Chapter 2 - Section 2-1 - Hw

7/25/2019 Chapter 2 - Section 2-1 - Hw

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Functions and Linear Functions

Avast expanse of open water at the top of our world was once covered with ice. The melting ofthe Arctic ice caps has forced polar bears to swim as far as 40 miles, causing them to drown insignificant numbers. Such deaths were rare in the past.

There is strong scientific consensus that human activities are changing the Earths climate. Scientists

now believe that there is a striking correlation between atmospheric carbon dioxide concentration and

global temperature. As both of these variables increase at significant rates, there are warnings of a planetaryemergency that threatens to condemn coming generations to a catastrophically diminished future.*

In this chapter, youll learn to approach our climate crisis mathematically by creating formulas, called

functions, that model data for average global temperature and carbon dioxide concentration over time.

Understanding the concept of a function will give you a new perspective on many situations, ranging

from global warming to using mathematics in a way that is similar to making a movie.

2CHAPTER

Mathematical models involving global

warming are developed in Example 10

and Check Point 10 in Section 2.4.

Using mathematics in a way that issimilar to making a movie is discussed

in the Blitzer Bonus on page 147.

*Sources: Al Gore, An Inconvenient Truth, Rodale, 2006; Time, April.3, 2006

103


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104 CHA PTE R 2 Functions and Linear Functions

Venus and Serena Williams

Objectives

1 Find the domain and

range of a relation.

2 Determine whether a

relation is a function.

3 Evaluate a function.

Introduction to Functions

SECTION

2.1Top U.S. Last Names

Name % of All Names

Smith 1.006%

Johnson 0.810%

Williams 0.699%

Brown 0.621%

Jones 0.621%

Source:Russell Ash, The Top 10 of Everything

1 Find the domain and

range of a relation.

Definition of a Relation

A relationis any set of ordered pairs. The set of all first components of the ordered

pairs is called the domainof the relation and the set of all second components iscalled the rangeof the relation.

EXAMPLE 1 Finding the Domain and Range of a Relation

Find the domain and range of the relation:

{(Smith, 1.006,), (Johnson, 0.810,), (Williams, 0.699,), (Brown, 0.621,), (Jones, 0.621,)}.

Solution The domain is the set of all first components. Thus, the domain is

{Smith, Johnson, Williams, Brown, Jones}.

The range is the set of all second components. Thus, the range is

Although Brown and Jones are both sharedby 0.621% of the U.S. population, it is not

necessary to list 0.621% twice.

{1.006%, 0.810%, 0.699%, 0.621%}.

CHECK POINT 1 Find the domain and the range of the relation:

{(0, 9.1), (10, 6.7), (20, 10.7), (30, 13.2), (38, 19.6)}.

The top five U.S. last names shown aboveaccount for nearly 4% of the entire population.The table indicates a correspondence between a last

name and the percentage of Americans who share thatname. We can write this correspondence using a set ofordered pairs:

{(Smith, 1.006%), (Johnson, 0.810%), (Williams, 0.699%),(Brown, 0.621%), (Jones, 0.621%)}.

These braces indicate we are representing a set.

The mathematical term for a set of ordered pairs is a relation.


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SECTION 2.1 Introduction to Functions 105

As you worked Check Point 1, did you wonder if there was a rule that assigned theinputs in the domain to the outputs in the range? For example, for the orderedpair (30, 13.2), how does the output 13.2 depend on the input 30? The ordered pairis based on the data in Figure 2.1(a), which shows the percentage of first-year U.S.college students claiming no religious affiliation.

Percentage of First-Year United States CollegeStudents Claiming No Religious Affiliation

Year

6%

3%

15%

24%

9%

12%

PercentageClaimingNo

ReligiousAffiliation

Women Men

18%

21%

1970 1980 1990 2000 2008

9.1

11.9

6.7

9.7 10.7

14.013.2

16.9

19.6

23.2

Figure 2.1(a) Data for women and men

Source:John Macionis, Sociology, 13th Edition, Prentice Hall, 2010.

