Intervals

Intervals

Certain sets of real numbers, called intervals, occur frequently in calculus

and correspond geometrically to line segments. For example, if a < b, the
open interval from a to b consists of all numbers between a and b and is
denoted by the symbol (a, b). Using set-builder notation, we can write

(a, b) = {x ∈ R a < x < b}

MAT 1001 Calculus I 1 / 79

 Intervals
A2 ■ APPENDIX A INTERVALS, INEQUALITIES, AND ABSOLUTE VAL

A Intervals, Inequalities, and Abso

Certain sets of real numb

(a, b) = {x|a < x < b}
spond geometrically to l
Notice that the endpoints of the interval - namely, a anda to consists
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ets 关 兴 and by the solid do
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in an interval, as shown
We also need to cons
MAT 1001 Calculus I 2 / 79
 Intervals
A Intervals, Inequalities, and Abso

Certain sets of real numb

spond geometrically to l
a to b consists of all num
Definition 1 Using set-builder notatio
The closed interval from a to b is the set

a {x ∈ R a ≤
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ets 关 兴 and by the solid do
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We also need to cons

▲ Table 1 lists the nine possible types This does not mean that
of intervals. When these intervals are
MAT 1001 Calculus I set of all numbers that
3 / 79 a
 Intervals

It is also possible to include only one endpoint in an interval, as shown in

Table 1.
We also need to consider infinite intervals such as

(a, ∞) = {x ∈ R x > a}

This does not mean that ∞ (“infinity”) is a number. The notation (a, ∞)
stands for the set of all numbers that are greater than a, so the symbol ∞
simply indicates that the interval extends indefinitely far in the positive
direction.

MAT 1001 Calculus I 4 / 79

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Calculus I 3. If a
b and c  0, then 5ac / 79

 Inequalities

Inequalities

Rules for Inequalities

1 If a < b then a + c < b + c,
2 If a < b and c < d then a + c < b + d,
3 If a < b and c > 0 then ac < bc,
4 If a < b and c < 0 then ac > bc,
1
5 If 0 < a < b or a < b < 0 then a > 1b .

MAT 1001 Calculus I 6 / 79

 Inequalities

Example 2
Solve the inequality x2 − 5x + 6 ≤ 0

Solution.
First we factor the left side:

(x − 2)(x − 3) ≤ 0

We know that the corresponding equation (x − 2)(x − 3) = 0 has the

solutions 2 and 3. The numbers 2 and 3 divide the real line into three
intervals:
(−∞, 2) (2, 3) (3, ∞)

MAT 1001 Calculus I 7 / 79

 Inequalities

Solution (cont.)
On each of these intervals we determine the signs of the factors.

x 2 3
(x − 2) − 0 + +
(x − 3) − − 0 +
(x − 2)(x − 3) + − +

Then we read from the chart that (x − 2)(x − 3) is negative when

2 < x < 3. Thus, the solution of the inequality (x − 2)(x − 3) ≤ 0 is

{x ∈ R 2 ≤ x ≤ 3} = [2, 3]

Notice that we have included the endpoints 2 and 3 because we are looking for values of
such x that the product is either negative or zero.

MAT 1001 Calculus I 8 / 79

 Inequalities

Example 3
Solve x3 + 3x2 > 4x.

Solution.
First we take all nonzero terms to one side of the inequality sign and factor
the resulting expression:

x3 + 3x2 − 4x > 0 x(x − 1)(x + 4) > 0

As in previous example we solve the corresponding equation

x3 + 3x2 − 4x = 0 and use the solutions x = 0, x = −4 and x = 1 to
divide the real line into four intervals (∞, −4), (−4, 0), (0, 1) and (1, ∞).

MAT 1001 Calculus I 9 / 79

 Inequalities

Solution (cont.)
On each of these intervals we determine the signs of the factors.

x −4 0 1
x − − 0 + +

(x − 1) − − − 0 +

(x + 4) − 0 + + +

x(x − 1)(x + 4) − + − +

Then we read from the chart that the solution set is

{x ∈ R − 4 < x < 0 or x > 1} = (−4, 0) ∪ (1, ∞).

MAT 1001 Calculus I 10 / 79

 Four Ways to Represent a Function

Functions and Models

Four Ways to Represent a Function

Functions arise whenever one quantity depends on another. Consider the

following four situations:
1 The area A of a circle depends on the radius r of the circle. The rule
that connects r and A is given by the equation A = πr2 . With each
positive number r there is associated one value of A, and we say that
A is a function of r.

MAT 1001 Calculus I 11 / 79

 Four Ways to Represent a Function

2 The human population of the world P depends on the time t. The

table gives estimates of the world population P (t) at time t for
certain years.
Year Population(billion)
1900 1650
1910 1750
For instance, 1920 1860
1930 2070
P (1950) ≈ 2.560.000.000 1940 2300
1950 2560
But for each value of the time t there
1960 3040
is a corresponding value of P and we
1970 3710
say that P is a function of t.
1980 4450
1990 5280
2000 6070

MAT 1001 Calculus I 12 / 79

 Four Ways to Represent a Function

3 The cost C of mailing a first-class letter depends on the weight w of

the letter. Although there is no simple formula that connects w and
C, the post office has a rule for determining C when w is known.

