore,

x = ± cos−1 (0.6) + 2nπ ≈ ±0.927 + 2nπ, where n = 0, ±1, ±2, ±3, . . .

c) Since the cosine is always −1 ≤ cos(x) ≤ 1, the cosine can never be

−3. Therefore, there is no solution to the equation cos(x) = −3. This can
also be seen from the graph which does not intersect with the horizontal line
y = −3.
y = cos(x)

1
x

−1

−3

d) A special solution of cos(x) = −1 is

cos−1 (−1) = π − cos−1 (1) = π − 0 = π,

274 SESSION 20. TRIGONOMETRIC EQUATIONS

so that the general solution is

x = ±π + 2nπ, where n = 0, ±1, ±2, ±3, . . .
However, since −π + 2π = +π, the solutions π + 2nπ and −π + 2nπ (for
n = 0, ±1, ±2, . . . ) can be identified with each other, and there is only one
solution in each period. Thus, the general solution can be written as
x = π + 2nπ, where n = 0, ±1, ±2, ±3, . . .
The graphs of y = cos(x) and y = −1 confirm these considerations.
y = cos(x)

1
−3π −π π 3π x

−1

The equation sin(x) = c

√
2
Example 20.7. Solve for x: sin(x) = 2
y = sin(x)

x
π π
4
π− 4

| {z }
one full period
√
Solution. First, we can find one obvious solution x = sin−1 ( 22 ) = π4 . Further-
more, from the top right equation in (17.3), we have that sin(π − x) = sin(x),
so that another solution is given by π − π4 :
 π  π  √2
sin π − = sin =
4 4 2
(This can also be seen by going back to the unit circle definition.) These are
all solutions within one period, as can be checked from the graph above. The
20.1. BASIC TRIGONOMETRIC EQUATIONS 275

function y = sin(x) is periodic with period 2π, so that

√
adding 2n · π for any
2
n = 0, ±1, ±2, . . . gives all solutions of sin(x) = 2 . This means that the
general solution is:
π π
x= + 2n · π, or x = (π − ) + 2n · π, for n = 0, ±1, ±2, ±3, . . .
4 4
We rewrite these solutions to obtain one single formula for the solutions.
Note, that π − π4 + 2n · π = − π4 + (2n + 1) · π. Therefore, all solutions are of
the form
π
x=± +k·π
4
where, for even k = 2n, the sign in front of π4 is “+,” and, for odd k = 2n + 1,
the sign in front of π4 is “−.” This can be summarized as:
π
x = (−1)k · +k·π
4
Renaming the indexing variable√
k to the usual n, we obtain the final version
2
for the solutions of sin(x) = 2 .
π
x = (−1)n · + n · π, where n = 0, ±1, ±2, ±3, . . .
4

We have the following general statement.

Observation 20.8. To solve sin(x) = c, we first determine one solution x =

sin−1 (c). Then the general solution is given by

x = (−1)n · sin−1 (c) + n · π where n = 0, ±1, ±2, ±3, . . . (20.3)

Example 20.9. Solve for x.

1
a) sin(x) = 2
b) sin(x) = − 12 c) sin(x) = − 75 d) sin(x) = 1

Solution. a) First, we calculate sin−1 ( 12 ) = π6 . The general solution is there-

fore,
π
x = (−1)n · + n · π, where n = 0, ±1, ±2, ±3, . . .
6
276 SESSION 20. TRIGONOMETRIC EQUATIONS

b) First, we need to find sin−1 (− 12 ). From equation (19.2) we know that

sin−1 (−c) = − sin−1 (c), so that sin−1 (− 12 ) = − sin−1 ( 12 ) = − π6 . We thus state
the general solution as
 π π π
x = (−1)n · − + n · π = −(−1)n · + n · π = (−1)n+1 · + n · π,
6 6 6
where n = 0, ±1, ±2, ±3, . . .
c) We do not have an exact value of sin−1 (− 57 ), so that we either need to
leave it as this, or approximate this with the calculator sin−1 (− 75 ) ≈ −0.796.
We get the solution:

x ≈ (−1)n · (−0.796) + n · π = (−1)n+1 · 0.796 + n · π,

where n = 0, ±1, ±2, ±3, . . .
d) We calculate sin−1 (1) = π2 . The general solution x can be written as
π
x = (−1)n · + n · π, where n = 0, ±1, ±2, ±3, . . . (20.4)
2
Now, if we look at the graph, we see that each period only has one solution.
y = sin(x)

x
−3π π 5π
2 2 2
−1

Algebraically, we can see this as follows. For an even number n, the solution
x = (−1)n · π2 + n · π coincides with the solution coming from the index n + 1,
that is:
π π
(−1)n · + n · π = + + n · π, and
2 2
n+1 π π π
(−1) · + (n + 1) · π = − + (n + 1) · π = − + n · π + π
2 2 2
π
= + +n·π
2
Therefore, we can write the solution more efficiently by removing the odd
solutions (since they coincide with the even solutions), and state this as
π
x = (−1)n · + n · π, where n = 0, ±2, ±4, . . .
2
20.1. BASIC TRIGONOMETRIC EQUATIONS 277

Since (−1)n = +1 for even n, we can just write this as

π
x= + 2n · π, where n = 0, ±1, ±2, ±3, . . . (20.5)
2
Writing the solutions as x = π2 + 2n · π as in (20.5) instead of the original
x = (−1)n · π2 + n · π from (20.4) for n = 0, ±1, ±2, . . . certainly does not
change the overall solution set. However, writing the solutions as in (20.5) is
more efficient, since it does not repeat any of the solutions, and is therefore
a simplified and preferred way of presenting the solutions.

Summary
We summarize the different formulas we used to solve the basic trigonometric
equations in the following table.

Solve: sin(x) = c Solve: cos(x) = c Solve: tan(x) = c

First, find one solution, First, find one solution, First, find one solution,
that is: sin−1 (c). Use: that is: cos−1 (c). Use: that is: tan−1 (c). Use:
sin−1 (−c) = − sin−1 (c) cos−1 (−c) = π − cos−1 (c) tan−1 (−c) = − tan−1 (c)
The general solution is: The general solution is: The general solution is:

x = (−1)n sin−1 (c) + nπ x = ± cos−1 (c) + 2nπ x = tan−1 (c) + nπ

where n = 0, ±1, ±2, . . . where n = 0, ±1, ±2, . . . where n = 0, ±1, ±2, . . .

Example 20.10. Find the general solution of the equation, and state at least
5 distinct solutions.
√
a) sin(x) = − 12 b) cos(x) = − 2
3

Solution. a) We already calculated the general solution in example 20.9 (b).

It is
π
x = (−1)n+1 · + n · π, where n = 0, ±1, ±2, ±3, . . .
6
We simplify the solutions for n = 0, 1, −1, 2, −2.
π π
n = 0 : x = (−1)0+1 · +0·π =−
6 6
π π π + 6π 7π
n = 1 : x = (−1)1+1 · + 1 · π = + π = =
6 6 6 6
278 SESSION 20. TRIGONOMETRIC EQUATIONS

π π π − 6π −5π
n = −1 : x = (−1)−1+1 · + (−1) · π = − π = =
6 6 6 6
π π −π + 12π 11π
n = 2 : x = (−1)2+1 · + 2 · π = − + 2π = =
6 6 6 6
π π −π − 12π −13π
n = −2 : x = (−1)−2+1 · + (−2) · π = − − 2π = =
6 6 6 6
√ √
−1
b) It is cos√ (− 23 ) = π − cos−1 ( 2
3
) =π− π
6
= 6π−π
6
= 5π
6
. The solutions
of cos(x) = − 23 are:
5π
x=± + 2nπ, where n = 0, ±1, ±2, ±3, . . .
6
We write the 6 solutions with n = 0, +1, −1, and for each use the two distinct
first terms + 5π
6
and − 5π
6
.
5π 5π
n=0: x= + +2·0·π =
6 6
5π 5π
n=0: x= − +2·0·π =−
6 6
5π 5π 5π + 12π 17π
n=1: x= + +2·1·π = + 2π = =
6 6 6 6
5π 5π −5π + 12π 7π
n=1: x= − +2·1·π =− + 2π = =
6 6 6 6
5π 5π 5π − 12π −7π
n = −1 : x = + + 2 · (−1) · π = − 2π = =
6 6 6 6
5π 5π −5π − 12π −17π
n = −1 : x = − + 2 · (−1) · π = − − 2π = =
6 6 6 6
Further solutions can be found by taking values n = +2, −2, +3, −3, . . . .

20.2 Equations involving trigonometric functions

The previous section showed how to solve the basic trigonometric equations
sin(x) = c, cos(x) = c, and tan(x) = c.
The next examples can be reduced to these basic equations.
Example 20.11. Solve for x.
√
a) 2 sin(x) − 1 = 0 b) sec(x) = − 2 c) 7 cot(x) + 3 = 0
20.2. EQUATIONS INVOLVING TRIGONOMETRIC FUNCTIONS 279

Solution. a) Solving for sin(x), we get

(+1) (÷2) 1
2 sin(x) − 1 = 0 =⇒ 2 sin(x) = 1 =⇒ sin(x) =
2
One solution of sin(x) = 1
2
is sin−1 ( 21 ) = π6 . The general solution is
π
x = (−1)n · + nπ, where n = 0, ±1, ±2, . . .
6
1
b) Recall that sec(x) = cos(x)
. Therefore,
√
√ 1 √ (reciprocal) 1 2
sec(x) = − 2 =⇒ =− 2 =⇒ cos(x) = − √ = −
cos(x) 2 2
√
A special solution of cos(x) = − 22 is
√ √
−1 2 −1 2 π 4π − π 3π
cos (− ) = π − cos ( )=π− = = .
2 2 4 4 4
The general solution is
3π
x=± + 2nπ, where n = 0, ±1, ±2, . . .
4
1
c) Recall that cot(x) = tan(x)
. So

(−3) (÷7) 3
7 cot(x) + 3 = 0 =⇒ 7 cot(x) = −3 =⇒ cot(x) = −
7
1 3 (reciprocal) 7
=⇒ =− =⇒ tan(x) = −
tan(x) 7 3
The solution is
 
−1 7
x = tan − + nπ ≈ −1.166 + nπ, where n = 0, ±1, ±2, . . .
3

For some of the more advanced problems it can be helpful to substitute

u for a trigonometric expression first, then solve for u, and finally apply the
rules from the previous section to solve for the wanted variable. This method
is used in the next example.
280 SESSION 20. TRIGONOMETRIC EQUATIONS

Example 20.12. Solve for x.

a) tan2 (x) + 2 tan(x) + 1 = 0 b) 2 cos2 (x) − 1 = 0

Solution. a) Substituting u = tan(x), we have to solve the equation

(factor) (−1)
u2 + 2u + 1 = 0 =⇒ (u + 1)(u + 1) = 0 =⇒ u + 1 = 0 =⇒ u = −1

Resubstituting u = tan(x), we have to solve tan(x) = −1. Using the fact that
tan−1 (−1) = − tan−1 (1) = − π4 , we have the general solution
π
x=− + nπ, where n = 0, ±1, ±2, . . .
4
b) We substitute u = cos(x), then we have

(+1) (÷2) 1
2u2 − 1 = 0 =⇒ 2u2 = 1 =⇒ u2 =
r 2
√
1 1 2
=⇒ u=± = ±√ = ±
2 2 2
√ √
2 2
=⇒ u=+ or u = −
2 2
For each of the two cases we need to solve the corresponding equation after
replacing u = cos(x).
√ √
cos(x) = 22 cos(x) = − 22
√   √  √ 
with cos−1 22 = π
4
with cos−1 − 22 = π − cos−1 22
= π − π4 = 3π
4

=⇒ x = ± π4 + 2nπ =⇒ x = ± 3π
4
+ 2nπ
where n = 0, ±1, ±2, . . . where n = 0, ±1, ±2, . . .

Thus, the general solution is

π 3π
x=± + 2nπ, or x=± + 2nπ, where n = 0, ±1, ±2, . . .
4 4
20.2. EQUATIONS INVOLVING TRIGONOMETRIC FUNCTIONS 281

Example 20.13. Solve the equation with the calculator. Approximate the
solution to the nearest thousandth.

a) 2 sin(x) = 4 cos(x) + 3 b) 5 cos(2x) = tan(x)

Solution. a) We rewrite the equation as 2 sin(x) − 4 cos(x) − 3 = 0, and use

the calculator to find the graph of the function f (x) = 2 sin(x) − 4 cos(x) − 3.
The zeros of the function f are the solutions of the initial equation. The graph
that we obtain is displayed below.

The graph indicates that the function f (x) = 2 sin(x)−4 cos(x)−3 is periodic.
This can be confirmed by observing that both sin(x) and cos(x) are periodic
with period 2π, and thus also f (x).

f (x + 2π) = 2 sin(x + 2π) − 4 cos(x + 2π) − 3 = 2 sin(x) − 4 cos(x) − 3 = f (x)

The solution
✞ of f (x)
☎✞ = 0 can☎✞ be obtained
☎ by finding the “zeros,” that is by
pressing lp2nd lp lptrace lp lp2 lp , then choosing a left- and right-bound,
✝ ✆✝ ✆✝ ✆
and making a guess for the zero. Repeating this procedure gives the following
two approximate solutions within one period.

The general solution is thus

x ≈ 1.842 + 2nπ or x ≈ 3.513 + 2nπ, where n = 0, ±1, ±2, . . .

b) We rewrite the equation as 5 cos(2x) − tan(x) = 0 and graph the

282 SESSION 20. TRIGONOMETRIC EQUATIONS

function f (x) = 5 cos(2x) − tan(x) in the standard window.

To get a better view of the function we zoom to a more appropriate window.

Note again that the function f is periodic. The period of cos(2x) is 2π

2
= π (see
definition 17.9 on page 246), and the period of tan(x) is also π (see equation
(17.6) on page 243). Thus, f is also periodic with period π. The solutions in
one period are approximated by finding the zeros with the calculator.

The general solution is given by any of these numbers, with possibly an

additional shift by any multiple of π.

x ≈ 1.788 + nπ or x ≈ 2.224 + nπ or x ≈ 3.842 + nπ,

where n = 0, ±1, ±2, ±3, . . .
20.3. EXERCISES 283

20.3 Exercises
Exercise 20.1. Find all solutions of the equation, and simplify as much as
possible. Do not approximate the solution.
√ √ √ √
a) tan(x) = 33 b) sin(x) = 23 c) sin(x) = − 22 d) cos(x) = 23
e) cos(x) = 0 f) cos(x) = −0.5 g) cos(x) = 1 √ h) sin(x) = 5
i) sin(x) = 0 j) sin(x) = −1 k) tan(x) = − 3 l) cos(x) = 0.2

Exercise 20.2. Find all solutions of the equation. Approximate your solution
with the calculator.
a) tan(x) = 6.2 b) cos(x) = 0.45 c) sin(x) = 0.91
d) cos(x) = −.772 e) tan(x) = −0.2 f) sin(x) = −0.06

Exercise 20.3. Find at least 5 distinct solutions of the equation.

√ √
a) tan(x) = −1 b) cos(x) = 22 c) sin(x) = − 23 d) tan(x) = 0
e) cos(x) = 0 f) cos(x) = 0.3 g) sin(x) = 0.4 h) sin(x) = −1

Exercise 20.4. Solve for x. State the general solution without approximation.

a) tan(x) − 1√ =0 b) 2 sin(x) = 1
c) 2 cos(x) −√ 3 = 0 d) sec(x) = −2
e) cot(x) = 3 f) tan2 (x) − 3 = 0
g) sin2 (x) − 1 = 0 h) cos2 (x) + 7 cos(x) + 6 = 0
i) 4 cos2 (x) − 4 cos(x) + 1 = 0 j) 2 sin2 (x) + 11 sin(x) = −5
k) 2 sin2 (x) + sin(x) − 1 = 0 l) 2 cos2 (x) − 3 cos(x) + 1 = 0
m) 2 cos2 (x) + 9 cos(x) = 5 n) tan3 (x) − tan(x) = 0

Exercise 20.5. Use the calculator to find all solutions of the given equation.
Approximate the answer to the nearest thousandth.

a) 2 cos(x) = 2 sin(x) + 1 b) 7 tan(x) · cos(2x) = 1

c) 4 cos2 (3x) + cos(3x) = sin(3x) + 2 d) sin(x) + tan(x) = cos(x)
Review of trigonometric functions

Exercise IV.1. Fill in all the trigonometric function values in the table below.
π π π π
0 6 4 3 2
sin(x)
cos(x)
tan(x)

Exercise IV.2. Find the exact values of

a) cos − π6 b) sin − π4 c) tan − π
3

Exercise IV.3. Find the exact value of the trigonometric function.

a) sin 5π
4
b) cos 11π
6

[Hint: Use the special 45◦ − 45◦ − 90◦ or 30◦ − 60◦ − 90◦ triangles to find the
solution.]
Exercise IV.4. Find the amplitude, period, and the phase shift of the given
function. Draw the graph over a one-period interval. Label all maxima, minima
and intercepts.



a) y = 3 cos 4x − π b) y = −5 sin x + π2

Exercise IV.5. Find the exact trigonometric function value.

π


a) cos 12 [Hint: Use the addition and subtraction of angles formulas.]


b) cos 3π8
[Hint: Use the half-angles formulas.]

Exercise IV.6. Let sin(α) = − 45 and let α be in quadrant III. Find sin(2α),
cos(2α), and tan(2α).

284
Exercise IV.7. Find the exact value of:

√
√

a) sin−1 − 21 b) cos−1 − 2
3
c) tan−1 − 3
3

√
Exercise IV.8. Solve for x: 2 sin(x) + 3=0

Exercise IV.9. Solve for x: tan2 (x) − 1 = 0

Exercise IV.10. Solve for x.

a) 2 cos2 (x) − 1 = 0 b) 2 sin2 (x) + 15 sin(x) + 7 = 0

285
 Part V

Complex numbers, sequences, and

the binomial theorem

286
Session 21

Complex numbers

21.1 Polar form of complex numbers

We now recall the definition of complex numbers and show how they are
represented on the complex plane.

Definition 21.1. We define the imaginary unit or complex unit to be

√
imaginary unit: i= −1 (21.1)

In other words, i is a solution of the equation:

i2 = −1 (21.2)

Definition 21.2. A complex number is a number of the form a + b · i. (Alter-

natively we also write this as a + bi without the dot for the multiplication.)
Here a and b are any real numbers. The number a is called the real part of
a + b · i, and b is called the imaginary part of a + b · i. The set of all complex
numbers is denoted by C.

Here are some examples of complex numbers:

√
3 + 2i, 1 − 1 · i, 2 + π · i, 5 + 0 · i, 0+3·i

Note that in the last two examples, we have 5 + 0 · i = 5, so that we see

that the real number 5 is also a complex number. Indeed, any real number
a = a + 0 · i is also a complex number. Similarly, 0 + 3 · i = 3i as well as any

287
288 SESSION 21. COMPLEX NUMBERS

multiple of i is also a complex number (these numbers are often called pure
imaginary numbers).
In analogy to section 1.1, where we represented the real numbers on the
number line, we can represent the complex numbers on the complex plane:
3
Im
2

0 Re
-3 -2 -1 0 1 2 3
-1

-2

-3

The complex number a + bi is represented as the point with coordinates (a, b)

in the complex plane.
4
Im
3

3 + 2i
2

−2 + i
1

0 Re
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2 3 + 0i

-3
−5 − 3i
-4

Example 21.3. Perform the operation.

5 + 4i
a) (2 − 3i) + (−6 + 4i) b) (3 + 5i) · (−7 + i) c)
3 + 2i
Solution. a) Adding real and imaginary parts, respectively, gives,
(2 − 3i) + (−6 + 4i) = 2 − 3i − 6 + 4i = −4 + i.
b) We multiply (using FOIL), and use that i2 = −1.
(3 + 5i) · (−7 + i) = −21 + 3i − 35i + 5i2 = −21 − 32i + 5 · (−1)
21.1. POLAR FORM OF COMPLEX NUMBERS 289

= −21 − 32i − 5 = −26 − 32i

c) Recall that we may simplify a quotient of complex numbers by multi-

plying numerator and denominator by the conjugate of the denominator.

5 + 4i (5 + 4i) · (3 − 2i) 15 − 10i + 12i − 8i2

= =
3 + 2i (3 + 2i) · (3 − 2i) 9 − 6i + 6i − 4i2
15 + 2i + 8 23 + 2i 23 2
= = = + i
9+4 13 13 13
23 2
The real part of the solution is 13
; the imaginary part is 13
.

There is also a notion of absolute value |a + bi| for any complex number
a + bi.

Definition 21.4. Let a+ bi be a complex number. The absolute value of a+ bi,

denoted by |a + bi| is the length between the point a + bi in the complex plane
and the origin (0, 0).
Im
a + bi
bi

|a + bi|

Re
a

Using the Pythagorean theorem, we can calculate |a + bi| as

|a + bi|2 = a2 + b2 ,

so that
√
|a + bi| = a2 + b2 (21.3)

Example 21.5. Find the absolute value of the complex numbers below.

a) 5 − 3i b) − 8 + 6i c) 7i
290 SESSION 21. COMPLEX NUMBERS

Solution. The absolute values are calculated as follows.

p √ √
a) |5 − 3i| = 52 + (−3)2 = 25 + 9 = 34
p √ √
b) | − 8 + 6i| = (−8)2 + 62 = 64 + 36 = 100 = 10
p √
c) |7i| = 02 + (7)2 = 0 + 49 = 7

We can write a complex number in the form a + bi, or alternatively, we

may use the absolute value and “angle” to write the number in the so-called
polar form, which we discuss now.
Observation 21.6. Let a + bi be a complex number. The coordinates in the
plane can be expressed in terms of the absolute value r = |a + bi| and the
angle θ with the real axis as shown below.
Im
a + bi
bi

r r · sin(θ)

θ Re
r · cos(θ) a

From the right triangle as shown in the complex plane above, we see that the
coordinates a and b in the plane are given by:
a = r · cos(θ) and b = r · sin(θ) (21.4)
Therefore, the complex number is a + bi = r · cos(θ) + r · sin(θ) · i, or factoring
r from this expression:


a + bi = r · cos(θ) + i · sin(θ) (21.5)

Writing the complex number in this way is called the polar form of the complex
number. Here, the number r is the absolute value, and θ is given by (21.4)
r·sin(θ) sin(θ)
via the calculation ab = r·cos(θ) = cos(θ) = tan(θ). Thus,

√ b
r= a2 + b2 and tan(θ) = (21.6)
a
21.1. POLAR FORM OF COMPLEX NUMBERS 291

When finding the directional angle θ of a complex number via equation

(21.6), we have to remember that tan(θ) may be obtained by the two angles
θ and π + θ; that is, we have

tan(π + θ) = tan(θ).

Therefore, we need to check that our answer for θ lies in the correct quadrant.
This is shown in the next example.
Example 21.7. Convert the complex number to polar form.
√
a) 2 + 3i, b) −2 − 2 3i, c) 4 − 3i, d) −4i
√ √
Solution. a) First, the absolute value is r = |2 + 3i| = 22 + 32 = 13.
Furthermore, since a = 2 and b = 3, we have tan(θ) = 23 . To obtain θ, we
calculate 3
tan−1 ≈ 56.3◦
2
Note that 56.3◦ is in the first quadrant, and so is the complex number 2 + 3i
4
Im
3

2 + 3i
2

56.3◦ Re
0
-4 -3 -2 -1 0 1 2 3 4
-1

-2

Therefore, θ ≈ 56.3◦, and we obtain our answer:

√ 
◦ ◦

2 + 3i ≈ 13 · cos(56.3 ) + i sin(56.3 ) .
√
b) For −2 − 2 3i, we first calculate the absolute value:
q √ √ √ √
r = (−2)2 + (−2 3)2 = 4 + 4 · 3 = 4 + 12 = 16 = 4.
√ √ √
Furthermore, tan(θ) = ab = −2−2 3 = 3. Now, tan−1 ( 3) = 60◦ = π3 . How-
√
ever, graphing the angle π3 = 60◦ and the number −2 − 2 3i, we see that 60◦
292 SESSION 21. COMPLEX NUMBERS
√
is in the first quadrant, whereas −2 − 2 3i is in the third quadrant.
4
Im
3

1
π
60◦ =
0
3 Re
-4 -3 -2 -1 0 1 2 3 4
-1

-2

√ -3
−2 − 2 3i
-4

√
Therefore, we have to add π to π3 to get the correct angle for −2 − 2 3i, that
is θ = π3 + π = π+3π
3
= 4π
3
. Our complex number in polar form is

√  4π
4π

−2 − 2 3i = 4 · cos + i sin .
3 3
p √ √
c) For 4 − 3i we calculate r = 42 + (−3)2 = 16 + 9 = 25 = 5.
The angle tan−1 ( −3
4
) ≈ −36.9◦ , which is in the fourth quadrant, just like the
number 4 − 3i.
2
Im
1

0 Re
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1 −36.9◦

-2

-3 4 − 3i

-4

Therefore, θ ≈ −36.9◦ , and we write

 
4 − 3i ≈ 5 · cos(−36.9◦ ) + i sin(−36.9◦ )

p √
d) We calculate the absolute value of 0 −4i as r = 02 + (−4)2 = 16 =
4. However, when calculating the angle θ of 0 − 4i, we are lead to consider
tan−1 ( −4
0
), which is undefined! The reason for this can be seen by plotting
21.1. POLAR FORM OF COMPLEX NUMBERS 293

the number −4i in the complex plane.

2
Im
1
3π
270◦ = 2
0 Re
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3

-4
−4i
-5

-6

The angle θ = 270◦ (or alternatively θ = −90◦ ), so that the complex number
is

−4i = 4 · (cos(270◦ ) + i sin(270◦))

 3π
3π

= 4 · cos + i sin
2 2

Note that we may write our answer either in degree or radian mode as we
did above.

We may also perform the reverse conversion of a complex number from

polar form into “standard form” a + bi.

Example 21.8. Convert the number from polar form into the standard form
a + bi.

a) 3 · (cos(117◦ ) + i sin(117◦ )) b) 4 · (cos( 5π

4
) + i sin( 5π
4
))

Solution. a) Since we don’t have an exact formula for cos(117◦) or sin(117◦ ),

we use the calculator to obtain approximate values.

3 · (cos(117◦) + i sin(117◦ )) ≈ 3 · (−0.454 + i · 0.891) = −1.362 + 2.673i

√ √
b) We recall that cos( 5π
4
) = − 22 and sin( 5π
4
) = − 22 . (This can be seen
as in Example 17.6 on page 237 by considering the point P (−1, −1) on the
294 SESSION 21. COMPLEX NUMBERS

5π 5·180◦
terminal side of the angle 4
= 4
= 225◦ .

−1 5π
4
= 225◦

√
2
−1
P

√ √
Therefore, cos( 5π
4
−1
)= √ 2
= − 22 and sin( 5π
4
)= √−1
2
= − 22 .)
With this, we obtain the complex number in standard form.
  √ √ !
5π 5π 2 2
4 · cos( ) + i sin( ) = 4· − −i
4 4 2 2
√ √
2 2 √ √
= −4 −i·4 = −2 2 − 2 2 · i
2 2

21.2 Multiplication and division of complex num-

bers in polar form
One interesting feature of the polar form of a complex number is that the
multiplication and division are very easy to perform.

Proposition 21.9. Let r1 (cos(θ1 ) + i sin(θ1 )) and r2 (cos(θ2 ) + i sin(θ2 )) be two

complex numbers in polar form. Then, the product and quotient of these are
given by

r1 (cos(θ1 ) + i sin(θ1 )) · r2 (cos(θ2 ) + i sin(θ2 ))

= r1 r2 · (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) (21.7)

r1 (cos(θ1 ) + i sin(θ1 )) r1
= · (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )) (21.8)
r2 (cos(θ2 ) + i sin(θ2 )) r2
21.2. MULTIPLICATION AND DIVISION OF COMPLEX NUMBERS 295

Proof. The proof uses the addition formulas for trigonometric functions sin(α + β) and cos(α + β) from
proposition 18.1 on page 252.

r1 (cos(θ1 ) + i sin(θ1 )) · r2 (cos(θ2 ) + i sin(θ2 ))

= r1 r2 · (cos(θ1 ) cos(θ2 ) + i cos(θ1 ) sin(θ2 ) + i sin(θ1 ) cos(θ2 ) + i2 sin(θ1 ) sin(θ2 ))
= r1 r2 · ((cos(θ1 ) cos(θ2 ) − sin(θ1 ) sin(θ2 )) + i(cos(θ1 ) sin(θ2 ) + sin(θ1 ) cos(θ2 )))
= r1 r2 · (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ))

For the division formula, note, that the multiplication formula (21.7) gives

1 1
r2 (cos(θ2 ) + i sin(θ2 )) · (cos(−θ2 ) + i sin(−θ2 )) = r2 (cos(θ2 − θ2 ) + i sin(θ2 − θ2 ))
r2 r2
= 1 · (cos 0 + i sin 0) = 1 · (1 + i · 0) = 1
1 1
=⇒ = (cos(−θ2 ) + i sin(−θ2 )),
r2 (cos(θ2 ) + i sin(θ2 )) r2

so that

r1 (cos(θ1 ) + i sin(θ1 )) 1
= r1 (cos(θ1 ) + i sin(θ1 )) ·
r2 (cos(θ2 ) + i sin(θ2 )) r2 (cos(θ2 ) + i sin(θ2 ))
1 r1
= r1 (cos(θ1 ) + i sin(θ1 )) · (cos(−θ2 ) + i sin(−θ2 )) = · (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )).
r2 r2

Example 21.10. Multiply or divide the complex numbers, and write your
answer in polar and standard form.

a) 5(cos(11◦) + i sin(11◦ )) · 8(cos(34◦ ) + i sin(34◦ ))

b) 3(cos( 5π
8
) + i sin( 5π
8
)) · 12(cos( 7π
8
) + i sin( 7π
8
))
32(cos( π4 ) + i sin( π4 )) 4(cos(203◦ ) + i sin(203◦))
c) d)
8(cos( 7π
12
) + i sin( 7π
12
)) 6(cos(74◦ ) + i sin(74◦ ))
Solution. We will multiply and divide the complex numbers using equations
(21.7) and (21.8), respectively, and then convert them to standard notation
a + bi.

a) 5(cos(11◦) + i sin(11◦ )) · 8(cos(34◦ ) + i sin(34◦ ))

= 5 · 8 · (cos(11◦ + 34◦ ) + i sin(11◦ + 34◦ )) = 40(cos(45◦ ) + i sin(45◦ ))
 √2 √ 
2
√
2
√
2 √ √
= 40 +i = 40 + i · 40 = 20 2 + 20 2i
2 2 2 2
b) Similarly, we obtain the next product.
296 SESSION 21. COMPLEX NUMBERS
 5π 5π   7π 7π 
3 cos( ) + i sin( ) · 12 cos( ) + i sin( )
8 8 8 8
 5π 7π 5π 7π 
= 36 cos( + ) + i sin( + )
8 8 8 8
5π 7π 5π + 7π 12π 3π
Now, + = = = , and cos( 3π
2
) = 0 and sin( 3π
2
) = −1.
8 8 8 8 2
Therefore, we obtain that the product is
 3π 3π 
36 cos( ) + i sin( ) = 36(0 + i · (−1)) = −36i
2 2
c) For the quotient, we use the subtraction formula (21.8).
32(cos( π4 ) + i sin( π4 )) 32  π 7π π 7π 
= cos( − ) + i sin( − )
8(cos( 7π
12
) + i sin( 7π
12
)) 8 4 12 4 12

Now, the difference in the argument of cos and sin is given by

π 7π 3π − 7π −4π −π
− = = = ,
4 12 12 12 3
√
and cos(− π3 ) = cos( π3 ) = 1
2
and sin(− π3 ) = − sin( π3 ) = − 2
3
. With this, we
obtain
32(cos( π4 ) + i sin( π4 ))  −π −π 
= 4 cos( ) + i sin( )
8(cos( 7π
12
) + i sin( 7π
12
)) 3 3
√ 
1 3 √
= 4· −i =2−2 3·i
2 2
d) Finally, we calculate
4(cos(203◦) + i sin(203◦ )) 2  ◦ ◦

= · cos(129 ) + i sin(129 ) .
6(cos(74◦ ) + i sin(74◦ )) 3
Since we do not have exact values of cos and sin for the angle 129◦ , we
approximate the complex number in standard form with the calculator.
2   2 2
· cos(129◦ ) + i sin(129◦) = · cos(129◦ ) + i · · sin(129◦ )
3 3 3
≈ −0.420 + 0.518 · i

Note that here we approximated the solution to the nearest thousandth.