Percentage of First-Year UnitedStates College Women Claiming

No Religious Affiliation

PercentageClaimingNo

ReligiousAffiliation

6%

3%

15%

24%

9%

12%

18%

21%

5 10 15 20 25 30 35 40

y

x

Years after 1970

(38, 19.6)

(30, 13.2)

(20, 10.7)

(10, 6.7)

(0, 9.1)

Figure 2.1(b) Visually representing the relation for womens data

In Figure 2.1(b), we used the data for college women to create the following orderedpairs:

percentage of first@year collegeyears after 1970, women claiming no religiousaffiliation

.Consider, for example, the ordered pair (30, 13.2).

(30, 13.2)

30 years after 1970,or in 2000,

13.2% of first-year college womenclaimed no religious affiliation.

The five points in Figure 2.1(b) visually represent the relation formed from thewomens data. Another way to visually represent the relation is as follows:

3020

38

100

13.210.7

19.6

6.79.1

Domain Range

Functions

Shown, again, in the margin are the top five U.S. last names and the percentage ofAmericans who share those names. Weve used this information to define two relations.

Figure 2.2(a)shows a correspondence between last names and percents sharing thosenames. Figure 2.2(b)shows a correspondence between percents sharing last names andthose last names.

2 Determine whether a

relation is a function.

Top U.S. Last Names

Name % of All Names

Smith 1.006%

Johnson 0.810%

Williams 0.699%

Brown 0.621%

Jones 0.621%

SmithJohnsonWilliams

BrownJones

1.006%0.810%0.699%

0.621%

Domain Range

Figure 2.2 (a)Names correspond

to percents.


BrownJones

1.006%0.810%0.699%0.621%

Domain Range

Figure 2.2 (b)Percents correspond

to names.


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A relation in which each member of the domain corresponds to exactly one memberof the range is a function. Can you see that the relation in Figure 2.2(a)is a function?Each last name in the domain corresponds to exactly one percent in the range. If weknow the last name, we can be sure of the percentage of Americans sharing that name.Notice that more than one element in the domain can correspond to the same elementin the range: Brown and Jones are both shared by 0.621% of Americans.

Is the relation in Figure 2.2(b) a function? Does each member of the domain

correspond to precisely one member of the range? This relation is not a functionbecause there is a member of the domain that corresponds to two different membersof the range:

(0.621,,Brown)(0.621,,Jones).

The member of the domain, 0.621%, corresponds to both Brown and Jones in therange. If we know the percentage of Americans sharing a last name, 0.621%, we cannotbe sure of that last name. Because a function is a relation in which no two ordered pairshave the same first component and different second components, the ordered pairs(0.621%, Brown) and (0.621%, Jones) are not ordered pairs of a function.

(0.621%, Brown) (0.621%, Jones)

Same first component

Different second components

Definition of a Function

A functionis a correspondence from a first set, called the domain, to a second set,called the range, such that each element in the domain corresponds to exactly oneelement in the range.

In Check Point 1, we considered a relation that gave a correspondence betweenyears after 1970 and the percentage of first-year college women claiming no religious

affiliation. Can you see that this relation is a function?

{(0, 9.1), (10, 6.7), (20, 10.7), (30, 13.2), (38, 19.6)}

corresponds to exactly one element in the range.

Each element in the domain

However, Example 2 illustrates that not every correspondence between sets is afunction.

EXAMPLE 2 Determining Whether a Relation Is a Function

Determine whether each relation is a function: a. {(1, 5), (2, 5), (3, 7), (4, 8)} b. {(5, 1), (5, 2), (7, 3), (8, 4)}.

Solution We begin by making a figure for each relation that shows the domain andthe range (Figure 2.3).

a. Figure 2.3(a)shows that every element in the domain corresponds to exactly oneelement in the range. The element 1 in the domain corresponds to the element 5 inthe range. Furthermore, 2 corresponds to 5, 3 corresponds to 7, and 4 correspondsto 8. No two ordered pairs in the given relation have the same first component anddifferent second components. Thus, the relation is a function.


