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Intervals

Intervals

Certain sets of real numbers, called intervals, occur frequently in calculusand correspond geometrically to line segments. For example, if a < b, theopen interval from a to b consists of all numbers between a and b and isdenoted 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

(a, b) = {x|a < x < b}

Notice that the endpoints of the interval - namely, a and b - are excluded.This is indicated by the round brackets ( ) and by the open dots in Figure.

Notation Set description Picture Notation Set description Picture

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

Intervals, Inequalities, and Absolute Values � � � � � � � �

Certain sets of real numbers, called intervals, occur frequently in calculus and corre-spond geometrically to line segments. For example, if , the open interval from

to consists of all numbers between and and is denoted by the symbol .Using set-builder notation, we can write

Notice that the endpoints of the interval—namely, and —are excluded. This isindicated by the round brackets and by the open dots in Figure 1. The closed inter-val from to is the set

Here the endpoints of the interval are included. This is indicated by the square brack-ets and by the solid dots in Figure 2. It is also possible to include only one endpointin an interval, as shown in Table 1.

We also need to consider infinite intervals such as

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

Table of Intervals

Inequalities

When working with inequalities, note the following rules.

Rules for Inequalities

1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b

ac � bcc � 0a � b


a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�

�a, b�babaa � b

A

A2 � APPENDIX A INTERVALS, INEQUALITIES, AND ABSOLUTE VALUES

� Table 1 lists the nine possible typesof intervals. When these intervals arediscussed, it is always assumed that

.a � b

a b

FIGURE 1Open interval (a, b)

a b

FIGURE 2Closed interval [a, b]

(set of all real numbers)��, ��

�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b


Intervals

Definition 1

The closed interval from a to b is the set

[a, b] = {x ∈ R∣∣ a ≤ x ≤ b}

Here the endpoints of the interval are included. This is indicated by thesquare brackets [ ] and by the solid dots in Figure.


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b


Intervals

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


Intervals

Notation Set description Picture

(a, b) {x : a < x < b}Notation Set description Picture Notation Set description Picture

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

[a, b] {x : a ≤ x ≤ b}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

[a, b) {x : a ≤ x < b}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

(a, b] {x : a < x ≤ b}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

(a,∞) {x : x > a}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

[a,∞) {x : x ≥ a}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

(−∞, b) {x : x < b}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b(−∞, b] {x : x ≤ b}


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

(−∞,∞) R


�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�

�x � a � x � b��a, b�








Table of Intervals

Inequalities



1. If , then .

2. If and , then .

3. If and , then .

4. If and , then .

5. If , then .1�a � 1�b0 � a � b



a � c � b � dc � da � b

a � c � b � ca � b

1

�a�a, ��

�a, �� x � x � a�

� �

�a, b� � �x � a � x � b�

ba� �

ba

�a, b� � �x � a � x � b�


A



.a � b

a b


a b



�x � x � b��, b�

�x � x � b��, b�

�x � x � a��a, ��

�x � x � a��a, ��a b

a b

a b

a b

a

a

b

b

Tablo 1: Table of intervals


Inequalities

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,

5 If 0 < a < b or a < b < 0 then 1a >

1b .


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 thesolutions 2 and 3. The numbers 2 and 3 divide the real line into threeintervals:

(−∞, 2) (2, 3) (3,∞)


Inequalities

Solution (cont.)

On each of these intervals we determine the signs of the factors.

x

(x − 2)

(x − 3)

(x − 2)(x − 3)

2 3

− 0 + +

− − 0 +

+ − +

Then we read from the chart that (x− 2)(x− 3) is negative when2 < 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 ofsuch x that the product is either negative or zero.


Inequalities

Example 3

Solve x3 + 3x2 > 4x.

Solution.

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

x3 + 3x2 − 4x > 0 x(x− 1)(x+ 4) > 0

As in previous example we solve the corresponding equationx3 + 3x2 − 4x = 0 and use the solutions x = 0, x = −4 and x = 1 todivide the real line into four intervals (∞,−4), (−4, 0), (0, 1) and (1,∞).


Inequalities

Solution (cont.)

On each of these intervals we determine the signs of the factors.

x

x

(x − 1)

(x + 4)

x(x − 1)(x + 4)

−4 0 1

− − 0 + +

− − − 0 +

− 0 + + +

− + − +

Then we read from the chart that the solution set is

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


Four Ways to Represent a Function

Functions and ModelsFour Ways to Represent a Function

Functions arise whenever one quantity depends on another. Consider thefollowing four situations:

1 The area A of a circle depends on the radius r of the circle. The rulethat connects r and A is given by the equation A = πr2. With eachpositive number r there is associated one value of A, and we say thatA is a function of r.



2 The human population of the world P depends on the time t. Thetable gives estimates of the world population P (t) at time t forcertain years.

For instance,

P (1950) ≈ 2.560.000.000

But for each value of the time t thereis a corresponding value of P and wesay that P is a function of t.

Year Population(billion)

1900 16501910 17501920 18601930 20701940 23001950 25601960 30401970 37101980 44501990 52802000 6070



3 The cost C of mailing a first-class letter depends on the weight w ofthe letter. Although there is no simple formula that connects w andC, the post office has a rule for determining C when w is known.



4 The vertical acceleration a of the ground as measured by aseismograph during an earthquake is a function of the elapsed time t.Figure shows a graph generated by seismic activity during theNorthridge earthquake that shook Los Angeles in 1994. For a givenvalue of t the graph provides a corresponding value of a.

Four Ways to Represent a Function � � � � � � � � � � �

Functions arise whenever one quantity depends on another. Consider the followingfour situations.

A. The area of a circle depends on the radius of the circle. The rule that con-nects and is given by the equation . With each positive number there is associated one value of , and we say that is a function of .

B. The human population of the world depends on the time . The table gives esti-mates of the world population at time for certain years. For instance,

But for each value of the time there is a corresponding value of and we saythat is a function of .

C. The cost of mailing a first-class letter depends on the weight of the letter.Although there is no simple formula that connects and , the post office has arule for determining when is known.

D. The vertical acceleration of the ground as measured by a seismograph duringan earthquake is a function of the elapsed time Figure 1 shows a graph gener-ated by seismic activity during the Northridge earthquake that shook Los Angelesin 1994. For a given value of the graph provides a corresponding value of .

Each of these examples describes a rule whereby, given a number ( , , , or ),another number ( , , , or ) is assigned. In each case we say that the second num-ber is a function of the first number.

aCPAtwtr

FIGURE 1Vertical ground acceleration during

the Northridge earthquake

{cm/s@}

(seconds)

Calif. Dept. of Mines and Geology

5

50

10 15 20 25

a

t

100

30

_50

at,

t.a

wCCw

wC

tPP,t

P�1950� � 2,560,000,000

t,P�t�tP

rAArA � �r 2Ar

rA

1.1

11

The fundamental objects that we deal with in calculusare functions. This chapter prepares the way for calcu-lus by discussing the basic ideas concerning functions,their graphs, and ways of transforming and combiningthem. We stress that a function can be represented indifferent ways: by an equation, in a table, by a graph,or in words. We look at the main types of functions

that occur in calculus and describe the process ofusing these functions as mathematical models of real-world phenomena. We also discuss the use of graph-ing calculators and graphing software for computersand see that parametric equations provide the bestmethod for graphing certain types of curves.

PopulationYear (millions)

1900 16501910 17501920 18601930 20701940 23001950 25601960 30401970 37101980 44501990 52802000 6070

Figure 1: Vertical ground acceleration during the Northridge earthquake



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 thatthe second number is a function of the first number.


Function

Function

Definition 4

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

We usually consider functions for which the sets A and B are sets of realnumbers.

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 throughoutthe domain.


Function

A symbol that represents an arbitrary number in the domain of a functionf is called an independent variable.

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


Function

The most common method for visualizing a function is its graph. If f is afunction with domain A, then its graph is the set of ordered pairs

{(x, f(x)) ∈ R2∣∣ x ∈ A}.