21.3. EXERCISES 297

21.3 Exercises
Exercise 21.1. Plot the complex numbers in the complex plane.
a) √
4 + 2i√ b) −3 − 5i c) 6 − 2i d) −5 + i e) −2i √
f) 2 − 2i g) 7 h) i i) 0 j) 2i − 3
Exercise 21.2. Add, subtract, multiply, and divide, as indicated.
a) (5 − 2i) + (−2 + 6i) b) (−9 − i) − (5 − 3i)
c) (3 + 2i) · (4 + 3i) d) (−2 − i) · (−1 + 4i)
e) 2+3i
2+i
f) (5 + 5i) ÷ (2 − 4i)

Exercise 21.3. Find the absolute value |a + bi| of the given complex number,
and simplify your answer as much as possible.
a) |4
√+ 3i| b) |1 − √2i| c) | − 3i| d) | − √
2 − 6i| √
e) | 8 − i| f) | − 2 3 − 2i| g) | − 5| h) | − 17 + 4 2i|
Exercise 21.4. Convert the complex number into polar form r(cos(θ)+i sin(θ)).
√
a) 2 + 2i b) 4 3 + 4i c) 3 −
√2i √ d) √
−5 + 5i
√
e) 4 − 3i f) −4 + 3i g) − 5 − 15i h) √ 7 − 21i
i) −5 − 12i j) 6i k) −10 l) − 3 + 3i
Exercise 21.5. Convert the complex number into the standard form a + bi.
π
a) 6(cos(134◦) + i sin(134◦ )) b) 21 (cos( 17 π
) + i sin( 17 ))
c) 2(cos(270◦ ) + i sin(270◦)) d) cos( π6 ) + i sin( π6 )
e) 10(cos( 7π
6
) + i sin( 7π
6
)) f) 6(cos(− 5π 12
) + i sin(− 5π12
))
Exercise 21.6. Multiply the complex numbers and write the answer in stan-
dard form a + bi.
a) 4(cos(27◦ ) + i sin(27◦ )) · 10(cos(33◦ ) + i sin(33◦ ))
b) 7(cos( 2π
9
) + i sin( 2π
9
)) · 6(cos( π9 ) + i sin( π9 ))
c) (cos( 13π
12
) + i sin( 13π
12
)) · (cos( −11π
12
) + i sin( −11π
12
))
d) 8(cos( 3π
7
) + i sin( 3π
7
)) · 1.5(cos( 4π
7
) + i sin( 4π
7
))
e) 0.2(cos(196◦) + i sin(196◦ )) · 0.5(cos(88◦ ) + i sin(88◦ ))
f) 4(cos( 7π
8
) + i sin( 7π
8
)) · 0.25(cos( −5π
24
) + i sin( −5π
24
))
298 SESSION 21. COMPLEX NUMBERS

Exercise 21.7. Divide the complex numbers and write the answer in standard
form a + bi.
18(cos( π2 ) + i sin( π2 )) 10(cos(254◦ ) + i sin(254◦))
a) b)
3(cos( π6 ) + i sin( π6 )) 15(cos(164◦ ) + i sin(164◦))
√
24(cos( 11π14
) + i sin( 11π
14
)) cos( 8π
5
) + i sin( 8π
5
)
c) √ 2π 2π
d) π π
6(cos( 7 ) + i sin( 7 )) 2(cos( 10 ) + i sin( 10 ))

42(cos( 7π
4
) + i sin( 7π
4
)) 30(cos(−175◦ ) + i sin(−175◦ ))
e) 5π 5π f)
7(cos( 12 ) + i sin( 12 )) 18(cos(144◦) + i sin(144◦ ))
Session 22

Vectors in the plane

22.1 Introduction to vectors

There is yet another way to discuss the 2-dimensional plane, which highlights
some other useful properties of the plane R2 . This is the notion of R2 as a
vector space. We start by defining vectors in the plane.

Definition 22.1. A geometric vector in the plane is a geometric object in the

plane R2 that is given by a direction and magnitude. We denote a vector by
~v (it is written by some authors as v), its magnitude is denoted by ||~v||, and
its directional angle by θ.
−→
Vectors are often represented by directed line segments ~v = P Q, where
two line segments represent the same vector, if one can be moved to the other
by parallel translation (without changing its direction or magnitute).
5 y

4 Q
~v ~v
3

P
2
~v
1 R
~v
O x
0
-2 -1 0 1 2 3 4 5 6 7 8 9 10
-1

-2

299
300 SESSION 22. VECTORS IN THE PLANE

−→
In particular, we can always represent a vector ~v by OR by arranging the
starting point of ~v to the origin O(0, 0). If R is given in coordinates by R(a, b),
−→
then we also write for ~v = OR,
 
a
~v = ha, bi or, alternatively, ~v = . (22.1)
b

−→
~ ~r, ~s, ~t in the plane, where ~v = P Q with
Example 22.2. Graph the vectors ~v , w,
P (6, 3) and Q(4, −2), and

w
~ = h3, −1i, ~r = h−4, −2i, ~s = h0, 2i, ~t = h−5, 3i.

Solution.
4 y

P
3

2
~t
~v
1
~s
0 x
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1
w
~
~r
-2

Q
-3

The formulas for the magnitude and the directional angle of a vector can be
obtained precisely the same way as the absolute value and angle of a complex
number. From equation (21.6) in Observation 21.6, we therefore obtain the
following analogue formulas.

Observation 22.3. Let ~v = ha, bi be a vector in the plane R2 . Then the

magnitude and angle of ~v are given by:

√ b
||~v|| = a2 + b2 and tan(θ) = (22.2)
a
22.1. INTRODUCTION TO VECTORS 301
y
R(a, b)
b
~v

θ x
a

Conversely, we can recover the coordinates of a vector ~v from its magnitude

||~v|| and angle θ by (see equation (21.4) from page 290):

~v = h ||~v|| · cos(θ) , ||~v|| · sin(θ) i (22.3)

Example 22.4. Find the magnitude and directional angle of the given vectors.
√
a) h−6, 6i b) h4, −3i c) h−2 3, −2i
√ −→
d) h8, 4 5i e) P Q, where P (9, 2) and Q(3, 10)

Solution. a) We use formulas (22.2), and the calculation is in analogy with

Example 21.7. The magnitude of ~v = h−6, 6i is
p √ √ √ √
||~v|| = (−6)2 + 62 = 36 + 36 = 72 = 36 · 2 = 6 2.
6
The directional angle θ is given by tan(θ) = −6 = −1. Now, since tan−1 (−1) =
− tan−1 (1) = −45◦ is in the fourth quadrant, but ~v = h−6, 6i drawn at the
origin O(0, 0) has its endpoint in the second quadrant, we see that the angle
θ = −45◦ + 180◦ = 135◦ .
y

x
−6 −45◦

b) The magnitude of ~v = h4, −3i is

p √ √
||~v|| = 42 + (−3)2 = 16 + 9 = 25 = 5.
302 SESSION 22. VECTORS IN THE PLANE

The directional angle is given by tan(θ) = −3 4

. Since tan−1 ( −3
4
) ≈ −36.9◦ is
in the fourth quadrant, and ~v = h4, −3i is in the fourth quadrant, we see that
 −3 
θ = tan−1 ≈ −36.9◦ .
4
√
c) The magnitude of ~v = h−2 3, −2i is
q √ √ √ √
||~v|| = (−2 3)2 + (−2)2 = 4 · 3 + 4 = 12 + 4 = 16 = 4.
√ √
−2
The directional angle is given by tan(θ) = = √13 = 33 . Now, tan−1 ( 33 ) =
√
−2 3
√
30◦ = π6 is in the first quadrant, whereas ~v = h−2 3, −2i is in the third quad-
rant. Therefore, the angle is given by adding an additional π to the angle.
π 6π + π 7π
θ=π+ + = .
6 6 6
√
d) The magnitude of ~v = h8, 4 5i is
q √ √ √ √
||~v|| = 82 + (4 5)2 = 64 + 16 · 5 = 64 + 80 = 144 = 12.
√ √
The directional
√
angle is given by the equation tan(θ) = 4 8 5 = 25 . Since
both tan−1 ( 25 ) ≈ 48.2◦ and the endpoint of ~v (represented with beginning
point at the origin) are in the first quadrant, we have:
 √5 
−1
θ = tan ≈ 48.2◦ .
2
−→
e) We first need to find the vector ~v = P Q in the form ~v = ha, bi. The
vector in the plane below shows that ~v is given by
~v = h3 − 9, 10 − 2i = h−6, 8i
11 y

Q
10

~v
6
10 − 2
5

P
1
3−9
0 x
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
22.2. OPERATIONS ON VECTORS 303

From this we calculate the magnitude to be

p √ √
||~v|| = (−6)2 + 82 = 36 + 64 = 100 = 10.
8
The directional angle is given by tan(θ) = −6 = − 43 . Now, tan−1 (− 34 ) ≈
−53.1◦ is in quadrant IV, whereas ~v = h−6, 8i has an endpoint in quadrant
II (when representing ~v with starting point at the origin O(0, 0)). Therefore,
the directional angle is
 4
◦ −1
θ = 180 + tan − ≈ 126.9◦
3

22.2 Operations on vectors

There are two basic operations on vectors, which are the scalar multiplication
and the vector addition. We start with the scalar multiplication.
Definition 22.5. The scalar multiplication of a real number r with a vector
~v = ha, bi is defined to be the vector given by multiplying r to each coordinate.
r · ha, bi := hr · a, r · bi (22.4)
We study the effect of scalar multiplication with an example.
Example 22.6. Multiply, and graph the vectors.
a) 4 · h−2, 1i b) (−3) · h−6, −2i
Solution. a) The calculation is straightforward.
4 · h−2, 1i = h4 · (−2), 4 · 1i = h−8, 4i
The vectors are displayed below. We see, that h−2, 1i and h−8, 4i both have
the same directional angle, and the latter stretches the former by a factor 4.
5 y

3
h−8, 4i
2

h−2, 1i x
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2
-1
304 SESSION 22. VECTORS IN THE PLANE

b) Algebraically, we calculate the scalar multiplication as follows:

(−3) · h−6, −2i = h(−3) · (−6), (−3) · (−2)i = h18, 6i

Furthermore, graphing the vectors gives:

y

h18, 6i

x
h−6, −2i

We see that the directional angle of the two vectors differs by 180◦. Indeed,
h18, 6i is obtained from h−6, −2i by reflecting it at the origin O(0, 0) and
then stretching the result by a factor 3.

We see from the above example, that scalar multiplication by a positive

number c does not change the angle of the vector, but it multiplies the mag-
nitude by c.

Observation 22.7. Let ~v be a vector with magnitude ||~v|| and angle θ~v . Then,
for a positive scalar, r > 0, the scalar multiple r · ~v has the same angle as ~v,
and a magnitude that is r times the magnitude of ~v :

for r > 0: ||r · ~v || = r · ||~v|| and θr·~v = θ~v

y

r · ~v

~v
θr·~v = θ~v x

Definition 22.8. A vector ~u is called a unit vector, if it has a magnitude of 1.

~u is a unit vector ⇐⇒ ||~u|| = 1

22.2. OPERATIONS ON VECTORS 305

There are two special unit vectors, which are the vectors pointing in the
x- and the y-direction.

~i := h1, 0i and ~j := h0, 1i (22.5)

Example 22.9. Find a unit vector in the direction of ~v .

√
a) h8, 6i b) h−2, 3 7i

Solution. a) Note that the magnitude of ~v = h8, 6i is

√ √ √
||h8, 6i|| = 82 + 62 = 64 + 36 = 100 = 10.
1 1 8 6
Therefore, if we multiply h8, 6i by 10 we obtain 10 · h8, 6i = h 10 , 10 i = h 45 , 53 i,
which (according to Observation 22.7 above) has the same directional angle
as h8, 6i, and has a magnitude of 1:

1 1 1
· h8, 6i = · ||h8, 6i|| = · 10 = 1
10 10 10
√
b) The magnitude of h−2, 3 7i is
√ q √ √ √ √
||h−2, 3 7i|| = (−2)2 + (3 7)2 = 4 + 9 · 7 = 4 + 63 = 67.

Therefore, we have the unit vector

√ √ √ √ √ √
1 √ −2 3 7 −2 67 3 7 67 −2 67 3 469
√ · h−2, 3 7i = h √ , √ i = h , i=h , i
67 67 67 67 67 67 67
√ √
which again has the same directional angle as h−2, 3 7i. Now, √167 ·h−2, 3 7i
√ √ √
is a unit vector, since || √167 · h−2, 3 7i|| = √167 · ||h−2, 3 7i|| = √167 · 67 =
1.

The second operation on vectors is vector addition, which we discuss now.

Definition 22.10. Let ~v = ha, bi and w~ = hc, di be two vectors. Then the
vector addition ~v + w
~ is defined by component-wise addition:

ha, bi + hc, di := ha + c, b + di (22.6)

306 SESSION 22. VECTORS IN THE PLANE

In terms of the plane, the vector addition corresponds to placing the vectors
~v and w~ as the edges of a parallelogram, so that ~v + w
~ becomes its diagonal.
This is displayed below.
y

~v

d
w
~
~v + w
~
w
~

b
~v x
c a

Example 22.11. Perform the vector addition and simplify as much as possible.
a) h3, −5i + h6, 4i b) 5 · h−6, 2i − 7 · h1, −3i c) 4~i + 9~j
~ for ~v = −6~i −
d) find 2~v + 3w ~ ~ = 10~i −
√4j and w
~
√7j
e) find −3~v + 5w~ for ~v = h8, 3i and w
~ = h0, 4 3i
Solution. We can find the answer by direct algebraic computation.
a) h3, −5i + h6, 4i = h3 + 6, −5 + 4i = h9, −1i
b) 5 · h−6, 2i − 7 · h1, −3i = h−30, 10i + h−7, 21i = h−37, 31i
c) 4~i + 9~j = 4 · h1, 0i + 9 · h0, 1i = h4, 0i + h0, 9i = h4, 9i
From the last calculation, we see that every vector can be written as a linear
combination of the vectors ~i and ~j.

ha, bi = a · ~i + b · ~j (22.7)

We will use this equation for the next example (d). Here, ~v = −6~i − 4~j =
~ = 10~i − 7~j = h10, −7i. Therefore, we obtain:
h−6, −4i and w
d) 2~v + 3w
~ = 2 · h−6, −4i + 3 · h10, −7i = h−12, −8i + h30, −21i = h18, −29i
Finally, we have:
√ √ √ √
e) − 3~v + 5w~ = −3 · h8, 3i + 5 · h0, 4 3i = h−24, −3 3i + h0, 20 3i
√
= h−24, 17 3i
√
~ = −24~i+ 17 3~j.
Note that the answer could also be written as −3~v + 5w
22.2. OPERATIONS ON VECTORS 307

In many applications in the sciences, vectors play an important role, since

many quantities are naturally described by vectors. Examples for this in
physics include the velocity ~v , acceleration ~a, and the force F~ applied to an
object.
Example 22.12. The forces F~1 and F~2 are applied to an object. Find the
resulting total force F~ = F~1 + F~2 . Determine the magnitude and directional
angle of the total force F~ . Approximate these values as necessary. Recall
that the international system of units for force is the newton [1N = 1 kg·ms2
].

a) F~1 has magnitude 3 newtons, and angle θ1 = 45◦ ,

F~2 has magnitude 5 newtons, and angle θ2 = 135◦
b) ||F~1 || = 7 newtons, and θ1 = π6 , and
||F~2 || = 4 newtons, and θ2 = 5π
3

Solution. a) The vectors F~1 and F~2 are given by their magnitudes and direc-
tional angles. However, the addition of vectors (in Definition 22.10) is defined
in terms of their components. Therefore, our first task is to find the vectors in
component form. As was stated in equation (22.3) on page 301, the vectors
are calculated by ~v = h ||~v|| · cos(θ) , ||~v|| · sin(θ) i. Therefore,
√ √ √ √
2 2 3 2 3 2
F~1 = h3 · cos(45 ), 3 · sin(45 )i = h3 ·
◦ ◦
,3 · i=h , i
2 √ 2 √ 2 2
√ √
~ ◦ ◦ − 2 2 −5 2 5 2
F2 = h5 · cos(135 ), 5 · sin(135 )i = h5 · ,5· i=h , i
2 2 2 2
The total force is the sum of the forces.
6 y

4
F~1 + F~2
3

2
F~2
1
F~1
0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

√ √ √ √
3 2 3 2 −5 2 5 2
F~ = F~1 + F~2 = h , i+h , i
2 2 2 2
308 SESSION 22. VECTORS IN THE PLANE
√ √ √ √ √ √ √ √
3 2 −5 2 3 2 5 2 3 2−5 2 3 2+5 2
= h + , + i=h , i
2√ √2 2 2 2 2
−2 2 8 2 √ √
= h , i = h− 2, 4 2i
2 2
√ √
The total force
q applied in components is F~ = h− 2, 4 2i. It has a magnitude
√ √ √ √
of ||F~ || = (− 2)2 + (4 2)2 = 2 + 16 · 2 = 34 ≈ 5.83 newton. The
√
directional angle is given by tan(θ) = −4 √22 = −4. Since tan−1 (−4) ≈ −76.0◦
√ √
is in quadrant IV, and F~ = h− 2, 4 2i is in quadrant II, we see that the
directional angle of F~ is

θ = 180◦ + tan−1 (−4) ≈ 180◦ − 76.0◦ ≈ 104◦.

b) We solve this example in much the same way we solved part (a). First,
F~1 and F~2 in components is given by
π  π  √ √
~ 3 1 7 3 7
F1 = h7 · cos , 7 · sin i = h7 · ,7 · i = h , i
6 6 2 2 √ 2 2
 5π   5π  1 − 3 √
F~2 = h4 · cos , 4 · sin i = h4 · , 4 · i = h2, −2 3i
3 3 2 2
The total force is therefore:
√ √
7 3 7 √ 7 3 7 √
~ ~
F = F1 + F2 = h~ , i + h2, −2 3i = h + 2, − 2 3i
2 2 2 2
≈ h8.06, 0.04i

The magnitude is approximately

p
||F~ || ≈ (8.06)2 + (0.04)2 ≈ 8.06 newton.

The directional angle is given by tan(θ) ≈ 0.04

8.06
. Since F~ is in quadrant I, we
0.04
see that θ ≈ tan−1 ( 8.06 ) ≈ 0.3◦ .

Remark 22.13. There is an abstract notion of a vector space. Although we do not use this structure for any
further computations, we will state its definition. A vector space is a set V , with the following structures
and properties. The elements of V are called vectors, denoted by the usual symbol ~v. For any vectors ~v
22.3. EXERCISES 309

and w~ there is a vector ~ v + w,

~ called the vector addition. For any real number r and vector ~v, there is a
vector r · ~v called the scalar product. These operations have to satisfy the following properties.

Associativity: (~
u+~ v) + w ~ =~ u + (~
v + w)~
Commutativity: ~v + w~ =w ~ + ~v
Zero element: there is a vector ~
o such that ~ o+~ v =~ v and ~ v +~ o=~v for every vector ~v
Negative element: for every ~v there is a vector −~ v such that ~ v + (−~ v) = ~
o and (−~v) + ~v=~ o
Distributivity (1): r · (~
v + w)~ = r ·~ v+r·w ~
Distributivity (2): (r + s) · ~v = r · ~v + s · ~v
Scalar compatibility: (r · s) · ~
v = r · (s · ~v)
Identity: 1 · ~v = ~
v

A very important example of a vector space is the 2-dimensional plane V = R2 as it was discussed in this
chapter.

22.3 Exercises
Exercise 22.1. Graph the vectors in the plane.
−→ −→
a) P Q with P (2, 1) and Q(4, 7) b) P Q with P (−3, 3) and Q(−5, −4)
−→
c) P Q with P (0, −4) and Q(6, 0) d) h−2,√4i
e) h−3, −3i f) h5, 5 2i

Exercise 22.2. Find the magnitude and directional angle of the vector.

a) h6, 8i b) h−2, 5i c) h−4,

√ −4i
d) h3,√−3i e) h2, −2i√ f) h4 √
3, 4i
g) h− 3, −1i h) h−4, 4 3i i) h−2 3, −2i

−→
j) P Q, where P (3, 1) and Q(7, 4)
−→
k) P Q, where P (4, −2) and Q(−5, 7)

Exercise 22.3. Perform the operation on the vectors.

a) 5 · h3, 2i b) 2 · h−1, 4i
c) (−10) · h− 23 , − 75 i d) h2, 3i + h6, 1i
e) h5, −4i − h−8, −9i f) 3 · h5, 3i + 4 · h2, 8i
g) (−2)h−5, −4i − 6h−1, −2i h) 32 h−3, 6i − 75 h10, −15i
√ √ √
i) 2 · h 68 , −512 2 i − 2h 23 , 53 i j) 6~i − 4~j
k) −5~i + ~j + 3~i l) 3 · h−4, 2i − 8~j + 12~i
310 SESSION 22. VECTORS IN THE PLANE
√
m) find 4~v + 7w~ for ~v = h2, 3i and w~ = h5, 1 3i
n) find ~v − 2w
~ for ~v = h−11, −6i and w ~ = h−3, 2i
~√for ~v = −4~i + 7~j and w
o) find 3~v − w ~ = 6~i +√
~j
p) find −~v − 5w ~ for ~v = 5~j and w~ = −8~i + 5~j
Exercise 22.4. Find a unit vector in the direction of the given vector.
√
a) h8, √
−6i√ b) h−3, −
√ √7i c) h9, 2i
d) h− 5, 31i e) h5 2, 3 10i f) h0, − 35 i

Exercise 22.5. The vectors v~1 and v~2 below are being added. Find the ap-
proximate magnitude and directional angle of sum ~v = v~1 + v~2 (see Example
22.12).
a) ||v~1 || = 6, and θ1 = 60◦ , and
||v~2 || = 2, and θ2 = 180◦
b) ||v~1 || = 3.7, and θ1 = 92◦ , and
||v~2 || = 2.2, and θ2 = 253◦
c) ||v~1 || = 8,√and θ1 = 3π
4
, and
||v~2 || = 8 2, and θ2 = 3π 2
Session 23

Sequences and series

23.1 Introduction to sequences and series

Definition 23.1. A sequence is an ordered list of numbers. We write a se-
quence in order as follows:

a1 , a2 , a3 , a4 , . . .

In short we write the above sequence as {an } or {an }n≥1 .

You might wonder how this topic fits into the general theme of functions. We can alternatively define a
sequence a to be a function a : N → R, where the domain is all natural numbers, and the range R is a
subset of the real or complex numbers. We write the nth term of the sequence as an = a(n).

Example 23.2. Here are some examples of sequences.

a) 4, 6, 8, 10, 12, 14, 16, 18, . . .

b) 1, 3, 9, 27, 81, 243, . . .

c) +5, −5, +5, −5, +5, −5, . . .

d) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .

√
e) 5, 8, −12, 4, 5.3, 7, −3, 2, 18, 32 , 9, . . .

For many of these sequences we can find rules that describe how to obtain
the individual terms. For example, in (a), we always add the fixed number 2
to the previous number to obtain the next, starting from the first term 4. This

311
312 SESSION 23. SEQUENCES AND SERIES

is an example of an arithmetic sequence, and we will study those in more

detail in section 23.2 below.
In (b), we start with the first element 1 and multiply by the fixed number
3 to obtain the next term. This is an example of a geometric sequence, and
we will study those in more detail in section 24 below.
The sequence in (c) alternates between +5 and −5, starting from +5.
Note, that we can get from one term to the next by multiplying (−1) to the
term. Therefore, this is another example of a geometric sequence.
The sequence in (d) is called the Fibonacci sequence. In the Fibonacci
sequence, each term is calculated by adding the previous two terms, starting
with the first two terms 1 and 1:

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, ...

Finally, the sequence in (e) does not seem to have any obvious rule by
which the terms are generated.

In many cases, the sequence is given by a formula for the nth term an .

Example 23.3. Consider the sequence {an } with an = 4n + 3. We can cal-

culate the individual terms of this sequence:

first term: a1 = 4 · 1 + 3 = 7,
second term: a2 = 4 · 2 + 3 = 11,
third term: a3 = 4 · 3 + 3 = 15,
fourth term: a4 = 4 · 4 + 3 = 19,
fifth term: a5 = 4 · 5 + 3 = 23
..
.

Thus, the sequence is: 7, 11, 15, 19, 23, 27, 31, 35, . . .
Furthermore, from the formula, we can directly calculate a higher term, for
example the 200th term is given by:

200th term: a200 = 4 · 200 + 3 = 803

Example 23.4. Find the first 6 terms of each sequence.

n
a) an = n2 b) an = n+1
c) an = (−1)n d) an = (−1)n+1 · 2n
23.1. INTRODUCTION TO SEQUENCES AND SERIES 313

Solution. a) We can easily calculate the first 6 terms of an = n2 directly:

1, 4, 9, 16, 25, 36, . . .

We can also use the calculator to produce the terms of a sequence. To this
end, we switch the✞calculator
☎ from function mode to sequence mode in the
mode menu (press lpmodelp ):
✝ ✆

✞ ☎
To enter the sequence, we need to use the LIST menu (press lp2nd lp and
✞ ☎ ✞ ☎ ✝ ✆
then lpstatlp ), then go to the OPS menu (press lp⊲ lp ), and use the fifth
✝ ✞✆ ☎ ✝ ✆
item (press lp5 lp ):
✝ ✆

✞ ☎✞ ☎ ✞ ☎ use fifth
✞ item☎
press lp2nd lp lpstatlp press lp⊲ lp (press lp5 lp )
✝ ✆✝ ✆ ✝ ✆ ✝ ✆

For the sequence, we need to specify

✞ four☎pieces of information, where the
index n can be entered via the lpX,T,θ,nlp key.
✝ ✆

1st entry is the given formula for an = n2

2nd entry is the index n of the sequence an
3rd and 4th entries are the first and last index 1 and
6 of the wanted sequence, here: a1 , . . . , a6
✄
Alternatively, we can enter the function in the lpy= lp menu, starting from the first index n = 1, and with
✂ ✁ ✄
nth term given by an = n2 . The values of the sequence are displayed in the table window (press lp2nd lp
✂ ✁
314 SESSION 23. SEQUENCES AND SERIES
✄
and lpgraph lp ).
✂ ✁

n
b) We calculate the lowest terms of an = n+1
:
1 1 2 2 3 3
a1 = = , a2 = = , a3 = = , ...
1+1 2 2+1 3 3+1 4
Identifying the pattern, we can simply write a1 , . . . , a6 as follows:
1 2 3 4 5 6
, , , , ,
2 3 4 5 6 7
We can also use the calculator to find the sequence:

Note 23.5.
✄ Fractions can be displayed with the calculator by adding “Frac” to it from the MATH menu
(using lpmathlp and the top item). For example, the fraction 72 is displayed below without approximation.
✂ ✁

Adding “Frac” to the sequence displays it more appropriately as a se-

quence of fractions.
23.1. INTRODUCTION TO SEQUENCES AND SERIES 315

c) Since (−1)n is +1 for even n, but is −1 for odd n, the sequence an =

(−1)n is:
−1, +1, −1, +1, −1, +1
d) Similar to part (c), (−1)n+1 is −1 for even n, and is +1 for odd n. This,
together with the calculation 21 = 2, 22 = 4, 23 = 8, 24 = 16, etc., we get the
first six terms of the sequence:

+2, −4, +8, −16, +32, −64

Another way to describe a sequence is by giving a recursive formula for

the nth term an in terms of the lower terms. Here are some examples.
Example 23.6. Find the first 6 terms in the sequence described below.

a) a1 = 4, and an = an−1 + 5 for n > 1

b) a1 = 3, and an = 2 · an−1 for n > 1
c) a1 = 1, a2 = 1, and an = an−1 + an−2 for n > 2

Solution. a) The first term is explicitly given as a1 = 4. Then, we can calcu-

late the following terms via an = an−1 + 5:

a2 = a1 + 5 = 4 + 5 = 9
a3 = a2 + 5 = 9 + 5 = 14
a4 = a3 + 5 = 14 + 5 = 19
a5 = a4 + 5 = 19 + 5 = 24
..
.

b) We have a1 = 3, and calculate a2 = 2 · a1 = 2 · 3 = 6, a3 = 2 · a2 =

2 · 6 = 12, a4 = 2 · a3 = 2 · 12 = 24, etc. We see that the effect of the recursive
relation an = 2 · an−1 is to double the previous number. The sequence is:

3, 6, 12, 24, 48, 96, 192, . . .

c) Starting from a1 = 1, and a2 = 1, we can calculate the higher terms:

a3 = a1 + a2 = 1 + 1 = 2
a4 = a2 + a3 = 1 + 2 = 3
316 SESSION 23. SEQUENCES AND SERIES

a5 = a3 + a4 = 2 + 3 = 5
a6 = a4 + a5 = 3 + 5 = 8
..
.