BrownJones

1.006%0.810%

0.699%0.621%

Domain Range

Figure 2.2(a) (repeated)


BrownJones

1.006%0.810%0.699%0.621%

Domain Range

Figure 2.2(b) (repeated)

4321

8

7

5

Domain Range

Figure 2.3(a)


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b. Figure 2.3(b) shows that 5 corresponds to both 1 and 2. If any element in thedomain corresponds to more than one element in the range, the relation is not afunction. This relation is not a function because two ordered pairs have the samefirst component and different second components.

Different second components

(5, 1) (5, 2)

Same first component

4321

8

7

5

Domain Range

Figure 2.3(b)

Look at Figure 2.3(a)again. The fact that 1 and 2 in the domain correspond to thesame number, 5, in the range does not violate the definition of a function. A functioncan have two different first components with the same second component.By contrast,a relation is not a function when two different ordered pairs have the same firstcomponent and different second components. Thus, the relation in Example 2(b) isnot a function.

CHECK POINT 2 Determine whether each relation is a function:

a. {(1, 2),(3, 4),(5, 6),(5, 7)}

b. {(1, 2),(3, 4),(6, 5),(7, 5)}.

Functions as Equations and Function Notation

Functions are usually given in terms of equations rather than as sets of ordered pairs.For example, here is an equation that models the percentage of first-year collegewomen claiming no religious affiliation as a function of time:

y = 0.014x2 - 0.24x + 8.8.

The variablexrepresents the number of years after 1970. The variableyrepresents thepercentage of first-year college women claiming no religious affiliation. The variableyis a function of the variable x. For each value of x, there is one and only one valueofy. The variablexis called the independent variablebecause it can be assigned anyvalue from the domain. Thus,xcan be assigned any nonnegative integer representingthe number of years after 1970. The variableyis called the dependent variablebecauseits value depends onx. The percentage claiming no religious affiliation depends on thenumber of years after 1970. The value of the dependent variable,y, is calculated afterselecting a value for the independent variable,x.

If an equation inxandygives one and only one value ofyfor each value ofx, thenthe variableyis a function of the variablex. When an equation represents a function,the function is often named by a letter such asf, g, h, F, G, or H. Any letter can be usedto name a function. Suppose thatfnames a function. Think of the domain as the setof the functions inputs and the range as the set of the functions outputs. As shown inFigure 2.4, the input is represented byxand the output byf(x). The special notationf(x), read fofx or fatx, represents the value of the function at the numberx.

Lets make this clearer by considering a specific example. We know that the equationy = 0.014x2 - 0.24x + 8.8

defines y as a function of x. Well name the function f. Now, we can apply our newfunction notation.

Input Output Equation

We read this equationas fof xequals

0.014x20.24x+8.8.

x f(x) f(x)=0.014x2-0.24x+8.8

3 Evaluate a function.

Great Question!

If I reverse a functions

components, will this new

relation be a function?

If a relation is a function,reversing the components ineach of its ordered pairs mayresult in a relation that is nota function.

Inputx

Outputf(x)

f

Figure 2.4 A function machine

with inputs and outputs

Great Question!

Doesnt f(x) indicate that I

need to multiply fandx?

The notationf(x) doesnotmean ftimesx. Thenotation describes the valueof the function atx.


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Suppose we are interested in findingf(30), the functions output when the input is 30.To find the value of the function at 30, we substitute 30 for x. We are evaluating thefunctionat 30.

f(x) = 0.014x2 - 0.24x + 8.8 This is the given function.

f(30) = 0.014(30)2 - 0.24(30) + 8.8 Replace each occurrence of xwith 30.

= 0.014(900) - 0.24(30) + 8.8 Evaluate the exponential expression:

302=

30#30

=

900. = 12.6 - 7.2 + 8.8 Perform the multiplications.