In other words, the graph of f con-sists of all points (x, y) in the co-ordinate plane such that y = f(x)and x is in the domain of f .

A function is a rule that assigns to each element in a set exactly oneelement, called , in a set .

We usually consider functions for which the sets and are sets of real numbers.The set is called the domain of the function. The number is the value of at and is read “ of .” The range of is the set of all possible values of as varies throughout the domain. A symbol that represents an arbitrary number in thedomain of a function is called an independent variable. A symbol that representsa number in the range of is called a dependent variable. In Example A, forinstance, r is the independent variable and A is the dependent variable.

It’s helpful to think of a function as a machine (see Figure 2). If is in the domainof the function then when enters the machine, it’s accepted as an input and themachine produces an output according to the rule of the function. Thus, we canthink of the domain as the set of all possible inputs and the range as the set of all pos-sible outputs.

The preprogrammed functions in a calculator are good examples of a function as amachine. For example, the square root key on your calculator is such a function. Youpress the key labeled (or ) and enter the input x. If , then is not in thedomain of this function; that is, is not an acceptable input, and the calculator willindicate an error. If , then an approximation to will appear in the display.Thus, the key on your calculator is not quite the same as the exact mathematicalfunction defined by .

Another way to picture a function is by an arrow diagram as in Figure 3. Eacharrow connects an element of to an element of . The arrow indicates that isassociated with is associated with , and so on.

The most common method for visualizing a function is its graph. If is a functionwith domain , then its graph is the set of ordered pairs

(Notice that these are input-output pairs.) In other words, the graph of consists of allpoints in the coordinate plane such that and is in the domain of .

The graph of a function gives us a useful picture of the behavior or “life history”of a function. Since the -coordinate of any point on the graph is , wecan read the value of from the graph as being the height of the graph above thepoint (see Figure 4). The graph of also allows us to picture the domain of on the -axis and its range on the -axis as in Figure 5.

FIGURE 4

{x, ƒ}

ƒ

f(1)f(2)

x

y

0 1 2 x

FIGURE 5

0 x

y � ƒ(x)

domain

range

y

yxffx

f �x�y � f �x��x, y�y

ffxy � f �x��x, y�

f

��x, f �x�� x � A�

Af

af �a�x,f �x�BA

f �x� � sxfsx

sxx � 0x

xx � 0sxs

f �x�xf,

x

ff

xf �x�fxfxff �x�A

BA

Bf �x�Axf

12 � CHAPTER 1 FUNCTIONS AND MODELS

FIGURE 2Machine diagram for a function ƒ

x(input)

ƒ(output)

f

fA B

ƒ

f(a)a

x

FIGURE 3Arrow diagram for ƒ


Function

The graph of f also allows us to picture the domain of f on the x−axisand its range on the y−axis as in Figure.

A function is a rule that assigns to each element in a set exactly oneelement, called , in a set .

We usually consider functions for which the sets and are sets of real numbers.The set is called the domain of the function. The number is the value of at and is read “ of .” The range of is the set of all possible values of as varies throughout the domain. A symbol that represents an arbitrary number in thedomain of a function is called an independent variable. A symbol that representsa number in the range of is called a dependent variable. In Example A, forinstance, r is the independent variable and A is the dependent variable.

It’s helpful to think of a function as a machine (see Figure 2). If is in the domainof the function then when enters the machine, it’s accepted as an input and themachine produces an output according to the rule of the function. Thus, we canthink of the domain as the set of all possible inputs and the range as the set of all pos-sible outputs.

The preprogrammed functions in a calculator are good examples of a function as amachine. For example, the square root key on your calculator is such a function. Youpress the key labeled (or ) and enter the input x. If , then is not in thedomain of this function; that is, is not an acceptable input, and the calculator willindicate an error. If , then an approximation to will appear in the display.Thus, the key on your calculator is not quite the same as the exact mathematicalfunction defined by .

Another way to picture a function is by an arrow diagram as in Figure 3. Eacharrow connects an element of to an element of . The arrow indicates that isassociated with is associated with , and so on.

The most common method for visualizing a function is its graph. If is a functionwith domain , then its graph is the set of ordered pairs

(Notice that these are input-output pairs.) In other words, the graph of consists of allpoints in the coordinate plane such that and is in the domain of .

The graph of a function gives us a useful picture of the behavior or “life history”of a function. Since the -coordinate of any point on the graph is , wecan read the value of from the graph as being the height of the graph above thepoint (see Figure 4). The graph of also allows us to picture the domain of on the -axis and its range on the -axis as in Figure 5.

FIGURE 4

{x, ƒ}

ƒ

f(1)f(2)

x

y

0 1 2 x

FIGURE 5

0 x

y � ƒ(x)

domain

range

y

yxffx

f �x�y � f �x��x, y�y

ffxy � f �x��x, y�

f

��x, f �x�� x � A�

Af

af �a�x,f �x�BA

f �x� � sxfsx

sxx � 0x

xx � 0sxs

f �x�xf,

x

ff

xf �x�fxfxff �x�A

BA

Bf �x�Axf


FIGURE 2Machine diagram for a function ƒ

x(input)

ƒ(output)

f

fA B

ƒ

f(a)a

x

FIGURE 3Arrow diagram for ƒ


Function

Example 5

EXAMPLE 1 The graph of a function is shown in Figure 6.(a) Find the values of and .(b) What are the domain and range of ?

SOLUTION(a) We see from Figure 6 that the point lies on the graph of , so the value of

at 1 is . (In other words, the point on the graph that lies above x � 1 isthree units above the x-axis.)

When x � 5, the graph lies about 0.7 unit below the x-axis, so we estimate that.

(b) We see that is defined when , so the domain of is the closedinterval . Notice that takes on all values from �2 to 4, so the range of is

EXAMPLE 2 Sketch the graph and find the domain and range of each function.(a) (b)

SOLUTION(a) The equation of the graph is , and we recognize this as being theequation of a line with slope 2 and y-intercept �1. (Recall the slope-intercept formof the equation of a line: . See Appendix B.) This enables us to sketchthe graph of in Figure 7. The expression is defined for all real numbers, sothe domain of is the set of all real numbers, which we denote by �. The graphshows that the range is also �.

(b) Since and , we could plot the points and , together with a few other points on the graph, and join them to producethe graph (Figure 8). The equation of the graph is , which represents aparabola (see Appendix B). The domain of t is �. The range of t consists of allvalues of , that is, all numbers of the form . But for all numbers x andany positive number y is a square. So the range of t is . This canalso be seen from Figure 8.

�y � y � 0� � �0, ��x 2 � 0x 2

t�x�

y � x 2��1, 1�

�2, 4�t��1� � ��1�2 � 1t�2� � 22 � 4

f2x � 1f

y � mx � b

y � 2x � 1

t�x� � x 2f�x� � 2x � 1

�y � �2 � y � 4� � ��2, 4�

ff�0, 7�f0 � x � 7f �x�

f �5� � �0.7

f �1� � 3ff�1, 3�

FIGURE 6

x

y

0

1

1

ff �5�f �1�

f

SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION � 13

FIGURE 7

x

y=2x-1

0

-1

12

y

� The notation for intervals is given inAppendix A.

(_1, 1)

(2, 4)

0

y

1

x1

y=≈

FIGURE 8

The graph of a function f is shown in Figure.

a) Find the values of f(1) and f(5).

b) What are the domain and range of f?


Function

Solution.

a) We see from Figure that the point (1, 3) lies on the graph of f , so thevalue of f at 1 is f(1) = 3. (In other words, the point on the graphthat 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 weestimate that f(5) ≈ −0.7.

b) We see that f(x) is defined when 0 ≤ x ≤ 7, so the domain of f isthe closed interval [0, 7]. Notice that f takes on all values from −2 to4, so the range of f is

{y| − 2 ≤ y ≤ 4} = [−2, 4].