Studying the sequence for a short while, we see that this is precisely the
Fibonacci sequence from example 23.2(d).

Note 23.7. There is no specific reason for using the indexing variable n in the sequence {an }. Indeed,
we may use as well any other variable. For example, if the sequence {an }n≥1 is given by the formula
an = 4n + 3, then we can also write this as ak = 4k + 3 or ai = 4i + 3. In particular, the sequences
{an }n≥1 = {ak }k≥1 = {ai }i≥1 = {aj }j≥1 are all identical as sequences.

We will be concerned with the task of adding terms in a sequence, such as

a1 + a2 + a3 + · · · + ak , for which we will use a standard summation notation.
Definition 23.8. A series is a sum of terms in a sequence. We denote the
sum of the first k terms in a sequence with the following notation:
k
X
ai = a1 + a2 + · · · + ak (23.1)
i=1
P
The summation symbol “ ” comes from the greek letter Σ, pronounced “sigma,”
which is the greek letter for “S.”
More generally, we denote the sum from the jth to the kth term (where j ≤ k) in a sequence with the
following notion:
Xk
ai = aj + aj+1 + · · · + ak
i=j

k
P Pk
Furthermore, the term ai is sometimes also written in the form i=j ai .
i=j

Example 23.9. Find the sum.

4
X
a) ai , for ai = 7i + 3
i=1
X6
b) aj , for an = (−2)n
j=1
5
X

c) 4 + k2
k=1
23.1. INTRODUCTION TO SEQUENCES AND SERIES 317

Solution. a) The first four terms a1 , a2 , a3 , a4 of the sequence {ai }i≥1 are:
10, 17, 24, 31
The sum is therefore:
4
X
ai = a1 + a2 + a3 + a4 = 10 + 17 + 24 + 31 = 82
i=1

We may
✞ also find
☎✞ the answer
☎✞ with
☎✞ the calculator.
☎ Before entering the sequence
(via lp2nd lp lpstatlp lp⊲ lp lp5 lp as in example 23.4(a) above), we have
✝ ✆✝ ✆✝ ✆✝ ✆ ✞ ☎
to put a sum symbol in the LIST menu and MATH submenu (with lp2nd lp
✞ ☎✞ ☎✞ ☎ ✝ ✆
lpstatlp lp⊳ lp lp5 lp ):
✝ ✆✝ ✆✝ ✆
✞ ☎
press lp2nd lp enter the sum and seq.
✞ ✝ ☎✞ ✆☎ ✞ ☎
lpstatlp lp⊳ lp press lp5 lp sum(seq(7n + 3, n, 1, 4))
✝ ✆✝ ✆ ✝ ✆

b) The first six terms of {an } with an = (−2)n are:

(−2)1 = −2, (−2)2 = 4, (−2)3 = −8, (−2)4 = 16, (−2)5 = −32, (−2)6 = 64
P
We calculate 6j=1 aj by adding these first six terms. (Note that the sum is
P
independent of the index j appearing in the sum 6j=1 aj , which we could as
P P
well replace by any other index 6j=1 aj = 6k=1 ak , etc.) We get:
6
X
aj = a1 + a2 + a3 + a4 + a5 + a6
j=1
= (−2) + 4 + (−8) + 16 + (−32) + 64 = 42
P
c) For the sum 5k=1 (4 + k 2 ) we need to add the first four terms of the
sequence ak = 4 + k 2 . Calculating and adding the terms of this sequence, we
obtain the sum:
5
X

4 + k2 = (4 + 12 ) + (4 + 22 ) + (4 + 32 ) + (4 + 42 ) + (4 + 52 )
k=1
318 SESSION 23. SEQUENCES AND SERIES

= (4 + 1) + (4 + 4) + (4 + 9) + (4 + 16) + (4 + 25)
= 5 + 8 + 13 + 20 + 29
= 75

This answer can of course also be confirmed with the calculator (after replac-
ing the index k with n) as we did in part (a).

23.2 The arithmetic sequence

We have already encountered examples of arithmetic sequences in the pre-
vious section. An arithmetic sequence is a sequence for which we add a
constant number to get from one term to the next, for example:

8, 11, 14, 17, 20, 23, . . .

+3 +3 +3 +3 +3

Definition 23.10. A sequence {an } is called an arithmetic sequence if any

two consecutive terms have a common difference d. The arithmetic sequence
is determined by d and the first value a1 . This can be written recursively as:

an = an−1 + d for n ≥ 2

Alternatively, we have the general formula for the nth term of the arithmetic
sequence:
an = a1 + d · (n − 1) (23.2)

Example 23.11. Determine if the sequence is an arithmetic sequence. If so,

then find the general formula for an in the form of equation (23.2).

a) 7, 13, 19, 25, 31, . . . b) 13, 9, 5, 1, −3, −7, . . .

c) 10, 13, 16, 20, 23, . . . d) an = 8 · n + 3

Solution. a) Calculating the difference between two consecutive terms always

gives the same answer 13 − 7 = 6, 19 − 13 = 6, 25 − 19 = 6, etc. Therefore
the common difference d = 6, which shows that this is an arithmetic sequence.
Furthermore, the first term is a1 = 7, so that the general formula for the nth
term is an = 7 + 6 · (n − 1).
b) One checks that the common difference is 9 − 13 = −4, 5 − 9 = −4,
etc., so that this is an arithmetic sequence with d = −4. Since a1 = 13, the
general term is an = 13 − 4 · (n − 1).
23.2. THE ARITHMETIC SEQUENCE 319

c) We have 13 − 10 = 3, but 20 − 16 = 4, so that this is not an arithmetic

sequence.
d) If we write out the first couple of terms of an = 8n + 3, we get the
sequence:
11, 19, 27, 35, 43, 51, . . .
From this it seems reasonable to suspect that this is an arithmetic sequence
with common difference d = 8 and first term a1 = 11. The general term of
such an arithmetic sequence is

a1 + d(n − 1) = 11 + 8(n − 1) = 11 + 8n − 8 = 8n + 3 = an .

This shows that an = 8n + 3 = 11 + 8(n − 1) is an arithmetic sequence.

Example 23.12. Find the general formula of an arithmetic sequence with the
given property.
a) d = 12, and a6 = 68
b) a1 = −5, and a9 = 27
c) a5 = 38, and a16 = 115

Solution. a) According to equation (23.2) the general term is an = a1 + d(n −

1). We know that d = 12, so that we only need to find a1 . Plugging a6 = 68
into an = a1 + d(n − 1), we may solve for a1 :
(−60)
68 = a6 = a1 + 12 · (6 − 1) = a1 + 12 · 5 = a1 + 60 =⇒ a1 = 68 − 60 = 8

The nth term is therefore, an = 8 + 12 · (n − 1).

b) In this case, we are given a1 = −5, but we still need to find the common
difference d. Plugging a9 = 27 into an = a1 + d(n − 1), we obtain
(+5)
27 = a9 = −5 + d · (9 − 1) = −5 + 8d =⇒ 32 = 8d
(÷8)
=⇒ 4=d

The nth term is therefore, an = −5 + 4 · (n − 1).

c) In this case we neither have a1 nor d. However, the two conditions
a5 = 38 and a16 = 115 give two equations in the two unknowns a1 and d.
 
38 = a5 = a1 + d(5 − 1) 38 = a1 + 4 · d
=⇒
115 = a16 = a1 + d(16 − 1) 115 = a1 + 15 · d
320 SESSION 23. SEQUENCES AND SERIES

To solve this system of equations, we need to recall the methods for doing
so. One convenient method is the addition/subtraction method. For this, we
subtract the top from the bottom equation:

115 = a1 +15 · d
−( 38 = a1 +4 · d )
(÷11)
77 = +11 · d =⇒ 7=d

Plugging d = 7 into either of the two equations gives a1 . We plug it into the
first equation 38 = a1 + 4d:
(−28)
38 = a1 + 4 · 7 =⇒ 38 = a1 + 28 =⇒ 10 = a1

Therefore, the nth term is given by an = 10 + 7 · (n − 1).

We can sum the first k terms of an arithmetic sequence using a trick, which,
according to lore, was found by the German mathematician Carl Friedrich
Gauss when he was a child at school.
Example 23.13. Find the sum of the first 100 integers, starting from 1. In
other words, we want to find the sum of 1 + 2 + 3 + · · · + 99 + 100. First,
note that he sequence 1, 2, 3, . . . is an arithmetic sequence. The main idea
for solving this problem is a trick, which will indeed work for any arithmetic
sequence:
Let S = 1 + 2 + 3 + · · · + 98 + 99 + 100 be what we want to find. Note that
1 + 2 + 3 · · · + 98 + 99 + 100
2S = .
+ 100 + 99 + 98 · · · + 3 + 2 + 1

Note that the second line is also S but is added in the reverse order. Adding
vertically we see then that

2S = 101 + 101 + 101 + · · · + 101 + 101 + 101,

where there are 100 terms on the right hand side. So

100 · 101
2S = 100 · 101 and therefore S = = 5050.
2
The previous example generalizes to the more general setting starting with
an arbitrary arithmetic sequence.
23.2. THE ARITHMETIC SEQUENCE 321

Observation 23.14. Let {an } be an arithmetic sequence, whose nth term is

given by the formula an = a1 +d(n−1). Then, the sum a1 +a2 +· · ·+ak−1 +ak
is given by adding (a1 + ak ) precisely k2 times:

k
X k
ai = · (a1 + ak ) (23.3)
i=1
2

In order to remember the formula above, it may be convenient to think of

the right hand side as k · a1 +a
2
k
(this is, k times the average of the first and
the last term).

Proof. For the proof of equation (23.3), we write S = a1 + a2 + · · · + ak−1 + ak . We then add it to itself
but in reverse order:

a1 + a2 + a3 ··· + ak−2 + ak−1 + ak

2S = .
+ ak + ak−1 + ak−2 ··· + a3 + a2 + a1

Now note that in general al + am = 2a1 + d(l + m − 2). We see that adding vertically gives

2S = k(2a1 + d(k − 1)) = k(a1 + (a1 + d(k − 1)) = k(a1 + ak ).

Dividing by 2 gives the desired result.

Example 23.15. Find the value of the arithmetic series.

a) Find the sum a1 + · · ·+ a60 for the arithmetic sequence an = 2 + 13(n − 1).
1001
P
b) Determine the value of the sum: (5 − 6j)
j=1

c) Find the sum of the first 35 terms of the sequence

4, 3.5, 3, 2.5, 2, 1.5, . . .

P
Solution. a) The sum is given by the formula (23.3): ki=1 ai = k2 · (a1 + ak ).
Here, k = 60, and a1 = 2 and a60 = 2+13·(60−1) = 2+13·59 = 2+767 = 769.
We obtain a sum of
60
X 60
a1 + · · · + a60 = ai = · (2 + 769) = 30 · 771 = 23130.
i=1
2
322 SESSION 23. SEQUENCES AND SERIES

We may confirm this with the calculator as described in example 23.9 (on
page 316) in the previous section.

Enter:
sum(seq((2 + 13 · (n − 1), n, 1, 60))

P
b) Again, we use the above formula kj=1 aj = k2 · (a1 + ak ), where the
arithmetic sequence is given by aj = 5 − 6j and k = 1001. Using the values
a1 = 5 − 6 · 1 = 5 − 6 = −1 and a1001 = 5 − 6 · 1001 = 5 − 6006 = −6001, we
obtain:
1001
X 1001 1001
(5 − 6j) = (a1 + a1001 ) = ((−1) + (−6001))
j=1
2 2
1001
= · (−6002) = 1001 · (−3001) = −3004001
2
c) First note that the given sequence 4, 3.5, 3, 2.5, 2, 1.5, . . . is an arithmetic
sequence. It is determined by the first term a1 = 4 and common difference
d = −0.5. The nth term is given by an = 4 − 0.5 · (n − 1), and summing the
first k = 35 terms gives:
35
X 35
ai = · (a1 + a35 )
i=1
2

We see that we need to find a35 in the above formula:

a35 = 4 − 0.5 · (35 − 1) = 4 − 0.5 · 34 = 4 − 17 = −13

This gives a total sum of

35
X 35 35 −315
ai = · (4 + (−13)) = · (−9) = .
i=1
2 2 2

The
P35 answer−315
may be written as a fraction or also as a decimal, that is:
i=1 ai = 2
= −157.5.
23.3. EXERCISES 323

23.3 Exercises
Exercise 23.1. Find the first seven terms of the sequence.

a) an = 3n 5n + 3 c) an = n2 + 2
b) an = √ d) an = n
e) an = (−1)n+1 f) an = n+1
n
g) ak = 10k h) ai = 5 + (−1)i

Exercise 23.2. Find the first six terms of the sequence.

a) a1 = 5, an = an−1 + 3 for n ≥ 2
b) a1 = 7, an = 10 · an−1 for n ≥ 2
c) a1 = 1, an = 2 · an−1 + 1 for n ≥ 2
d) a1 = 6, a2 = 4, an = an−1 − an−2 for n ≥ 3

Exercise 23.3. Find the value of the series.

P P
a) 4n=1 an , where an = 5n, b) 5k=1 ak , where ak = k,
P P
c) 4i=1 ai , where an = n2 , d) 6n=1 (n − 4),
P P
e) 3k=1(k 2 + 4k − 4), f) 4j=1 j+1
1
.

Exercise 23.4. Is the sequence below part of an arithmetic sequence? In the

case that it is part of an arithmetic sequence, find the formula for the nth term
an in the form an = a1 + d · (n − 1).

a) 5, 8, 11, 14, 17, . . . b) −10, −7, −4, −1, 2, . . .

c) −1, 1, −1, 1, −1, 1, . . . d) 18, 164, 310, 474, . . .
e) 73.4, 51.7, 30, . . . f) 9, 3, −3, −8, −14, . . .
g) 4,
√ 4, √4, 4,√
4, . .√
. h) −2.72, −2.82, −2.92, −3.02, −3.12, . . .
−3 −1 2
i) 2, 5, 8, 11, . . . j) 5
, 10 , 5 , . . .
k) an = 4 + 5 · n l) aj = 2 · j − 5
m) an = n2 + 8n + 15 n) ak = 9 · (k + 5) + 7k − 1

Exercise 23.5. Determine the general nth term an of an arithmetic sequence

{an } with the data given below.

a) d = 4, and a8 = 57 b) d = −3, and a99 = −70

c) a1 = 14, and a7 = −16 d) a1 = −80, and a5 = 224
e) a3 = 10, and a14 = −23 f) a20 = 2, and a60 = 32
324 SESSION 23. SEQUENCES AND SERIES

Exercise 23.6. Determine the value of the indicated term of the given arith-
metic sequence.
a) if a1 = 8, and a15 = 92, find a19
b) if d = −2, and a3 = 31, find a81
c) if a1 = 0, and a17 = −102, find a73
d) if a7 = 128, and a37 = 38, find a26
Exercise 23.7. Determine the sum of the arithmetic sequence.
a) Find the sum a1 + · · · + a48 for the arithmetic sequence ai = 4i + 7.
P
b) Find the sum 21 i=1 ai for the arithmetic sequence an = 2 − 5n.
99
P
c) Find the sum: (10 · i + 1)
i=1

200
P
d) Find the sum: (−9 − n)
n=1

e) Find the sum of the first 100 terms of the arithmetic sequence:
2, 4, 6, 8, 10, 12, . . .

f) Find the sum of the first 83 terms of the arithmetic sequence:

25, 21, 17, 13, 9, 5, . . .

g) Find the sum of the first 75 terms of the arithmetic sequence:

2012, 2002, 1992, 1982, . . .

h) Find the sum of the first 16 terms of the arithmetic sequence:

−11, −6, −1, . . .

i) Find the sum of the first 99 terms of the arithmetic sequence:

−8, −8.2, −8.4, −8.6, −8.8, −9, −9.2, . . .

j) Find the sum

7 + 8 + 9 + 10 + · · · + 776 + 777
k) Find the sum of the first 40 terms of the arithmetic sequence:
5, 5, 5, 5, 5, . . .
Session 24

The geometric series

24.1 Finite geometric series

We now study another sequence, the geometric sequence, which will be anal-
ogous to our study of the arithmetic sequence in section 23.2. We have
already encountered examples of geometric sequences in Example 23.2(b). A
geometric sequence is a sequence for which we multiply a constant number
to get from one term to the next, for example:
5, 20, 80, 320, 1280, . . .
×4 ×4 ×4 ×4
Definition 24.1. A sequence {an } is called a geometric sequence, if any
two consecutive terms have a common ratio r. The geometric sequence is
determined by r and the first value a1 . This can be written recursively as:
an = an−1 · r for n ≥ 2
Alternatively, we have the general formula for the nth term of the geometric
sequence:
an = a1 · r n−1 (24.1)
Example 24.2. Determine if the sequence is a geometric, or arithmetic se-
quence, or neither or both. If it is a geometric or arithmetic sequence, then
find the general formula for an in the form (24.1) or (23.2).
a) 3, 6, 12, 24, 48, . . . b) 100, 50, 25, 12.5, . . .
c) 700, −70, 7, −0.7, 0.07, . . . d) 2, 4, 16, 256, . . .
e) 3, 10, 17,
24, . . . f) −3, −3, −3, −3, −3, . . .
n
g) an = 73 h) an = n2

325
326 SESSION 24. THE GEOMETRIC SERIES

Solution. a) Calculating the quotient of two consecutive terms always gives

the same number 6÷3 = 2, 12÷6 = 2, 24÷12 = 2, etc. Therefore the common
ratio is r = 2, which shows that this is a geometric sequence. Furthermore,
the first term is a1 = 3, so that the general formula for the nth term is
an = 3 · 2n−1 .
b) We see that the common ratio between two terms is r = 21 , so that this
is a geometric sequence. Since the first term is a1 = 100, we have the general
term an = 100 · ( 21 )n−1 .
1
c) Two consecutive terms have a ratio of r = − 10 , and the first term is
1 n−1
a1 = 700. The general term of this geometric sequence is an = 700 · (− 10 ) .
d) The quotient of the first two terms is 4 ÷ 2 = 2, whereas the quotient of
the next two terms is 16 ÷ 4 = 4. Since these quotients are not equal, this is
not a geometric sequence. Furthermore, the difference between the first two
terms is 4 − 2 = 2, and the next two terms have a difference 16 − 4 = 12.
Therefore, this is also not an arithmetic sequence.
e) The quotient of the first couple of terms is not equal 10
3
17
6= 10 , so that this
is not a geometric sequence. The difference of any two terms is 7 = 10 − 3 =
17 − 10 = 24 − 17, so that this is part of an arithmetic sequence with common
difference d = 7. The general formula is an = a1 + d · (n − 1) = 3 + 7 · (n − 1).
f) The common ratio is r = (−3) ÷ (−3) = 1, so that this is a geometric
sequence with an = (−3) · 1n−1 . On the other hand, the common difference
is (−3) − (−3) = 0, so that this is also an arithmetic sequence with an =
(−3) + 0 · (n − 1). Of course, both formulas reduce to the simpler expression
an = −3.
g) Writing the first couple of terms in the sequence {( 37 )n }, we obtain:
 1  2  3  4  5
3 3 3 3 3
, , , , ,...
7 7 7 7 7

Thus, we get from one term to the next by multiplying r = 37 , so that this is

n−1
a geometric sequence. The first term is a1 = 37 , so that an = 37 · 73 . This
is clearly the given sequence, since we may simplify this as
 n−1  1+n−1  n
3 3 3 3
an = · = =
7 7 7 7
h) We write the first terms in the sequence {n2 }n≥1 :

1, 4, 9, 16, 25, 36, 49, . . .

24.1. FINITE GEOMETRIC SERIES 327

Calculating the quotients of consecutive terms, we get 4 ÷ 1 = 4 and 9 ÷

4 = 2.25, so that this is not a geometric sequence. Also the difference of
consecutive terms is 4 − 1 = 3 and 9 − 4 = 5, so that this is also not an
arithmetic sequence.

Example 24.3. Find the general formula of a geometric sequence with the
given property.

a) r = 4, and a5 = 6400
b) a1 = 52 , and a4 = − 27
20
c) a5 = 216, a7 = 24, and r is positive

Solution. a) Since {an } is a geometric sequence, it is an = a1 · r n−1. We

know that r = 4, so we still need to find a1 . Using a5 = 64000, we obtain:

(÷256) 6400
6400 = a5 = a1 · 45−1 = a1 · 44 = 256 · a1 =⇒ a1 = = 25
256
The sequence is therefore given by the formula, an = 25 · 4n−1 .
b) The geometric sequence an = a1 · r n−1 has a1 = 52 . We calculate r
using the second condition.

27 2 (× 52 ) 27 5 27 1 −27
− = a4 = a1 · r 4−1 = · r 3 =⇒ r3 = − · = − · =
20 5 r 20 2 √ 4 2 8
(take √
3 )
3 −27
3
−27 −3
=⇒ r= = √ =
8 3
8 2

n−1
Therefore, an = 52 · −32
.
c) The question does neither provide a1 nor r in the formula an = a1 · r n−1 .
However, we obtain two equations in the two variables a1 and r:
 
216 = a5 = a1 · r 5−1 216 = a1 · r 4
=⇒
24 = a7 = a1 · r 7−1 24 = a1 · r 6

In order to solve this, we need to eliminate one of the variables. Looking at

the equations on the right, we see dividing the top equation by the bottom
equation cancels a1 .

216 a1 · r 4 9 1 (take reciprocal) 1 r2 1

= =⇒ = 2 =⇒ = =⇒ r 2 =
24 a1 · r 6 1 r 9 1 9
328 SESSION 24. THE GEOMETRIC SERIES

To obtain r we have to solve this quadratic equation. In general, there are in

fact two solutions: r
1 1
r=± =±
9 3
Since the problem states that r is positive, we see that we need to take the
positive solution r = 13 . Plugging r = 31 back into either of the two equations,
we may solve for a1 . For example, using the first equation a5 = 216, we
obtain:
 5−1  4
1 1 1 1
216 = a5 = a1 · = a1 · = a1 · 4 = a1 ·
3 3 3 81
(×81)
=⇒ a1 = 81 · 216 = 17496

So, we finally arrive at the general formula for the nth term of the geometric
sequence, an = 17496 · ( 31 )n−1 .

We can find the sum of the first k terms of a geometric sequence using
another trick, which is very different from the one we used for the arithmetic
sequence.

Example 24.4. Consider the geometric sequence an = 8 · 5n−1 , that is the

sequence:
8, 40, 200, 1000, 5000, 25000, 125000, . . .
We want to add the first 6 terms of this sequence.

8 + 40 + 200 + 1000 + 5000 + 25000 = 31248

In general, it may be much more difficult to simply add the terms as we did
above, and we need to use a better general method. For this, we multiply
(1 − 5) to the sum (8 + 40 + 200 + 1000 + 5000 + 25000) and simplify this
using the distributive law:

(1 − 5) · (8 + 40 + 200 + 1000 + 5000 + 25000)

= 8 − 40 + 40 − 200 + 200 − 1000 + 1000 − 5000
+5000 − 25000 + 25000 − 125000
= 8 − 125000
24.1. FINITE GEOMETRIC SERIES 329

In the second and third lines above, we have what is called a telescopic sum,
which can be canceled except for the very first and last terms. Dividing by
(1 − 5), we obtain:
8 − 125000 −124992
8 + 40 + 200 + 1000 + 5000 + 25000 = = = 31248
1−5 −4
The previous example generalizes to the more general setting starting with
an arbitrary geometric sequence.
Observation 24.5. Let {an } be a geometric sequence, whose nth term is given
by the formula an = a1 · r n−1. We furthermore assume that r 6= 1. Then, the
sum a1 + a2 + · · · + ak−1 + ak is given by:
k
X 1 − rk
ai = a1 · (24.2)
i=1
1−r

Proof. We multiply (1 − r) to the sum (a1 + a2 + · · · + ak−1 + ak ):

(1 − r) · (a1 + a2 + · · · + ak )
= (1 − r) · (a1 · r 0 + a1 · r 1 + · · · + a1 · r k−1 )
= a1 · r 0 − a1 · r 1 + a1 · r 1 − a1 · r 2 + · · · + a1 · r k−1 − a1 · r k
= a1 · r 0 − a1 · r k = a1 · (1 − r k )

Dividing by (1 − r), we obtain

a1 · (1 − r k ) 1 − rk
a1 + a2 + · · · + ak = = a1 ·
(1 − r) 1−r
This is the formula we wanted to prove.
Example 24.6. Find the value of the geometric series.
6
P
a) Find the sum an for the geometric sequence an = 10 · 3n−1 .
n=1

5
P
k
b) Determine the value of the geometric series: − 12
k=1

c) Find the sum of the first 12 terms of the geometric sequence

−3, −6, −12, −24, . . .
330 SESSION 24. THE GEOMETRIC SERIES

Solution. a) We need to find the sum a1 + a2 + a3 + a4 + a5 + a6 , and we will

do so using the formula provided in equation (24.2). Since an = 10 · 3n−1 , we
have a1 = 10 and r = 3, so that
6
X 1 − 36 1 − 729 −728
an = 10 · = 10 · = 10 · = 10 · 364 = 3640
n=1
1−3 1−3 −2

P
b) Again, we use the formula for the geometric series nk=1 ak = a1 · 1−r
n
1−r
,
1 k
since ak = (− 2 ) is a geometric series. We may calculate the first term
a1 = − 21 , and the common ratio is also r = − 21 . With this, we obtain:

5  k     5
X 1 1 1 − (− 12 )5 1 1 − ((−1)5 215 )
− = − · = − · =
k=1
2 2 1 − (− 12 ) 2 1 − (− 12 )
  1   1   32+1
1 1 − (− 32 ) 1 1 + 32 1 32
= − · 1 = − · 1 = − · 2+1
2 1 − (− 2 ) 2 1+ 2 2 2
  33  
1 32 1 33 2 1 11 11
= − · 3 = − · · =− · =−
2 2
2 32 3 2 16 32

c) Our first task is to find the formula for the provided geometric series
−3, −6, −12, −24, . . . . The first term is a1 = −3 and the common ratio is
r = 2, so that an = (−3) · 2n−1. The sum of the first 12 terms of this sequence
is again given by equation (24.2):

12
X 1 − 212 1 − 4096 −4095
(−3) · 2i−1 = (−3) · = (−3) · = (−3) ·
i=1
1−2 1−2 −1
= (−3) · 4095 = −12285

24.2 Infinite geometric series

In some cases, it makes sense to add not only finitely many terms of a geo-
metric sequence, but all infinitely many terms of the sequence! An informal
and very intuitive infinite geometric series is exhibited in the next example.
24.2. INFINITE GEOMETRIC SERIES 331

Example 24.7. Consider the geometric sequence

1 1 1 1
1, , , , , . . .
2 4 8 16

Here, the common ratio is r = 21 , and the first term is a1 = 1, so that the

n−1
formula for an is an = 12 . We are interested in summing all infinitely
many terms of this sequence:

1 1 1 1
1+ + + + + ...
2 4 8 16
We add these terms one by one, and picture these sums on the number line:

0 1 1.5 1.75 2

1 = 1
1
1+ = 1.5
2
1 1
1+ + = 1.75
2 4
1 1 1
1+ + + = 1.875
2 4 8
1 1 1 1
1+ + + + = 1.9375
2 4 8 16
We see that adding each term takes the sum closer and closer to the number
2. More precisely, adding a term an to the partial sum a1 + · · · + an−1 cuts
the distance between 2 and a1 + · · · + an−1 in half. For this reason we can,
in fact, get arbitrarily close to 2, so that it is reasonable to expect that

1 1 1 1
1+ + + + + ··· = 2
2 4 8 16
In the next definition and observation, this equation will be justified and made
more precise.

We start by providing the definition of an infinite series.

332 SESSION 24. THE GEOMETRIC SERIES

Definition 24.8. An infinite series is given by the

∞
X
ai = a1 + a2 + a3 + . . . (24.3)
i=1

∞ k
 
P P
To be more precise, the infinite sum is defined as the limit ai := lim ai . Therefore, an infinite
i=1 k→∞ i=1
sum is defined, precisely when this limit exists.

Observation 24.9. Let {an } be a geometric sequence with an = a1 · r n−1 .

Then the infinite geometric series is defined whenever −1 < r < 1. In this
case, we have:
∞
X 1
ai = a1 · (24.4)
i=1
1 − r

P
Proof. Informally, this follows from the formula ki=1 ai = a1 · 1−r k
1−r
and the
fact that r k approaches zero when k increases without bound.
More formally, the proof uses the notion of limits, and goes as follows:

∞ k 1 − lim (r k )
1 − rk
X   
X k→∞ 1
ai = lim ai = lim a1 · = a1 · = a1 ·
i=1
k→∞
i=1
k→∞ 1−r 1−r 1−r

Example 24.10. Find the value of the infinite geometric series.