MAT 1001 Calculus I 13 / 79

 Four Ways to Represent a Function
C. The cost C of mailing a first-class letter depends on the weight w of the letter.
4 The Although
vertical there is no simplea formula
acceleration of the that connects
ground and C, the by
as wmeasured postaoffice has a
rule for determining C when w is known.
seismograph during an earthquake is a function of the elapsed time t.
D. The vertical acceleration a of the ground as measured by a seismograph during
Figure shows a graph generated by seismic activity during the
an earthquake is a function of the elapsed time t. Figure 1 shows a graph gener-
Northridge earthquake
ated by seismic activity that
duringshook Los Angeles
the Northridge in 1994.
earthquake ForLos
that shook a given
Angeles
value of t the graph provides a corresponding value of a.
in 1994. For a given value of t, the graph provides a corresponding value of a.
a
{cm/s@}

100

50

5 10 15 20 25 30 t (seconds)

FIGURE 1 _50
ion during
arthquake Calif. Dept. of Mines and Geology

Each of these examples describes a rule whereby, given a number (r, t, w, or t),
another
Figure 1: number
Vertical( Aground
, P, C, oracceleration
a) is assigned.during
In eachthe
case we say thatearthquake
Northridge the second num-
ber is a function of the first number.
MAT 1001 Calculus I 14 / 79
 Four Ways to Represent a Function

Each of these examples describes a rule whereby, given a number (r, t, w,

or t), another number (A, P, C, or a) is assigned. In each case we say that
the second number is a function of the first number.

MAT 1001 Calculus I 15 / 79

 Function

Function

Definition 4
A function f is a rule that assigns to each element in a set A exactly one
element, called f (x), in a set B.

We usually consider functions for which the sets A and B are sets of real
numbers.
The set A is called the domain of the function.

The number f (x) is called the value of f at x.

The range of f is the set of all possible values of f as x varies throughout

the domain.

MAT 1001 Calculus I 16 / 79

 Function

A symbol that represents an arbitrary number in the domain of a function

f is called an independent variable.

A symbol that represents a number in the range of is called a dependent

variable.

MAT 1001 Calculus I 17 / 79

 Function
兵共x, f 共x兲兲 ⱍ x

(Notice that these are input-output pairs.) In o

points 共x, y兲 in the coordinate plane such that
The graph of a function f gives us a usefu
The most common method for visualizing a function is its graph. If f is a
of a function. Since the y-coordinate of any p
function with domain A, then its graph is thethe
can read setvalue
of ordered
of f 共x兲 pairs
from the graph as
point
2
x (see Figure 4). The graph of f also all
{(x, f (x)) ∈ R ∈ A}.
x-axisxand its range on the y-axis as in Figure

y { x, ƒ}

In other words, the graph of f con-

sists of all points (x, y) in the co- ran
ƒ
ordinate plane such that y = f (x)
f (2)
and x is in the domain of f . f (1)

0 1 2 x x

FIGURE 4 FIG

MAT 1001 Calculus I 18 / 79

 any point 共x, y兲 on the graph is y 苷 f 共x兲, we
y-coordinate of Function
x兲 from the graph as being the height of the graph above the
he graph of f also allows us to picture the domain of f on the
he The
y-axis as of
graph in fFigure 5. us to picture the domain of f on the x−axis
also allows
and its range on the y−axis as in Figure.

ƒ
range y ⫽ ƒ(x)

x x 0 x
domain
FIGURE 5

MAT 1001 Calculus I 19 / 79

 Function

EXAMPLE 1 The graph of a function f is shown in Figure 6.

(a) Find the values of f 共1兲 and f 共5兲.
Example 5 the domain and range of f ?
(b) What are
y

0 1 x

SOLUTION
The graph of a function f is shown in Figure.
(a) We see from Figure 6 that the point 共1, 3兲 lies on the graph of f , so the value of
f a) is f 共1兲
at 1Find the苷values of f (1)
3. (In other andthe
words, point on the graph that lies above x 苷 1 is
f (5).
three units above the x-axis.)
b) What are the domain and range of f ?
When x 苷 5, the graph lies about 0.7 unit below the x-axis, so we estimate that
f 共5兲 ⬇ ⫺0.7.
(b) We see that f 共x兲 is defined when 0 艋 x 艋 7, so the domain of f is the closed
MAT 1001 Calculus I 20 / 79
 Function

Solution.
a) We see from Figure that the point (1, 3) lies on the graph of f , so the
value of f at 1 is f (1) = 3. (In other words, the point on the graph
that lies above x = 1 is 3 units above the x-axis.)
When x = 5, the graph lies about 0.7 unit below the x−axis, so we
estimate that f (5) ≈ −0.7.

b) We see that f (x) is defined when 0 ≤ x ≤ 7, so the domain of f is

the closed interval [0, 7]. Notice that f takes on all values from −2 to
4, so the range of f is

{y| − 2 ≤ y ≤ 4} = [−2, 4].