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)



Example 6

A rectangular storage container with an open top has a volume of 10m3.The length of its base is twice its width. Material for the base costs 10 TLper 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 introducenotation by letting w and 2w be the width andlength of the base, respectively, and h be theheight.

A more accurate graph of the function in Example 3 could be obtained by using athermometer to measure the temperature of the water at 10-second intervals. In gen-eral, scientists collect experimental data and use them to sketch the graphs of func-tions, as the next example illustrates.

EXAMPLE 4 The data shown in the margin come from an experiment on the lactoni-zation of hydroxyvaleric acid at 25 C. They give the concentration of this acid(in moles per liter) after minutes. Use these data to draw an approximation to thegraph of the concentration function. Then use this graph to estimate the concentra-tion after 5 minutes.

SOLUTION We plot the five points corresponding to the data from the table in Fig-ure 14. The curve-fitting methods of Section 1.2 could be used to choose a modeland graph it. But the data points in Figure 14 look quite well behaved, so we simplydraw a smooth curve through them by hand as in Figure 15.

Then we use the graph to estimate that the concentration after 5 minutes is

moleliter

In the following example we start with a verbal description of a function in a phys-ical situation and obtain an explicit algebraic formula. The ability to do this is a use-ful skill in solving calculus problems that ask for the maximum or minimum values ofquantities.

EXAMPLE 5 A rectangular storage container with an open top has a volume of 10 m .The length of its base is twice its width. Material for the base costs $10 per squaremeter; material for the sides costs $6 per square meter. Express the cost of materialsas a function of the width of the base.

SOLUTION We draw a diagram as in Figure 16 and introduce notation by letting andbe the width and length of the base, respectively, and be the height.The area of the base is , so the cost, in dollars, of the material for

the base is . Two of the sides have area and the other two have area , so the cost of the material for the sides is . The total cost is

therefore

To express as a function of alone, we need to eliminate and we do so by usingthe fact that the volume is 10 m . Thus

w�2w�h � 10

3hwC

C � 10�2w2 � � 6�2�wh� � 2�2wh�� 20w2 � 36wh

6�2�wh� � 2�2wh��2whwh10�2w2 �

�2w�w � 2w2h2w

w

3

C�5� � 0.035

FIGURE 14

C(t )

0.08

0.06

0.04

0.02

0 1 2 3 4 5 6 7 8 t t

0.02

0.04

0.06

C(t )

0.08

1 2 30 4 5 6 7 8

FIGURE 15

t C�t�


t

0 0.08002 0.05704 0.04086 0.02958 0.0210

C�t�

w

2w

h

FIGURE 16



Solution (cont.)

The area of the base is (2w)w = 2w2 ⇒ the cost, in TL, of the materialfor the base is 10(2w2).Two of the sides have area wh and the other two have area 2wh, so thecost 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.



Solution (cont.)

To express C as a function of w alone, we need to eliminate h and we doso by using the fact that the volume is 10m3. Thus

w(2w)h = 10

which gives

h =10

2w2=

5

w2.

Substituting this into the expression for C, we have

C = 20w2 + 36w

(5

w2

)= 20w2 +

180

w.

Therefore, the equation

C(w) = 20w2 +180

w, w > 0

expresses C as a function of w.


Function Vertical Line Test

Vertical Line Test

The graph of a function is a curve in the xy−plane. But the questionarises: Which curves in the xy−plane are graphs of functions? This isanswered by the following test.


Function Vertical Line Test

A curve in the xy−plane is the graph of a function of x if and only if novertical line intersects the curve more than once.

If each vertical line x = a intersects a curve only once, at (a, b), thenexactly one functional value is defined by f(a) = b. But if a line x = aintersects the curve twice, at (a, b) and (a, c), then the curve can’trepresent a function because a function can’t assign two different values toa.

which gives

Substituting this into the expression for , we have


expresses as a function of .

EXAMPLE 6 Find the domain of each function.

(a) (b)

SOLUTION(a) Because the square root of a negative number is not defined (as a real number),the domain of consists of all values of x such that . This is equivalent to

, so the domain is the interval .

(b) Since

and division by is not allowed, we see that is not defined when or. Thus, the domain of is

which could also be written in interval notation as

The graph of a function is a curve in the -plane. But the question arises: Whichcurves in the -plane are graphs of functions? This is answered by the following test.

The Vertical Line Test A curve in the -plane is the graph of a function of ifand only if no vertical line intersects the curve more than once.

The reason for the truth of the Vertical Line Test can be seen in Figure 17. If eachvertical line intersects a curve only once, at , then exactly one functionalvalue is defined by . But if a line intersects the curve twice, at and , then the curve can’t represent a function because a function can’t assigntwo different values to .

FIGURE 17xa

y

(a, c)

(a, b)

x=a

0xa

yx=a

(a, b)

0

a�a, c�

�a, b�x � af �a� � b�a, b�x � a

xxy

xyxy

��, 0� � �0, 1� � �1, ��

�x � x � 0, x � 1�

tx � 1x � 0t�x�0

t�x� �1

x 2 � x�

1

x�x � 1�

��2, ��x � �2x � 2 � 0f

t�x� �1

x 2 � xf �x� � sx � 2

wC

w 0C�w� � 20w2 �180

w

C � 20w2 � 36w 5

w2� � 20w2 �180

w

C

h �10

2w2 �5

w2


� In setting up applied functions as inExample 5, it may be useful to reviewthe principles of problem solving as dis-cussed on page 88, particularly Step 1:Understand the Problem.

� If a function is given by a formulaand the domain is not stated explicitly,the convention is that the domain is theset of all numbers for which the formulamakes sense and defines a real number.

which gives

Substituting this into the expression for , we have


expresses as a function of .

EXAMPLE 6 Find the domain of each function.

(a) (b)

SOLUTION(a) Because the square root of a negative number is not defined (as a real number),the domain of consists of all values of x such that . This is equivalent to

, so the domain is the interval .

(b) Since

and division by is not allowed, we see that is not defined when or. Thus, the domain of is

which could also be written in interval notation as

The graph of a function is a curve in the -plane. But the question arises: Whichcurves in the -plane are graphs of functions? This is answered by the following test.

The Vertical Line Test A curve in the -plane is the graph of a function of ifand only if no vertical line intersects the curve more than once.

The reason for the truth of the Vertical Line Test can be seen in Figure 17. If eachvertical line intersects a curve only once, at , then exactly one functionalvalue is defined by . But if a line intersects the curve twice, at and , then the curve can’t represent a function because a function can’t assigntwo different values to .

FIGURE 17xa

y

(a, c)

(a, b)

x=a

0xa

yx=a

(a, b)

0

a�a, c�

�a, b�x � af �a� � b�a, b�x � a

xxy

xyxy

��, 0� � �0, 1� � �1, ��

�x � x � 0, x � 1�

tx � 1x � 0t�x�0

t�x� �1

x 2 � x�

1

x�x � 1�

��2, ��x � �2x � 2 � 0f

t�x� �1

x 2 � xf �x� � sx � 2

wC

w 0C�w� � 20w2 �180

w

C � 20w2 � 36w 5

w2� � 20w2 �180

w

C

h �10

2w2 �5

w2


� In setting up applied functions as inExample 5, it may be useful to reviewthe principles of problem solving as dis-cussed on page 88, particularly Step 1:Understand the Problem.

� If a function is given by a formulaand the domain is not stated explicitly,the convention is that the domain is theset of all numbers for which the formulamakes sense and defines a real number.


Function Piecewise Defined Function

Piecewise Defined Function

The functions in the following examples are defined by different formulasin different parts of their domains.

Example 7

A function f is defined by

f(x) =

{1− x, x ≤ 1

x2, x > 1

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



Solution.

If x ≤ 1 then the value of f(x) is 1− x. On the other hand, if x > 1, thenthe value of f(x) is x2.