P∞

1 j−1
a) j=1 aj , for aj = 5 · 3
P∞ n
b) n=1 3 · (0.71)

c) 500 − 100 + 20 − 4 + . . .
d) 3 + 6 + 12 + 24 + 48 + . . .

Solution. a) We use formula (24.4) for the geometric series an = 5 · ( 13 )n−1 ,

that is a1 = 5 · ( 31 )1−1 = 5 · ( 13 )0 = 5 · 1 = 5 and r = 31 . Therefore,
∞
X 1 1 1 1 3 15
aj = a1 · =5· 1 =5· 3−1 =5· 2 =5· =
j=1
1−r 1− 3 3 3
2 2
24.2. INFINITE GEOMETRIC SERIES 333

b) In this case, an = 3 · (0.71)n , so that a1 = 3 · 0.711 = 3 · 0.71 = 2.13

and r = 0.71. Using again formula (24.4), we can find the infinite geometric
series as
∞
X 1 1 1 2.13 213
3 · (0.71)n = a1 · = 2.13 · = 2.13 · = =
n=1
1−r 1 − 0.71 0.29 0.29 29

In the last step we simplified the fraction by multiplying both numerator and
denominator by 100, which had the effect of eliminating the decimals.
c) Our first task is to identify the given sequence as an infinite geometric
sequence:
{an } is given by 500, −100, 20, −4, . . .
Notice that the first term is 500, and each consecutive term is given by di-
viding by −5, or in other words, by multiplying by the common ratio r = − 15 .
Therefore, this is an infinite geometric series, which can be evaluated as
∞
X 1 1
500 − 100 + 20 − 4 + . . . = an = a1 · = 500 ·
n=1
1−r 1 − (− 15 )
1 500 500 5
= 500 · 1 = 1+5 = 6 = 500 ·
1+ 5 5 5
6
2500 1250
= =
6 3
d) We want to evaluate the infinite series 3 + 6 + 12 + 24 + 48 + . . . . The
sequence 3, 6, 12, 24, 48, . . . is a geometric sequence, with a1 = 3 and common
ratio r = 2. Since r ≥ 1, we see that formula (24.4) cannot be applied, as
(24.4) only applies to −1 < r < 1. However, since we add larger and larger
terms, the series gets larger than any possible bound, so that the whole sum
becomes infinite.
3 + 6 + 12 + 24 + 48 + · · · = ∞

Example 24.11. The fraction 0.55555 . . . may be written as:

0.55555 · · · = 0.5 + 0.05 + 0.005 + 0.0005 + 0.00005 + . . .

Noting that the sequence

0.5, 0.05, 0.005, 0.0005, 0.00005, . . .

×0.1 ×0.1 ×0.1 ×0.1
334 SESSION 24. THE GEOMETRIC SERIES

is a geometric sequence with a1 = 0.5 and r = 0.1, we can calculate the

infinite sum as:
∞
X 1 1 0.5 5
0.55555 · · · = 0.5 · (0.1)i−1 = 0.5 · = 0.5 · = = ,
i=1
1 − 0.1 0.9 0.9 9

Here we multiplied numerator and denominator by 10 in the last step in order

to eliminate the decimals.

24.3 Exercises
Exercise 24.1. Which of these sequences is geometric, arithmetic, or neither
or both. Write the sequence in the usual form an = a1 + d(n − 1) if it is an
arithmetic sequence and an = a1 · r n−1 if it is a geometric sequence.

a) 7, 14, 28, 56, . . . b) 3, −30, 300, −3000, 30000, . . .

c) 81, 27, 9, 3, 1, 31 , . . . d) −7, −5, −3, −1, 1, 3, 5, 7, . . .

2
3
e) −6, 2, − 23 , 29 , − 27
2
,... f) −2, −2 · 23 , −2 · 32 , −2 · 23 , . . .
1 1 1 1
g) , , , ,...
2 4 8 16
h) 2, 2, 2, 2, 2, . . .
i) 5, 1, 5, 1, 5, 1, 5, 1, . . . j) −2, 2, −2, 2, −2, 2, . . .
k) 0, 5, 10, 15, 20, . . . l) 5, 53 , 352 , 353 , 354 , . . .
1 1 1 1
m) , , , ,...
2 4 8 16
n) log(2), log(4), log(8), log(16), . . .
o) an = −4n p) an = −4n

n
q) an = 2 · (−9)n r) an = 31

n
n
s) an = − 75 t) an = − 57
u) an = n2 v) an = 3n + 1

Exercise 24.2. A geometric sequence, an = a1 · r n−1, has the given properties.

Find the term an of the sequence.

a) a1 = 3, and r = 5, find a4
b) a1 = 200, and r = − 21 , find a6
c) a1 = −7, and r = 2, find an (for all n)
d) r = 2, and a4 = 48, find a1
e) r = 100, and a4 = 900, 000, find an (for all n)
24.3. EXERCISES 335

f) a1 = 20, a4 = 2500, find an (for all n)

5
g) a1 = 18 , and a6 = 836 , find an (for all n)
h) a3 = 36, and a6 = 972, find an (for all n)
i) a8 = 4000, a10 = 40, and r is negative, find an (for all n)
Exercise 24.3. Find the value of the finite geometric series using formula
(24.2). Confirm the formula either by adding the the summands directly, or
alternatively by using the calculator.
4
P
a) Find the sum aj for the geometric sequence aj = 5 · 4j−1.
j=1

7

1 n
P
b) Find the sum ai for the geometric sequence an = 2
.
i=1

5
P
m
c) Find: − 15
m=1

6
P
d) Find: 2.7 · 10k
k=1

e) Find the sum of the first 5 terms of the geometric sequence:

2, 6, 18, 54, . . .

f) Find the sum of the first 6 terms of the geometric sequence:

−5, 15, −45, 135, . . .

g) Find the sum of the first 8 terms of the geometric sequence:

−1, −7, −49, −343, . . .

h) Find the sum of the first 10 terms of the geometric sequence:

600, −300, 150, −75, 37.5, . . .

i) Find the sum of the first 40 terms of the geometric sequence:

5, 5, 5, 5, 5, . . .
336 SESSION 24. THE GEOMETRIC SERIES

Exercise 24.4. Find the value of the infinite geometric series.

P∞

2 j−1
P∞
j
a) j=1 aj , for aj =3· 3
b) j=1 7 · − 15
P∞ P∞
c) 1
j=1 6 · 3j d) n=1 −2 · (0.8)n
P∞ n
e) n=1 (0.99) f) 27 + 9 + 3 + 1 + 31 + . . .
g) −2 + 1 − 21 + 41 − . . . h) −6 − 2 − 32 − 29 − . . .
i) 100 + 40 + 16 + 6.4 + . . . j) −54 + 18 − 6 + 2 − . . .

Exercise 24.5. Rewrite the decimal using an infinite geometric sequence, and
then use the formula for infinite geometric series to rewrite the decimal as a
fraction (see example 24.11).

a) 0.44444 . . . b) 0.77777 . . . c) 5.55555 . . . d) 0.2323232323 . . .

e) 39.393939 . . . f) 0.248248248 . . . g) 20.02002 . . . h) 0.5040504 . . .
Session 25

The binomial theorem

25.1 The binomial theorem

Recall the well-known binomial formula:

(a + b)2 = a2 + 2ab + b2

(since, using “FOIL,” we have: (a + b)2 = (a + b) · (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 )

In this section we generalize this to find similar expressions for (a + b)n

for any natural number n. This is the content of the (generalized) binomial
theorem below. Before we can state the theorem, we need to define the notion
of a factorial and combinations.
Definition 25.1. For a natural number n = 1, 2, 3, . . . , we define n! to be the
number
n! = 1 · 2 · 3 · · · · · n
The number n! is called n factorial.
To make the formulas below work nicely, we also define 0! to be 0! = 1.
It is easy to calculate some examples of factorials.
Example 25.2.

4! = 1 · 2 · 3 · 4 = 24
7! = 1 · 2 · 3 · 4 · 5 · 6 · 7 = 5040

337
338 SESSION 25. THE BINOMIAL THEOREM

2! = 1 · 2 = 2

To calculate
✞ factorials
☎ with the calculator, we have to ✞
use the
☎ MATH menu
(press lpmathlp ), then move to the PRB menu (press lp⊳ lp ), and use the
✝ ✆✞ ☎ ✝ ✆
fourth item (press lp4 lp ).
✝ ✆

✞ ☎ ✞ ☎ use fourth
✞ item☎
press lpmathlp press lp⊳ lp (press lp4 lp )
✝ ✆ ✝ ✆ ✝ ✆

For example, we can calculate 12! = 479001600 by entering 12 and the fac-
torial symbol as described above.

Note that the factorial becomes very large even for relatively small integers.
For example 17! ≈ 3.557 · 1014 as shown above.

The next concept that we introduce is that of the binomial coefficient.

Definition 25.3. Let n = 0, 1, 2, . . . and r = 0, 1, 2, . . . , n be natural numbers

or zero, so that 0 ≤ r ≤ n. Then we define the binomial coefficient as
 
n n!
=
r r! · (n − r)!


The binomial coefficient is also written as n Cr = nr , and we read them as
“n-choose-r.”

n
n


Note 25.4. • The binomial coefficient r
should not be confused with the fraction r
.
25.1. THE BINOMIAL THEOREM 339

• A subset of the set {1, 2, . . . , n} with r elements is called an r-combination. The binomial coeffi-
cient can be interpreted as counting the number of distinct r-combinations. More precisely, there
are exactly n


r
distinct r-combinations of the set {1, . . . , n}.

Example 25.5. Calculate the binomial coefficients.

a) 64 b) 85 c) 25
23
d) 7
1
e) 11
11
f) 11
0

Solution. a) Many binomial coefficients may be calculated by hand, such as:

 
6 6! 6! 1·2·3·4·5·6 5·6
= = = = = 15
4 4!(6 − 4)! 4!2! 1·2·3·4·1·2 2
b) Again, we can calculate this by hand
 
8 8! 1·2·3·4·5·6·7·8 6·7·8
= = = = 7 · 8 = 56
5 5!3! 1·2·3·4·5·1·2·3 1·2·3
However, we✞can also☎use the calculator to find
✞ the ☎
answer. Using the MATH
and (press lpmathlp ) PRB menus (press lp⊳ lp ) as above, we use the
✝ ✞ ✆ ☎ ✝ ✆
third item (press lp3 lp ). The answer is obtained by pressing the following
✝ ✆
sequence of keys:
✞ ☎ ✞ ✞
☎ ✞
☎ ☎ ✞ ☎ ✞ ☎
lp8 lp lpmathlp lp⊳ lp lp3 lp lp5 lp lpenter lp
✝ ✆ ✝ ✆
✝ ✆
✝ ✆ ✝ ✆ ✝ ✆

We also calculate the remaining binomial coefficients (c)-(f), which can

also be confirmed with the calculator.
 
25 25! 23! · 24 · 25 24 · 25
c) = = = = 300
23 23! · 2! 23! · 1 · 2 2
 
7 7! 6! · 7 7
d) = = = =7
1 1! · 6! 1 · 6! 1
 
11 11! 1
e) = = =1
11 11! · 0! 1·1
 
11 11! 1
f) = = =1
0 0! · 11! 1·1
Note that in the last two equations we needed to use the fact that 0! = 1.
340 SESSION 25. THE BINOMIAL THEOREM

We state some useful facts about the binomial coefficient, that can already
be seen in the previous example.

Observation 25.6. For all n = 0, 1, 2, . . . and r = 0, 1, 2, . . . , n, we have:

           
n n n n n n
= = =1 = =n
n−r r 0 n 1 n−1

Proof. We have:
   
n n! n! n
= = =
n−r (n − r)! · (n − (n − r))! (n − r)! · r! r
   
n n n! 1
= = = =1
0 n 0! · n! 1
   
n n n! n
= = = =n
1 n−1 1! · (n − 1)! 1

Note 25.7.
n

• We can obtain all binomial coefficients r
for fixed n from the calculator by using the function and table
menus.

• The binomial coefficients are found in what is known as Pascal’s triangle. For this, calculate the lowest
binomial coefficients and write them in a triangular arrangement:

0

0 1
1
1

0 1 1 1
2
2
2

0 1 2 1 2 1
3
3
3
3
1 3 3 1
4

0
4

1
4

2
4

3
4

=
1 4 6 4 1
0 1 2 3 4
5
5
5
5
5
5
1 5 10 10 5 1
0 1 2 3 4 5
.. ..
. .

The triangle on the right is known as Pascal’s triangle. Each entry in the triangle is obtained by adding
the two entries right above it.
25.1. THE BINOMIAL THEOREM 341

The binomial coefficients appear in the expressions for (a + b)n as we will

see in the next example.
Example 25.8. We now calculate some simple examples.
(a + b)3 = (a + b) · (a + b) · (a + b)
= (a2 + 2ab + b2 ) · (a + b)
= a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3
= a3 + 3a2 b + 3ab2 + b3
Note that the coefficients 1, 3, 3, and 1 in front of

a3
, a2 b,

ab2 , and

b3 , respec-
tively, are precisely the binomial coefficients 30 , 31 , 32 , and 33 .
We also calculate the fourth power.
(a + b)4 = (a + b) · (a + b) · (a + b) · (a + b)
= (a3 + 3a2 b + 3ab2 + b3 ) · (a + b)
= a4 + 3a3 b + 3a2 b2 + ab3 + a3 b + 3a2 b2 + 3ab3 + b4
= a4 + 4a3 b + 6a2 b2 + 4ab3 + b4
Again,

4
the
numbers

1,
4, 6, 4, and 1 are precisely the binomial coefficients
4
0
, 1 , 2 , 3 , and 44 .
4 4

We are now ready to state the general binomial theorem:

Theorem 25.9 (Binomial theorem). The nth power (a + b)n can be expanded
as:
         
n n n n n−1 1 n n−2 2 n 1 n−1 n n
(a + b) = a + a b + a b + · · ·+ ab + b
0 1 2 n−1 n
Using the summation symbol, we may write this in short:
n  
n
X n
(a + b) = · an−r · br (25.1)
r=0
r

Example 25.10. Expand (a + b)5 .

Solution.
           
5 5 5 5 4 1 5 3 2 5 2 3 5 1 4 5 5
(a + b) = a + ab + ab + ab + ab + b
0 1 2 3 4 5
= a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
342 SESSION 25. THE BINOMIAL THEOREM

25.2 Binomial expansion

Using the binomial theorem, we can also expand more general powers of sums
or differences.

Example 25.11. Expand the expression.

4 3
√
a) (x2 + 2y 3 )5 b) (2xy 2 − y2
) c) ( 2 + 1)6 d) (i − 3)4

Solution. a) We use the binomial theorem with a = x2 and b = 2y 3:

   
2 3 5 5 2 4
2 5 3 5
(x + 2y ) = (x ) + (x ) (2y ) + (x2 )3 (2y 3)2
1 2
   
5 2 2 3 3 5
+ (x ) (2y ) + (x2 )(2y 3 )4 + (2y 3)5
3 4
= x10 + 5x8 · 2y 3 + 10x6 · 4y 6 + 10x4 · 23 y 9 + 5x2 · 24 y 12 + 25 y 15
= x10 + 10x8 y 3 + 40x6 y 6 + 80x4 y 9 + 80x2 y 12 + 32y 15

b) For part (b), it is a = 2xy 2 and b = − y42 .

4 3
(2xy 2 − )
y2
   
3
2 3 2 2 4 3 4 4
= (2xy ) + (2xy ) (− 2 ) + (2xy 2 )(− 2 )2 + (− 2 )3
1 y 2 y y
2
4 4 43
= 23 x3 y 6 + 3 · 22 x2 y 4 (− 2 ) + 3(2xy 2)(−1)2 4 + (−1)3 6
y y y
1 1
= 8x3 y 6 − 48x2 y 2 + 96x · 2 − 64 · 6
y y
√
c) Similarly, for part (c), we now have a = 2 and b = 1:
     
√ 6
√ 6 6 √ 5 6 √ 4 2 6 √ 3 3
( 2 + 1) = ( 2) + ( 2) · 1 + ( 2) · 1 + ( 2) · 1
1 2 3
   
6 √ 2 4 6 √
+ ( 2) · 1 + ( 2) · 15 + 16
4 5
√ √ √ √ √ √
= 64 + 6 · 32 + 15 · 16 + 20 · 8 + 15 · 4 + 6 · 2 + 1
√ √ √
= 8 + 6 · 16 · 2 + 15 · 4 + 20 · 4 · 2 + 15 · 2 + 6 · 2 + 1
25.2. BINOMIAL EXPANSION 343
√ √ √
= 8 + 24 2 + 60 + 40 2 + 30 + 6 2 + 1
√
= 99 + 70 2

Note that the last expression cannot be simplified any further (due to the
order of operations).
d) Finally, we have a = i and b = −3, and we use the fact that i2 = −1,
and therefore, i3 = −i and i4 = +1:
     
4 4 4 3 4 2 2 4
(i − 3) = i + · i · (−3) + · i · (−3) + · i · (−3)3 + (−3)4
1 2 3
= 1 + 4 · (−i) · (−3) + 6 · (−1) · 9 + 4 · i · (−27) + 81
= 1 + 12i − 54 − 108i + 81
= 28 − 96i

In some instances it is not necessary to write the full binomial expansion,

but it is enough to find a particular term, say the kth term of the expansion.

Observation 25.12. Recall, for example, the binomial expansion of (a + b)6 :

6 6 0


6 5 1


6 4 2


6 3 3


6 2 4


6 1 5


6 0 6
0
a b + 1
a b + 2
a b + 3
a b + 4
a b + 5
a b + 6
ab

first second third fourth fifth sixth seventh

term term term term term term term

Note that the exponents of the a’s and b’s for each term always add up to
6, and that the exponents of a decrease from 6 to 0, and the exponents of b
increase from
0 to 6. Furthermore observe that in the above expansion the
fifth term is 64 a2 b4 .
In general, we define the kth term by the following formula:
 
n n
The kth term in the expansion of (a + b) is: an−k+1bk−1 (25.2)
k−1

Note in particular, that the kth term
has a power of b given by bk−1 (and not
n
bk ), it has a binomial coefficient k−1 , and the exponents of a and b add up
to n.
344 SESSION 25. THE BINOMIAL THEOREM

Example 25.13. Determine:

a) the 4th term in the binomial expansion of (p + 3q)5
b) the 8th term in the binomial expansion of (x3 y − 2x2 )10
c) the 12th term in the binomial expansion of (− 5a
b7
− b)15
Solution. a) We have a = p and b = 3q,
and n = 5 and k3 = 4. Thus, the
5
binomial coefficient of the 4th term is 3 , the b-term is (3q) , and the a-term
is p2 . The 4th term is therefore given by
 
5
· p2 · (3q)3 = 10 · p2 · 33 q 3 = 270p2 q 3
3
b) In this case, a = x3 y and b = −2x2 , and
furthermore, n = 10 and k = 8.
The binomial coefficient of the 8th term is 10 7
, the b-term is (−2x2 )7 , and the
a-term is (x3 y)3 . Therefore, the 8th term is
 
10
· (x3 y)3 · (−2x2 )7 = 120 · x9 y 3 · (−128)x14 = −15360 · x23 y 3
7
c) Similarly, we obtain the 12th term of (− 5a
b7
− b)15 as
  
15 5a 4 11 54 a4 11 625 · a4 · (−b11 )
· − 7 · (−b) = 1365 · 28 · (−b ) = 1365 ·
11 b b b28
a4
= −853125 · 17
b

Here is a variation of the above problem.

Example 25.14. Determine:
a) the x4 y 12 -term in the binomial expansion of (5x2 + 2y 3)6
b) the x15 -term in the binomial expansion of (x3 − x)7
c) the real part of the complex number (3 + 2i)4

Solution. a) In this case we have a = 5x2 and b = 2y 3. The term x4 y 12 can

n−k+1
4 12 2 2 3 4 n
be rewritten as x y = (x ) · (y ) , so that the full term k−1 a bk−1
(including the coefficients) is
 
6
· (5x2 )2 · (2y 3)4 = 15 · 25x4 · 16y 12 = 6000 · x4 y 12
4
25.3. EXERCISES 345

b) The various powers of x in (x3 − x)7 (in the order in which they appear
in the binomial expansion) are:
(x3 )7 = x21 , (x3 )6 · x1 = x19 , (x3 )5 · x2 = x17 , (x3 )4 · x3 = x15 , ...
The last term is precisely the x15 -term, that is we take the fourth term, k = 4.
We obtain a total term (including the coefficients) of
 
7
· (x3 )4 · (−x)3 = 35 · x12 · (−x)3 = −35 · x15
3
c) Recall that in is real when n is even, and imaginary when n is odd:
i1 =i
i2 = −1
i3 = −i
i4 =1
i5 =i
i6 = −1
..
.
The real part of (3 + 2i)4 is therefore given by the first, third, and fifth term
of the binomial expansion:
     
4 4 0 4 2 2 4
real part = · 3 · (2i) + · 3 · (2i) + · 30 · (2i)4
0 2 4
= 1 · 81 · 1 + 6 · 9 · 4i2 + 1 · 1 · 16i4 = 81 + 216 · (−1) + 16 · 1
= 81 − 216 + 16 = −119
The real part of (3 + 2i)4 is −119.

25.3 Exercises
Exercise 25.1. Find the value of the factorial or binomial coefficient.
a) 5!
b) 3!
c) 9!
d) 2!
e) 0!
f) 1!
g) 19!
h) 64!

i) 52 j) 96 k) 12 1
l) 12
0
m) 2322
n) 19
12
o) 13
11
p) 16
5

Exercise 25.2. Expand the expression via the binomial theorem.

a) (m + n)4 b) (x + 2)5 c) (x − y)6 d) (−p − q)5
346 SESSION 25. THE BINOMIAL THEOREM

Exercise 25.3. Expand the expression.

a) (x − 2y)3 b) (x − 10)4 c) (x2 y + y 2 )5 d) (2y 2 − 5x4 )4

√ 2 3 √ √
e) (x + x )3 f) (−2 xy − yx )5 g) ( 2 − 2 3)3 h) (1 − i)3

Exercise 25.4. Determine:

a) the first 3 terms in the binomial expansion of (xy − 4x)5
b) the first 2 terms in the binomial expansion of (2a2 + b3 )9
c) the last 3 terms in the binomial expansion of (3y 2 − x2 )7
d) the first 3 terms in the binomial expansion of ( xy − xy )10
e) the last 4 terms in the binomial expansion of (m3 n + 21 n2 )6

Exercise 25.5. Determine:

a) the 5th term in the binomial expansion of (x + y)7
b) the 3rd term in the binomial expansion of (x2 − y)9
c) the 10th term in the binomial expansion of (2 − w)11
d) the 5th term in the binomial expansion of (2x + xy)7
e) the 7th term in the binomial expansion of (2a + 5b)6
f) the 6th term in the binomial expansion of (3p2 − q 3 p)7
g) the 10th term in the binomial expansion of (4 + 2b )13

Exercise 25.6. Determine:

a) the x3 y 6 -term in the binomial expansion of (x + y)9
b) the r 4 s4 -term in the binomial expansion of (r 2 − s)6
c) the x4 -term in the binomial expansion of (x − 1)11
d) the x3 y 6 -term in the binomial expansion of (x3 + 5y 2 )4
e) the x7 -term in the binomial expansion of (2x − x2 )5
f) the imaginary part of the number (1 + i)3
Review of complex numbers,
sequences, and the binomial
theorem

Exercise V.1. Multiply and write the answer in standard form:

(−4)(cos(207◦) + i sin(207◦ )) · 2(cos(33◦ ) + i sin(33◦ ))

Exercise V.2. Divide and write the answer in standard form:

3(cos( π3 ) + i sin( π3 ))
15(cos( π2 ) + i sin( π2 ))
Exercise V.3. Find the magnitude and directional angle of the vector
√
~v = h−7, −7 3i.

Exercise V.4. Determine if the sequence is an arithmetic or geometric se-

quence or neither. If it is one of these, then find the general formula for the
nth term an of the sequence.

a) 54, −18, 6, −2, 32 , . . .

b) 2, 4, 8, 10, . . .
c) 9, 5, 1, −3, −7, . . .

Exercise V.5. Find the sum of the first 75 terms of the arithmetic sequence:

−30, −22, −14, −6, 2, . . .

Exercise V.6. Find the sum of the first 8 terms of the geometric series:

−7, −14, −28, −56, −112, . . .

347
Exercise V.7. Find the value of the infinite geometric series:

80 − 20 + 5 − 1.25 + . . .

Exercise V.8. Expand the expression via the binomial theorem.

(3x2 − 2xy)3

Exercise V.9. Write the first 3 terms of the binomial expansion:

 10 9
ab2 +
a
Exercise V.10. Find the 6th term of the binomial expansion:

(5p − q 2 )8

348
Appendix A

Introduction to the TI-84

A.1 Basic calculator functions

In this section we explain the basic functions of the calculator. There are
various buttons and screens. We will describe below those which are most
useful for us.

The main screen The main screen is the screen that appears when you first
turn on the calculator. It is the screen where you can perform operations such
as
✞ adding☎✞numbers. ☎You can always return to the main screen by pressing
lp2nd lp lpmodelp . Other screens will be discussed below according to
✝ ✆✝ ✆
their use.

The buttons There are three levels of buttons. The physical buttons together
with the green and blue print above each button,
✞ for example
☎✞ ☎. To select
the items in green (blue) print first select lpalpha lp ( lp2nd lp ) then the
✝ ✆✝ ✆
button
✞ beneath
☎ the
✞ printed
☎ item. For example to select the letter K press
lpalpha lp then lp(lp.
✝ ✆ ✝ ✆
From now on we will use ’button’ to mean any of the the physical buttons or
the green or blue pseudo-buttons.

The clear✞button ☎ Whenever you want to erase a line of input or a screen

press the lpclearlp button (once or twice).
✝ ✆

349
350 APPENDIX A. INTRODUCTION TO THE TI-84
✞ ☎
The delete and insert buttons The ’DEL’ button ( lpdellp ) and the ’INS’
✞ ☎✞ ☎ ✝ ✆
button ( lp2nd lp lpdellp ) are handy when editing an expression. The ’DEL’
✝ ✆✝ ✆
button just deletes the character that is at the location of the cursor. The
insert button is inserts the entered numbers or letters at the location of the
cursor. For example, if you type 5+5 and you realize you wanted 15+5
instead,
✞ you
☎✞ should☎✞use the
☎ left arrow to move the cursor to the first✞5, press
☎
lp2nd lp lpdellp lp1 lp . What happens if you just had pressed lp1 lp ?
✝ ✆✝ ✆✝ ✆ ✝ ✆
Note that the calculator stays in ’insert mode’ until the cursor is moved with
the error or enter is pressed. So you could have ✞ changed ☎ the
✞ first☎✞5 to a☎
115 by moving the cursor to the first 5 and pressing lp2nd lp lpdellp lp1 lp
✞ ☎ ✝ ✆✝ ✆✝ ✆
lp1 lp .
✝ ✆

The ANS button The answer button is used when you want to use ✞ the☎
answer above in an expression. For example suppose you had entered lp3 lp
✞ ☎✞ ☎✞ ☎ ✝ ✆
lp+ lp lp5 lp lpenter lp . Now you want to compute 7 minus the result of
✝ ✆✝ ✆✝ ✆ ✞ ☎✞ ☎✞ ☎✞ ☎
that computation. You can press lp7 lp lp− lp lp2nd lp lp(−) lp .
✝ ✆✝ ✆✝ ✆✝ ✆

✞ ☎
The ENTRY button The ’ENTRY’ button (the blue selection above lpenter lp
✞ ☎✞ ☎ ✝ ✆
which is entered by pressing lp2nd lp lpenter lp ) displays the last entry.
✝ ✆✝ ✆
This is a particularly useful button when you have entered an expression then
realized there is an error. You can recall the expression then edit it. Unlike
the ’ANS’ button, using ’ENTRY’ twice gives the entry before the last. For
example:
✞ ☎✞ ☎✞ ☎✞ ☎
Example A.1. We first prepare the screen: lp5 lp lp+ lp lp5 lp lpenter lp
✞ ☎✞ ☎✞ ☎✞ ☎ ✝ ✆✝ ✆✝ ✆✝ ✆
lp2 lp lp+ lp lp2 lp lpenter lp . Now press the ’ENTRY’ button. 2 + 2
✝ ✆✝ ✆✝ ✆✝ ✆
appears in the window. Press the ’ENTRY’ button again. 5 + 5 replaces the
2 + 2.