MAT 1001 Calculus I 21 / 79

 Function Representations of Functions

Representations of Functions

Representations of Functions
• verbally (by a description in words)
• numerically (by a table of values)
• visually (by a graph)
• algebraically (by an explicit formula)

MAT 1001 Calculus I 22 / 79

 Function Representations of Functions

Example 6
A rectangular storage container with an open top has a volume of 10 m3 .
The length of its base is twice its width. Material for the base costs 10 TL
per square meter; material for the sides costs 6 TL per square meter.
Express the cost of materials as a function of the width of the base.

Solution.

We draw a diagram as in Figure and introduce

notation by letting w and 2w be the width and
length of the base, respectively, and h be the h

height. w
2w

FIGURE 16

MAT 1001 Calculus I 23 / 79

 Function Representations of Functions

Solution (cont.)
The area of the base is (2w)w = 2w2 ⇒ the cost, in TL, of the material
for the base is 10(2w2 ).
Two of the sides have area wh and the other two have area 2wh, so the
cost of the material for the sides is 6[2(wh) + 2(2wh)].
The total cost is therefore

C = 10(2w2 ) + 6[2(wh) + 2(2wh)] = 20w2 + 36wh.

MAT 1001 Calculus I 24 / 79

 Function Representations of Functions

Solution (cont.)
To express C as a function of w alone, we need to eliminate h and we do
so by using the fact that the volume is 10 m3 . Thus

w(2w)h = 10

which gives
10 5
h= 2
= 2.
2w w
Substituting this into the expression for C, we have
 
2 5 180
C = 20w + 36w 2
= 20w2 + .
w w

Therefore, the equation

180
C(w) = 20w2 + , w>0
w
expresses C as a function of w.
MAT 1001 Calculus I 25 / 79
 Function Vertical Line Test

Vertical Line Test

The graph of a function is a curve in the xy−plane. But the question

arises: Which curves in the xy−plane are graphs of functions? This is
answered by the following test.

MAT 1001 Calculus I 26 / 79

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0 0 a a x x 0 0 a x a x
E 17

MAT 1001 Calculus I 27 / 79

 Function Piecewise Defined Function

Piecewise Defined Function

The functions in the following examples are defined by different formulas

in different parts of their domains.
Example 7
A function f is defined by
(
1 − x, x ≤ 1
f (x) =
x2 , x>1

Evaluate f (0), f (1) and f (2) and sketch the graph.

MAT 1001 Calculus I 28 / 79

 Function Piecewise Defined Function

Evaluate f 共0兲, f 共1兲,

SOLUTION Remember
the following: First l
Solution. value of f 共x兲 is 1 ⫺
If x ≤ 1 then the value of f (x) is 1 − x. On the other hand, if x > 1, then
the value of f (x) is x2 .

How do we draw
so the part of the gra
1
cide with the line y 苷
f 共x兲 苷 x 2, so the par
1 x coincide with the gra
graph in Figure l9. T
FIGURE 19 graph; the open dot i

MAT 1001 Calculus I 29 / 79

 Function Piecewise Defined Function

Example 8
Consider the cost C(w) of mailing a first-class letter with weight w. In
effect, this is a piecewise defined function because, from the table of
values, we have the following.
C


0.39, if 0 < w ≤ 1 1


0.63, if 1 < w ≤ 2

C(w) =

 0.87, if 2 < w ≤ 3
..


.
0 1 2 3 4 5 w

FIGURE 22

MAT 1001 Calculus I 30 / 79

 Function Absolute Values

Absolute Values

The absolute value of a number a, denoted by |a|, is the distance from a

to 0 on the real number line.

Distances are always positive or zero, so we have

|a| ≥ 0 for every number a.

For example,

|3| = 3 | − 3| = 3 |0| = 0
√ √
| 2 − 1| = 2 − 1 |3 − π| = π − 3

MAT 1001 Calculus I 31 / 79

 Function Absolute Values

In general, we have

|a| = a if a ≥ 0
|a| = −a if a < 0.
√
Recall that the symbol √ means “the positive square root of”.
Therefore, the equation a2 = a is not always
√ true. It is true only when
a ≥ 0. If a < 0, then −a > 0, so we have a2 = −a. Then we have the
equation √
a2 = |a|
which is true for all values of a.

MAT 1001 Calculus I 32 / 79

 Function Properties of Absolute Values

Properties of Absolute Values

Properties of Absolute Values

Suppose a and b are any real numbers and n is an integer. Then
1 |ab| = |a|.|b|
a |a|
2 = (b 6= 0)
b |b|
3 |an | = |a|n

Suppose a > 0. Then

4 |x| = a if and only if x = ∓a
5 |x| < a if and only if −a < x < a
6 |x| > a if and only x > a or x < −a

MAT 1001 Calculus I 33 / 79

 Function Properties of Absolute Values

Example 9
Solve |3x + 2| ≥ 4.

Solution.
By Properties 4 and 6 of absolute values, |3x + 2| ≥ 4 is equivalent to

3x + 2 ≥ 4 or 3x + 2 ≤ −4

In the first case 3x ≥ 2 which gives x ≥ 32 . In the second case 3x ≤ −6,

which gives x ≤ −2. So the solution set is
   
2 2
x ∈ R x ≤ −2 or x ≥ = (−∞, −2] ∪ , ∞ .
3 3

MAT 1001 Calculus I 34 / 79

 Function Symmetric Function

Symmetric Function

Definition 10
If a function f satisfies f (−x) = f (x) for every number x in its domain,
then is f called an even function.