For example, the parabola shown in Figure 18(a) is not the graph of afunction of because, as you can see, there are vertical lines that intersect the parabolatwice. The parabola, however, does contain the graphs of two functions of . Noticethat implies , so . So the upper and lowerhalves of the parabola are the graphs of the functions [from Example6(a)] and . [See Figures 18(b) and (c).] We observe that if we reversethe roles of and , then the equation does define as a functionof (with as the independent variable and as the dependent variable) and theparabola now appears as the graph of the function .

Piecewise Defined Functions

The functions in the following four examples are defined by different formulas in dif-ferent parts of their domains.

EXAMPLE 7 A function is defined by

Evaluate , , and and sketch the graph.

SOLUTION Remember that a function is a rule. For this particular function the rule isthe following: First look at the value of the input . If it happens that , then thevalue of is . On the other hand, if , then the value of is .

How do we draw the graph of ? We observe that if , then ,so the part of the graph of that lies to the left of the vertical line must coin-cide with the line , which has slope and -intercept 1. If , then

, so the part of the graph of that lies to the right of the line mustcoincide with the graph of , which is a parabola. This enables us to sketch thegraph in Figure l9. The solid dot indicates that the point is included on thegraph; the open dot indicates that the point is excluded from the graph.�1, 1�

�1, 0�y � x 2

x � 1ff �x� � x 2x 1y�1y � 1 � x

x � 1ff �x� � 1 � xx � 1f

Since 2 1, we have f �2� � 22 � 4.

Since 1 � 1, we have f �1� � 1 � 1 � 0.

Since 0 � 1, we have f �0� � 1 � 0 � 1.

x 2f �x�x 11 � xf �x�� 1xx

f �2�f �1�f �0�

f �x� � �1 � x

x 2

if x � 1

if x 1

f

FIGURE 18

(_2, 0)

(a) x=¥-2

0 x

y

(c) y=_œ„„„„x+2

_2

0 x

y

(b) y=œ„„„„x+2

_2 0 x

y

hxyy

xx � h�y� � y 2 � 2yxt�x� � �sx � 2

f �x� � sx � 2y � �sx � 2y 2 � x � 2x � y 2 � 2

xx

x � y 2 � 2


FIGURE 19

x

y

1

1



Example 8

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

C(w) =

0.39, if 0 < w ≤ 1

0.63, if 1 < w ≤ 2

0.87, if 2 < w ≤ 3...

We also see that the graph of coincides with the -axis for . Putting thisinformation together, we have the following three-piece formula for :

EXAMPLE 10 In Example C at the beginning of this section we considered the costof mailing a first-class letter with weight . In effect, this is a piecewise

defined function because, from the table of values, we have

The graph is shown in Figure 22. You can see why functions similar to this one arecalled step functions—they jump from one value to the next. Such functions will bestudied in Chapter 2.

Symmetry

If a function satisfies for every number in its domain, then iscalled an even function. For instance, the function is even because

The geometric significance of an even function is that its graph is symmetric withrespect to the -axis (see Figure 23). This means that if we have plotted the graph of

for , we obtain the entire graph simply by reflecting about the -axis. If satisfies for every number in its domain, then is called an

odd function. For example, the function is odd because

The graph of an odd function is symmetric about the origin (see Figure 24). If wealready have the graph of for , we can obtain the entire graph by rotatingthrough about the origin.

EXAMPLE 11 Determine whether each of the following functions is even, odd, or neither even nor odd.(a) (b) (c)

SOLUTION

(a)

Therefore, is an odd function.

(b)

So is even.t

t��x� � 1 � ��x�4 � 1 � x 4 � t�x�

f

� �f �x�

� �x 5 � x � ��x 5 � x�

f ��x� � ��x�5 � ��x� � ��1�5x 5 � ��x�

h�x� � 2x � x 2t�x� � 1 � x 4f �x� � x 5 � x

180x � 0f

f ��x� � ��x�3 � �x 3 � �f �x�

f �x� � x 3fxf ��x� � �f �x�f

yx � 0fy

f ��x� � ��x�2 � x 2 � f �x�

f �x� � x 2fxf ��x� � f �x�f

0.34

0.56

0.78

1.00

if 0 � w � 1

if 1 � w � 2

if 2 � w � 3

if 3 � w � 4

C�w� �

wC�w�

f �x� � �x

2 � x

0

if 0 � x � 1

if 1 � x � 2

if x 2

fx 2xf


FIGURE 22

C

1

1

0 2 3 4 5 w

x0

y

x_x

f(_x) ƒ

FIGURE 23An even function

x0

y

x

_x ƒ

FIGURE 24An odd function


Function Absolute Values

Absolute Values

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


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.


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|

2

∣∣∣ab

∣∣∣ = |a||b| (b 6= 0)

3 |an| = |a|n

Suppose a > 0. Then

4 |x| = a if and only if x = ∓a5 |x| < a if and only if −a < x < a

6 |x| > a if and only x > a or x < −a


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 ≥ 23 . In the second case 3x ≤ −6,

which gives x ≤ −2. So the solution set is{x ∈ R

∣∣ x ≤ −2 or x ≥ 2

3

}= (−∞,−2] ∪

[2

3,∞).


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).



The geometric significance of an even function is that its graph issymmetric with respect to the y−axis. This means that if we have plottedthe graph of f for x ≥ 0, we obtain the entire graph simply by reflectingabout the y−axis.





Symmetry







SOLUTION

(a)


(b)

So is even.t

t��x� � 1 � ��x�4 � 1 � x 4 � t�x�

f

� �f �x�

� �x 5 � x � ��x 5 � x�

f ��x� � ��x�5 � ��x� � ��1�5x 5 � ��x�

h�x� � 2x � x 2t�x� � 1 � x 4f �x� � x 5 � x

180x � 0f

f ��x� � ��x�3 � �x 3 � �f �x�

f �x� � x 3fxf ��x� � �f �x�f

yx � 0fy

f ��x� � ��x�2 � x 2 � f �x�

f �x� � x 2fxf ��x� � f �x�f

0.34

0.56

0.78

1.00

if 0 � w � 1

if 1 � w � 2

if 2 � w � 3

if 3 � w � 4

C�w� �

wC�w�

f �x� � �x

2 � x

0

if 0 � x � 1

if 1 � x � 2

if x 2

fx 2xf


FIGURE 22

C

1

1

0 2 3 4 5 w

x0

y

x_x

f(_x) ƒ


x0

y

x

_x ƒ

FIGURE 24An odd function



Definition 11

If f satisfies f(−x) = −f(x) for every number x in its domain, then f iscalled an odd function.

For example, the function f(x) = x3 is odd because

f(−x) = (−x)3 = −x3 = −f(x).



The graph of an odd function is symmetric about the origin. If we alreadyhave the graph of f for x ≥ 0, we can obtain the entire graph by rotatingthrough 180◦ about the origin.





Symmetry







SOLUTION

(a)


(b)

So is even.t

t��x� � 1 � ��x�4 � 1 � x 4 � t�x�

f

� �f �x�

� �x 5 � x � ��x 5 � x�

f ��x� � ��x�5 � ��x� � ��1�5x 5 � ��x�

h�x� � 2x � x 2t�x� � 1 � x 4f �x� � x 5 � x

180x � 0f

f ��x� � ��x�3 � �x 3 � �f �x�

f �x� � x 3fxf ��x� � �f �x�f

yx � 0fy

f ��x� � ��x�2 � x 2 � f �x�

f �x� � x 2fxf ��x� � f �x�f

0.34

0.56

0.78

1.00

if 0 � w � 1

if 1 � w � 2

if 2 � w � 3

if 3 � w � 4

C�w� �

wC�w�

f �x� � �x

2 � x

0

if 0 � x � 1

if 1 � x � 2

if x 2

fx 2xf


FIGURE 22

C

1

1

0 2 3 4 5 w

x0

y

x_x

f(_x) ƒ


x0

y

x

_x ƒ

FIGURE 24An odd functionMAT 1001 Calculus I 38 / 79


Example 12

Determine whether each of the following functions is even, odd or neithereven 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)5x5 + (−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) andh(−x) 6= −h(x), we conclude that h is neither even nor odd.