✞ ☎
Press lpenter lp (so that 10 is the answer). To multiply the 2+2 by 7 we
✝ ✆
A.1. BASIC CALCULATOR FUNCTIONS 351
✞ ☎✞ ☎✞ ☎✞ ☎
can enter: lp2nd lp lpenter lp lp2nd lp lpenter lp . Then move the cursor
✝ ✆✝ ✆✝ ✆✝ ✆ ✞ ☎✞ ☎
using the left arrow to the position of the first 2 and press lp2nd lp lpdellp
✞ ☎ ✝ ✆✝ ✆
lp(lp . Then move the cursor using the right button to after the second 2 and
✝ ✆ ✞ ☎✞ ☎✞ ☎
press the sequence lp)lp lp× lp lp7 lp .
✝ ✆✝ ✆✝ ✆

✞ ☎
What happens if we leave out the parentheses? Can we enter lp(lp before
✞ ☎✞ ☎ ✝ ✆
lp2nd lp lpenter lp ?
✝ ✆✝ ✆
✞ ☎✞ ☎✞ ☎✞ ☎
Example A.2. First type lp5 lp lp+ lp lp2 lp lpenter lp .
✞ ☎✞✝ ✆✝
☎✞ ✆✝
☎✞ ✆✝
☎✞ ✆☎
Then type lp2nd lp lp(−) lp lp+ lp lp5 lp lpenter lp ,
✞ ☎✞✝ ✆
☎✝
✞ ✆✝☎ ✆✝ ✆✝ ✆
lp2nd lp lpenter lp lpenter lp ,
✝
✞ ✆
☎✝
✞ ✆✝
☎✞ ✆
☎
lp2nd lp lpenter lp lpenter lp .
✝ ✆✝ ✆✝ ✆

What would have happened if you had used the ’ANS’ button instead of the
’ENTRY’ button? ✞ ☎✞ ☎✞ ☎
Note that instead of the second lp2nd lp lpenter lp lpenter lp you could
✞ ☎ ✝ ✆✝ ✆✝ ✆
have just pressed lpenter lp .
✝ ✆

✞ ☎
− vs (−) There are two minus signs on the keyboard: lp− lp which is
✞ ☎ ✞ ☎ ✞ ☎ ✝ ✆
between lp× lp and lp+ lp , and lp(−) lp which is to the lower left of
✞ ✝☎ ✆ ✝ ✆ ✝ ✆ ✞ ☎
lpenter lp . What is the difference between the two? lp− lp means subtract.
✝ ✆ ✝ ✆
✞ ☎
So this is the button you use to compute 7 − 5. The button lp(−) lp means
✝ ✆
352 APPENDIX A. INTRODUCTION TO THE TI-84

’negative’ or
✞ ’the opposite
☎✞ of’
☎✞ so, for
☎✞example,
☎ if you want to compute −5 + 7
you enter lp(−) lp lp5 lp lp+ lp lp7 lp . Notice what happens if you use
✞ ☎ ✝ ✆✝ ✆✝ ✆✝ ✆ ✞ ☎
lp− lp instead (what will you subtract from?)! The lp(−)lp button is also
✝ ✆ ✝ ✆
what you use to enter expressions like −x + 7 or exp(−x).

The MODE button When calculating using the trigonometric functions (sin,
cos, tan, and their inverses) it is important that you distinguish between
degrees and radians and that you are in sync with your calculator. For
that
✞ we need
☎ to look at the MODE screen. To view the Mode screen press
lpmodelp . The third line has the words Radian and Degree. If the word
✝ ✆
’radian’ is highlighted and you want to compute in degrees you should press
the down arrow twice (so that the highlighted
✞ word
☎ on the third line blinks)
and then the right arrow and press lpenter lp . The word ’degree’ should
✝ ✆
then be highlighted.

You can also verify✞your mode

☎✞ by evaluating
☎ sin(180) on the main screen (you
can press ’quit’ - lp2nd lp lpmodelp to return to the main screen). If you
✝ ✆✝ ✆
get 0 as an answer then you are in degree mode. What do you get if you are
in radian mode?

The STO button When evaluating complicated expressions, sometimes it is

handy to use the storage feature. For example

2
Example A.3. Suppose you want to evaluate (1−(1−1/2))
1−1/2
. You can first assign
a letter, V✞say, the
☎✞ value☎✞
(1 − 1/2)
☎✞ since☎✞
it appears
☎✞ twice.
☎✞You can do ☎✞this by
☎
pressing lp1 lp lp− lp lp1 lp lp÷ lp lp2 lp lpstolp lpalpha lp lp6 lp .
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✞ ✆✝ ☎✞ ✆✝
☎✞ ☎✞ ✆✝ ✆☎
Then to evaluate the original expression type lp(lp lp1 lp lp− lp lpalpha lp
✝ ✆✝ ✆✝ ✆✝ ✆
A.2. GRAPHING A FUNCTION 353
✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎
lp6 lp lp)lp lpx2 lp lp÷ lp lpalpha lp lp6 lp .
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆

Example A.4. Note that you can use this feature with the ’ENTRY’ button to
evaluate a function at different places. Let’s evaluate x2 + x at x = 0 and at
x = 1. ✞ ☎✞ ☎✞ ☎✞ ☎
First set x = 0: lp0 lp lpstolp lpX,T,θ,nlp lpenter lp .
✝ ✆✝ ✞ ✆✝ ☎✞ ✆✝ ☎✞ ✆
☎✞ ☎✞ ☎
2
Then evaluate the function: lpX,T,θ,nlp lpx lp lp+ lp lpX,T,θ,nlp lpenter lp
✞ ☎✞ ✝ ☎✞ ✆✝ ☎ ✆✝ ✆✝ ✆✝ ✆
Now set x = 1: lp1 lp lpstolp lpX,T,θ,nlp
✝ ✆✝ ✆✝ ✞ ✆
☎✞ ☎✞ ☎✞ ☎
Now evaluate the function again: lp2nd lp lpenter lp lp2nd lp lpenter lp
✞ ☎ ✝ ✆✝ ✆✝ ✆✝ ✆
lpenter lp .
✝ ✆

Note that there is another way to evaluate a function at several points

which will be introduced below in the graphing section.

A.2 Graphing a function

The Y= screen This ✞ is the☎screen that we will need to graph functions. It
is seen by pressing lpy= lp .
✝ ✆
354 APPENDIX A. INTRODUCTION TO THE TI-84
✞ ☎
Press the lpy= lp button. This brings you to the screen where you can enter
✝ ✆
functions that you want to graph. Let us suppose that you want to graph
y = sin(x).
✞ You
☎✞ would first
☎✞ move the☎ cursor (using
✞ the arrow☎ keys) to ’Y1’
enter lpsinlp lpX,T,θ,nlp lpenter lp . The press lpgraph lp . This will show
✝ ✆✝ ✆✝ ✆ ✝ ✆
a portion of the graph of the function (naturally, it can not generally show
you the whole graph as this is often ’infinite’).

A.2.1 Rescaling the graphing window

The graphing window is what determines what portion of the graph is seen.
There are various reasons you may want to have a different window. For
example, the standard view for graphing the sin function (as we will show
in the next paragraph) the setting for the y-axis is too large and makes it
difficult to see what is going on. We describe here a few using the example
of sin(x) (in degree mode) and encourage experimenting.

The standard window This sets the x-axis from −10 to 10 and the y-axis
from −10 to 10. To choose this view ✞you must ☎
choose
✞ the standard window
☎
from the ’ZOOM’ screen by pressing lpzoom lp lp6 lp . This is the window
✝ ✆✝ ✆
shown above.

Notice that in this case you can’t really see what you want. Don’t worry,
there are other zooming options!

Zoom fit One easy way to ’choose’ an appropriate ✞ window ☎is

✞ to ask
☎ the
calculator to choose it using the ’zoom fit’ windows ( lpzoom lp lp0 lp ). You
✝ ✆✝ ✆
A.2. GRAPHING A FUNCTION 355
✞ ☎
can also press lpzoom lp then press the down arrow until you arrive at the
✝ ✆
bottom of the list with is ’ZoomFit’.

Custom window Maybe the zoom fit choice of the calculator isn’t what
you
✞ wanted ☎to see. Then you can choose your own window. Press the
lpwindow lp key. Move the cursor to Xmin. This is the smallest number
✝ ✆
on the x-axis
✞ in your
✞ custom
☎ ☎ window. let’s say you want this to be −5. Then
you press lp(−)lp lp5 lp . Then move the cursor to Xmax and enter your de-
✝ ✝ ✞
✆ ✆ ☎
sired maximum, say, lp5 lp. You can also adjust the scale (where the marks
✝ ✆
on the x-axis are given) by setting Xscl equal to whatever is convenient. In
this example we will set it to 1. Now
✞ you can
✞ do ☎
☎ the
✞ same
☎ for ✞
the y-axis
☎✞ with ☎
changing Ymin and Ymax to say, lp(−)lp lp. lp lp5 lp and lp. lp lp5 lp ,
✝ ✝
✆ ✆✝ ✆ ✝ ✆✝ ✆
respectively. Press graph again and observe the window.

✞ ☎
Zooming in and out You can zoom in and out also. Press the lpzoom lp
✞ ☎ ✝ ✆
lp2 lp . You will now be looking at the graph. Move your cursor to what you
✝ ✆ ✞ ☎
want to be the center and press lpenter lp . You can zoom out as well.
✝ ✆
356 APPENDIX A. INTRODUCTION TO THE TI-84

ZTrig
✞ The☎✞best ☎
choice for our current Y1=sin(X) function is ZTrig (press
lpzoom lp lp7 lp ).
✝ ✆✝ ✆

Zbox You should experiment with ZBox, which is the first option on the zoom
menu. You first move the cursor to where you want the upper left point of
your box to be and press enter. Then you move the cursor to where you want
your lower right corner to be and press enter. Our result is

ZSquare Finally, we point out another zoom option, the zoom square option,
which scales the x- and y-axes to the same pixel size. To see the issue recall
from example 4.4 on page 52,√that the circle x2 + y 2 = 16 may be graphed by
entering the functions y = ± 16 − x2 .

Note, that the obtained graph in the above window (with −6 ≤ x ≤ 6 and
−6 ≤ y ≤✞6) looks more
☎✞ like☎ an ellipse rather than a circle. To rectify this,
we press lpzoom lp lp5 lp , and obtain a graph which now does have the
✝ ✆✝ ✆
A.3. GRAPHING MORE THAN ONE FUNCTION 357

shape of a circle.

A.3 Graphing more than one function

Let us suppose that you want to graph three functions: y = sin(360x), y = x
3
and y ✞= x − x6☎✞. You would
☎✞
first
☎✞
move☎✞the cursor ☎
(using
✞
the arrow keys)
☎✞ ☎
to ’Y1’
enter lpsinlp lp3 lp lp6 lp lp0 lp lpX,T,θ,nlp lp)lp lpenter lp . Note that
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆
if you need to clear the field you can use the✞ clear button.☎✞ Then you
☎ would
move the cursor to the ’Y2’ position and enter lpX,T,θ,nlp lpenter lp . Finally
✝ ✆✝
✞ ✆☎✞ ☎
you would move the cursor to the ’Y3’ position and enter lpX,T,θ,nlp lp− lp
✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎✞ ✝☎ ✆✝ ✆
lpX,T,θ,nlp lp∧lp lp3 lp lp⊲ lp lp÷ lp lp6 lp lpenter lp . To graph these
✝ ✆✝ ✞ ✆✝ ✆✝☎ ✆✝ ✆✝ ✆✝ ✆
functions press lpgraph lp . The three graphs will be drawn one after the
✝ ✆
other.

If you would like to just graph the first two but do not want to erase the
work you put in to enter the third function, you can move back to the screen
and toggle the third✞ function ☎
off (or on) by moving the cursor to the ’=’ sign
after Y3 and press lpenter lp . This equal sign should now be unhighlighted.
✝ ✆
358 APPENDIX A. INTRODUCTION TO THE TI-84

You may also want to have the calculator distinguish between the graphs
by the type of line. Let us suppose you want Y1 bold, Y2 normal, and Y3
dashed.
✞ Look at the
☎ ’t́o the left of Y1. This is what indicates the type of line.
Press lpenter lp until the line type you want is displayed. Move the cursor
✝ ✆
down and do the same for Y2 and then for Y3.

✞ ☎✞ ☎
The Calc menu With the help of the Calc menu ( lp2nd lp lptrace lp ) you
✝ ✆✝ ✆
can find the minima of a function, the maxima of a function, intersections of
functions, and other similar things. We will describe the intersection feature
in detail.

Intersection of functions When solving equations graphically, you will want

to find the intersection of functions. For example graph y = x2 and y = x + 2
on the same screen (zoom standard).

✞ ☎
To find the left most intersection of the two functions first press lp2nd lp
✞ ☎✞ ☎ ✝ ✆
lptrace lp lp5 lp . The calculator shows you the graphing screen and will
✝ ✆✝ ✆
ask you which is the first curve. Move
✞ the cursor
☎ (using the arrow keys) until
it is on the first curve and press lpenter lp . It will ask you which is the
✝ ✆ ✞ ☎
second curve. Move the cursor to the second curve and press lpenter lp .
✝ ✆
A.4. GRAPHING A PIECEWISE DEFINED FUNCTION 359

It will ask you for a guess.

✞ Move the
☎ cursor to near the intersection point you
want to know and press lpenter lp . The intersection will then be displayed.
✝ ✆

A.4 Graphing a piecewise defined function

Suppose you wanted to graph the piecewise defined function
n 2
x, x<2
f (x) =
2x − 7, x ≥ 2
✞ ☎
Go to the lpy= lp menu. We need to enter the function in the following way:
✝ ✆
y = x2 (x < 2) + (2x − 7)(x ≥ 2)
The only new entries are ✞
in the inequality
☎✞ signs,
☎ which can be accessed
through the “TEST” menu ( lp2nd lp lpmathlp ).
✝ ✆✝ ✆

✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎
Therefore, we press the keys: lpX,T,θ,nlp lpx2 lp lp(lp lpX,T,θ,nlp lp2nd lp
✞ ☎✞ ☎✞ ☎✞ ✝
✞
☎ ☎✞ ✆✝☎✞ ✆✝☎✞ ✆✝ ☎✞ ✆✝ ☎✞ ✆☎
lpmathlp lp5 lp lp2 lp lp)lp lp+ lp lp(lp lp2 lp lpX,T,θ,nlp lp− lp lp7 lp
✝
✞ ☎✞ ✆✝☎✞ ✆✝ ✆✝
☎✞ ✆✝ ☎✞ ✆✝ ✆✝
☎✞ ✆✝
☎✞ ☎✞ ✆✝ ☎✞ ✆✝ ☎✆
lp)lp lp(lp lpX,T,θ,nlp lp2nd lp lpmathlp lp4 lp lp2 lp lp)lp lpenter lp
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆
. The graph is displayed on the right:
360 APPENDIX A. INTRODUCTION TO THE TI-84

The meaning of y = (x < 2) as a function is that it is equal to 1 when

it is true and 0 when it is false. In other words, the expression (x < 2)
is 1 when x < 2 and 0 otherwise. You can use this idea to graph more
complicated functions. For example to enter x + 2 for 0 < x < 3 you could
enter (x + 2)(0 < x)(x < 3).
Note: When graphing piecewise defined functions the graph is sometimes
different from what is expected. See the common errors section to interpret
the resulting graph below.

Graphing with polar✞ coordinates

☎ Suppose you want to graph the curve
r = θ + sin θ. Press lpmodelp . On the fourth line highlight ’Pol’ and press
✞ ☎ ✝ ✆
lpenter lp (make sure you in radian mode also)
✝ ✆

✞ ☎
and press lpy= lp . Your menu will have a list r1 , r2 , .... After r1 type
✞ ✝
☎✞ ✆✞
☎ ☎✞ ☎✞ ☎
lpX,T,θ,nlp lp+ lp lpsinlp lpX,T,θ,nlp lp)lp . Then to see the graph type
✝
✞ ✆✝
☎ ✆✝ ✆✝ ✆✝ ✆
lpgraph lp . You may have to adjust your window (zoom fit) to see the
✝ ✆
resulting spiral.

Graphing with parametric coordinates Suppose you want to graph the curve
given by y(t) = 3 sin(t),✞x(t) = 4☎cos(t). First we need to set the mode to
parametric mode: press lpmodelp and highlight ’PAR’ in the fourth line and
✝ ✆
A.5. USING THE TABLE 361
✞ ☎
press lpenter lp .
✝ ✆

✞ ☎
Now go to the graphing screen lpy= lp . You will see a list that begins with
✝ ✞✆ ☎✞ ☎✞ ☎
X1T and Y1T . To the right of X1T type lp4 lp lpcoslp lpX,T,θ,nlp and to the
✞ ☎✞ ☎✞ ✝ ✆
☎✝ ✆✝ ✆
right of Y1T type lp3 lp lpsinlp lpX,T,θ,nlp . Then to see the graph press
✞ ☎ ✝ ✆✝ ✞ ✆✝ ☎✞ ✆ ☎
lpgraph lp . You may need to lpzoom lp lp0 lp (zoomfit) to see the resulting
✝ ✆ ✝ ✆✝ ✆
ellipse.
✞ (Also,
☎✞ be aware
☎ that the scale may be misleading– you may want to
lpzoom lp lp5 lp for a clearer understanding of the graph).
✝ ✆✝ ✆

A.5 Using the table

✞ ☎✞ ☎
The table ( lp2nd lp lpgraph lp ) is useful as it gives a table representation
✝ ✆✝ ✞ ☎✆
of a function (from the lpy= lp screen). Make sure there is a function typed
✝ ✆
in the Y1 postion and look at the table. If you want to know the value of
the function
✞ for ☎
particular
✞ values
☎ of x then we must use the ’TBLSET’ screen.
Press lp2nd lp lpwindow lp . Move the cursor to the ’Indpnt’ line then to
✝ ✆✝ ✆ ✞ ☎
the right to highlight ask and press lpenter lp .
✝ ✆
362 APPENDIX A. INTRODUCTION TO THE TI-84
✞ ☎✞ ☎
Go back to the table lp2nd lp lpgraph lp . Now your table is empty. Evalu-
✝ ✆✝ ✆ ✞ ☎
ate your function at 2.5 by typing 2.5 and pressing lpenter lp . You will see
✝ ✆
the value of the function in Y1 at 2.5 in the table now. You can evaluate at
any number of points.

You can also evaluate at every✞half integer

☎✞ starting at
☎ −1 for example by
going to the ’TBLSET’ screen ( lp2nd lp lpwindow lp ) changing ’TblStart’
✞ ☎
✞ ☎ ✝ ✆✝ ✆
to lp(−)lp lp1 lp and ∆Tbl= .5. Also change the ‘Indpnt’ setting to ‘Auto’
✝ ✝
✆ ✆
again.

✞ ☎✞ ☎
View the table ( lp2nd lp lpgraph lp ) to see what happened.
✝ ✆✝ ✆

A.6 Solving an equation using the solver

We now solve the equation x2 = 2. The first thing we must do in order to use
the equation solver is to rewrite the equation as an expression equal to zero,
that is we need to write it in the format ‘something= 0.’ In our✞example☎this
will be x2 − 2 = 0. Now, let’s go to the equation solver. Press lpmathlp and
✝ ✆
use the cursor to go to the bottom of the list and press enter when ’Solver...’
✞ is☎
highlighted. Various things appear on the screen. Press the up arrow lp△ lp
✝ ✆
until ’EQUATION SOLVER’ appears on the top line. Edit the right hand side
of the equation on the second line so that it reads ’0 = x2 − 2’ and press
A.7. SPECIAL FUNCTIONS (ABSOLUTE VALUE, N-TH ROOT, ETC.) 363
✞ ☎
lpenter lp .
✝ ✆

Next, on the second line pick a number near where you expect the answer to
be, for example, ’x = 1’.

✞ ☎✞ ☎
Now solve by pressing lpalpha lp lpenter lp . The second line gives an
✝ ✆✝ ✆
answer (check that it is a square root of two!).

✞ ☎
What happens if you had entered ‘x = −1’ on the second line (use lp(−)lp
✝ ✆
here)?

A.7 Special functions (absolute value, n-th root,

etc.)
✞ ☎
We now point out some important functions in the MATH menu ( lpmathlp )
✞ ☎✞ ☎ ✝ ✆
and the LIST menu ( lp2nd lp lpstatlp ), that are used in this course.
✝ ✆✝ ✆

Fractions Suppose in the main screen you evaluate 4/6. Your calculator
will display
✞ .66666666667.
☎✞ ☎✞If you wanted
☎ to see this as a fraction you would
type lpmathlp lpenter lp lpenter lp . You will now see 2/3 displayed.
✝ ✆✝ ✆✝ ✆
364 APPENDIX A. INTRODUCTION TO THE TI-84
✞ ☎
Absolute value To evaluate |−4| you would type in the main screen lpmathlp
✞ ☎✞ ☎✞ ☎
✞ ☎✞ ☎ ✝ ✆
lp⊲ lp lpenter lp lp(−)lp lp4 lp lpenter lp . The answer 4 should be dis-
✝ ✆✝ ✆✝ ✝
✆ ✆✝ ✆✞ ☎
played. To graph |x − 2|, go to the Y= screen ( lpy= lp ) and enter in the Y1
✞ ☎✞ ☎✞ ✝ ☎✞ ✆ ☎✞ ☎✞ ☎
(for example) space lpmathlp lp⊲ lp lpenter lp lpX,T,θ,nlp lp− lp lp2 lp
✞ ☎✞ ☎ ✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆
lp⊲ lp lpenter lp .
✝ ✆✝ ✆

Now you can go to the graphing screen to see the graph.

✞ ☎
n-th root To calculate the cube root of 8. Press on the main screen lp3 lp
✞ ☎✞ ☎ ✞ ☎✞ ☎ ✝ ✆
lpmathlp lp5 lp (or select the x-root) lp8 lp lpenter lp .
✝ ✆✝ ✆ ✝ ✆✝ ✆

You can compute any root ✞ this way

☎✞ (e.g.
☎ the 5th root of 7). For the cube
root you can also use lpmathlp lp4 lp instead. Recall also the algebraic
✝ ✆✝ ✆
definition of the nth root,
√
n
1
x = xn,
✞ ☎✞ ☎
so that the 5th root of 7 can also be computed by pressing lp7 lp lp∧lp
✞ ☎✞ ☎✞ ☎✞ ☎✞ ☎ ✝ ✆✝ ✆
lp(lp lp1 lp lp÷ lp lp5 lp lp)lp .
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆

✞ ☎
Factorials To compute 5!, for example, in the main screen type lp5 lp
✞ ☎ ✝ ✆
lpmathlp then move the cursor to the right three times (or left once) so
✝ ✆ ✞ ☎
that the ’PRB’ menu is displayed. Option 4 is ’!’ so press lp4 lp or move the
✞ ☎ ✝ ✆✞ ☎
cursor to highlight this option and press lpenter lp . Finally press lpenter lp
✝ ✆ ✝ ✆
A.7. SPECIAL FUNCTIONS (ABSOLUTE VALUE, N-TH ROOT, ETC.) 365

and the answer 120 will be displayed.

 
5
Combinations and permutations To calculate 5 C2 = , in the main
✞ ☎
2
menu type lp5 lp . Then go to the math-probability menu by pressing
✞ ☎ ✝ ✆
lpmathlp then the right (or left) arrow until ’PRB’ is highlighted. You will
✝ ✆ ✞ ☎
see that the third option is n Cr so press lp3 lp . This will return you to the
✞ ☎✞ ☎ ✝ ✆
main screen. Press lp2 lp lpenter lp . The answer 10 will be displayed.
✝ ✆✝ ✆

Permutations n Pr are handled the same way except it is the second option
under the math-probability menu instead of the third.

Sequences and series A sequence a1 , a2✞, a3 , . . . that ☎✞ is given

☎ by a formula
can be added via the ‘LIST’ menu (press lp2nd lp lpstatlp ). For example,
✝ ✆✝ ✆
to enter the first 10 terms a1 , a2 , . . . , a10 of the sequence an = n2 + 1, we
have to write seq(x2 + 1, x, 1, 10). Here, the ‘sequence’ command takes four
inputs, first the assignment x2 + 1, second the independent variable x, third
the starting index 1, and fourth✞the final☎index✞ 10.☎✞This expression
☎✞ ☎✞is entered☎
to the calculator by pressing lp2nd lp lpstatlp lp⊲ lp lp5 lp lpX,T,θ,nlp
✞ ☎✞ ☎✞ ☎✞ ☎✞✝ ✆✝
☎✞ ☎✞✆✝ ✆✝ ☎✞ ✆✝ ☎✞
☎✞ ✆
☎
2
lpx lp lp+ lp lp1 lp lp,lp lpX,T,θ,nlp lp,lp lp1 lp lp,lp lp1 lp lp0 lp
✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆✝ ✆
366 APPENDIX A. INTRODUCTION TO THE TI-84
✞ ☎ ✞ ☎
lp)lp and then confirmed by pressing lpenter lp . We obtain:
✝ ✆ ✝ ✆

To add the ten numbers a1 + · · · + a10 of

✞ the sequence,
☎✞ we☎✞use the
☎✞ ‘sum’☎
command in the LIST-MATH menu. Press lp2nd lp lpstatlp lp⊳ lp lp5 lp
✝ ✆✝ ✆✝ ✆✝ ✆
to enter the
✞ ‘sum(’
☎✞expression.
☎ Then using the previously
✞ ☎✞ entered answer
☎ by
pressing lp2nd lp lp(−) lp , and finishing with lp)lp lpenter lp , we now
✝ ✆✝ ✆ ✝ ✆✝ ✆
calculate the wanted sum.

A.8 Programming the calculator

This section is more advanced and is only recommended for students who are
familiar with programming languages.

Hello world A ‘Hello World’ program is a program that only displays the
words “Hello World.” It is a starting point to understand the procedure of
creating and running a program before going to more advanced programming
techniques. ✞ ☎ ✞ ☎
To create a new program press lpprgmlp , move to the right lp⊲ lp to
✝ ✆ ✝ ✆
‘NEW,’ and give it the name ‘HELLO.’
A.8. PROGRAMMING THE CALCULATOR 367
✞ ☎
Next, we need to add the Display command. Press lpprgmlp ; you see a
✝ ✆
list of commands. We move to the right to ‘I/O’ for input/output commands.
Select the third item ‘Disp’ for display.

To finish,✞we need to☎✞add ‘”HELLO

☎ WORLD.”’ (Quotation
✞ marks
☎✞obtained
☎ by
pressing lpalpha lp lp+ lp ; space is obtained by lpalpha lp lp0 lp .) The
✝ ✆✝ ✆ ✝ ✆✝ ✆
complete program is shown below.

✞ ☎✞ ☎ ✞ ☎
To execute the program, quit lp2nd lp lpmodelp , press lpprgmlp and confirm
✞ ☎ ✝ ✞ ✆✝ ☎ ✆ ✝ ✆
with lpenter lp . After another lpenter lp , the program should run as shown
✝ ✆ ✝ ✆
below.

The quadratic formula We next implement the quadratic formula

√
2 −b ± b2 − 4ac
ax + bx + c = 0 =⇒ x= .
2a

Create a new program ‘QUADRATI.’ We now need a new command, the Prompt
command which will ask for three input values and places them in the variables
368 APPENDIX A. INTRODUCTION TO THE TI-84

A, B, C. The Prompt command is the second item in the ‘I/O’ menu.

After executing the new program, enter any input values (for example A= 3,
B= −10, C= −8). The answer should appear below.

Directional angle Our third and last program calculates the directional an-
gle of a vector ~v = ha, bi. Recall that the angle is tan−1 (b/a) or tan−1 (b/a) +
180◦ depending on the quadrant in which ~v is pointing. In quadrant II and
III, we need to add 180◦ degrees. This is exactly the case when a < 0. Thus,
we prompt for input values A and B, and we calculate tan−1 (b/a) ✞ and☎place
it into the new variable G. (The arrow is obtained by pressing lpstolp .) We
✝ ✆
then check if a < 0, in which case we add 180 to G. (Here, we are assuming
that the calculator is in degree mode).

✞ ☎
Note that the ‘If’ command is obtained by pressing lpprgmlp , and ‘<’ is
✞ ☎✞ ✝ ☎ ✆ ✞ ☎
obtained from the TEST menu, press lp2nd lp lpmathlp and then lp5 lp .
✝ ✆✝ ✆ ✝ ✆
A.8. PROGRAMMING THE CALCULATOR 369

We may now run the program and check its correctness in some examples.

Other projects Here is a list of projects that you may want to try to imple-
ment on your calculator.

1. Display the final amount of an investment for a given principal P , annual

interest rate r, number of compoundings per year n, and number of years
t (see Observation 16.8 on page 222).

2. In the program above for the directional angle, check for the cases when
a = 0, and display the correct answers.

3. Implement the long division (see Session 8).

✞ ☎
Erasing a program To erase a program that you have written, press lp2nd lp
✞ ☎ ✞ ☎ ✞ ☎ ✝ ✆
lp+ lp , then lp2 lp , then lp1 lp . A list of all your programs appears.
✝ ✆ ✝ ✆ ✝ ✆

✞ ☎✞ ☎
Select the program you want to delete with lp△ lp lp▽ lp and press the
✞ ☎ ✞ ✝☎ ✆✝ ✆
lpdellp key. Confirm your selection with lp2 lp ‘Yes.’
✝ ✆ ✝ ✆
370 APPENDIX A. INTRODUCTION TO THE TI-84

A.9 Common errors

The following shows many common errors and mistakes that may produce
a false result or an error. If none of the items below help to resolve your
problem, then you can also reset the calculator to factory setting (see section
A.10 on page 373).

Reset window to standard window Many problems with graphing can be

resolved
✞ by changing
☎✞ the
☎ window back to the standard window. To do so,
press lpzoom lp lp6 lp .
✝ ✆✝ ✆

Note, that the standard window has the following settings:

PLOT marked Sometimes an error is produced when graphing a function

due to having any of the PLOTs marked in the ‘Y=’ menu.

Make sure that all of the PLOTs are unmarked!

A.9. COMMON ERRORS 371

1
Errors in graphing functions Graph the functions y = ln(x−3) or y = x3 −9x
.

The calculator graphic is not really the graph, but has extra parts or missing
parts of the graph (depending on the version of the calculator: TI-83 is much
worse than TI-84). The reason for this is that the calculator just approximates
the graph pixel by pixel and does not represent the exact graph. In particular,
you should not just copy the graph from the calculator onto your paper, but
interpret what you see and draw the interpreted graph!

Fractions need parenthesis for numerator and denominator When entering

a rational function, (or any fraction), the numerator and denominator has to
be entered with parenthesis. For example, y = xx−32 −1 is entered as follows.

Radian versus degree Trigonometric functions should generally be graphed

in radians. For example, the cosine function in radian gives the following.
372 APPENDIX A. INTRODUCTION TO THE TI-84

On the other hand, the cosine in degrees displays as follows.

The reason is that one period is now 360 on the x-axis. Rescaling the cal-
culator to a wider x-scale shows the effect more clearly.