For instance, the function f (x) = x2 is even because

f (−x) = (−x)2 = x2 = f (x).

MAT 1001 Calculus I 35 / 79

 1 2 3 4 5
Function Symmetric Function
called step
FIGURE of
The geometric significance 22an even function is that its graph studied
is in C
symmetric with respect to the y−axis. This means that if we have plotted
the graph of f for x ≥ 0, we obtain the entire graph simply by reflecting
about the y−axis. Sym
y If a functio
called an e
f (_x) ƒ
_x 0 x x
The geome
respect to t
f for x 艌 0
If f sati
FIGURE 23 odd functi
MAT 1001 An even function
Calculus I 36 / 79
 Function Symmetric Function

Definition 11
If f satisfies f (−x) = −f (x) for every number x in its domain, then f is
called an odd function.
For example, the function f (x) = x3 is odd because

f (−x) = (−x)3 = −x3 = −f (x).

MAT 1001 Calculus I 37 / 79

 Function Symmetric Function
f for x 艌 0
If f satis
FIGURE
The graph of an odd 23 is symmetric about the origin.
function If odd functi
we already
have the graph ofAn x ≥function
even
f for 0, we can obtain the entire graph by rotating
◦
through 180 about the origin.

y The graph
already hav
through 18
_x 0
ƒ
x x
EXAMPLE 11
neither eve
(a) f 共x兲 苷
SOLUTION
FIGURE 24 (a)
MAT 1001 An odd functionCalculus I 38 / 79
 Function Symmetric Function

Example 12
Determine whether each of the following functions is even, odd or neither
even nor odd.
(a) f (x) = x5 + x (b) g(x) = 1 − x4 (c) h(x) = 2x − x2

Solution.
(a) f (−x) = (−x)5 + (−x) = (−1)5 x5 + (−x)
= −x5 − x = −(x5 + x)
= −f (x)
Therefore, f is an odd function.
(b) g(−x) = 1 − (−x)4 = 1 − x4 = g(x). So g is even.
(c) h(−x) = 2(−x) − (−x)2 = −2x − x2 . Since h(−x) 6= h(x) and
h(−x) 6= −h(x), we conclude that h is neither even nor odd.

MAT 1001 Calculus I 39 / 79

 Function Symmetric Function
Since h共⫺x兲 苷 h共x兲 and h共⫺x兲 苷 ⫺h共x兲, we conclude that h is neither even nor
odd.

The graphs of the functions in Example 11 are shown in Figure 25. Notice that the
graph of h is(cont.)
Solution symmetric neither about the y-axis nor about the origin.
y y y
1
1 f g 1 h

1
_1 1 x x 1 x

_1

(a) (b) (c)

Increasing and Decreasing Functions

The graph shown in Figure 26 rises from A to B, falls from B to C, and rises again
from C to D. The function f is said to be increasing on the interval 关a, b兴, decreasing
on 关b, c兴, and
MAT increasing
1001 again on 关c, d兴. Calculus
NoticeI that if x and x are any two numbers
40 / 79
 Mathematical Models

Mathematical Models

A mathematical model is a mathematical description (often by means of

a function or an equation) of a real-world phenomenon such as the size of
a population, the demand for a product, the speed of a falling object, the
concentration of a product in a chemical reaction, the life expectancy of a
person at birth, or the cost of emission reductions.

The purpose of the model is to understand the phenomenon and perhaps

to make predictions about future behavior.

MAT 1001 Calculus I 41 / 79

ata. The graph might even suggest a suitable algebraic fo
Mathematical Models

Real-world Formulate Mathematical

problem model

Test Solve

Real-world Mathematical
predictions Interpret conclusions

d stage is to apply the mathematics that we know (such a

MAT 1001 Calculus I 42 / 79
 Mathematical Models Linear Models

Linear Models

When we say that y is a linear function of x, we mean that the graph of

the function is a line.

Therefore we can use the slope-intercept form of the equation of a line to

write a formula for the function as

y = f (x) = mx + b

where m is the slope of the line and b is the y−intercept.

MAT 1001 Calculus I 43 / 79

 Mathematical Models Linear Models

NS A
AND MODELS
characteristic feature of linear functions is that they grow at a
constant
MODELS rate.
For instance,A characteristic
Figure shows feature
a graphof linear
of thefunctions is that fthey
linear function (x) grow
= 3x at
− a2 consta
and instance, Figure
a table of sample 2 shows a graph of the linear function f 共x兲
by 0.1, ⫺
苷 3x 2 an
A characteristic featurevalues. Notice
of linear that is
functions whenever
that theyxgrow
increases
at a constant the
rate. F
sample values. Notice that whenever x increases by 0.1, the value of f 共x兲 i
instance,
value ofFigure 2 shows
increases a graph
by 0.3. So fof
(x)theincreases
linear function f 共x兲 as
three times 苷 3xfast⫺as2 x.
and a table
0.3. So f 共x兲 increases three times as fast as x. Thus, the slope of the graph y
sample
Thus,values. Notice
the slope thatgraph
of the whenevery = x3x − 2, namely
increases by 0.1,
3,the
canvalue of f 共x兲 increases
be interpreted as
namely 3, can be interpreted as the rate of change of y with respect to x.
0.3.the f 共x兲ofincreases
Sorate change three timesrespect
of y with tox.x.Thus, the slope of the graph y 苷 3x ⫺
as fast as
namely 3, can be interpreted y as the rate of change of y with respect to x.
x f 共x兲 苷 3x ⫺ 2
y
y=3x-2
x 共x兲 苷 3x ⫺ 2
f1.0 1.0
1.1 1.3
y=3x-2 1.0 1.0
1.2 1.6
1.1 1.3
1.3 1.9
0 x 1.2 1.6
1.4 2.2
1.3 1.9
0 _2 x 1.5 2.5
1.4 2.2
_2 1.5 2.5
GURE 2
MAT 1001 Calculus I 44 / 79
 Mathematical Models Linear Models