(30) = 14.2 Subtract and add from left to right.

The statement f(30) = 14.2, read fof 30 equals 14.2, tells us that the value of thefunction at 30 is 14.2. When the functions input is 30, its output is 14.2. Figure 2.5illustrates the input and output in terms of a function machine.

f(30)=14.2

30 years after1970, or in 2000,

14.2% of first-year college womenclaimed no religious affiliation.

We have seen that in 2000, 13.2% actually claimed nonaffiliation, so our function

that models the data overestimates the data value for 2000 by 1%.

Inputx30

Outputf(30)14.2

f(x) 0.014x20.24x8.8

0.014(30)20.24(30)8.8

Figure 2.5 A function machine

at work

Graphing utilities can be usedto evaluate functions. Thescreens on the right show theevaluation of

f

(x) = 0.014x2 - 0.24x + 8.8

at 30 on a TI-84 Plus graphingcalculator. The functionfisnamed Y1 .

Using Technology

We used f(x) = 0.014x2 - 0.24x + 8.8 to findf(30). To find other function values,such asf(40) orf(55), substitute the specified input value, 40 or 55, forxin the functionsequation.

If a function is namedfandxrepresents the independent variable, the notationf(x)corresponds to they-value for a givenx. Thus,

f(x) = 0.014x2 - 0.24x + 8.8andy = 0.014x2 - 0.24x + 8.8

define the same function. This function may be written as

y = f(x) = 0.014x2 - 0.24x + 8.8.

EXAMPLE 3 Using Function Notation

Find the indicated function value: a. f(4) for f(x) = 2x + 3 b. g(-2) forg(x) = 2x2 - 1

c. h(-5) for h(r) = r3 - 2r2 + 5 d. F(a + h) for F(x) = 5x + 7.

Solution

a. (x) = 2x + 3 This is the given function.

(4) = 2 # 4 + 3 To find fof 4, replace xwith 4.

= 8 + 3 Multiply: 2 #4 = 8.

fof 4 is 11.f(4)=11

Add.


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b. g(x) = 2x2 - 1 This is the given function.

g(-2) = 2(-2)2 - 1 To find gof -2, replace xwith -2.

= 2(4) - 1 Evaluate the exponential expression: (-2)2 = 4.

= 8 - 1 Multiply: 2 4) = 8.

gof 2 is 7.g(2)=7 Subtract.

c. h(r) = r3 - 2r2 + 5 The functions name is hand rrepresentsthe independent variable.

h(-5) = (-5)3 - 2(-5)2 + 5 To find hof -5, replace each occurrence of rwith -5.

= -125 - 2(25) + 5 Evaluate exponential expressions.

= -125 - 50 + 5 Multiply.

hof 5 is 170.h(5)=170 - 125 - 50 = -175 and -175 + 5 = - 170.

d. F(x) = 5x + 7 This is the given function.

F(a + h) = 5(a + h) + 7 Replace xwith a + h .

Fof a+his 5a+5h+7.

F(a+h)=5a+5h+7

Apply the distributive property.

CHECK POINT 3 Find the indicated function value: a. f(6) forf(x) = 4x + 5

b. g(-5) forg(x) = 3x2 - 10

c. h(-4) for h(r) = r2 - 7r + 2

d. F(a + h) for F(x) = 6x + 9.

Great Question!

In Example 3 and Check Point 3, finding some of the function values involved

evaluating exponential expressions. Cant this be a bit tricky when such functions

are evaluated at negative numbers?

Yes. Be particularly careful if there is a term with a coefficient of-1. Notice the followingdifferences:

Replace xwith 4.

f(4)=(4)2

f(x)=x2

=16

Replace xwith 4.

g(4)=((4))2

g(x)=(x)2

=42=16

Functions Represented by Tables and Function Notation

Function notation can be applied to functions that are represented by tables.