Solution (cont.)

(c)

Since and , we conclude that is neither even norodd.

The graphs of the functions in Example 11 are shown in Figure 25. Notice that thegraph of h is symmetric neither about the y-axis nor about the origin.

Increasing and Decreasing Functions

The graph shown in Figure 26 rises from to , falls from to , and rises againfrom to . The function is said to be increasing on the interval , decreasingon , and increasing again on . Notice that if and are any two numbersbetween and with , then . We use this as the defining prop-erty of an increasing function.

A function is called increasing on an interval if

It is called decreasing on if

In the definition of an increasing function it is important to realize that the inequal-ity must be satisfied for every pair of numbers and in with

.You can see from Figure 27 that the function is decreasing on the inter-

val and increasing on the interval .�0, ��, 0�f �x� � x 2

x1 � x2

Ix2x1f �x1 � � f �x2 �

whenever x1 � x2 in If �x1 � f �x2 �

I

whenever x1 � x2 in If �x1 � � f �x2 �

If

A

B

C

D

y=ƒ

f(x¡)

f(x™)

a

y

0 xx¡ x™ b c dFIGURE 26

f �x1 � � f �x2 �x1 � x2bax2x1�c, d ��b, c�

�a, b�fDCCBBA

1

1 x

y

h1

1

y

x

g1

_1

1

y

x

f

_1

(a) (b) (c)FIGURE 25

hh��x� � �h�x�h��x� � h�x�

h��x� � 2��x� � ��x�2 � �2x � x 2


0

y

x

y=≈

FIGURE 27


Mathematical Models

Mathematical Models

A mathematical model is a mathematical description (often by means ofa function or an equation) of a real-world phenomenon such as the size ofa population, the demand for a product, the speed of a falling object, theconcentration of a product in a chemical reaction, the life expectancy of aperson at birth, or the cost of emission reductions.

The purpose of the model is to understand the phenomenon and perhapsto make predictions about future behavior.


Mathematical Models

a chemical reaction, the life expectancy of a person at birth, or the cost of emissionreductions. The purpose of the model is to understand the phenomenon and perhapsto make predictions about future behavior.

Figure 1 illustrates the process of mathematical modeling. Given a real-world prob-lem, our first task is to formulate a mathematical model by identifying and naming theindependent and dependent variables and making assumptions that simplify the phe-nomenon enough to make it mathematically tractable. We use our knowledge of thephysical situation and our mathematical skills to obtain equations that relate the vari-ables. In situations where there is no physical law to guide us, we may need to collectdata (either from a library or the Internet or by conducting our own experiments) andexamine the data in the form of a table in order to discern patterns. From this numeri-cal representation of a function we may wish to obtain a graphical representation byplotting the data. The graph might even suggest a suitable algebraic formula in somecases.

The second stage is to apply the mathematics that we know (such as the calculusthat will be developed throughout this book) to the mathematical model that we haveformulated in order to derive mathematical conclusions. Then, in the third stage, wetake those mathematical conclusions and interpret them as information about the orig-inal real-world phenomenon by way of offering explanations or making predictions.The final step is to test our predictions by checking against new real data. If the pre-dictions don’t compare well with reality, we need to refine our model or to formulatea new model and start the cycle again.

A mathematical model is never a completely accurate representation of a physicalsituation—it is an idealization. A good model simplifies reality enough to permitmathematical calculations but is accurate enough to provide valuable conclusions. Itis important to realize the limitations of the model. In the end, Mother Nature has thefinal say.

There are many different types of functions that can be used to model relationshipsobserved in the real world. In what follows, we discuss the behavior and graphs of these functions and give examples of situations appropriately modeled by suchfunctions.

Linear Models

When we say that y is a linear function of x, we mean that the graph of the functionis a line, so we can use the slope-intercept form of the equation of a line to write a for-mula for the function as

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

y � f �x� � mx � b

� The coordinate geometry of lines isreviewed in Appendix B.

FIGURE 1The modeling process

Real-worldproblem

Mathematicalmodel

Real-worldpredictions

Mathematicalconclusions

Formulate

Interpret

SolveTest

SECTION 1.2 MATHEMATICAL MODELS � 25


Mathematical Models Linear Models

Linear Models

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

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



A characteristic feature of linear functions is that they grow at aconstant rate.

For instance, Figure shows a graph of the linear function f(x) = 3x− 2and a table of sample values. Notice that whenever x increases by 0.1, thevalue of increases by 0.3. So f(x) increases three times as fast as x.Thus, the slope of the graph y = 3x− 2, namely 3, can be interpreted asthe rate of change of y with respect to x.

A characteristic feature of linear functions is that they grow at a constant rate. Forinstance, Figure 2 shows a graph of the linear function and a table ofsample values. Notice that whenever x increases by 0.1, the value of increases by0.3. So increases three times as fast as x. Thus, the slope of the graph ,namely 3, can be interpreted as the rate of change of y with respect to x.

EXAMPLE 1(a) As dry air moves upward, it expands and cools. If the ground temperature is

and the temperature at a height of 1 km is , express the temperature T(in °C) as a function of the height h (in kilometers), assuming that a linear model isappropriate.(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?

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

We are given that when , so

In other words, the y-intercept is .We are also given that when , so

The slope of the line is therefore and the required linear func-tion is

(b) The graph is sketched in Figure 3. The slope is , and this repre-sents the rate of change of temperature with respect to height.

(c) At a height of , the temperature is

If there is no physical law or principle to help us formulate a model, we constructan empirical model, which is based entirely on collected data. We seek a curve that“fits” the data in the sense that it captures the basic trend of the data points.

T � �10�2.5� � 20 � �5 C

h � 2.5 km

m � �10 Ckm

T � �10h � 20

m � 10 � 20 � �10

10 � m � 1 � 20

h � 1T � 10b � 20

20 � m � 0 � b � b

h � 0T � 20

T � mh � b

10 C20 C

FIGURE 2

x

y

0

y=3x-2

_2

y � 3x � 2f �x�f �x�

f �x� � 3x � 2


x

1.0 1.01.1 1.31.2 1.61.3 1.91.4 2.21.5 2.5

f �x� � 3x � 2

FIGURE 3

T

h0

10

20

1 3

T=_10h+20











T � �10�2.5� � 20 � �5 C

h � 2.5 km

m � �10 Ckm

T � �10h � 20

m � 10 � 20 � �10

10 � m � 1 � 20

h � 1T � 10b � 20

20 � m � 0 � b � b

h � 0T � 20

T � mh � b

10 C20 C

FIGURE 2

x

y

0

y=3x-2

_2

y � 3x � 2f �x�f �x�

f �x� � 3x � 2


x

1.0 1.01.1 1.31.2 1.61.3 1.91.4 2.21.5 2.5

f �x� � 3x � 2

FIGURE 3

T

h0

10

20

1 3

T=_10h+20



Example 13

(a) As dry air moves upward, it expands and cools. If the groundtemperature is 20◦C and the temperature at a height of 1 km is10◦C, express the temperature T (in ◦C) as a function of the heighth (in kilometers), assuming that a linear model is appropriate.

(b) Draw the graph of the function in part (a). What does the sloperepresent?

(c) What is the temperature at a height of 2.5 km?



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 therequired linear function is

T = −10h+ 20.



Solution (cont.)

(b) The graph is sketched in Figure. The slope is m = −10◦C/km, andthis represents the rate of change of temperature with respect toheight.