Example 13
(a) As dry air moves upward, it expands and cools. If the ground
temperature is 20◦ C and the temperature at a height of 1 km is
10◦ C, express the temperature T (in ◦ C) as a function of the height
h (in kilometers), assuming that a linear model is appropriate.
(b) Draw the graph of the function in part (a). What does the slope
represent?
(c) What is the temperature at a height of 2.5 km?

MAT 1001 Calculus I 45 / 79

 Mathematical Models Linear Models

Solution.
(a) Because we are assuming that T is a linear function of h, we can write

T = mh + b

We are given that T = 20 when h = 0, so

20 = m · 0 + b = b

In other words, the y−intercept is b = 20.

We are also given that T = 10 when h = 1, so

10 = m · 1 + 20

The slope of the line is therefore m = 10 − 20 = −10 and the

required linear function is

T = −10h + 20.

MAT 1001 Calculus I 46 / 79

 Mathematical Models Linear Models
We are given that T
Solution (cont.)
(b) The graph is sketched in Figure. The slope is m = −10◦ C/km, and
In other to
this represents the rate of change of temperature with respect words, the
We are also give
height.
T

20 The slope of the lin

tion is
T=_10h+20
10

0 h (b) The graph is sk

1 3
sents the rate of cha
(c) At a height of h
(c) At a height of h = 2.5 FIGURE
km, the
3 temperature is

T = −10(2.5) + 20 = −5◦ C.
If there is no ph
an empirical mode
MAT 1001 Calculus I “fits” the data47in
/ 79th
 Mathematical Models Polynomials

Polynomials

Definition 14
A function P is called a polynomial if

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

where n is a nonnegative integer and the numbers a0 , a1 , a2 , . . . , an are

constants, which are called the coefficients of the polynomial.

The domain of any polynomial is R = (−∞, ∞).

If the leading coefficient an 6= 0, then the degree of the polynomial is n.

MAT 1001 Calculus I 48 / 79

 Mathematical Models Polynomials

For example, the function

2 √
P (x) = 2x6 − x4 + x3 + 2
5
is a polynomial of degree 6.

A polynomial of degree 1 is of the form P (x) = mx + b and so it is a

linear function.

A polynomial of degree 2 is of the form P (x) = ax2 + bx + c and is called

a quadratic function.

MAT 1001 Calculus I 49 / 79

 ynomial
A polynomial
of degree of 1degree
Mathematical is of the
Models 1 isform
of the form
P共x兲 苷 P共x兲
mx ⫹苷bmx and⫹sobitand
Polynomials is asolinear
it is afunc
line
polynomial
ion. A polynomial
of degree of 2degree
is of the 2 isform
of the 苷 P共x兲
form
P共x兲 ax 2 ⫹苷bxax⫹ 2
⫹c bx and⫹isccalled
and isa
quadratic
ic The
function.
graph function.
The
of P graph Theofgraph
is always aPparabola
is of
always
P obtained
is always
a parabola
byashifting
parabola
obtainedtheobtained
by shifting
parabola by shithe
y y苷=axaxy, ,as
parabola 22
苷aswe
ax 2
we,will
as we
will seewill
see inthe
in the
see next
in section.
next the
section.
nextThe
section.
Theparabola
parabola
Theopensparabola
opens upward
upwardopensif ui
and⬎downward
a 0>and
0 and if a ⬍ 0if. if(See
downward
downward a a⬍<Figure
. (See 7.)
00. Figure 7.)
y y y y

2 2 2 2

0 10 x1 x 1 1x x

Figure 2: y = x2 + x + 1 y = −2x2 + 3x + 1

(a) y=≈+x+1
(a) y=≈+x+1 (b) y=_2≈+3x+1
(b) y=_2≈+3x+1

ynomial
A polynomial
ofMATdegree
1001 of 3
degree
is of the
3 isform
of the form
Calculus I 50 / 79
 Mathematical Models Polynomials

A polynomial of degree 3 is of the form

ax3 + bx2 + cx + d

and is called a cubic function.

MAT 1001 Calculus I 51 / 79

 Mathematical Models Power Functions

Power Functions

Definition 15
A function of the form
f (x) = xa
where a is a constant, is called a power function.