EXAMPLE 4 Using Function Notation

Functionfis defined by the following table:

x f(x)

-2 5

-1 0

0 3

1 1

2 4

a. Explain why the table defines a function.

b. Find the domain and the range of the function.

Find the indicated function value:

c. f(-1)

d. f(0)

e. Findxsuch that f(x) = 4.


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Solution

a. Values in the first column of the table make up the domain, or input values. Valuesin the second column of the table make up the range, or output values. We seethat every element in the domain corresponds to exactly one element in the range,shown in Figure 2.6. Therefore, the relation given by the table is a function.

The voice balloons pointing to appropriate parts of the table illustrate the solution toparts (b)(e).

f(x)

2

1

0

1

2

5

0

3

1

4

c.

b. The domain is theset of inputs:

{2, 1, 0, 1, 2}.

b. The range is theset of outputs:{5, 0, 3, 1, 4}.

d.

e.

f(1) =0: When the input is 1, the output is 0.

f(0) =3: When the input is 0, the output is 3.

f(x) =4 whenx = 2: The output,f(x), is 4 when the input,x, is 2.

x

CHECK POINT 4 Functiongis defined by the following table:

x g(x)

0 3

1 0

2 1

3 2

4 3

12

012

14

305

Domain Range

Figure 2.6

a. Explain why the table defines a function.

b. Find the domain and the range of the function.

Find the indicated function value:

c. g(1)

d. g(3)

e. Findxsuch thatg(x) = 3.

Fill in each blank so that the resulting statement is true.

1. Any set of ordered pairs is called a/an ______________. The set of all first components of the ordered pairs is called the______________. The set of all second components of the ordered pairs is called the ______________.

2. A set of ordered pairs in which each member of the set of first components corresponds to exactly one member of the set ofsecond components is called a/an ______________.

3. The notationf(x) describes the value of ______________ at ______________.

4. If h(r) = -r2 + 4r - 7, we can find h(2) by replacing each occurrence of ______________ by ______________.

CONCEPT AND VOCABULARY CHECK

Check out Professor Dan MillersLearning Guidethat accompanies this textbook.

Benefits of using the Learning Guide include:

It will help you become better organized. This includes organizing your class notes,

assigned homework, quizzes, and tests.

It will enable you to use your textbook more efficiently.

It will bring together the learning tools for this course, including the textbook, the

Video Lecture Series, and the PowerPoint Presentation. It will help increase your study skills.

It will help you prepare for the chapter tests.

Ask your professor about the availability of this textbook supplement.

Achieving Success

x f(x)

-2 5

-1 0

0 3

1 1

2 4

The table defining f(repeated)


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Practice Exercises

In Exercises 18, determine whether each relation is a function.

Give the domain and range for each relation.

1. {(1, 2), (3, 4), (5, 5)}

2. {(4, 5), (6, 7), (8, 8)}

3. {(3, 4), (3, 5), (4, 4), (4, 5)}

4. {(5, 6), (5, 7), (6, 6), (6, 7)}

5. {(-3, -3),(-2, -2),(-1, -1),(0, 0)}

6. {(-7, -7), (-5, -5), (-3, -3), (0, 0)}

7. {(1, 4), (1, 5), (1, 6)}

8. {(4, 1), (5, 1), (6, 1)}

In Exercises 924, find the indicated function values. 9. f(x) = x + 1

a. f(0) b. f(5) c. f(-8)

d. f(2a) e. f(a + 2)

10. f(x) = x + 3

a. f(0) b. f(5) c. f(-8)

d. f(2a) e. f(a + 2)

11. g(x) = 3x - 2

a. g(0) b. g(-5) c. ga23b

d. g(4b) e. g(b + 4)

12. g(x) =

4x -

3 a. g(0) b. g(-5) c. ga3

4b

d. g(5b) e. g(b + 5)

13. h(x) = 3x2 + 5

a. h(0) b. h(-1) c. h(4)

d. h(-3) e. h(4b)