T � �10�2.5� � 20 � �5 C

h � 2.5 km

m � �10 Ckm

T � �10h � 20

m � 10 � 20 � �10

10 � m � 1 � 20

h � 1T � 10b � 20

20 � m � 0 � b � b

h � 0T � 20

T � mh � b

10 C20 C

FIGURE 2

x

y

0

y=3x-2

_2

y � 3x � 2f �x�f �x�

f �x� � 3x � 2


x

1.0 1.01.1 1.31.2 1.61.3 1.91.4 2.21.5 2.5

f �x� � 3x � 2

FIGURE 3

T

h0

10

20

1 3

T=_10h+20

(c) At a height of h = 2.5 km, the temperature is

T = −10(2.5) + 20 = −5◦C.


Mathematical Models Polynomials

Polynomials

Definition 14

A function P is called a polynomial if

P (x) = anxn + an−1x

n−1 + . . .+ a2x2 + a1x+ a0

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



For example, the function

P (x) = 2x6 − x4 + 2

5x3 +

√2

is a polynomial of degree 6.

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

A polynomial of degree 2 is of the form P (x) = ax2 + bx+ c and is calleda quadratic function.



The graph of P is always a parabola obtained by shifting the parabolay = ax2, as we will see in the next section. The parabola opens upward ifa > 0 and downward if a < 0.

We therefore predict that the level will exceed 400 ppm by the year 2020. This prediction is somewhat risky because it involves a time quite remote from ourobservations.

Polynomials

A function is called a polynomial if

where is a nonnegative integer and the numbers are constants,which are called the coefficients of the polynomial. The domain of any polynomial is

If the leading coefficient , then the degree of the polynomial is . For example, the function

is a polynomial of degree 6.A polynomial of degree 1 is of the form and so it is a linear func-

tion. A polynomial of degree 2 is of the form and is called aquadratic function. The graph of P is always a parabola obtained by shifting theparabola , as we will see in the next section. The parabola opens upward if

and downward if . (See Figure 7.)

A polynomial of degree 3 is of the form

and is called a cubic function. Figure 8 shows the graph of a cubic function in part(a) and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see laterwhy the graphs have these shapes.

FIGURE 8 (a) y=˛-x+1 (b) y=x$-3≈+x (c) y=3x%-25˛+60x

x

1

y

10 x

2

y

1

x

20

y

1

P�x� � ax 3 � bx 2 � cx � d

FIGURE 7The graphs of quadratic functions are parabolas.

y

2

x1

(b) y=_2≈+3x+1

0

y

2

x1

(a) y=≈+x+1

a � 0a 0y � ax 2

P�x� � ax 2 � bx � cP�x� � mx � b

P�x� � 2x 6 � x 4 �25 x 3 � s2

nan � 0� � ��, ��.

a0, a1, a2, . . . , ann

P�x� � an xn � an�1xn�1 � � � � � a2x 2 � a1x � a0

P

CO2


We therefore predict that the level will exceed 400 ppm by the year 2020. This prediction is somewhat risky because it involves a time quite remote from ourobservations.

Polynomials

A function is called a polynomial if

where is a nonnegative integer and the numbers are constants,which are called the coefficients of the polynomial. The domain of any polynomial is

If the leading coefficient , then the degree of the polynomial is . For example, the function

is a polynomial of degree 6.A polynomial of degree 1 is of the form and so it is a linear func-

tion. A polynomial of degree 2 is of the form and is called aquadratic function. The graph of P is always a parabola obtained by shifting theparabola , as we will see in the next section. The parabola opens upward if

and downward if . (See Figure 7.)


and is called a cubic function. Figure 8 shows the graph of a cubic function in part(a) and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see laterwhy the graphs have these shapes.

FIGURE 8 (a) y=˛-x+1 (b) y=x$-3≈+x (c) y=3x%-25˛+60x

x

1

y

10 x

2

y

1

x

20

y

1

P�x� � ax 3 � bx 2 � cx � d

FIGURE 7The graphs of quadratic functions are parabolas.

y

2

x1

(b) y=_2≈+3x+1

0

y

2

x1

(a) y=≈+x+1

a � 0a 0y � ax 2

P�x� � ax 2 � bx � cP�x� � mx � b

P�x� � 2x 6 � x 4 �25 x 3 � s2

nan � 0� � ��, ��.

a0, a1, a2, . . . , ann

P�x� � an xn � an�1xn�1 � � � � � a2x 2 � a1 x � a0

P

CO2


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




ax3 + bx2 + cx+ d

and is called a cubic function.


Mathematical Models Power Functions

Power Functions

Definition 15

A function of the formf(x) = xa

where a is a constant, is called a power function.



We consider several cases:

i) a = n, where n is a positive integer. The graphs of f(x) = xn forn = 3, 4, and 5 are shown in Figures below. (These are polynomialswith only one term.)

The general shape of the graph of depends on whether is even orodd. If is even, then is an even function and its graph is similar to theparabola . If is odd, then is an odd function and its graph is simi-lar to that of . Notice from Figure 12, however, that as increases, the graphof becomes flatter near 0 and steeper when . (If is small, then issmaller, is even smaller, is smaller still, and so on.)

( i i ) , where n is a positive integerThe function is a root function. For it is the square rootfunction , whose domain is and whose graph is the upper half ofthe parabola . [See Figure 13(a).] For other even values of n, the graph of

is similar to that of . For we have the cube root functionwhose domain is (recall that every real number has a cube root) and

whose graph is shown in Figure 13(b). The graph of for n odd issimilar to that of .

(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)

FIGURE 13Graphs of root functions

y � s3 x

�n 3�y � sn x

�f �x� � s3 x

n � 3y � sxy � sn x

x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈

FIGURE 12Families of power functions

x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2

f �x� � xnnnf �x� � xn

Graphs of ƒ=xn for n=1, 2, 3, 4, 5

x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11






(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11






(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11




Notice from the figure below, however, that as n increases, the graphof f(x) = xn becomes flatter near 0 and steeper when |x| ≥ 1.





(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11






(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11


(If x is small, then x2 is smaller, x3 is even smaller, x4 is smaller still,and so on.)



ii) a =1

n, where n is a positive integer. The function

f(x) = x1/n = n√x is a root function. For n = 2 it is the square

root function f(x) =√x, whose domain is [0,∞) and whose graph is

the upper half of the parabola x = y2.





(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11


For other even values of n, the graphof y = x1/n is similar to that of y =√x.



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 graphis shown below.





(b) ƒ=#œ„x

x

y

0

(1, 1)

(a) ƒ=œ„x

x

y

0

(1, 1)


y � s3 x

�n 3�y � sn x

�f �x� � s3 x


x � y 2�0, ��f �x� � sx

n � 2f �x� � x 1n � sn x

a � 1n

0

y

x

y=x$

(1, 1)(_1, 1)

y=x^

y=≈


x

y

0

y=x#

y=x%

(_1, _1)

(1, 1)

x 4x 3x 2x� x � � 1y � xn

ny � x 3f �x� � xnny � x 2



x

1

y

10

y=x%

x

1

y

10

y=x#

x

1

y

10

y=≈

x

1

y

10

y=x

x

1

y

10

y=x$

FIGURE 11


The graph of y = n√x for n odd

(n > 3) is similar to that of y =3√x.



Let a = −1. The graph of the reciprocal function f(x) = x−1 =1

x

is shown in Figure. Its graph has the equation y =1

xor xy = 1, and

is a hyperbola with the coordinate axes as its asymptotes.

( i i i )The graph of the reciprocal function is shown in Figure 14. Itsgraph has the equation , or , and is a hyperbola with the coordinateaxes as its asymptotes.

This function arises in physics and chemistry in connection with Boyle’s Law,which says that, when the temperature is constant, the volume of a gas is inverselyproportional to the pressure:

where C is a constant. Thus, the graph of V as a function of P (see Figure 15) hasthe same general shape as the right half of Figure 14.

Another instance in which a power function is used to model a physical phenom-enon is discussed in Exercise 20.