The calculator also gives different values for the trigonometric functions
in degrees or radians. This is shown below.

gives

gives

Table setup The table can be setup to generate a list of outputs, or to take
an input value and generate its output value. This depends on the independent
variable in the TABLE SETUP being set to ‘Auto’ or ‘Ask,’ respectively.
A.10. RESETTING THE CALCULATOR TO FACTORY SETTINGS 373

Using✞ the wrong

☎ sign Often
minus ✞ ☎ a syntax✞error is
☎ due to using the minus
sign lp− lp instead of lp(−)lp. Note that lp− lp is used to subtract two
✝ ✆ ✞ ✝ ☎ ✆ ✝ ✆
numbers, whereas lp(−) lp gives the negative of a number.
✝ ✆ ✞ ☎✞ ☎✞ ☎✞ ☎
For example, to calculate −3 − 5, press lp(−) lp lp3 lp lp− lp lp5 lp .
✝ ✆✝ ✆✝ ✆✝ ✆

A.10 Resetting the calculator to factory settings

To reset everything
✞ and
☎✞ return
☎ the calculator to factory settings, first✞press☎
the buttons lp2nd lp lp+ lp to get to the MEMORY menu, and then lp7 lp
✝ ✆✝ ✆ ✝ ✆
(‘Reset...’).

✞ ☎
Then move the cursor to the right (with lp⊲ lp ) until ‘ALL’ is highlighted.
✞ ☎ ✞ ✝ ☎ ✆
Press lpenter lp (‘All Memory...’) and lp2 lp (‘Reset’).
✝ ✆ ✝ ✆

(With some of the other options in the MEMORY/Reset... menu you can also
reset selective parts of the calculator only.)
Answers to exercises

Here are the answers to the exercises given in each sections and in the
reviews for each part.
Session 1 (exercises starting on page 11):
√ √
Exercise 1.1 Examples: a) 2, 3, 5, b) −3, 0, 6, c) −3, −4, 0, d) 23 , −4
7 , 8, e) 5, π, 3 31,
f) 12 , 25 , 0.75
√ √
Exercise 1.2 natural: 17000, 12 4√, 25, integer: −5, 0, 17000, 12 4 , 25, rational:
7 12
3 , −5, 0, 17000, √4 , 25, real: all of the given numbers are real num-
bers, irrational: 7
√
Exercise 1.3 a) 8, b) 10, c) 99, d) −3, e) −2, f) 6, g) 7, h) 7, i) 5.4, j) 32 , k) 25 , l) −2
Exercise 1.4 a) S = {−8, 8}, b) S = {0}, c) S = {}, d) S = {−13, 7},
e) S = {−7, 2}, f) S = {−4, 24 5 }, g) S = {}, h) S = { 7 }, i) S =
−3
−4 8
{−10, 12}, j) S = { 3 , 3 }, k) S = {−8, 1}, l) S = {−10, 2}
Exercise 1.5 a) S = {−8, 8}, b) S = {0}, c) S = {}, d) S = {2, 6}, e) S = {−14, 4},
f) S = {−3, 7}
Exercise 1.6
Inequality Number line Interval
2≤x<5 2 5 [2, 5)
x≤3 3 (−∞, 3]
12 < x ≤ 17 12 17 (12, 17]
x < −2 −2 (−∞, −2)
−2 ≤ x ≤ 6 −2 6 [−2, 6]
x<0 0 (−∞, 0)
4.5 ≤ x 4.5 [4.5, ∞)
√ √ √
5 < x ≤ 30 5 30 (5, 30]
13 π
( 13
13
7 <x<π 7
7 , π)

374
 375

Exercise 1.7 a) S = [−3, 11], b) S = (−∞, −3]∪[11, ∞), c) S = (−∞, −3)∪(11, ∞),
d) S = [−10, 3], e) S = (−∞, − 25 ) ∪ ( 23 , ∞), f) S = ( −19 15
4 , 4 ), g)
2 2
S = (−∞, 3] ∪ [7, ∞), h) S = (−∞, 15 ) ∪ ( 5 , ∞), i) S = [−1, 3], j)
S = {}, k) S = (−∞, ∞) = R, l) S = (−∞, −1)∪(−1, ∞) = R−{−1}
Session 2 (exercises starting on page 28):
Exercise 2.1 a) y = 2x − 4, b) y = −x + 3, c) y = −2x − 2, d) y = 52 x + 3, e)
y = −x + 0 or y = −x, f) y = 32 x + 4
Exercise 2.2 a) y = −2x + 4, b) y = 3x + 5, c) y = − 21 x − 2, d) y = 35 x − 57 , e)
y = 23 x − 15 8
2 , f) y = 9 x
Exercise 2.3 a) y − 3 = 31 · (x − 5), b) y − 1 = − 23 · (x − 4), c) y + 2 = − 21 · (x − 3),
d) y − 1 =2 1 · (x + 1) 6
y y

1 5 4 y

0 x 4 3
-3 -2 -1 0 1 2 3 4
-1 3 2

-2 2 1

-3 1 0 x
-4 -3 -2 -1 0 1 2 3 4
-4 0 x -1
-2 -1 0 1 2 3 4 5
-5 -1 -2

Exercise 2.4 a) -6
4 y
b) -2 c) 3 y
-3

3 2

2 1

1 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
0 x -1
-4 -3 -2 -1 0 1 2 3 4
-1 -2

-2 -3

-3 -4

d) -4 e) y = 2x − 3 3 y
-5

0 x
-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3

f) y = − 31 x − 2 -4

Exercise 2.5 a) this is a function with domain D = {−5, −1, 0, 3, 6} and range
R = {2, 3, 5, 7, 8}, for example: the input x = −5 gives output y = 5,
etc., b) not a function since for x = 4 we have both y = 0 and y = −1,
c) this is a function with D = {−11, −2, √ 3, 6, 7, 19}, R = {3}, d) this is
a function with D = {1, 2, 3, 4, 5}, R = { 19, 5.33, 9, 13, 17}, e) this is
not a function
Exercise 2.6 a) yes, b) no, c) the domain for the function in (a) is the set of all
children, d) the range for the function in (a) is the set of all mothers
Exercise 2.7 a) a given cash amount x determines the interest amount y, b) i) $0,
ii) $100, 000, iii) $0, iv) $300, 000, v) $200, 000, vi) $40, 000
Exercise 2.8 a) C = 2πr, b) P = 3a, c) P = 2a + 6, d) V = a3
376 ANSWERS TO EXERCISES

Session 3 (exercises starting on page 44):

√ √
Exercise 3.1 a): i) 10, ii) 16, iii) −5, iv) 1, v) 3 13 + 1, vi) 3 2 + 10, vii) −3x + 1,
viii) 3x + 7, ix) 3x + 1 + h, x) 3x + 3h + 1
√ √
b): i) 6, ii) 20, iii) 6, iv) 0, v) 13 − 13, vi) 8 + 5 2, vii) x2 + x, viii)
x2 + 3x + 2, ix) x2 − x + h, x) x2 + 2xh + h2 − x − h
p √
√ i) 0, ii) 4,√iii) undefined, iv)√undefined, v) 2,√vi) 2 + 6 2, vii)
c):
x2 − 9, viii) x2 + 4x − 5, ix) x2 − 9 + h, x) x2 + 2xh + h2 − 9
√ √
13 3− 2
d): i) 13 , ii) 51 , iii) − 12 , iv) undefined, v) 13 , vi) 7 , vii) − x1 , viii)
1 1+xh 1
x+2 , ix) x , x) x+h
√ √ √
e): i) √13−5 = 23−7 13 , √2−2
5 , ii) 0, iii) undefined, iv) , v) vi) =
−2 −5
2 13+2 9 2+5
√
−12+7 2 x−3 x−5+hx+2h x+h−5
23 , vii) −x+2 ,
−x−5
viii) x+4 , ix) x+2 , x) x+h+2
√ √
f): i) −27, ii) −125, iii) 8, iv) 0, v) − 2197, vi) −45 − 29 2, vii)
x3 , viii) f (x + 2) = −(x + 2)3 or in descending order f (x + 2) =
−x3 −6x2 −12x−8, ix) −x3 +h, x) −(x+h)3 or −x3 −3x2 h−3xh2 −h3
Exercise 3.2 a) D = (−4, 6], b) −3, c) 25, d) −8, e) 9
Exercise 3.3 a) D = (−∞, 5) ∪ (5, ∞) or, alternatively, D = R − {5}, b) 0, c) −2,
d) 7, e) 7, f) undefined, g) 22
Exercise 3.4 a) 5, b) 2, c) 2x + h, d) 2x + 5 + h, e) 2x + 3 + h, f) −2x − h, g)
2x + 4 + h, h) 6x − 2 + 3h, i) 3x2 + 3xh + h2
Exercise 3.5 a) 3, b) 4, c) x + a − 3, d) −1
ax

Exercise 3.6 a) D = R all real numbers, b) D = R, c) D = [2, ∞), d) D = (−∞, 4],

e) D = R, f) D = R − {−6}, g) D = R − {7}, h) D = R − {2, 5},
i) D = R − {2}, j) D = (1, 2) ∪ [3, ∞), k) D = [0, 9) ∪ (9, ∞), l)
D = (−4, ∞)
Exercise 3.7 a) domain Df = [1, 3) ∪ [4, 6] and range Rf = [1, 3], b) Dg = R and
Rg = [2, 3], c) Dh = (−2, 0) ∪ (0, 2) ∪ (2, 3) and Rh = {−1} ∪ (0, 1], d)
1, e) 3, f) undefined, g) 2, h) 2, i) 3, j) undefined, k) 2, l) 2, m) 3, n) 2.5,
o) 2, p) 2, q) 2, r) undefined, s) 1, t) undefined, u) −1, v) undefined, w)
undefined, x) −1
Exercise 3.8 a) not a function, b) this is a function, c) not a function, d) not a function
Exercise 3.9 a) D = (−3, 4) ∪ (4, 7], b) R = (−2, 2], c) x = −2 or x = 0 or x = 7,
d) x ∈ (4, 5], e) x ∈ (−3, −1] ∪ [0, 4) ∪ [6, 7], f) x ∈ (−2, 0) ∪ (4, 7), g)
f (2) = −1, f (5) = 2, h) f (2)+f (5) = 1, i) f (2)+5 = 4, j) f (2+5) = 0
Exercise 3.10 a) Approximately 3, 900 students were admitted in the year 2000, b)
The most students were admitted in 2007. c) In 2000, the number
of admitted students rose fastest. d) In 2003 the number of admitted
students declined.
 377

Exercise 3.11 domain D = (−2, 5], graph:

5 y

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6
-1

Session 4 (exercises starting on page 61):

Exercise 4.1 a) b) c)

d) e) f)
√ √
Exercise 4.2 a) y1 = 4 − x2p , y2 = − 4 − x2 (The graphsp are displayed be-
low.)p b) y1 = 15 − (x +p5)2 , y2 = − 15 − (x + 5)2 , c) y1 =
√
2 + 9 −√(x − 1)2 , y2 = 2 − 9 − (x − 2 2
√1) , d) y1 = −x √+ 8x + 14,
y2 =√− −x2 + 8x + 14,√ e) y1 = x2 + 3, y2 = − x2 + 3, f)
y1 = −x2 + 77, y2 = − −x2 + 77

a) b) c)

d) e) f)
√
Exercise 4.3 a) x ≈ −2.62, x ≈ −0.38, b) x = ±1, x = ± 2 ≈ ±1.41, c) x ≈ 1.46,
d) x ≈ −2.83, x ≈ 0.01, x ≈ 2.82, x ≈ 4.01, x ≈ 7.00, e) x ≈ −0.578,
x ≈ 1.187, x ≈ 4.388, f) x ≈ 1.61, x = 2, x ≈ 6.91, g) x ≈ −4.00, h)
x = −4, x = 2.25, i) x ≈ −4.00, x ≈ 2.22, x ≈ 2.28,
378 ANSWERS TO EXERCISES

Exercise 4.4 a) x ≈ −1.488, b) x ≈ −1.764, x ≈ −0.416, x ≈ 0.681, c) x ≈ 5.220, d)

x ≈ −1.431, x ≈ 0.038, e) x ≈ −1.247, x = 0, x ≈ 0.445, x ≈ 1.802,
f) x = 0, x = 1, x = 3
Exercise 4.5 a) There is one minimum. Zoom out for the graph (see below). b) There
is one maximum. Resize the window to Ymin= −100. c) There is one
local maximum and one local minimum. The graph with Xmin=−4,
Xmax= 4, Ymin= −2, Ymax= 2 is below. d) Zooming into the graph
reveals two local minima and one local maximum. We graph the func-
tion with Xmin=−2, Xmax= 4, Ymin= −1.3, Ymax= 0.5.

a) b) c)

d)
(x, y) ≈ (−1.5, 3.6) (x, y) ≈ (0.2, 0.9) (x, y) ≈ (1.3, 1.1)

Exercise 4.6 a) b)

(x, y) = (2, 1) (x, y) ≈ (2.5, 16.1) (x, y) ≈ (−0.2, 2.0)

c)

(x, y) = (0, 2) (x, y) ≈ (−1.4, −7.7) (x, y) ≈ (0.8, 4.1)

d)

(x, y) ≈ (1.4, 3.7) (x, y) ≈ (−1.5, −10.6)

e)

(there is only one minimum and no maximum in part (e))

 379

(x, y) ≈ (−1.5, −9.1) (x, y) = (1, 5) (x, y) ≈ (1.2, 5.0)

f)

Session 5 (exercises starting on page 73):

√ 1
Exercise 5.1 a) y = |x| + 1, b) y = − x, c) y = (x − 2)2 + 1, d) y = x+2 + 2, e)
y = −(x + 2)3 , f) y = −|x − 3| + 2
2 y 3 y 1 y

1 2 0 x
-3 -2 -1 0 1 2 3
0 x 1 -1
-4 -3 -2 -1 0 1 2 3 4
-1 0 x -2
-6 -5 -4 -3 -2 -1 0 1 2
-2 -1 -3

-3 -2 -4

Exercise 5.2 a) 3 y
-4 b) 5 y
-3 c) 5 y
-5

2 4 4

1 3 3

0 x 2 2
-2 -1 0 1 2 3 4
-1 1 1

-2 0 x 0 x
-6 -5 -4 -3 -2 -1 0 1 2 -2 -1 0 1 2 3 4 5 6
d) 3 y
-3 e) -1
7 y
f) -1

6
2
5
1
4
0 x
-1 0 1 2 3 4 5 6 7 3

-1
2

-2 1

g) -3 h) -5 -4 -3 -2 -1
0
0 1 2 3 4
x
5

Exercise 5.3 a) y = x − 7x − 3, b) y = (x + 3) − 7 · (x + 3) + 1 = x2 − x − 11,

2 2

c) y = −x2 + 7x − 1, d) y = x2 + 7x + 1, e) y = 19 x2 − 73 x + 1, f)
y = 4x2 − 14x + 1
Exercise 5.4 a) shift to the right by 5, b) stretched away from the x-axis by a factor
2, c) shift to the right by 4, d) compressed towards the y-axis by a
factor 2, e) shifted to the right by 3 (to get the graph of y = x1 ) and
then reflected about the x-axis, f) compressed towards the x-axis by
a factor 2 (you get y = |x|) then shifted to the left by 1 and up by 1
Exercise 5.5 a) odd, b) even, c) even, d) neither, e) even, f) odd, g) even, h) neither,
i) odd
4 y 4 y 3 y

3 3 2

2 2 1

1 1 0 x
-4 -3 -2 -1 0 1 2 3 4
0 x 0 x -1
-4 -3 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 4 5
-1 -1 -2

Exercise 5.6 a) -2 b) -2 c) -3
380 ANSWERS TO EXERCISES
4 y 4 y 4 y

3 3 3

2 2 2

1 1 1

0 x 0 x 0 x
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 -1 -1

d) -2 e) -2 f) -2

Session 6 (exercises starting on page 83):

Exercise 6.1 a) (f + g)(x) = x2 + 9x − 5 with domain Df +g = R, (f − g)(x) =
x2 + 3x + 5 with domain Df −g = R, (f · g)(x) = 3x3 + 13x2 − 30x
with domain Df ·g = R
b) (f + g)(x) = x3 + 5x2 + 12, Df +g = R, (f − g)(x) = x3 − 5x2 − 2,
Df −g = R, (f · g)(x) = 5x5 + 7x3 + 25x2 + 35, Df ·g = R
√
g)(x) = 2x2 + 3x + 12 x, Df +g = [0, ∞), (f√− g)(x) √
c) (f +√ = −2x2 +
3 2
3x + 2 x, Df −g = [0, ∞), (f · g)(x) = 6x + 14x x + 15x x + 35x,
Df ·g = [0, ∞)
d) (f + g)(x) = 5x+1 1−5x
x+2 , Df +g = R − {−2}, (f − g)(x) = x+2 , Df −g =
5x
R − {−2}, (f · g)(x) = (x+2) 2 , Df ·g = R − {−2}
√ √
e) (f + g)(x) = 3 x − 3, Df +g√ = [3, ∞), (f − g)(x) = − x − 3,
Df −g = [3, ∞), (f · g)(x) = 2 · ( x − 3)2 = 2 · (x − 3), Df ·g = [3, ∞)
f) (f + g)(x) = x2 + 5x − 1, Df +g = R, (f − g)(x) = x2 − x + 11,
Df −g = R, (f · g)(x) = 3x3 + 3x − 30, Df ·g = R
g) (f + g)(x) = 3x2 + 6x + 4, Df +g = R, (f − g)(x) = −x2 − 4,
Df −g = R, (f · g)(x) = 2x4 + 9x3 + 13x2 + 12x, Df ·g = R



Exercise 6.2 a) fg (x) = 3x+62x−8 with domain D gf = R − {4},
g
f (x) = 3x+6
2x−8


with domain D fg = R − {−2}, b) fg (x) = x2 −5x+4 x+2 x+2
= (x−4)(x−1) ,
g

2
x −5x+4
D f = R − {1, 4}, f (x) = x+2 , D f = R − {−2}, c)
g
g
f

x+3 g

(x−5)(x−2)
g (x) = (x−5)(x−2) , D fg = R − {−3, 2, 5}, f (x) = x+3 ,

√
D fg = R − {−3, 5}, d) fg (x) = 2x+5 x+6
, D f = [−6, − 25 ) ∪ (− 52 , ∞),
g
g

2x+5 f

x2 +8x−33
f (x) = √
x+6
, D g = (−6, ∞), e)
f g (x) = √
x
, D f = (0, ∞),
g

√
g x
f (x) = 2
x +8x−33 , D g = [0, 3) ∪ (3, ∞)
f

Exercise 6.3 a) 37, b) 7, c) 11, d) 147, e) −1, f) 81, g) 12x2 + 20x + 7, h) −4x − 9,
i) −141, j) −5, k) 2x + 2h − 3, l) 3x2 + 6xh + 3h2 + 4x + 4h
Exercise 6.4 a) (f ◦g)(x) = 6x+4, b) (f ◦g)(x)√= x2 +6x+11, c) (f ◦g)(x) = 4x2 −
2
2x, d) (f ◦g)(x) = x4 +4x3 +4x2 + x2 + 2x + 3, e) (f ◦g)(x) = x+h+4 ,
2 2
f) (f ◦ g)(x) = x + 2xh + h + 4x + 4h + 3
Exercise 6.5 a) (f ◦ g)(x) = 2x − 6, (g ◦ f )(x) = 2x − 1, (f ◦ f )(x) = 4x + 12,
(g ◦ g)(x) = x − 10
b) (f ◦ g)(x) = x2 − 2x + 3, (g ◦ f )(x) = x2 + 4x + 3, (f ◦ f )(x) = x + 6,
(g ◦ g)(x) = x4 − 4x3 + 2x2 + 4x
 381
√ √
c) (f ◦ g)(x) = 6x − 2 − 3x + 2, (g ◦ f )(x) = p6x2 − 3x − 16,
√
(f ◦ f )(x) = 8x4 − 8x3 − 48x2 + 25x + 72, (g ◦ g)(x) = 3 3x + 2 + 2
x+3
d) (f ◦g)(x) = x, (g ◦f )(x) = x, (f ◦f )(x) = 3x+10 , (g ◦g)(x) = 10x−3
1−3x
e) (f ◦ g)(x) = x, (g ◦ f )(x) = x, (f ◦ f )(x) = (2(2x − 7)2 − 7)2 or
expanded in descending degrees: (f ◦f )(x) = 64x4 −896x3 +4592x2 −
q√
x+7
√ √
+7 14+ 14+2 x
10192x + 8281, (g ◦ g)(x) = 2
2 = 4
Exercise 6.6
x 1 2 3 4 5 6 7
f (x) 4 5 7 0 −2 6 4
g(x) 6 −8 5 2 9 11 2
f (x) + 3 7 8 10 3 1 9 7
4g(x) + 5 29 −28 25 13 41 49 13
g(x) − 2f (x) −2 −18 −9 2 13 −1 −6
f (x + 3) 0 −2 6 4 undef. undef. undef.

Note, however, that the complete table for y = f (x + 3) is given by:

x −2 −1 0 1 2 3 4
f (x + 3) 4 5 7 0 −2 6 4

Exercise 6.7
x 1 2 3 4 5 6
f (x) 3 1 2 5 6 3
g(x) 5 2 6 1 2 4
(g ◦ f )(x) 6 5 2 2 4 6
(f ◦ g)(x) 6 1 3 3 1 5
(f ◦ f )(x) 2 3 1 6 3 2
(g ◦ g)(x) 2 2 4 5 2 1

Exercise 6.8
x 0 2 4 6 8 10 12
f (x) 4 8 5 6 12 −1 10
g(x) 10 2 0 −6 7 2 8
(g ◦ f )(x) 0 7 undef. −6 8 undef. 2
(f ◦ g)(x) −1 8 4 undef. undef. 8 12
(f ◦ f )(x) 5 12 undef. 6 10 undef. −1
(g ◦ g)(x) 2 2 10 undef. undef. 2 7

Session 7 (exercises starting on page 95):

Exercise 7.1 a) no (that is: the function is not one-to-one), b) yes, c) no, d) no, e)
no, f) no, g) yes, h) no
382 ANSWERS TO EXERCISES

Exercise 7.2 a) f −1 (x) = x−9 x+3

4 , b) f
(x) = − 8 , c) f
−1 −1
(x) = x2 − 8, d) f −1 (x) =
2
x −7 2 2 √
3 , e) f
−1
(x) = − x6 −2 = −x36−72 , f) f −1 (x) = 3 x, g) f −1 (x) =
√3
q
x−5
2 , h) f
−1
(x) = 3 x−5 2 , i) f
−1
(x) = x1 , j) f −1 (x) = x1 + 1 = 1+x x ,

2 2
k) f −1 (x) = x1 + 2 = 1+2x x2 , l) f
−1
(x) = y5 + 4 = 5+4y y , m)
f (x) = 1−x , n) f (x) = x−3 , o) f (x) = 2−3x
−1 2x −1 6x −1
x−1 , p) f −1
(x) = 5x+7
x+1
q)
x 3 7 1 8 5 2
f −1 (x) 2 4 6 8 10 12
√
Exercise 7.3 a) restricting to the domain D = [0, ∞) gives the inverse f −1 (x) = x,
b)
√ restricting to the domain D = [−5, ∞) gives the inverse f (x) =
−1

x − 1 − 5, c) restricting to the domain D = [0, ∞) gives the inverse

f −1 (x) = x, d) restricting to the domain D = [4, ∞) gives the inverse
f −1 (x) = x + 6, q e) restricting to the domain D = (0, ∞) gives the
1
inverse f −1 (x) = x,
f) restricting to the domain D = (−7, ∞) gives
q
the inverse f −1 (x) = − x3 −7, g) restricting to the domain D = [0, ∞)
√
gives the inverse f −1 (x) = 4 x, h)√ restricting to the domain D = [3, ∞)
gives the inverse f −1 (x) = 3 + 4 10x
Exercise 7.4 a) yes (that is: the functions f and g are inverses of each other), b)
no, c) no, d) yes, e) no, f) yes 5 y

5 y 4

4 3

3 2

2 1

dashed = original function x

1 0
solid = inverse function -2 -1 0 1 2 3 4 5
0 (dotten = diagonal line) x -1
-1 0 1 2 3 4 5 6 7 8
Exercise 7.5 a) -1
6 y
b) -2

3
7 y
2
6

1 5

0 x 4
-4 -3 -2 -1 0 1 2 3 4 5 6
3
-1
2
-2
1

-3 x
0
-1 0 1 2 3 4 5 6 7
c) -4 d) -1
 383
4 y 6 y

3 5

2 4

1 3

0 x 2
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4
-1 1

-2 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
-3 -1

-4 -2

-5 -3

-6 -4

e) 5 y
-7 f) -5

4
3 y
3 6 y
2
2 5

1
1 4

0 x
0 x 3
-3 -2 -1 0 1 2 3 4
-4 -3 -2 -1 0 1 2 3
-1 2 -1

-2 1 -2

-3 0 x -3
-1 0 1 2 3 4 5 6
g) -4 h) -1 i) -4

Review of part I (exercises starting on page 98):

Exercises I 1. x = 1 or x = 5, 2. y = − 21 x + 1, 3. x ≈ 2.52, 4. 2x − 2 + 2h, 5.
domain D = [2, 7], range R = (1, 4], f (3) = 2, f (5) = 2, f (7) = 4, f (9)
is undefined, 6. f (x) = −x2 + 2, 7. ( fg )(x) = x25x+4 5x+4
+8x+7 = (x+7)(x+1) has
2
√
domain D = R − {−7, −1}, 8. (f ◦ g)(x) = 4x − 12x + 9 + 2x − 6 has
domain D = [3, ∞), 9. f and g are both functions, and the composition
is given by the table:
x 2 3 4 5 6
f (x) 5 0 2 4 2
g(x) 6 2 3 4 1
(f ◦ g)(x) 2 5 0 2 undef.
1
10. f −1 (x) = 2x − 52
Session 8 (exercises starting on page 111):
3 4 10
Exercise 8.1 a) x2 − 2x − 2 − x−2 , b) x2 + 3x − 2 + x+3 , c) x + 6 − x+1 , d)
2 5 2 11 3 2 373
x + x + x+2 , e) 2x + 3x + 6 + x−1 , f) 2x − 3x + 15x − 74 + x+5 ,
3 2
g) 2x3 + 8x2 + x + 4 + x−4 , h) x2 − 3x + 9, i) x3 + 2x2 + x + 2 + 3x+1 ,
2 7x+6 3 2 9x+7
j) 4x + 3x + 6, k) x + 1 − x2 +2x+1 , l) x + 3x − 3x − 9 + x2 +3
Exercise 8.2 a) remainder r = 15, b) r = 20, c) r = −2, d) r = 12
Exercise 8.3 a) yes, g(x) is a factor of f (x), the root of f (x) is x = −3, b) g(x)
is not a factor of f (x), c) g(x) is a factor of f (x), the root of f (x) is
x = −7, d) g(x) is a factor of f (x), the root of f (x) is x = −1
Exercise 8.4 a) f (x) = (x−2)(x−1)(x+1), b) f (x) = (x−1)(x−2)(x−3), c) f (x) =
(x−3)(x−i)(x+i), d) f (x) = (x+2)3 , e) f (x) = (x+2)(x+4)(x+7), f)
f (x) = (x−4)(x+3)(x+4), g) f (x) = (x−2)(x−1)(x+1)(x+2)(x+5)
384 ANSWERS TO EXERCISES

25 18 15
Exercise 8.5 a) 2x2 + 7x + 9 + x−2 , b) 4x2 − 9x + 12 − x+3 , c) x2 + 2x − 7 + x+2 , d)
3 2 3 4 3 2 2 5
x + 2x + 2x + 2 + x−1 , e) x − 2x + 4x − 8x + 16, f) x − 3 + x+5 .