14. h(x) = 2x2 - 4

a. h(0) b. h(-1) c. h(5)

d. h(-3) e. h(5b)

15. f(x) = 2x2 + 3x - 1

a. f(0) b. f(3) c. f(-4)

d. f(b)

e. f(5a)

16. f(x) = 3x2 + 4x - 2

a. f(0) b. f(3) c. f(-5)

d. f(b)

e. f(5a)

17. f(x) = (-x)3 - x2 - x + 7

a. f(0) b. f(2)

c. f(2) d. f(1) + f(-1)

18. f(x) = (-x)3 - x2 - x + 10

a. f(0) b. f(2)

c. f(2) d. f(1) + f(-1)

19. f(x) = 2x -

3x - 4

a. f(0) b. f(3) c. f(-4)

d. f(-5) e. f(a + h)

f. Why must 4 be excluded from the domain off?

20. f(x) =3x - 1

x - 5

a. f(0) b. f(3) c. f(-3)

d. f(10) e. f(a + h)

f. Why must 5 be excluded from the domain off?

21.

2.1 EXERCISE SET Download theMyDashBoard AppWatch the videos

in MyMathLab

x f(x)

-4 3

-2 6

0 9

2 12

4 15

a. f(-2)

b. f(2)

c. For what value ofxis

f(x) = 9?

22.x f(x)

-5 4

-3 8

0 123 16

5 20

a. f(-3)

b. f(3)

c. For what value of x is

f(x) = 12?

23. x h(x)

-2 2

-1 1

0 0

1 1

2 2

a. h(-2)

b. h(1)

c. For what values of x is

h(x) = 1?

24.x h(x)

-2 -2

-1 -1

0 0

1 -1

2 -2

a. h(-2)

b. h(1)

c. For what values of x is

h(x) = -1?

Practice PLUS

In Exercises 2526, let f(x) = x2 - x + 4and g(x) = 3x - 5.

25. Findg(1) andf(g(1)).

26. Findg(-1) andf(g(-1)).


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In Exercises 2728, let f and g be defined by the following table:

x f(x) g(x)

-2 6 0

-1 3 4

0 -1 1

1 -4 -32 0 -6

27. Find 2f(-1) - f(0) - [g(2)]2 + f(-2) , g(2) #g(-1).

28. Find 0f(1) - f(0) 0 - [g(1)]2 + g(1) , f(-1) #g(2).

In Exercises 2930, find f(-x) - f(x)for the given function f.Then simplify the expression.

29. f(x) = x3 + x - 5

30. f(x) = x2 - 3x + 7

In Exercises 3132, each function is defined by two equations.

The equation in the first row gives the output for negative

numbers in the domain. The equation in the second row givesthe output for nonnegative numbers in the domain. Find the

indicated function values.

31. f(x) =b 3x + 5 ifx 6 04x + 7 ifx 0

a. f(-2) b. f(0)

c. f(3) d. f(-100) + f(100)

32. f(x) =b 6x - 1 ifx 6 07x + 3 ifx 0

a. f(-3) b. f(0)

c. f(4) d. f(-100) + f(100)

Application Exercises

The Corruption Perceptions Index uses perceptions of the

general public, business people, and risk analysts to rate

countries by how likely they are to accept bribes. The ratings

are on a scale from 0 to 10, where higher scores represent less

corruption. The graph shows the corruption ratings for the

worlds least corrupt and most corrupt countries. (The rating for

the United States is 7.6.) Use the graph to solve Exercises 3334.

10

Country

9

8

7

6

54

3

2

1

Top Four Least Corrupt and Most Corrupt Countries

9.7

Most corrupt

countries

Iceland

1.7

Bangladesh

9.6

Finland

1.7

Chad

9.6

NewZealand

1.8

Haiti

9.5

Denmark

Corruptio

nRating

1.8

Myanmar

Least corruptcountries

Source:Transparency International, Corruption Perceptions Index

33. a. Write a set of four ordered pairs in which each of theleast corrupt countries corresponds to a corruptionrating. Each ordered pair should be in the form

(country,

corruption

rating).

b. Is the relation in part (a) a function? Explain your

answer.

c. Write a set of four ordered pairs in whichcorruption ratings for the least corrupt countriescorrespond to countries. Each ordered pair shouldbe in the form

(corruption

rating,

country).

d. Is the relation in part (c) a function? Explain youranswer.