MAT 1001 Calculus I 52 / 79

 Mathematical Models Power Functions

We consider several cases:

i) a = n, where n is a positive integer. The graphs of f (x) = xn for
n = 3, 4, and 5 are shown in Figures below. (These are polynomials
SECTION
SECTION
1.2
SECTION
1.2 MATHEMATICAL
1.2 MATHEMATICAL
MATHEMATICAL MODELS
MODELS ◆
MODELS ◆
31 ◆3
with only one term.)

y=x#y=x#
y=x# y=x$y=x$
y=x$ y=x%y=xy
y y y y y y y y y

1 1 1 1 1 1 1 1 1

0 0 10 x 1 x1 x 0 0 10 x 1 x1 x 0 0 10 x 1 x

MAT 1001 Calculus I 53 / 79

 odd. If nlar
is to
even, of y f苷
thatthen 共x兲x 苷
Mathematical Models3
x n is anfrom
. Notice evenFigure
function
12, and Power Functions
its graph
however, thatisassimilar to the the gra
n increases,
parabolaofy y苷苷x x. Ifbecomes
2 n
Notice from
n is odd,flatter
the figure below,
共x兲 苷
then fnear 0
n
x issteeper
and an oddwhen ⱍ ⱍ
function
x 艌
and1.its
(Ifgraph
however, that as n increases, the graph x is is simi-
small, then x
lar to that of y 苷xx3 3is
smaller, .nNotice from Figure
even smaller, 12, however,
x 4 is smaller thatso
still, and as on.)
n increases, the graph
of y 苷ofx nfbecomes
3
(x) = xflatterbecomes
near 0 flatter
4
nearwhen
and steeper ⱍ ⱍ
0 andx steeper
艌 1. (Ifwhen |x| ≥then
x is small,
y
1. x 2 is
smaller, x is even smaller, x is smaller still, and so on.)
y y
y=x $
y (1, 1)
y=x
y=x $ ^ y=x #
y=≈ (1, 1)
y=x ^ y=x
(_1, 1) (1, 1) y=x #
y=≈ 0 y=x %
(_1, 1) (1, 1)
0 x
0 x (_1, _1)

GURE 12 0 x (_1, _1)

unctions

(If(ii)x is small, then x2 is smaller, x3 is even smaller, x4 is smaller still,

a 苷 1兾n, where n is a positive integer
andThesofunction
on.) f 共x兲 苷 x 1兾n 苷 s
n
x is a root function. For n 苷 2 it is the square roo
(ii) a 苷function
MAT, 1001
1兾n wheref 共x兲n 苷
is sx
a positive
, whose integer
domain isI 关0, ⬁兲 and whose graph is the upper54half
Calculus / 79 o
 x 苷 y 2. [See Figure 13(a).] For other even values of n, the g
Mathematical Models Power Functions

milar to that of y 苷 sx. For n 苷 3 we have the cube root fun

whose domain is ⺢ (recall that every real number has a cube
h is ii)
shown 1
a = ,inwhere
Figure
n is 13(b). The
a positive graphThe
integer. y苷s n
of function x for n odd 共n
n3
苷=
at of yf (x) √
. = x is a root function. For n = 2 it is the square
sxx1/n n
√
root function f (x) = x, whose domain is [0, ∞) and whose graph is
the upper half of the parabola x = y 2 .
y y
For other even values of n, the graph
of y = x1/n is similar to that of y =
√
(1, 1) x. (1, 1)
0 x 0

(a) ƒ=œ„
x (b) ƒ=#œx„
MAT 1001 Calculus I 55 / 79
.] For other even values
Mathematical Models of n, the graph of Power Functions

or n 苷 3 we have the cube root function

that every real number has a cube root) and
The graph of y 苷 s n
x for n odd 共n ⬎ 3兲 is
√
For n = 3 we have the cube root function f (x) = 3 x whose domain
is R (recall that every real number has a cube root) and whose graph
is shown below.
y

√
The graph of y = n x for n odd
(1, 1)
(n > 3) is similar to that of y =
0 x √3
x.

(b) ƒ=#œx„

MAT 1001 Calculus I 56 / 79

 Mathematical Models Power Functions

1
Let a = −1. The graph of the reciprocal function f (x) = x−1 =
x
32 ■ CHAPTER 1 1
FUNCTIONS AND MODE
is shown in Figure. Its graph has the equation y = or xy = 1, and
x
is a hyperbola with the coordinate axes as its asymptotes.
y (iii)
The
y=∆
grap
1 axe
T
0 1 x
whi
prop

MAT 1001 FIGURE 14

Calculus I 57 / 79
 Mathematical Models Rational Functions

Rational Functions

Definition 16
A rational function f is a ratio of two polynomials:

P (x)
f (x) =
Q(x)

where P and Q are polynomials.

The domain consists of all values of x such that Q(x) 6= 0.

A simple example of a rational function is the function f (x) = 1/x, whose

domain is {x|x 6= 0}.

MAT 1001 Calculus I 58 / 79

 Mathematical Models Rational Functions
Ano
enon is
The function
2x4 − x2 + 1
f (x) =
x2 − 4
is a rational function with domain {x|x 6= ±2}.
R

y A ratio

20

0 2 x where
Q共x兲 苷
domain

FIGURE 16
MAT 1001 Calculus I 59 / 79
 Mathematical Models Algebraic Functions

Algebraic Functions

Definition 17
A function is called an algebraic function if it can be constructed using
algebraic operations (such as addition, subtraction, multiplication, division,
and taking roots) starting with polynomials.