Session 9 (exercises starting on page 127):

Exercise 9.1 a) yes, b) no (due to the discontinuity), c) no (due to horizontal asymp-
tote), d) no (due to corner), e) yes (polynomial of degree 1), f) yes

Exercise 9.2 a) corresponds to (iii), b) corresponds to (v), c) corresponds to (vi), d)

corresponds to (ii), e) corresponds to (iv)
Exercise 9.3 a) corresponds to (iii), b) corresponds to (i), c) corresponds to (ii)

Exercise 9.4 The windows and graphs are displayed below for all parts (a)-(f).

a) b) c)

d) e) f)

Exercise 9.5 a) x = 2, x = 3, or x = 5, b) x = −1,

√ c) x = 11, d) x = −8, x = −3,
x = −2, or x = 4, e) x = −3 ± 6 (use the quadratic formula), f)
x=0
y y

−4 5 x

−2 0 4

Exercise 9.6 a) b)
 385
y y

−4 −3 −2 −1 2

3 5 7

c) d)

Session 10 (exercises starting on page 144):

Exercise 10.1 a) x = −1, x = 2, x = 21 , b) x = 31 , c) x = −32 , x = −1, x =
4
3, d)
√ √
x = 21 , x = −2
3 , x = −2 + 3, x = −2 − 3, e) x = − 14

Exercise 10.2 a) x = 1, b) x = 1 or x = −1, c) x = 3, d) x = −10, e) x = 3 or

x = −3, f) x = 5, g) x = −2, h) x = 1, i) x = 8i or x = −8i

Exercise 10.3 a) f (x) = (x − 2)(x − 1)(x + 3), b) f (x) = (x − 5)(x √ + 2)2 , c) f (x)
√
=
−3+ 5 −3− 5
(x− 1)(x+ 1)(x− 2)(x+ 2), d) f (x) = (x− 2)(x− 2 )(x− 2 ),
e) f (x) = 2(x + 32 )(x + 1)(x − 2), f) f (x) = 12(x − 32 )(x + 34 )(x + 4), g)
f (x) = (x − 1)(x + 1)(x − √ i)(x + i), h) f (x) = x(x − 1)(x + 1)(x − 3)2 ,
−3+3 3·i
√
−3−3 3·i
√
i) f (x) = (x − 3)(x − )(x − ), j) f (x) = (x − 3)(x +
√ √ √ 2 2
3)(x − 5 · i)(x + 5 · i)
y y

x x
√ √
−2 −4 −1− 5 −1+ 5
1 3 2 2

Exercise 10.4 a) y
b) y

x −1 x
√ √ 5 √ √
− 7 7 −3 − 2 −3 + 2

c) y
d) y

√ √
− 3 −1 1 3

e) f)
386 ANSWERS TO EXERCISES
y y

x x

−4 −2 2 5 −1 4 2
7

g) y
h) y

x x
√ √
− 53 1
3
3 2− 6 0 2 2+ 6

i) j)
Exercise 10.5 a) f (x) = 2(x−2)(x−3)(x−4), b) f (x) = (−1)·x(x−2)(x+1)(x+3), c)
f (x) = (− 25 )·(x−2)(x+2)(x+1), d) f (x) = −2·x(x−2)(x+1)(x+4),
e) f (x) = 3(x − 7)(x − (2 + 5i))(x − (2 − 5i)), f) f (x) = (−2) · (x −
i)(x + i)(x − 3), g) f (x) = 74 · (x − (5 + i))(x − (5 − i))(x − 3)2 ,
h) f (x) = (x − i)(x + i)(x − (3 + 2i))(x − (3 − 2i)) (other correct
answers are possible, depending on the choice of the first coefficient),
i) f (x) = (x − (1 + i))(x − (2 + i))(x − (4 − 3i))(x + 2)3 (other correct
answers are possible, depending on the choice of the first coefficient),
j) f (x) = (x − i)(x − 3)(x + 7)2 (other correct answers are possible,
depending on the choice of the first coefficient and the fourth root),
k) f (x) = (x − 2)(x − 3)(x − 4) (other correct answers are possible,
depending on the choice of the first coefficient), l) f (x) = (x − 1)2 (x −
3)2 , m) f (x) = −x(x − 1)(x − 3)(x − 4) (other correct answers are
possible, depending on the choice of the first coefficient)
Session 11 (exercises starting on page 168):
Exercise 11.1 a) domain D = R − {2}, vertical asymptote at x = 2, no removable
discontinuities, b) D = R−{2, 4}, vertical asympt. at x = 2 and x = 4,
no removable discont., c) D = R − {−2, 0, 2}, vertical asympt. at x = 0
and x = 2, removable discont. at x = −2, d) D = R − {−3, 2, 5},
vertical asympt. at x = 2 and x = 5, removable discont. at x = −3,
e) D = R − {1}, no vertical asympt., removable discont. at x = 1, f)
D = R − {−1, 1, 2}, vertical asympt. at x = −1 and x = 1 and x = 2,
no removable discont.
Exercise 11.2 a) y = 4, b) y = 0, c) no horizontal asymptote (asymptotic behavior
y = x + 4), d) y = −4
Exercise 11.3 a) x-intercept at x = 3, y-intercept at y = 3, b) x-intercepts at x = 0
and x = −2 and x = 2, y-intercept at y = 0, c) x-intercepts at x = −4
and x = 1 and x = 3, y-intercept at y = 56 , d) x-intercept at x = −3
(but not at x = −2 since f (−2) is undefined), no y-intercept since
f (0) is undefined
 387

Exercise 11.4 a) D = R − {−2}, horizontal asympt. y = 3, vertical asympt. x = −2,

no removable discont., x-intercept at x = 31 , y-intercept at y = −1
2 ,
graph:

−10 ≤ x ≤ 10, 8 y

−10 ≤ y ≤ 10 : 7

−2 ≤ x ≤ 2, 1

−2 ≤ y ≤ 2 : 0
1
3 x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
−1
-1 2

-2

-3

-4

x−3
b) f (x) = (x−4)(x−1)(x+2) has domain D = R − {−2, 1, 4}, horizontal
asympt. y = 0, vertical asympt. x = −2 and x = 1 and x = 4, no
removable discont., x-intercept at x = 3, y-intercept at y = −38 =
−0.375, graph:

4 y

2
−6 ≤ x ≤ 6,
−2 ≤ y ≤ 2 : 1

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3

-4

c) f (x) = (x−3)(x+3)(x−1)(x+1)
(x−2)(x−1) has domain D = R − {1, 2}, no hori-
zontal asympt., vertical asympt. x = 2, removable discont. at x = 1, x-
intercept at x = −3 and x = −1 and x = 3, y-intercept at y = 29 = 4.5,
graph:
388 ANSWERS TO EXERCISES
18 y

17

−10 ≤ x ≤ 10, 16

15

−10 ≤ y ≤ 10 : 14

13

12

11

10

0 x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

d) f (x) = (x−3)(x−1)(x+1)
x2 (x−2) has domain D = R − {0, 2}, horizontal
asympt. y = 1, vertical asympt. x = 0 and x = 2, no removable
discont., x-intercepts at x = −1 and x = 1 and x = 3, no y-intercept
since f (0) is undefined, graph:
4 y
−4 ≤ x ≤ 6,
−5 ≤ y ≤ 5 : 3

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3

-4

Note that the graph intersects the horizontal asymptote y = 1 at

approximately x ≈ −2.3 and approaches the asymptote from above.
1 5x2
Exercise 11.5 a) for example f (x) = x−4 , b) for example f (x) = x2 −5x+6 , c) for
x2 −x
example f (x) = x−1
Session 12 (exercises starting on page 179):
Exercise 12.1 a) x ≤ 3, b) 21 > x, c) −4 ≥ x, d) x > 22 5 , e) −5 ≤ x ≤ 7, f)
−4 < x ≤ 3, g) x ≥ 2 (this then also implies x ≥ − 72 ), h) no solution
Exercise 12.2 a) (−∞,
√
−2) √∪ (7, ∞), b) (−∞, −5] ∪ [7, ∞), c) [−2, 2], d)
−3− 21 −3+ 21



2 , 2 , e) [−3, 2], f) − 13 , 1 , g) R−{2}, h) [−2, 1]∪[3, ∞),
i) (−∞, −4), j) (0, 2) ∪ (2, ∞), k) [−3, −1] ∪ [1, 3], l) (−1, 1) ∪ (2, 3), m)
(−∞, 0) ∪ (0, 2) ∪ (3, ∞), n) (−∞, −2] ∪ [0, 1] ∪ [2, 5], o) [1, 2] ∪ [3, 4] ∪
[5, ∞), p) (−∞, 1]
Exercise 12.3 a) D = (−∞, 3] ∪ [5, ∞), b) D = (−∞, −3] ∪ [0, 3], c) D = [1, 4], d)
D = [2, 5] ∪ [6, ∞), e) D = (−∞, 3), f) D = (−∞, −1) ∪ (7, ∞)
 389

Exercise 12.4 a) (2, 5), b) (−2, 1] ∪ (2, ∞), c) (−∞, −1) ∪ (2, 5), d) (−∞, −3] ∪
(−2, 2) ∪ [3, ∞), e) (−∞, −5] ∪ (−3, ∞), f) (−10, −9.8), g) (−1, 2) ∪
[4, ∞), h) (−∞, −4) ∪ [0, ∞)
1 1
Exercise 12.5 a) (−∞, −8) ∪ (1, ∞) , b) ( −5 6 , 6 ), c) (−∞, 3 ] ∪ [3, ∞), d) [−12, −2],
1 1
e) (−∞, − 4 ] ∪ [ 2 , ∞), f) (−15, −5)
Review of part II (exercises starting on page 181):
1
Exercises II 1. x2 − x − 3 + 2x+3 , 2. 21, 3. x − 1 is a factor, x + 1 is not a factor,
x − 0 is not a factor, 4. a) ↔ iii), b) ↔ iv), c) ↔ i), d) ↔ ii),

5.

6. f (x) = (x − 1)(x + 3)(x + 4), 7. f (x) = (−5) · x(x − 1)(x − 3), 8.

f (x) = (x + 2)(x − 5)(x − (3 − 2i))(x − (3 + 2i)) (other correct answers
are possible, depending on the choice of the first coefficient),
9. f (x) = 3(x−2)(x+2)
(x−3)(x+1) has domain D = R−{−1, 3}, horizontal asympt.
y = 3, vertical asympt. x = −1 and x = 3, no removable discont., x-
intercepts at x = −2 and x = 2 and x = 3, y-intercept at y = 4,
graph: 7 y

0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1

-2

-3
√   √

−3− 37
10. a) (−1, 0)∪(1, 2), b) −∞, 2 ∪ −3+2 37 , ∞ , c) (−∞, −7]∪
(−4, ∞)
390 ANSWERS TO EXERCISES

Session 13 (exercises starting on page 197):

−200 ≤ x ≤ 200 :

Exercise 13.1 a) , b) , c) ,

−200 ≤ x ≤ 200 :

d) , e) same as c) since y = ( 13 )x = 3−x ,

y = ( 13 )−x = 3x : −1 ≤ x ≤ 1 :

f) , g) , h) ,

y = 1x = 1 :

i) , j) , k) ,

−2 ≤ y ≤ 2 :

l)

Exercise 13.2 a) y = 4x is compressed towards the x-axis by the factor 0.1 (graph
below), b) y = 2x stretched away from x-axis, c) y = 2x reflected
about the x-axis, d) y = 2x compressed towards the x-axis, e) y = ex
reflected about the y-axis, f) y = ex reflected about the y-axis and
shifted up by 1, g) y = ( 21 )x shifted up by 3, h) y = 2x shifted to the
right by 4, i) y = 2x shifted to the left by 1 and down by 6
 391
5 y 5 y 1 y

4 4 0 x
-4 -3 -2 -1 0 1 2
3 3 -1

2 2 -2

1 1 -3

0 x 0 x -4
-3 -2 -1 0 1 2 3 -4 -3 -2 -1 0 1 2
a) -1 b) -1 c) -5

0.05 y 5 y 5 y

0.04 4 4

0.03 3 3

0.02 2 2

0.01 1 1

0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
d) -1 e) -1 f) -1

6 y 6 y 0 y x
-4 -3 -2 -1 0 1 2
5 5 -1

4 4 -2

3 3 -3

2 2 -4

1 1 -5

g) -3 -2 -1
0
0 1 2
x
h)0 0
3 1 2 3 4 5
x
i)
6
-6

Exercise 13.3 a) log4 (16) = 2, b) log2 (256) = 8, c) ln(7) = x, d) log(0.1) = −1, e)

log3 (12) = x, f) log5 (12) = 7x, g) log3 (44) = 2a + 1, h) log 21 (30) = xh

Exercise 13.4 a) 2, b) 4, c) 6, d) 2, e) −2, f) 3, g) 4, h) 1, i) −1, j) −2, k) 0, l) −3

Exercise 13.5 a) 3.561, b) 2.262, c) −0.290, d) −4.911

2 y D = (0, ∞) D = (−7, ∞) 2 y

1 1

0 x 0 x
-2 -1 0 1 2 3 4 5 6 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2
Exercise 13.6 a) -1 b) -1

-2 -2

-3 -3

-4 -4
392 ANSWERS TO EXERCISES

D = (−5, ∞) 2 y 3 y D = (2, ∞) D = (−4, ∞)3 y

1 2 2

0x 1 1
-6 -5 -4 -3 -2 -1 0
c) -1 d)0 x e) 0 x
0 1 2 3 4 5 6 -5 -4 -3 -2 -1 0 1
-2 -1 -1

-3 -2 -2

-4 -3 -3

D
2 y= (−2, ∞) 2 y D = (5, ∞)
1 1

0 x 0 x
-3 -2 -1 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9
f) -1 g)-1
-2 -2

-3 -3

-4 -4

6 y D = (1, ∞) 3 y D = (0, ∞)
5 2

4 1

h)3 i) 0 x
-2 -1 0 1 2 3 4 5 6 7
2 -1

1 -2

0 x -3
0 1 2 3 4 5 6 7 8 9
D = (−∞, 0)1 y
R −2 {0}
y
R − {−2} 2 y

0 x 1 1
-5 -4 -3 -2 -1 0 1
-1 0 x 0) x
-3 -2 -1 0 1 2 3 -5 -4 -3 -2 -1 0 1
j) -2 k) -1 l) -1

-3 -2 -2

-4 -3 -3

-5 -4 -4

Session 14 (exercises starting onpage 206):

  √   
3 xz 4
Exercise 14.1 a) ln(x3 · y), b) log x2 = log √ 3
x
2
, c) log y , d) log xz3 y ,
y
√ √ 
4 x 3 z2
y 3 

1 10
e) ln √
y , f) ln x+1 , g) ln(x ), h) log5 25 · (a + 1)
Exercise 14.2 a) 3u + v, b) 23 u + 74 v, c) 21 u + 16 v, d) 3u − 4v, e) 2u − 12 v − 2w, f)
1 3 1 3 1 1 1 4 2
2 u + 2 v − 4 w, g) 4 u − 3v + 4 w, h) 2 − 2v + 5 w, i) 6 v + 3 w − 3
6
Exercise 14.3 a) x = 4, b) x = 4, c) x = 9, d) x = −2, e) x = − 5 , f) x = 2, g)
x = −2, h) x = 11 10
2 , i) x = 17 , j) x = 6
7
5
Exercise 14.4 a) x = 3, b) x = 6, c) x = 31 , d) x = 76, e) x = 1, f) x = 8, g) x = 18,
h) x = 25, i) x = 2, j) x = −1
 393

log 57 log 7 log 31

Exercise 14.5 a) x = log 4 ≈ 2.92, b) x = log 9 + 2 ≈ 2.89, c) x = log 2 − 1 ≈ 3.94,
d) x = log(63)−7 log(3.8)
2 log(3.8)
5·log(5)
≈ −1.95, e) x = log(8)−log(5) ≈ 17.12, f)
2·log(3) 9 log(1.02)
x = log(0.4)−log(3) ≈ −1.09, g) x = 2 log(1.02)−log(4.35) ≈ −0.12, h)
x = log(4)−2 log(5)
log(5)−log(4) ≈ −8.21, i) x =
3 log(9)+6 log(4)
log(9)+log(4) ≈ 4.16, j) x =
7 log(2.4)−4 log(3.8) 4 log(9)−2 log(4)
2 log(2.4)+3 log(3.8) ≈ 0.14, k) x = 2 log(9)−9 log(4) ≈ −0.74, l) x =
4 log(1.2)+4 log(1.95)
7 log(1.2)−3 log(1.95) ≈ −4.68
Session 15 (exercises starting on page 214):
Exercise 15.1 a) f (x) = 4·3x, b) f (x) = 5·2x, √
c) f (x) = 3200·0.1x, d) f (x) = 1.5·2x,
x 3 x
e) f (x) = 20 · 5 , f) f (x) = 5 · 5
Exercise 15.2 a) y = 8.4·1.101t with t = 0 corresponding to the year 2005, b) approx.
22.0 million, c) It will reach 20 million in the year 2014.
Exercise 15.3 a) y = 79000 · 1.037t with t = 0 corresponding to the year 1990, b)
approx. 163, 400, c) approx. 195, 900, d) The city will reach maximum
capacity in the year 2061.
Exercise 15.4 The city will be at 90% of its current size after approximately 4.5 years.
Exercise 15.5 It will take the company 4.87 years.
Exercise 15.6 The ant colony has doubled its population after approximately 23.4
weeks.
Exercise 15.7 It will take 4.27 months for the beehive to have decreased to half its
current size.
Exercise 15.8 It will take 139.0 years until the world population has doubled.
Session 16 (exercises starting on page 226):
Exercise 16.1 It takes 49.262 minutes until 2 mg are left of the element.
Exercise 16.2 2.29 grams are left after 1 year.
Exercise 16.3 The half-life of fermium-252 is 25.38 minutes.
Exercise 16.4 You have to wait approximately 101.3 days.
Exercise 16.5 67.8% of the carbon-14 is left in the year 2000.
Exercise 16.6 The wood is approximately 3323 years old.
Exercise 16.7 The bone is approximately 3952 years old.
Exercise 16.8 a) $7, 346.64, b) It takes approximately 18 years.
Exercise 16.9 a) $862.90, b) $1, 564.75, c) $1, 566.70, d) $541.46, e) $6, 242.86, f)
$1, 654.22, g) $910.24
Exercise 16.10 a) P = $1, 484.39, b) P = $2, 938.67, c) P = $709.64, d) r = 4.23%,
e) r = 4.31%, f) t ≈ 1.69 years, g) t ≈ 3.81 years, h) t ≈ 10.27 years,
i) t ≈ 13.73 years
Review of part III (exercises starting on page 229):
√
√ 3 
xz
x2 y 4 , b) log2 √
3
Exercises III 1. 37.5 million, 2. 19.1 hours, 3. a) ln 4 3
, 4. a)
y
4
3u − 2v, b) u + 52 v, c) 51 u + 54 v, 5. x = 9, 6. a) x = log27−log
log 6
6 , b)
x ≈ 23.25, 7. 1.82 hours, 8. 3561 years, 9. $1, 828.48, 10. r = 6.5%
394 ANSWERS TO EXERCISES

Session 17 (exercises starting on page 249):

√ √
Exercise 17.1 a) sin(120◦ ) = 23 , cos(120◦ ) = − 21 , tan(120◦ ) = − 3, b)
√ √
sin(390◦ ) = 12 , cos(390◦ ) = 23 , tan(390◦ ) = 33 , c) sin(−150◦ ) = − 21 ,
√ √ √
cos(−150◦ ) = − 23 , tan(−150◦ ) = 33 , d) sin(−45◦ ) = − 22 ,
√
cos(−45◦ ) = 22 , tan(−45◦ ) = −1, e) sin(1050◦) = − 12 , cos(1050◦) =
√ √
3 3
2 , tan(1050 ) = − 3 , f) sin(−810 ) = √ −1, cos(−810◦ ) = 0,
◦ ◦
√
tan(−810 ) is undefined, g) sin( 4 ) = − 22 , cos( 5π
◦ 5π
4 ) = − 22 ,
√ √
tan( 5π4 ) = 1, √ h) sin( 5π 1 5π
6 ) = 2 , cos( 6 ) = − 2 √
3
, tan( 5π 3
6 ) = − 3 , i)
sin( 10π 3 10π 1
3 ) = − 2 , cos( 3 ) = − 2 , tan( 3 ) =
10π
3, j) sin( 15π
2 ) = −1,
cos( 2 ) = 0, tan( 2 ) is undefined, k) sin( 6 ) = − 21 , cos( −π
15π 15π −π
6 )√=
√ √ √
3 3 2
2 , tan( −π
6 ) = − 3 , l) sin( −54π
8 ) = − 2 , cos( −54π
8 ) = − 22 ,
tan( 8 ) = 1
−54π

Exercise 17.2 a) shift y = sin(x) up by 2 (see graph below), b) y = cos(x) shifted to

the right by π, c) y = tan(x) shifted down by 4, d) y = sin(x) stretched
away from the x-axis by a factor 5, e) y = cos(x) compressed towards
the y-axis by a factor 2, f) y = sin(x) shifted to the right by 2 and
down by 5
4 y 2 y

3 1

2 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5
1 -1

a) -5 -4 -3 -2 -1
0
0 1 2 3 4
x
b)
5
-2
3 y 5 y

2 4

1 3

0 x 2
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1 1

-2 0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-3 -1

-4 -2

-5 -3

-6 -4

c) -7 d) -5

3 y 0 y x
-5 -4 -3 -2 -1 0 1 2 3 4 5
2 -1

1 -2

0 x -3
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1 -4

-2 -5

e) -3 f) -6

Exercise 17.3 a) g(x), b) h(x), c) j(x), d) k(x), e) i(x), f) f (x)

Exercise 17.4 a) y = 5 cos(x), b) y = −5 cos(x), c) y = −5 sin(x), d) y = cos(x) + 5,
e) y = sin(x) + 5, f) y = 2 sin(x) + 3
 395

Exercise 17.5 a) amplitude 5, period π, phase-shift −3 2 , b) amplitude 1, period 2,

phase-shift π5 , c) amplitude 6, period π2 , phase-shift 0, d) amplitude
2, period 2π, phase-shift −π4 , e) amplitude 8, period π, phase-shift 3,
f) amplitude 3, period 8π, phase-shift 0, g) amplitude 1, period 2π,
phase-shift −2, h) amplitude 7, period 5, phase-shift 3, i) amplitude 1,
period π, phase-shift 0

Exercise 17.6 a) amplitude 5, period π, phase-shift 0 (graph see below), b) ampli-

tude 4, period 2, phase-shift 0, c) amplitude 2, period 3, phase-shift
0, d) amplitude 1, period π, phase-shift π2 , e) amplitude 1, period 2,
phase-shift 1, f) amplitude 6, period 8π, phase-shift 0, g) amplitude 1,
period π2 , phase-shift −π
4 , h) amplitude 7, period 2π, phase-shift 4 ,
−π

i) amplitude 5, period 2π, phase-shift 2 , j) amplitude 4, period 2π

−3π
5 ,
phase-shift π5 , k) amplitude 3, period 1, phase-shift π2 , l) amplitude 7,
period 8π, phase-shift −π, m) amplitude 1, period 2π 4π
3 , phase-shift 3 ,
n) amplitude 2, period 10π, phase-shift 2 , o) amplitude 3 , period 5π
π 1
7 ,
phase-shift 3π7
y y

5 4
π
2 1.5
x x

π 3π
4 4
π 0.5 1 2
−5 −4

a) b)

y y

2 1
2.25 π
x x

π 3π 5π 3π
0.75 1.5 3 2 4 4 2
−2 −1

c) d)

y y y

1
1 6
−π π
2 8π 8 8 π
x x −π
4 4 x

3 5
1 2 2 3 2π 4π 6π
−1 −6
−1
e) f) g)

y y

7 5
5π −π
4 2
x x

π 3π 7π π
−π
4 4 4 4
−3π
2 −π 2
−7 −5

h) i)
396 ANSWERS TO EXERCISES
y y
2 1 2 3
π + 4 π + 4
4 3
π 2 2
2 π π +1
x x

π 3π 2π 3π 2 1
5 10 5 5 π + 2
−4 −3

j) k)

y y

7 1
5π
5π 3
x x

4π 3π 11π
−π π 3π 7π 3 2 6 2π
−7 −1

l) m)

y y

1
2 3
11π
8π 14
x x

π 11π 21π 3π 17π 27π 8π

2 3π 2 2 7 28 28 7
−2 − 31

n) o)

Session 18 (exercises starting on page 261):

√ √ √ √ √ √ √ √ √ √
Exercise 18.1 a) 2+ 6
4 , b) 6− 4
2
, c) 2 − 3, d) 2+ 4
6
, e) −( 2+ 4
6)
, f) 23 , g) 21 ,
√
− 2
√ √ √ √ √ √ √
h) 2 , i) 2 − 3, j) 2+ 4
6
, k) 6− 4
2
, l) 2+ 4
6

√
Exercise 18.2 a) sin(x + π2 ) = cos(x), b) cos x − π4 = 22 (sin(x) + cos(x)),

√
c) tan (π − x) = − tan(x), d) sin π6 − x = 21 cos(x) − 23 sin(x),

√ √ √ √
6)


e) cos x + 11π 12 = −( 2+ 4 · cos(x) − 6− 2
4 · sin(x), f)

−1 √
2π 3
cos 3 − x = 2 · cos(x) + 2 · sin(x)
√ √ √ √ √ √ √ √ √ √ √
Exercise 18.3 a) 2+ 3
= 2+ 6
, b) 2 − 1, c) 8−2 2−2 6
, d) 8+2 2+2 6
, e)
√ 2√ 4
√ √ √
q √ 4
√
4
2+ 2+ 3
− 2− 2
2
, f) 1 + 2, g) − 2− 2
2
, h) 2
√ √
Exercise 18.4 α 5 α 2 5
a) sin( 2 ) = 5 , cos( 2 ) = 5 , tan( α2 ) = 21 , sin(2α) = 24 ,
√ √ 25
α 39 α − 130
cos(2α) = 25 , tan(2α) = 7 , b) sin( 2 ) = 13 , cos( 2 ) = 13 ,
−7 −24
√ √ √
tan( α2 ) = −1030 , sin(2α) = −28 169 √
30
, cos(2α) = −71
169 , tan(2α) = 28 30
71 ,
√
α 3 10 α − 10
c) sin( 2 ) = 10 , cos( 2 ) = 10 , tan( 2 ) = −3, sin(2α) = 24 α
25 ,
√ √
7 2 5 − 5
cos(2α) = 25 , tan(2α) = 24 7 , d) sin( α
2 ) = 5 , cos( α
2 ) = 5 ,
24
tan( α2 ) = −2, sin(2α) = 25 , cos(2α) = −7 25 , tan(2α) = −24
7 , e)
√ √
α 5 26 α 26 α
sin( 2 ) = 26 , cos( 2 ) = 26 , tan( 2 ) = 5, sin(2α) = 169 , −120
√ √
cos(2α) = 119 , tan(2α) = −120
, f) sin( α
) = 30
, cos( α
) = 6
6 ,
α
√ 169 √ 119
−4 5
2 6 2
√
tan( 2 ) = 5, sin(2α) = 9 , cos(2α) = 9 , tan(2α) = 4 5 −1

Session 19 (exercises starting on page 269):

 397

−2 ≤ x ≤ 2 −2 ≤ x ≤ 2 −10 ≤ x ≤ 10
−2 ≤ y ≤ 2 −1 ≤ y ≤ 4 −2 ≤ y ≤ 2

Exercise 19.1 a) , b) , c)

3π 5π
Exercise 19.2 a) π3 , b) π
6, c) π
3, d) 0, e) π
4, f) 4 , g) − π2 , h) − π3 , i) 6 , j) − π4 , k) − π3 ,
l) − π6
Exercise 19.3 a) 1.37, b) −0.85, c) 1.23, d) 1.57, e) −1.11, f) undefined, g) 47.16◦ , h)
−45◦, i) 75◦ , j) 90.00◦, k) 67.5◦, l) −7.5◦
Session 20 (exercises starting on page 283):
Exercise 20.1 a) x = π6 + nπ, where n = 0, ±1, . . . , b) x = (−1)n π3 + nπ, where
n = 0, ±1, . . . , c) x = (−1)n+1 π4 + nπ, where n = 0, ±1, . . . , d)
x = ± π6 + 2nπ, where n = 0, ±1, . . . , e) x = ± π2 + 2nπ, where
n = 0, ±1, . . . , f) x = ± 2π
3 + 2nπ, where n = 0, ±1, . . . , g) x = 2nπ,
where n = 0, ±1, . . . , h) there is no solution (since −1 ≤ sin(x) ≤
1), i) x = nπ, where n = 0, ±1, . . . , j) x = (−1)n+1 π2 + nπ, where
n = 0, ±1, . . . , (since each solution appears twice, it is enough to
take n = 0, ±2, ±4, . . . ), k) x = −π3 + nπ, where n = 0, ±1, . . . , l)
x = ± cos−1 (0.2) + 2nπ, where n = 0, ±1, . . .
Exercise 20.2 a) x ≈ 1.411 + nπ, where n = 0, ±1, . . . , b) x ≈ ±1.104 + 2nπ, where
n = 0, ±1, . . . , c) x ≈ (−1)n 1.143 + nπ, where n = 0, ±1, . . . , d)
x ≈ ±2.453 + 2nπ, where n = 0, ±1, . . . , e) x ≈ −0.197 + nπ, where
n = 0, ±1, . . . f) x ≈ (−1)n+1 0.06 + nπ, where n = 0, ±1, . . . ,
−π 3π 7π −5π −9π π −π 9π −9π 17π −17π
Exercise 20.3 a) 4 , 4 , 4 , 4 , 4 , b) 4 , 4 , 4 , 4 , 4 , 4 , c) 3 ,
−π
4π 5π −2π −7π π −π 3π −3π 5π
3 , 3 , 3 , 3 , d) 0, π, 2π, −π, −2π, e) 2 , 2 , 2 , 2 , 2 ,
−5π
2 , f) cos−1
(0.3), − cos−1
(0.3), cos−1
(0.3) + 2π, − cos−1
(0.3) + 2π,
cos (0.3) − 2π, − cos (0.3) − 2π, g) sin (0.4), − sin (0.4) + π,
−1 −1 −1 −1

− sin−1 (0.4) − π, sin−1 (0.4) + 2π, sin−1 (0.4) − 2π, h) 3π 7π 11π −π

2 , 2 , 2 , 2 ,
−5π
2
Exercise 20.4 a) x = π4 + nπ, where n = 0, ±1, . . . , b) x = (−1)n π6 + nπ, where n =
0, ±1, . . . , c) x = ± π6 + 2nπ, where n = 0, ±1, . . . , d) x = ± 2π
3 + 2nπ,
where n = 0, ±1, . . . , e) x = π6 + nπ, where n = 0, ±1, . . . , f) x =
± π3 + nπ, where n = 0, ±1, . . . , g) x = ± π2 + nπ, where n = 0, ±1, . . . ,
h) x = π + 2nπ, where n = 0, ±1, . . . (Note: The solution given by
the formula from page 277 is x = ±π + 2nπ with n = 0, ±1, . . . .
Since every solution appears twice in this expression, we can reduce
this to x = π + 2nπ.), i) x = ± π3 + 2nπ, where n = 0, ±1, . . . , j)
x = (−1)n+1 π6 + nπ, where n = 0, ±1, . . . , k) x = (−1)n+1 π2 + nπ, or
x = (−1)n π6 + nπ, where n = 0, ±1, . . . , l) x = 2nπ, or x = ± π3 + 2nπ,
where n = 0, ±1, . . . , m) x = ± π3 + nπ, where n = 0, ±1, . . . , n)
x = ± π4 + nπ, or x = nπ, where n = 0, ±1, . . . ,
398 ANSWERS TO EXERCISES

Exercise 20.5 a) x ≈ −1.995 + 2nπ, or x ≈ 0.424 + 2nπ, where n = 0, ±1, . . . , b)

x ≈ −0.848 + nπ, or x ≈ 0.148 + nπ, or x ≈ 0.700 + nπ, where n =
0, ±1, . . . , c) x ≈ 0.262 + n 2π 2π 2π
3 , or x ≈ 0.906 + n 3 , or x ≈ 1.309 + n 3 ,
2π
or x ≈ 1.712 + n 3 , where n = 0, ±1, . . . , d) x ≈ 0.443 + 2nπ, or
x ≈ 2.193 + 2nπ, where n = 0, ±1, . . .
Review of part IV (exercises starting on page 284):
Exercises IV 1.
π π π π
x 0 = 0◦ 6 = 30◦ 4 = 45◦ 3 = 60◦ 2 = 90◦
√ √
1 2 3
sin(x) 0 1
√2 √2 2
3 2 1
cos(x) 1 0
√2 2
√2
3
tan(x) 0 3 1 3 undef.