34. a. Write a set of four ordered pairs in which each of themost corrupt countries corresponds to a corruption

rating. Each ordered pair should be in the form

(country,

corruption

rating).

b. Is the relation in part (a) a function? Explain youranswer.

c. Write a set of four ordered pairs in which corruptionratings for the least corrupt countries correspond tocountries. Each ordered pair should be in the form

(corruption

rating,

country).

d.

Is the relation in part (c) a function? Explain youranswer.

Writing in Mathematics

35. What is a relation? Describe what is meant by its domainand its range.

36. Explain how to determine whether a relation is a function.What is a function?

37. Doesf(x) meanftimesxwhen referring to functionf? Ifnot, what doesf(x) mean? Provide an example with yourexplanation.

38. For people filing a single return, federal income tax is afunction of adjusted gross income because for each valueof adjusted gross income there is a specific tax to be paid.By contrast, the price of a house is not a function of the lotsize on which the house sits because houses on same-sized

lots can sell for many different prices.

a. Describe an everyday situation between variables thatis a function.

b. Describe an everyday situation between variables thatis not a function.


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Critical Thinking Exercises

Make Sense? In Exercises 3942, determine whether eachstatement makes sense or does not make sense and explain

your reasoning.

39. Todays temperature is a function of the time of day.

40. My height is a function of my age.

41. Although I presented my function as a set of orderedpairs, I could have shown the correspondences using atable or using points plotted in a rectangular coordinatesystem.

42. My function models how the chance of divorce dependson the number of years of marriage, so the range is{x xis the number of years of marriage}.

In Exercises 4348, determine whether each statement is true or

false. If the statement is false, make the necessary change(s) to

produce a true statement.

43. All relations are functions.

44. No two ordered pairs of a function can have the samesecond components and different first components.

Using the tables that define f and g, determine whether each

statement in Exercises 4548 is true or false.

x f(x)

-4 -1

-3 -2

-2 -3

-1 -4

51. If f(x + y) = f(x) + f(y) and f(1) = 3, find f(2), f(3),and f(4). Is f(x + y) = f(x) + f(y) for all functions?

Review Exercises

52. Simplify: 24 ,

4[2 -

(5 -

2)]2-

6.(Section 1.2, Example 7)

53. Simplify: 3x2y-2y3

-2. (Section 1.6, Example 9)54. Solve:

x

3 =

3x

5 + 4. (Section 1.4, Example 4)

Preview Exercises

Exercises 5557 will help you prepare for the material covered

in the next section.

55. Graphy = 2x. Select integers forx, starting with -2 andending with 2.

56. Graphy = 2x + 4. Select integers forx, starting with -2and ending with 2.

57. Use the following graph to solve this exercise.

1

1

2

3

4

5

6

1 2 3 4 512345

y

x

a. What is the y-coordinate when the x-coordinate is2?

b. What are thex-coordinates when they-coordinate is 4?

c. Use interval notation to describe thex-coordinates ofall points on the graph.

d. Use interval notation to describe they-coordinates ofall points on the graph.

x g(x)

-1 -4

-2 -3

-3 -2

-4 -1

45.The domain off

=

the range off46. The range off = the domain ofg

47. f(-4) - f(-2) = 2

48. g(-4) + f(-4) = 0

49. If f(x) = 3x + 7, findf(a + h) - f(a)

h.

50. Give an example of a relation with the followingcharacteristics: The relation is a function containing twoordered pairs. Reversing the components in each orderedpair results in a relation that is not a function.

Chapter 2 - Section 2-1 - Hw

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