Any rational function is automatically an algebraic function.

Here are two more examples:

p x4 − 16x2 √
f (x) = x2 + 1 g(x) = √ + (x − 2) 3 x + 1
x+ x

MAT 1001 Calculus I 60 / 79

 New Functions from Old Functions

New Functions from Old Functions

In this section we start with the basic functions we discussed in previous

section and obtain new functions by shifting, stretching, and reflecting
their graphs. We also show how to combine pairs of functions by the
standard arithmetic operations and by composition.

MAT 1001 Calculus I 61 / 79

 New Functions from Old Functions Transformations of Functions

Transformations of Functions

By applying certain transformations to the graph of a given function we

can obtain the graphs of certain related functions.
This will give us the ability to sketch the graphs of many functions quickly
by hand.
It will also enable us to write equations for given graphs.

MAT 1001 Calculus I 62 / 79

 New Functions from Old Functions Transformations of Functions

Let’s first consider translations.

If c is a positive number, then the graph of y = f (x) + c is just the graph
of y = f (x) shifted upward a distance of c units (because each
y−coordinate is increased by the same number c).
Likewise, if g(x) = f (x − c) , where c > 0, then the value of g at x is the
same as the value of f at x − c (c units to the left of x). Therefore, the
graph of y = f (x − c) is just the graph of y = f (x) shifted c units to the
right.

MAT 1001 Calculus I 63 / 79

 New Functions from Old Functions Transformations of Functions

Vertical and Horizontal Shifts

Suppose c > 0. To obtain the graph of
• y = f (x) + c, shift the graph y = f (x) a distance c units upward.
• y = f (x) − c, shift the graph y = f (x) a distance c units downward.
• y = f (x − c), shift the graph y = f (x) a distance c units to the right.
• y = f (x + c), shift the graph y = f (x) a distance c units to the left.

MAT 1001 Calculus I 64 / 79

 New Functions from Old Functions Transformations of Functions SECT
Suppose c > 0.
y

y=ƒ+c

y=f(x+c) c y =ƒ y=f(x-c)

c c

0 c x

y=ƒ-c

MAT 1001 Calculus I 65 / 79

 New Functions from Old Functions Transformations of Functions

Now let’s consider the stretching and reflecting transformations.

If c > 1, then the graph of y = cf (x) is the graph of y = f (x) stretched
by a factor of c in the vertical direction (because each y−coordinate is
multiplied by the same number c).
The graph of y = −f (x) is the graph of y = f (x) reflected about the
x−axis because the point (x, y) is replaced by the point (x, −y).

MAT 1001 Calculus I 66 / 79

 New Functions from Old Functions Transformations of Functions

Vertical and Horizontal Stretching and Reflecting

Suppose c > 1. To obtain the graph of
• y = cf (x), stretch the graph of y = f (x) vertically by a factor of c.
• y = (1/c)f (x), compress the graph of y = f (x) vertically by a factor
of c.
• y = f (cx), compress the graph of y = f (x) horizontally by a factor of
c.
• y = f (x/c), stretch the graph of y = f (x) horizontally by a factor of
c.
• y = −f (x), reflect the graph of y = f (x) about the x-axis.
• y = f (−x), reflect the graph of y = f (x) about the y-axis.

MAT 1001 Calculus I 67 / 79

 SECTION 1.3 from
New Functions NEW FUNCTIONS FROM OLD FUNCTIONS
Old Functions ◆ 39
Transformations of Functions

Suppose c > 1.
y

y=cƒ
(c>1)

y=f(_x)
y=ƒ

y= 1c ƒ

x 0 x

y=_ƒ

MAT 1001 Calculus I 68 / 79

 New Functions from Old Functions Transformations of Functions

Example 18
√ √
Given√the graph of y = x, use transformations
√ to graph y = x − 2,
√ √
y = x − 2, y = − x, y = 2 x and y = −x

Solution.
√
The graph of the square root function y = x is:

0
4 x

-2

MAT 1001 Calculus I 69 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
in the figure we sketch y = x − 2 by shifting 2 units downward:

0
4 x

-2

MAT 1001 Calculus I 70 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
in the figure we sketch y = x − 2 by shifting 2 units downward:

0
4 x

-2

MAT 1001 Calculus I 70 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
in the figure we sketch y = x − 2 by shifting 2 units downward:

0
4 x

-2

MAT 1001 Calculus I 70 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= x − 2 by shifting 2 units to the right:

0 x
2

MAT 1001 Calculus I 71 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= x − 2 by shifting 2 units to the right:

0 x
2

MAT 1001 Calculus I 71 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= x − 2 by shifting 2 units to the right:

2
0 x
2

MAT 1001 Calculus I 71 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = − x by reflecting about the x−axis:

0
x

MAT 1001 Calculus I 72 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = − x by reflecting about the x−axis:

0
x

MAT 1001 Calculus I 72 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = − x by reflecting about the x−axis:

a
0
x
a

MAT 1001 Calculus I 72 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = 2 x by stretching vertically by a factor of 2:

0
x

MAT 1001 Calculus I 73 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = 2 x by stretching vertically by a factor of 2:

0
x

MAT 1001 Calculus I 73 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y = 2 x by stretching vertically by a factor of 2:

2a
a

0
x

MAT 1001 Calculus I 73 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= −x by reflecting about the y−axis:

MAT 1001 Calculus I 74 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= −x by reflecting about the y−axis:

MAT 1001 Calculus I 74 / 79

 New Functions from Old Functions Transformations of Functions

Solution (cont.)
√
y= −x by reflecting about the y−axis:

a a

MAT 1001 Calculus I 74 / 79

 New Functions from Old Functions Transformations of Functions

Example 19
Sketch the graph of the function f (x) = x2 + 6x + 10.