Rational Functions

A rational function is a ratio of two polynomials:

where and are polynomials. The domain consists of all values of such that. A simple example of a rational function is the function , whose

domain is ; this is the reciprocal function graphed in Figure 14. The function

is a rational function with domain . Its graph is shown in Figure 16.

Algebraic Functions

A function is called an algebraic function if it can be constructed using algebraicoperations (such as addition, subtraction, multiplication, division, and taking roots)starting with polynomials. Any rational function is automatically an algebraic func-tion. Here are two more examples:

t�x� �x 4 � 16x 2

x � sx� �x � 2�s

3 x � 1f �x� � sx 2 � 1

f

�x � x � �2�

f �x� �2x 4 � x 2 � 1

x 2 � 4

�x � x � 0�f �x� � 1xQ�x� � 0

xQP

f �x� �P�x�Q�x�

f

P

V

0

FIGURE 15Volume as a function of pressure

at constant temperature

V �C

P

xy � 1y � 1xf �x� � x�1 � 1x

a � �1


FIGURE 14The reciprocal function

x

1

y

10

y=∆

FIGURE 16

ƒ=2x$-≈+1

≈-4

x

20

y

20


Mathematical Models Rational Functions

Rational Functions

Definition 16

A rational function f is a ratio of two polynomials:

f(x) =P (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, whosedomain is {x|x 6= 0}.


Mathematical Models Rational Functions

The function

f(x) =2x4 − x2 + 1

x2 − 4

is a rational function with domain {x|x 6= ±2}.

( i i i )The graph of the reciprocal function is shown in Figure 14. Itsgraph has the equation , or , and is a hyperbola with the coordinateaxes as its asymptotes.

This function arises in physics and chemistry in connection with Boyle’s Law,which says that, when the temperature is constant, the volume of a gas is inverselyproportional to the pressure:

where C is a constant. Thus, the graph of V as a function of P (see Figure 15) hasthe same general shape as the right half of Figure 14.

Another instance in which a power function is used to model a physical phenom-enon is discussed in Exercise 20.

Rational Functions

A rational function is a ratio of two polynomials:

where and are polynomials. The domain consists of all values of such that. A simple example of a rational function is the function , whose

domain is ; this is the reciprocal function graphed in Figure 14. The function

is a rational function with domain . Its graph is shown in Figure 16.

Algebraic Functions

A function is called an algebraic function if it can be constructed using algebraicoperations (such as addition, subtraction, multiplication, division, and taking roots)starting with polynomials. Any rational function is automatically an algebraic func-tion. Here are two more examples:

t�x� �x 4 � 16x 2

x � sx� �x � 2�s

3 x � 1f �x� � sx 2 � 1

f

�x � x � �2�

f �x� �2x 4 � x 2 � 1

x 2 � 4

�x � x � 0�f �x� � 1xQ�x� � 0

xQP

f �x� �P�x�Q�x�

f

P

V

0

FIGURE 15Volume as a function of pressure

at constant temperature

V �C

P

xy � 1y � 1xf �x� � x�1 � 1x

a � �1


FIGURE 14The reciprocal function

x

1

y

10

y=∆

FIGURE 16

ƒ=2x$-≈+1

≈-4

x

20

y

20


Mathematical Models Algebraic Functions

Algebraic Functions

Definition 17

A function is called an algebraic function if it can be constructed usingalgebraic 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:

f(x) =√x2 + 1 g(x) =

x4 − 16x2

x+√x

+ (x− 2) 3√x+ 1


New Functions from Old Functions

New Functions from Old Functions

In this section we start with the basic functions we discussed in previoussection and obtain new functions by shifting, stretching, and reflectingtheir graphs. We also show how to combine pairs of functions by thestandard arithmetic operations and by composition.


New Functions from Old Functions Transformations of Functions

Transformations of Functions

By applying certain transformations to the graph of a given function wecan obtain the graphs of certain related functions.

This will give us the ability to sketch the graphs of many functions quicklyby hand.

It will also enable us to write equations for given graphs.



Let’s first consider translations.

If c is a positive number, then the graph of y = f(x) + c is just the graphof y = f(x) shifted upward a distance of c units (because eachy−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 thesame as the value of f at x− c (c units to the left of x). Therefore, thegraph of y = f(x− c) is just the graph of y = f(x) shifted c units to theright.



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.



Suppose c > 0.

of is the graph of reflected about the -axis because the pointis replaced by the point . (See Figure 2 and the following chart, where

the results of other stretching, compressing, and reflecting transformations are alsogiven.)

Vertical and Horizontal Stretching and Reflecting Suppose . To obtain the graph of

Figure 3 illustrates these stretching transformations when applied to the cosinefunction with . For instance, to get the graph of we multiply the y-coordinate of each point on the graph of by 2. This means that the graphof gets stretched vertically by a factor of 2.

FIGURE 3

x

1

2

y

0

y=Ł x

y=Ł 2x

y=Ł 2 x12

x

1

2

y

0

y=2 Ł x

y=Ł x

y= Ł x12

y � cos xy � cos x

y � 2 cos xc � 2

y � f ��x�, reflect the graph of y � f �x� about the y-axis

y � �f �x�, reflect the graph of y � f �x� about the x-axis

y � f �xc�, stretch the graph of y � f �x� horizontally by a factor of c

y � f �cx�, compress the graph of y � f �x� horizontally by a factor of c

y � �1c� f �x�, compress the graph of y � f �x� vertically by a factor of c

y � cf �x�, stretch the graph of y � f �x� vertically by a factor of c

c 1

�x, �y��x, y�xy � f �x�y � �f �x�

FIGURE 2Stretching and reflecting the graph of ƒ

y= ƒ1c

x

y

0

y=f(_x)

y=ƒ

y=_ƒ

y=cƒ(c>1)

FIGURE 1Translating the graph of ƒ

x

y

0

y=f(x-c)y=f(x+c) y =ƒ

y=ƒ-c

y=ƒ+c

c

c

c c

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS � 39

In Module1.3 you can see theeffect of combining the transfor-

mations of this section.



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) stretchedby a factor of c in the vertical direction (because each y−coordinate ismultiplied by the same number c).

The graph of y = −f(x) is the graph of y = f(x) reflected about thex−axis because the point (x, y) is replaced by the point (x,−y).



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 factorof c.

• y = f(cx), compress the graph of y = f(x) horizontally by a factor ofc.

• y = f(x/c), stretch the graph of y = f(x) horizontally by a factor ofc.

• 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.



Suppose c > 1.

of is the graph of reflected about the -axis because the pointis replaced by the point . (See Figure 2 and the following chart, where

the results of other stretching, compressing, and reflecting transformations are alsogiven.)

Vertical and Horizontal Stretching and Reflecting Suppose . To obtain the graph of

Figure 3 illustrates these stretching transformations when applied to the cosinefunction with . For instance, to get the graph of we multiply the y-coordinate of each point on the graph of by 2. This means that the graphof gets stretched vertically by a factor of 2.

FIGURE 3

x

1

2

y

0

y=Ł x

y=Ł 2x

y=Ł 2 x12

x

1

2

y

0

y=2 Ł x

y=Ł x

y= Ł x12

y � cos xy � cos x

y � 2 cos xc � 2

y � f ��x�, reflect the graph of y � f �x� about the y-axis

y � �f �x�, reflect the graph of y � f �x� about the x-axis

y � f �xc�, stretch the graph of y � f �x� horizontally by a factor of c

y � f �cx�, compress the graph of y � f �x� horizontally by a factor of c

y � �1c� f �x�, compress the graph of y � f �x� vertically by a factor of c

y � cf �x�, stretch the graph of y � f �x� vertically by a factor of c

c 1

�x, �y��x, y�xy � f �x�y � �f �x�

FIGURE 2Stretching and reflecting the graph of ƒ

y= ƒ1c

x

y

0

y=f(_x)

y=ƒ

y=_ƒ

y=cƒ(c>1)

FIGURE 1Translating the graph of ƒ

x

y

0

y=f(x-c)y=f(x+c) y =ƒ

y=ƒ-c

y=ƒ+c

c

c

c c

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS � 39

In Module1.3 you can see theeffect of combining the transfor-

mations of this section.