√ √ √ √ √
2. a) 23 , b) −2 2 , c) − 3, 3. a) −2 2 , b) 23 , 4. a) amplitude 3, period
π π
2 , phase-shift 4 , b) amplitude 5, period 2π, phase-shift 2 (graphs
−π

below:)
y y

3 5
π
2
x x

π 3π 5π 3π π 3π
4 8 8 4
−π
2 2 π 2
−3
−5

a) b)

√ √ √ √
2− 2
5. a) 2+ 4
6
, b) 2
24
6. sin(2α) = 25 , cos(2α) = −725 , tan(2α) =
5π n+1 π
−24
7 , 7. a) −π
6 , b) 6 , c) −π
6 , 8. x = (−1) 3 + nπ, where n =
0, ±1, . . . , 9. x = ± π4 +nπ, where n = 0, ±1, . . . , 10. a) x = ± π4 +2nπ,
or x = ± 3π 4 + 2nπ, where n = 0, ±1, . . . , b) (−1)
n+1 π
6 + nπ, where
n = 0, ±1, . . .
Session 21 (exercises starting on page 297):
Exercise 21.1 3
Im
a)
2
d)
1 h)
g)
0 i) Re
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-1
f)
j)
-2 e)
c)
-3

-4

-5
b)
-6
 399

Exercise 21.2 a) 3 + 4i, b) −14 + 2i, c) 6 + 17i, d) 6 − 7i, e) 57 + 54 i, f) − 21 + 23 i

√ √
Exercise 21.3 a) 5, b) 5, c) 3, d) 2 10, e) 3, f) 4, g) 5, h) 7
√
Exercise 21.4 a) 2 2(cos( π4 ) + i sin( π4 ), b) 8(cos( π6 ) + i sin( π6 )), c) approximately
√ √
13(cos(−.588)
√ + i sin(−.588)) or 13(cos(−.187π) + i sin(−.187π)),
d) 5 2(cos( 3π 3π
4 ) + i sin( 4 )), e) approximately 5(cos(−.644) +
i sin(−.644)) or 5(cos(−.205π) + i sin(−.205π)), f) approximately
5(cos(2.498)
√ + i sin(2.498)) or √ 5(cos(.795π) + i sin(.795π)), g)
2 5(cos( 4π 4π π π
3 ) + i sin( 3 )), h) 2 7(cos(− 3 ) + i sin(− 3 )) i) approxi-
mately 13(cos(4.318) + i sin(4.318)) or 13(cos(1.374π) +√ i sin(1.374π))
j) 6(cos( π2 ) + i sin( π2 )), k) 10(cos(π) + i sin(π)), l) 2 3(cos( 2π 3 ) +
i sin( 2π
3 ))
Exercise 21.5 a) approximately −4.168 + 4.316i, b) approximately .491 + 0.0919i, c)
√
3 1
√
−2i, d) 2 + 2 i, e)−5 3 − 5i, f) approximately 1.553 − 5/796i
√
Exercise 21.6 a) 40(cos(60◦ ) + i sin(60◦ )) = 20 + 20 3i, b) 42(cos( π3 ) + i sin( π3 ) =
√ √
21 + 21 3i, c) cos( π6 ) + i sin( π6 ) = 23 + 12 i, d) 12(cos(π) + i sin(π)) =
−12, e) .1(cos(284◦) + i sin(284◦ )) ≈ .0242 − .0970i, f) cos( 2π 3 ) +
√
2π 1 3
i sin( 3 ) = − 2 + 2 i
√
Exercise 21.7 a) 6(cos(π/3)+i sin(π/3)) = 3+3 3i, b) 32 (cos(90◦ )+i sin(90◦ )) = 23 i,
c) 2(cos(π/2) + i sin(π/2)) = 2i, d) 21 (cos( 3π 3π 1
2 ) + i sin( 2 )) = − 2 i, e)
4π 4π
√ 5
6(cos( 3 )+i sin( 3 )) = −3−3 3i, f) 3 (cos(−319 )+i sin(−319 )) ≈
◦ ◦

1.258 + 1.093i
Session 22 (exercises starting on page 309):
Exercise 22.1 8 y

7 f)
a)
6

4
d)

3
b)
2

0 x
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1

-2

-3
e)
-4 c)

-5
400 ANSWERS TO EXERCISES
√ √ √ √
Exercise 22.2 a) 10, 53◦, b) 29, 112◦ , c) 4 2, 225◦, d) 3 2, −45◦√ , e) 2 2, −45◦ , f)
8, 30◦ , g) 2, 210◦, h) 8, 120◦ , i) 4, 210◦, j) 5, 37◦ , k) 9 2, 135◦
Exercise 22.3 a) h15, 10i, b)h−2, 8i, c) h15, 14i, d) h8, 4i, e) h13, 5i, f) h23, 41i, g)
h16, 20i, h) h−16, 25i, i) h− 32 , − 25
6 i, j) h6, −4i, k) √h−2, 1i, l) h0, −2i, m)
√
h43, 12 + 7 3i, n) h−5, −10i, o) h−18, 20i, p) h8 5, −10i
√ √ √ √ √
Exercise 22.4 a) h 45 , − 35 i, b) h− 34 , − 47 i, c) h 9 8585 , 2 8585 i, d) h− 65 , 631 i, e)
√ √
h 5 7070 , 3 1414 i, f) h0, −1i
√ √
Exercise 22.5 a) ~v = h1, 3 3i, k~v k = 2√ 7, θ ≈ √ 79 b) ~v ≈ h−.772,◦1.594i,
◦
k~vk ≈
7.63, θ ≈ 116 c) ~v = h−4 2, −4 2i, kvk = 8, θ = 225 = 5π
◦
4
Session 23 (exercises starting on page 323):
Exercise 23.1 a) 3, 6, 9, 12, 15, 18, 21, b) 8, 13, 18, 23, 28, 33, 38, c) 3, 6, 11, 18, 27,
√ √3 2
38, 51, d) 1, 2, 3, 4, 5, 6, 7, e) 1, −1, 1, −1, 1, −1, 1, f) 2, 2 , 3 ,
√ √ √ √
5 6 7 8
4 , 5 , 6 , 7 , g) 10, 100, 1000, 10000, 100000, 1000000, 10000000,
h) 4, 6, 4, 6, 4, 6, 4
Exercise 23.2 a) 5, 8, 11, 14, 17, b) 7, 70, 700, 7000, 70000, c) 1, 3, 7, 15, 31, d) 6,
4, −2, −6, −4,
77
Exercise 23.3 a) 50, b) 15, c) 30, d) −3, e) 26, f) 60
Exercise 23.4 For the convenience of those who prefer to use an = a + b · n as
standard form we have provided answers also in that form.
a) 5 + 3(n − 1) = 2 + 3n, b) −10 + 3(n − 1) = −13 + 3n c) no, d) no, e)
73.4 − 21.7(n − 1) = 95.1 − 21.7n, f) no, g) 4 + 0 · (n − 1) = 4 + 0 · n, h)
−2.72 − .1(n − 1) = −2.62 − .1n, i) no, j) − 35 + 12 (n − 1) = − 10 11
+ 12 n,
k) 9 + 5(n − 1) = 4 + 5n, l) −3 + 2(j − 1) = −5 + 2j, m) no, n)
29 + 16(k − 1) = 13 + 16k
Exercise 23.5 a) 57 + 4(n − 8) = 29 + 4(n − 1) = 25 + 4n, b) −70 − 3(n − 99) =
224 − 3(n − 1) = 227 − 3n, c) 14 − 5(n − 1) = 19 − 5n, d) −80 +
76(n − 1) = −156 + 76n, e) 10 − 3(n − 3) = 16 − 3(n − 1) = 19 − 3n,
f)2 + 43 (n − 20) = −49/4 + 43 (n − 1) = −13 + 43 n
Exercise 23.6 a) 116, b) 187, c) − 3621
8 , d) 71
Exercise 23.7 a) 5, 040, b) −1, 113, c) 49, 599, d) −21, 900, e) 10, 100, f) −11, 537, g)
123, 150, h) 424, i) −1762.2, j) 302, 232, k) 200
Session 24 (exercises starting on page 334):
Exercise 24.1 a) geometric, 7·2n−1 , b) geometric, 3·(−10)n−1, c) geometric, 81( 31 )n−1 ,
d) arithmetic, −7+2(n−1) = −9+2n, e) geometric, −6(− 31 )n−1 , f) ge-
ometric, −2( 32 )n−1 , g) geometric, 21 ( 12 )n−1 , h) both, 2 = 2 + 0(n − 1) =
2(1)n−1 , i) neither, j) geometric, −2(−1)n−1 , k) arithmetic, 5(n − 1),
l) geometric, 5( 13 )n−1 , m) geometric, 12 ( 12 )n−1 , n) neither, o) geomet-
ric, −4(4)n−1 , p) arithmetic, −4 − 4(n − 1) = −4n, q) geometric,
−18(−9)n−1 , r) geometric, 13 ( 13 )n−1 , s) geometric, − 75 ( 57 )n−1 , t) geo-
metric, − 75 (− 57 )n−1 , u) neither, v) arithmetic, 4 + 3(n − 1) = 1 + 3n

9
Exercise 24.2 a) 375, b) 6.25, c) −7 · 2n−1 , d) 6, e) 10 (100)n−1 , f) 20 · (5)n−1 , g)
1 3 n−1 n−1 1 n−1
(
8 8 ) , h) 4 · 3 , i) −40000000000(− 10 )
 401

127 521
Exercise 24.3 a) 425, b) 128 , c) − 3125 , d) 2999997, e) 242, f) 910, g) −960, 800, h)
25,575
64 , i) 200
Exercise 24.4 a) 9, b) − 67 , c) 3, d) −8, e) 99, f) 81 4 500 81
2 , g) − 3 , h) −9, i) 3 , j) − 2
Exercise 24.5 a) 49 , b) 79 , c) 50 23 1300 248 20000 560
9 , d) 99 , e) 33 , f) 999 , g) 999 , h) 1111
Session 25 (exercises starting on page 345):
Exercise 25.1 a) 120, b) 6, c) 362880, d) 2, e) 1, f) 1, g) ≈ 1.216·1017, h) ≈ 1.269·1089,
i) 10, j) 84, k) 12, l) 1, m) 23, n) 50388, o) 78, p) 4368
Exercise 25.2 a) m4 + 4m3 n + 6m2 n2 + 4mn3 + n4 , b) x5 + 10x4 + 40x3 + 80x2 +
80x + 32, c) x6 − 6x5 y + 15x4 y 2 − 20x3 y 3 + 15x2 y 4 − 6xy 5 + y 6 , d)
−p5 − 5p4 q − 10p3 q 2 − 10p2 q 3 − 5pq 4 − q 5
Exercise 25.3 a) x3 − 6x2 y + 12xy 2 − 8y 3 , b) x4 − 40x3 + 600x2 − 4000x + 10000, c)
x10 y 5 + 5x8 y 6 + 10x6 y 7 + 10x4 y 8 + 5x2 y 9 + y 10 , d) 16y 8 − 160x4 y 6 +
5 3 10
600x8 y 4 − 1000x12 y 2 + 625x16 , e) x3 + 3x 2 + 3x2 + x 2 , f) −32 xy5 −
7 11 15 √ √
80 xy − 80x4 y 3 − 40xy 7 − 10 yx2 − yx5 , g) 38 2 − 36 3, h) −2 − 2i
Exercise 25.4 a) x5 y 5 − 20x5 y 4 + 160x5 y 3 , b) 512a18 + 2304a16b3 , c) −189x10 y 4 +
10 8 6
21x12 y 2 −x14 , d) xy10 −10 yx8 +45 xy6 , e) 25 m9 n9 + 15 6 10 3 3 11
16 m n + 16 m n +
1 12
64 n
Exercise 25.5 a) 35x3 y 4 , b) 36x14 y 2 , c) −220w9 , d) 280x7 y 4 , e) 15625b6, f)
−189p9q 15 , g) 7152 b
9

Exercise 25.6 a) 84x y , b) 15r s , c) −330x4 , d) 500x3 y 6 , e) 80x7 , f) 2i

3 6 4 4

Review of part V (exercises

√ √
starting on page 347): √
3 1
Exercises V 1. 4 + i4 3, 2. 10 − i 10 , 3. magnitude ||h−7, −7 3i|| = 14, directional
angle θ = 4π3 = 240 , 4. a) geometric an = 54 · (− 3 )
◦ 1 n−1
, b) neither,
c) arithmetic an = 9 − 2 · (n − 1), 5. 19950, 6. −1785, 7. 64, 8.
27x6 − 54x5 y + 36x4 y 2 − 8x3 y 3 , 9. a9 b18 + 90a7 b16 + 3600a5b14 , 10.
−7000p3q 10
References

The topics in this book are all standard and can be found in many precalculus
textbooks. In particular, the precalculus textbooks below can be consulted for
further reading.
1. Robert F. Blitzer, Precalculus, 4th edition, Prentice Hall
2. Thomas W. Hungerford, Douglas J. Shaw, Contemporary Precalculus: A
Graphing Approach, 5th edition, Brooks Cole
3. Ron Larson, Precalculus, 8th edition, Brooks Cole
4. David Lippman, Melonie Rasmussen, Precalculus: An Investigation of
Functions, available on www.opentextbookstore.com/precalc
5. Revathi Narasimhan, Precalculus: Building Concepts and Connections,
Instructor’s edition, Houghton Mifflin
6. Fred Safier, Schaum’s Outline of PreCalculus, 2nd edition, Schaum’s Out-
line Series, McGraw-Hill
7. Karl J. Smith, Precalculus: A Functional Approach To Graphing And Prob-
lem Solving, 6th edition, Jones & Bartlett Learning
8. James Stewart, Lothar Redlin, Saleem Watson, Precalculus: Mathematics
for Calculus, 6th edition, Brooks Cole
Here are some books on calculus, which should be accessible to the reader
after finishing this book.
1. Ron Larson, Bruce H. Edwards, Calculus, 9th edition, Brooks Cole
2. Jon Rogawski, Calculus, 2nd edition, W. H. Freeman
3. James Stewart, Calculus, 6th edition, Brooks Cole

402
Important formulas used in
precalculus

Algebraic formulas
Quadratic formula: The solutions of ax2 + bx + c = 0 are:
√
−b ± b2 − 4ac
x1/2 =
2a

Remainder theorem:
Dividing a polynomial f (x) by (x − c) has a remainder of r = f (c).

In particular: g(x) = x − c is a factor of f (x) ⇐⇒ f (c) = 0

Rational root theorem: The rational solutions of an xn + an−1 xn−1 + · · · +

a1 x + a0 = 0 (with a0 6= 0 and an 6= 0) are of the form x = pq where p is a
factor of a0 , and q is a factor of an .

Fundamental theorem of algebra:

Every non-constant polynomial has a root.

Exponential and logarithmic formulas:

logb (x · y) = logb (x) + logb (y)

bx · by = bx+y
bx logb ( xy ) = logb (x) − logb (y)
by
= bx−y
logb (xn ) = n · logb (x)
(bx )n = bn·x logb (x) = log(x)
log(b)

403
404 IMPORTANT FORMULAS USED IN PRECALCULUS

Applications of exponential and logarithmic functions:

Rate of growth: y = c · bx where b = 1 + r

x
Half-life: y = c · 21 h where h is the half-life

n·t
Compound interest: A = P · 1 + nr (compounded n times per year)
Compound interest: A = P · er·t (continuous compounding)

Complex numbers z = a + bi: i2 = −1

√
Absolute value: |a + bi| = a2 + b2
Polar form: a + bi = r · (cos(θ) + i · sin(θ))
√ b
where r = a2 + b2 and tan(θ) =
a
Multiplication: r1 (cos(θ1 ) + i sin(θ1 )) · r2 (cos(θ2 ) + i sin(θ2 ))
= r1 r2 · (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ))
r1 (cos(θ1 )+i sin(θ1 ))
Division: r2 (cos(θ2 )+i sin(θ2 ))
= rr12 · (cos(θ1 − θ2 ) + i sin(θ1 − θ2 ))

Vectors ~v = ha, bi:

√
Magnitude: ||ha, bi|| = a2 + b2
Directional angle: tan(θ) = ab
Scalar multiplication: r · ha, bi = hr · a, r · bi
Vector addition: ha, bi + hc, di = ha + c, b + di

Arithmetic and geometric series:

arithmetic sequence geometric sequence
nth term an = a1 + d · (n − 1) an = a1 · r n−1
Pk k
P k
series a1 + · · · + ak ai = k2 · (a1 + ak ) ai = a1 · 1−r
1−r
i=1 i=1
∞
P 1
infinite series – ai = a1 · 1−r
i=1

Binomial formula:
n  

X n where nr = r!·(n−r)!
n!
,
(a + b)n = · an−r · br
r=0
r and k! = 1 · 2 · · · · · k
n

n−k+1 k−1
The kth term is k−1 a b .
 405

Graphs of basic functions

Lines:
Slope-intercept form of the line: Slope:
y2 − y1
y =m·x+b m=
x2 − x1
y y

e
slo p y2
m= P2

b = y-intercept y1 P1

x x
x1 x2

y y y

x x x
Line y = mx + b:

m>0 m=0 m<0

Absolute value and square root:

y y

√
y = |x| y= x
x x

Polynomials and rational functions:

1 1
y = x2 y = x3 y= x
y= x2
3 y 3 y 3 y 3 y

2 2 2 2

1 1 1 1

0 x 0 x 0 x 0 x
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
-1 -1 -1 -1

-2 -2 -2 -2

-3 -3 -3 -3
406 IMPORTANT FORMULAS USED IN PRECALCULUS

Exponential and logarithmic functions:

y = ex y = log(x)
4 y 4 y

3 3

2 2

1 1

0 x 0 x
-5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5
-1 -1

-2 -2

-3 -3

Trigonometric and inverse trigonometric functions:

y = sin(x) y = sin−1 (x)
2 y 2 y

1 1

0 x 0 x
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3
-1 -1

-2 -2

y = cos(x) y = cos−1 (x)

4 y

2 y 3

1 2

0 x 1
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1 0 x
-3 -2 -1 0 1 2 3
-2 -1

y = tan(x) y = tan−1 (x)

3 y 3 y

2 2

1 1

0 x 0 x
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -4 -3 -2 -1 0 1 2 3 4
-1 -1

-2 -2

-3 -3
 407

f (x) = a · sin(b · x + c) or f (x) = a · cos(b · x + c)

y = a · sin(b · x + c)

|a|
|a| = amplitude
−c −c 2π
+ 2π
b b b x
b
= period
−c
b
= phase-shift
−|a|

| {z }
period| 2πb |

Trigonometric formulas
Basic facts:
π
2
= 90◦ π
2π
3
= 120◦ 3
= 60◦
π
3π
4
= 135◦ 4
= 45◦
π
5π
6
= 150◦ 6
= 30◦

π = 180◦ π = 180◦ 0 = 0◦

7π 11π
6
= 210◦ 6
= 330◦
5π ◦ 7π
4
= 225 4
= 315◦
4π ◦ 5π
3
= 240 3π 3
= 300◦
2
= 270◦

terminal side of x
a2 + b2 = r 2
P (a, b) √
b
=⇒ r = a2 + b2

r
b r
sin(x) = r
csc(x) = b
x
a r
cos(x) = r
sec(x) = a
a
b a
tan(x) = a
cot(x) = b
408 IMPORTANT FORMULAS USED IN PRECALCULUS

Quadrant II Quadrant I

sin(x) is positive sin(x) is positive

cos(x) is negative cos(x) is positive

tan(x) is negative tan(x) is positive

Quadrant III Quadrant IV

sin(x) is negative sin(x) is negative

cos(x) is negative cos(x) is positive

tan(x) is positive tan(x) is negative

1 1 sin(x) cos(x)
csc(x) = , sec(x) = , tan(x) = , cot(x) =
sin(x) cos(x) cos(x) sin(x)

sin2 (x) + cos2 (x) = 1, sec2 (x) = 1 + tan2 (x)

sin(x + 2π) = sin(x), sin(−x) = − sin(x), sin(π − x) = sin(x)
cos(x + 2π) = cos(x), cos(−x) = cos(x), cos(π − x) = − cos(x)

45◦ − 45◦ − 90◦ 30◦ − 60◦ − 90◦

√
2 45◦ 2 60◦
1 1

45◦ 90◦ 90◦

◦
30
√
1 3
π π π π
x 0 = 0◦ 6
= 30◦ 4
= 45◦ 3
= 60◦ 2
= 90◦
√ √
1 2 3
sin(x) 0 2 2 2
1
√ √
3 2 1
cos(x) 1 2 2 2
0
√
3
√
tan(x) 0 3
1 3 undef.
 409

Solving trigonometric equations:

Solve: sin(x) = c Solve: cos(x) = c Solve: tan(x) = c

First, find one solution, First, find one solution, First, find one solution,
that is: sin−1 (c). Use: that is: cos−1 (c). Use: that is: tan−1 (c). Use:
sin−1 (−c) = − sin−1 (c) cos−1 (−c) = π − cos−1 (c) tan−1 (−c) = − tan−1 (c)
The general solution is: The general solution is: The general solution is:

x = (−1)n sin−1 (c) + nπ x = ± cos−1 (c) + 2nπ x = tan−1 (c) + nπ

where n = 0, ±1, ±2, . . . where n = 0, ±1, ±2, . . . where n = 0, ±1, ±2, . . .

Addition and subtraction of angles:

sin(α + β) =
sin α cos β + cos α sin β
sin(α − β) =
sin α cos β − cos α sin β
cos(α + β) =
cos α cos β − sin α sin β
cos(α − β) =
cos α cos β + sin α sin β
tan α + tan β
tan(α + β) =
1 − tan α tan β
tan α − tan β
tan(α − β) =
1 + tan α tan β

Half-angles and multiple angles:

r
α 1 − cos α
sin = ±
2 2
r
α 1 + cos α
cos = ±
2 2
r
α 1 − cos α sin α 1 − cos α
tan = = = ±
2 sin α 1 + cos α 1 + cos α
sin(2α) = 2 sin α cos α
cos(2α) = cos2 α − sin2 α = 1 − 2 sin2 α = 2 cos2 α − 1
2 tan α
tan(2α) =
1 − tan2 α

The signs “±” in the half-angle formulas above are determined by the quad-
rant in which the angle α2 lies.
410 IMPORTANT FORMULAS USED IN PRECALCULUS

Geometric formulas
Pythagorean Theorem:

hypotenuse
c
a a2 + b2 = c2

2-dimensional (planar) geometric shapes:

Circle Square Rectangle

r a b
a a
Area A = πr 2 Area A = a2 Area A = a · b
Circumference C = 2πr Perimeter P = 4a Perimeter P = 2a + 2b

Triangle Parallelogram

a c a h
h
b b
Area A = 21 bh Area A = bh
Perimeter P = a + b + c Perimeter P = 2a + 2b
Index

10n , 203 {an }n≥1 , 311

30◦ − 60◦ − 90◦ triangle, 235 e, 187
45◦ − 45◦ − 90◦ triangle, 235 n!, 337
n Cr , 338 n factorial, 337
C, 287 n-choose-r, 338
N, 2 nth root, 225
Q, 2
R, 2 absolute value, 3, 62, 63, 289
R2 , 299 addition of angles formula for trigono-
Z, 2 metric functions, 252
arccos(x), 267 angle, 232
arcsin(x), 265 arccosine, 267
arctan(x), 263 arcsine, 265
n


, 338 arctangent, 263
r
cos(x), 233 arithmetic sequence, 318
arithmetic series, 321
cos−1 (x), 267
asymptote, 118, 147, 148
cosh, 188
cot(x), 233, 263 base, 184, 191
csc(x), 233 binomial coefficient, 338
⌊x⌋, 27 binomial theorem, 341
sec(x), 233
sin(x), 233 carbon dating method, 219
sin−1 (x), 265 carbon-14, 219
sinh,
P 197 codomain, 22
, 316 coefficients, 101
tan(x), 233 combination, 339
tan−1 (x), 263 common difference, 318
tanh, 197 common ratio, 325
~v , 299 complex conjugate, 140
{an }, 311 complex number, 287

411
412 INDEX

absolute value, 289 function, 22

imaginary part, 287 arccosine, 267
polar form, 290 arcsine, 265
real part, 287 arctangent, 263
complex numbers, 134 composition, 78
complex plane, 288 domain, 22, 36
complex unit, 287 exponential, 184
compound interest, 216 floor, 27
compounded n times per year, 222 graph, 37
compounded quarterly, 221 injective, 86
compounding, 222 inverse cosine, 267
conjugate, 140 inverse function, 89
continuous compounding, 222 inverse sine, 265
cosecant csc(x), 233 inverse tangent, 263
cosine cos(x), 233 logarithm, 191
cotangent cot(x), 233, 263 one-to-one, 86
piecewise defined, 33
degree, 101
range, 23, 36
degree measure, 232
fundamental theorem of algebra, 135
dependent variable, 28
difference quotient, 34 geometric sequence, 325
directional angle, 299 geometric series, 329
dividend, 104 infinite, 332
divisor, 104 geometric vector, 299
domain, 22, 36 graphs
double angle formula for trigonomet- y = arccos(x), 267
ric functions, 257 y = arcsin(x), 265
Euler number, 187 y = arctan(x), 263
even function, 71, 242 y = cos(x), 241
exponential function, 184 y = cos−1 (x), 267
extrema, 114 y = x1 , 64
extremum, 53, 114 y = x1n , 148
y = ln(x), 193, 197
factor theorem, 107 y = log(x), 193, 197
factorial, 337 y = logb (x), 197
Fibonacci sequence, 312 y = sin(x), 241
floor function, 27 y = sin−1 (x), 265
√
frequency, 246 y = x, 64
INDEX 413

y = tan(x), 243 imaginary, 134

y = tan−1 (x), 263 imaginary unit, 287
y = bx , 186 independent variable, 28
y = c · 2x , 189 inequality, 10, 169
y = ex , 187 infinite geometric series, 332
y = x2 , 64 infinite series, 332
y = x3 , 64 injective, 86
y = xn , 114 input, 22
absolute value, 63 integer, 2
arccosine, 267 interest rate, 222
arcsine, 265 interval, 5
arctangent, 263 inverse function, 89
hole, 151 irrational number, 3
inverse cosine, 267
leading coefficient, 101
inverse sine, 265
less than, 4
inverse tangent, 263
line, 13
maximum, 53
x-intercept, 15
minimum, 53
y-intercept, 15
polynomial degree 2, 114
graphing the line, 16
polynomial degree 3, 115
point-slope form of the line, 15
polynomial degree 4, 116
slope, 14
polynomial degree 5, 116
slope-intercept form of the line,
rational function, 150
16
reflect, 67
logarithm, 191
shift, 65, 66
logarithm change of base, 200
stretch, compress, 66, 67
logarithm properties, 199
greater than, 4
long division, 102
half angle formula for trigonometric magnitude, 299
functions, 257 maximum, 53
half-life, 217 minimum, 53
carbon-14, 219 monomial, 101
horizontal asymptote, 147, 151 monthly, 223
horizontal line test, 87 multiplicity, 122, 140
hyperbolic cosine, 188
hyperbolic sine, 197 natural logarithm, 191
hyperbolic tangent, 197 natural number, 2
hypotenuse, 410 newton, 307
414 INDEX

odd function, 71, 242, 243 resetting the calculator to factory set-
one-to-one, 86 tings, 373
output, 22 root, 53, 101

parabola, 114 scalar multiplication, 303

Pascal’s triangle, 340 secant sec(x), 233
period, 246 sequence, 311
periodic function, 240 series, 316
phase shift, 246 arithmetic, 321
piecewise defined function, 33 geometric, 329
point-slope form of the line, 15 infinite, 332
polynomial, 101 infinite geometric, 332
degree, 101 Sigma Σ, 316
polynomial division, 102 sine sin(x), 233
powers of 10, 203 slant asymptote, 150
present value, 224 slope-intercept form of the line, 16
principal, 221, 222 solving trigonometric equations, 277
programming the TI-84 calculator, 366 standard convention of domain, 36
Pythagorean theorem, 410 subtraction of angles formula for trigono-
metric functions, 252
quadrant, 234 summation symbol, 316
quadratic formula, 124 synthetic division, 109
quotient, 104
tangent tan(x), 233
radian measure, 232 telescopic sum, 329
range, 23, 36 TI-84, 49
rate, 222 (−), 351
rate of growth, 212 e, 188
rational function, 102, 147 nth root, 225, 364
asymptotic behaviour, 152 absolute value, 62, 63, 364
rational number, 2 alpha key, 349
rational root theorem, 131 calculate function values, 54
real number, 2 combination, 365
real part, 134 degree, 352
recursive, 315 equation solver, 362
remainder, 104 exponential function, 188
remainder theorem, 107 factorial, 364
removable discontinuity, 151 graphing, 49
INDEX 415

graphing functions, 353, 357 vector space, 308

graphing polar coordinates, 360 vertical asymptote, 147, 151
insert, 350 vertical line test, 40
intersecting functions, 358
intersection points, 61 zero, 53
line style, 186
maximum, 53
minimum, 53
minus vs. subtract, 50, 351
permutation, 365
piecewise defined functions, 359
prior answer, 350
prior entry, 350, 352
programming the calculator, 366
radian, 352
rescaling window, 354
resetting to factory settings, 373
sequences, 365
table, 361
zeros, 53
zoom window, 354–356
trigonometric equations, 277
trigonometric functions
addition of angles formula, 252
double angle formula, 257
exact values, 236
half angle formula, 257
positive/negative, 234
subtraction of angles formula, 252

union of sets, 8
unit vector, 304

vector, 299
directional angle, 299
magnitude, 299
unit vector, 304
vector addition, 305