Solution.
y Completing the square, we write
the equation of the graph as

y = x2 + 6x + 10 = (x + 3)2 + 1

This means we obtain the de-

sired graph by starting with the
1
parabola y = x2 and shifting 3
-3 0 x units to the left and then 1 unit
upward.

MAT 1001 Calculus I 75 / 79

 New Functions from Old Functions Transformations of Functions

Example 19
Sketch the graph of the function f (x) = x2 + 6x + 10.

Solution.
y Completing the square, we write
the equation of the graph as

y = x2 + 6x + 10 = (x + 3)2 + 1

This means we obtain the de-

sired graph by starting with the
1
parabola y = x2 and shifting 3
-3 0 x units to the left and then 1 unit
upward.

MAT 1001 Calculus I 75 / 79

 New Functions from Old Functions Transformations of Functions

Example 19
Sketch the graph of the function f (x) = x2 + 6x + 10.

Solution.
y Completing the square, we write
the equation of the graph as

y = x2 + 6x + 10 = (x + 3)2 + 1

This means we obtain the de-

sired graph by starting with the
1
parabola y = x2 and shifting 3
-3 0 x units to the left and then 1 unit
upward.

MAT 1001 Calculus I 75 / 79

 New Functions from Old Functions Transformations of Functions

Example 20
Sketch the graph of the function y = |x2 − 1|.

Solution.
y

We first graph the parabola y =

x2 −1 by shifting the parabola y =
x2 downward 1 unit. We see that
1 the graph lies below the x−axis
when −1 < x < 1, so we reflect
that part of the graph about the
-1 0 1 x x−axis to obtain the graph of y =
|x2 − 1|.

MAT 1001 Calculus I 76 / 79

 New Functions from Old Functions Transformations of Functions

Example 20
Sketch the graph of the function y = |x2 − 1|.

Solution.
y

We first graph the parabola y =

x2 −1 by shifting the parabola y =
x2 downward 1 unit. We see that
1 the graph lies below the x−axis
when −1 < x < 1, so we reflect
that part of the graph about the
-1 0 1 x x−axis to obtain the graph of y =
|x2 − 1|.

MAT 1001 Calculus I 76 / 79

 New Functions from Old Functions Transformations of Functions

Example 20
Sketch the graph of the function y = |x2 − 1|.

Solution.
y

We first graph the parabola y =

x2 −1 by shifting the parabola y =
x2 downward 1 unit. We see that
1 the graph lies below the x−axis
when −1 < x < 1, so we reflect
that part of the graph about the
-1 0 1 x x−axis to obtain the graph of y =
|x2 − 1|.

MAT 1001 Calculus I 76 / 79

 New Functions from Old Functions Algebra of Functions

Algebra of Functions

Algebra of Functions
Let f and g be functions with domains A and B.
Then the functions f + g, f − g, f g, and f /g are defined as follows:

(f + g)(x) = f (x) + g(x) domain = A ∩ B

(f − g)(x) = f (x) − g(x) domain = A ∩ B
(f g)(x) = f (x)g(x) domain = A ∩ B
(f /g)(x) = f (x)/g(x) domain = {x ∈ A ∩ B : g(x) 6= 0}

MAT 1001 Calculus I 77 / 79

 New Functions from Old Functions Algebra of Functions

Example 21
√ √
If f (x) = x and g(x) = 4 − x2 , find the functions f + g, f − g, f g,
and f /g.

Solution.
√
The domain of f (x) = x is [0, ∞).
√
The domain of g(x) = 4 − x2 consists of all numbers x such that
4 − x2 ≥ 0, that is, x2 ≤ 4.
Taking square roots of both sides, we get |x| ≤ 2, or −2 ≤ x ≤ 2, so the
domain of g is the interval [−2, 2].
The intersection of the domains of f and g is

[0, ∞) ∩ [−2, 2] = [0, 2].

MAT 1001 Calculus I 78 / 79

 New Functions from Old Functions Algebra of Functions

Solution (cont.)
Thus, according to the definitions, we have
√ p
(f + g)(x) = x + 4 − x2 0≤x≤2
√ p
(f − g)(x) = x − 4 − x2 0≤x≤2
√ p p
(f g)(x) = x 4 − x2 = 4x − x3 0≤x≤2
  √ r
f x x
(x) = √ = 0≤x<2
g 4−x 2 4 − x2

Notice that the domain of f /g is the interval [0,2) because we must

exclude the points where g(x) = 0, that is, x = ±2.

MAT 1001 Calculus I 79 / 79