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:

EXAMPLE 1 Given the graph of , use transformations to graph ,, , , and .

SOLUTION The graph of the square root function , obtained from Figure 13 in Section 1.2, is shown in Figure 4(a). In the other parts of the figure we sketch

by shifting 2 units downward, by shifting 2 units to theright, by reflecting about the -axis, by stretching vertically by afactor of 2, and by reflecting about the -axis.

EXAMPLE 2 Sketch the graph of the function

SOLUTION Completing the square, we write the equation of the graph as

This means we obtain the desired graph by starting with the parabola andshifting 3 units to the left and then 1 unit upward (see Figure 5).

EXAMPLE 3 Sketch the graphs of the following functions.(a) (b)

SOLUTION(a) We obtain the graph of from that of by compressing hori-zontally by a factor of 2 (see Figures 6 and 7). Thus, whereas the period of is , the period of is .

FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x

y � sin xy � sin 2x

y � 1 � sin xy � sin 2x

FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �

(a) y=œ„x (b) y=œ„-2x (c) y=œ„„„„x-2 (d) y=_œ„x (e) y=2œ„x (f ) y=œ„„_x

0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y

yy � s�xy � 2sxxy � �sx

y � sx � 2y � sx � 2

y � sx

y � s�xy � 2sxy � �sxy � sx � 2y � sx � 2y � sx


FIGURE 4



Solution (cont.)

in the figure we sketch y =√x− 2 by shifting 2 units downward:









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4MAT 1001 Calculus I 70 / 79


Solution (cont.)

y =√x− 2 by shifting 2 units to the right:









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4 MAT 1001 Calculus I 71 / 79


Solution (cont.)

y = −√x by reflecting about the x−axis:









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx





Solution (cont.)

y = 2√x by stretching vertically by a factor of 2:









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx





Solution (cont.)

y =√−x by reflecting about the y−axis:









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx





Example 19

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

Solution.

Completing the square, we write the equation of the graph as

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

This means we obtain the desired graph by starting with the parabolay = x2 and shifting 3 units to the left and then 1 unit upward.









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4









FIGURE 6

x0

y

1

π2

π

y=sin x

FIGURE 7

x0

y

1

π2

π4

π

y=sin 2x

2�2 � �y � sin 2x2�y � sin x



FIGURE 5 (a) y=≈ (b) y=(x+3)@+1

x0_1_3

1

y

(_3, 1)

x0

y

y � x 2

y � x 2 � 6x � 10 � �x � 3�2 � 1

x 2 � 6x � 10.f �x� �


0 x

y

0 x

y

0 x

y

20 x

y

_2

0 x

y

1

10 x

y


y � sx � 2y � sx � 2

y � sx



FIGURE 4



Example 20

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

Solution.

We first graph the parabola y = x2 − 1 by shifting the parabola y = x2

downward 1 unit. We see that the graph lies below the x−axis when−1 < x < 1, so we reflect that part of the graph about the x−axis toobtain the graph of y = |x2 − 1|.

and when . This tells us how to get the graph offrom the graph of : The part of the graph that lies above the -axis

remains the same; the part that lies below the -axis is reflected about the -axis.

EXAMPLE 5 Sketch the graph of the function .

SOLUTION We first graph the parabola in Figure 10(a) by shifting theparabola downward 1 unit. We see that the graph lies below the x-axis when

, so we reflect that part of the graph about the x-axis to obtain the graphof in Figure 10(b).

Combinations of Functions

Two functions and can be combined to form new functions , , , andin a manner similar to the way we add, subtract, multiply, and divide real numbers.

If we define the sum by the equation

then the right side of Equation 1 makes sense if both and are defined, that is,if x belongs to the domain of and also to the domain of . If the domain of is Aand the domain of is B, then the domain of is the intersection of thesedomains, that is, .

Notice that the sign on the left side of Equation 1 stands for the operation ofaddition of functions, but the sign on the right side of the equation stands for addi-tion of the numbers and .

Similarly, we can define the difference and the product , and their domainsare also . But in defining the quotient we must remember not to divide by 0.

Algebra of Functions Let and be functions with domains and . Then thefunctions , and are defined as follows:

f

t��x� �

f �x�t�x�

domain � �x � A � B � t�x� � 0�

� ft� �x� � f �x�t�x� domain � A � B

� f � t� �x� � f �x� � t�x� domain � A � B


ftf � t, f � t, ftBAtf

ftA � Bftf � t

t�x�f �x��

�A � B

f � tt

ftft�x�f �x�

� f � t��x� � f �x� � t�x�1

f � t

ft

ftf � tf � ttf

0 x

y

_1 1 0 x

y

_1 1

FIGURE 10 (a) y=≈-1 (b) y=| ≈-1 |

y � � x 2 � 1��1 � x � 1

y � x 2y � x 2 � 1

y � � x 2 � 1 �

xxxy � f �x�y � � f �x��

f �x� � 0y � �f �x�f �x� � 0


and when . This tells us how to get the graph offrom the graph of : The part of the graph that lies above the -axis

remains the same; the part that lies below the -axis is reflected about the -axis.

EXAMPLE 5 Sketch the graph of the function .

SOLUTION We first graph the parabola in Figure 10(a) by shifting theparabola downward 1 unit. We see that the graph lies below the x-axis when

, so we reflect that part of the graph about the x-axis to obtain the graphof in Figure 10(b).

Combinations of Functions

Two functions and can be combined to form new functions , , , andin a manner similar to the way we add, subtract, multiply, and divide real numbers.

If we define the sum by the equation

then the right side of Equation 1 makes sense if both and are defined, that is,if x belongs to the domain of and also to the domain of . If the domain of is Aand the domain of is B, then the domain of is the intersection of thesedomains, that is, .

Notice that the sign on the left side of Equation 1 stands for the operation ofaddition of functions, but the sign on the right side of the equation stands for addi-tion of the numbers and .

Similarly, we can define the difference and the product , and their domainsare also . But in defining the quotient we must remember not to divide by 0.

Algebra of Functions Let and be functions with domains and . Then thefunctions , and are defined as follows:

f

t��x� �

f �x�t�x�

domain � �x � A � B � t�x� � 0�

� ft� �x� � f �x�t�x� domain � A � B



ftf � t, f � t, ftBAtf

ftA � Bftf � t

t�x�f �x��

�A � B

f � tt

ftft�x�f �x�

� f � t��x� � f �x� � t�x�1

f � t

ft

ftf � tf � ttf

0 x

y

_1 1 0 x

y

_1 1

FIGURE 10 (a) y=≈-1 (b) y=| ≈-1 |

y � � x 2 � 1��1 � x � 1

y � x 2y � x 2 � 1

y � � x 2 � 1 �

xxxy � f �x�y � � f �x��

f �x� � 0y � �f �x�f �x� � 0



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, fg, 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

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



Example 21

If f(x) =√x and g(x) =

√4− x2, find the functions f + g, f − g, fg,

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 thedomain of g is the interval [−2, 2].

The intersection of the domains of f and g is

[0,∞) ∩ [−2, 2] = [0, 2].



Solution (cont.)

Thus, according to the definitions, we have

(f + g)(x) =√x+

√4− x2 0 ≤ x ≤ 2

(f − g)(x) =√x−

√4− x2 0 ≤ x ≤ 2

(fg)(x) =√x√

4− x2 =√4x− x3 0 ≤ x ≤ 2(

f

g

)(x) =

√x√

4− x2=

√x

4− x20 ≤ x < 2

Notice that the domain of f/g is the interval [0,2) because we mustexclude the points where g(x) = 0, that is, x = ±2.


intervals - deukisi.deu.edu.tr/cem.celik/files/1001_week_01.pdf · the closed inter-val from to is...

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