Algebraic, Exponential, and Logarithmic Functions · ˜ is unit focuses on a study of functions, establishing the algebraic basis for precalculus. Linear, exponential, power, logarithmic,

Chapter 1: Functions and Mathematical Models

Chapter 2: Properties of Elementary Functions

Chapter 3: Fitting Functions to Data

Chapter 4: Polynomial and Rational Functions

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Algebraic, Exponential, andLogarithmic FunctionsAlgebraic, Exponential, andAlgebraic, Exponential, and

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Unit Overview� is unit focuses on a study of functions, establishing the algebraic basis for precalculus. Linear, exponential, power, logarithmic, logistic, polynomial and rational functions are investigated and expanded using graphical, numerical, and algebraic approaches. In Chapter 1, students learn overarching concepts such as transformations, composition, and inverses. An introduction to parametric equations strengthens the concept of inverse functions. In Chapter 2, students approach functions by examining shapes of graphs and numerical patterns of evenly spaced data. � ey explore characteristics and signi� cant features of graphs and make generalizations about di� erent functions. Chapter 3 focuses on � tting functions to data using re-expression and regression. Students make decisions about which function best � ts the data by examining the shape of the scatter and residual plots, the end behavior, and the correlation coe� cient. Chapter 4 completes the unit, focusing on polynomial and rational functions.

Using this UnitEach chapter in this unit examines functions from di� erent viewpoints. Students need a strong understanding of functions to be successful in precalculus, calculus, and courses beyond calculus. You may decide to teach Unit 1 and then branch into other parts of the text. If you would like to study trigonometry and periodic functions early in the course, start with Chapter 1 and then go directly to Unit 2: Trigonometric and Periodic Functions.

Depending on the level of preparation your students have, you may be able to omit a portion of Chapter 1 with a cautionary note that the topics in this chapter form a critical foundation for the rest of the course. Consider omitting Chapter 3 if your students have taken or will be taking statistics. It provides an introduction to � tting functions to data, but time considerations may make it necessary to study this concept in a di� erent course. Finally, if your students studied polynomial and rational functions in second-year algebra, you may prefer to delay Chapter 4 until later in the course, as it provides a solid foundation for the discussion of limits in Chapter 16. You may also want to omit the sections on partial fractions, which will be covered in calculus when students study integration techniques.

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If you shoot an arrow into the air, its height above the ground depends on the number of seconds since you released it. In this chapter you will learn ways to express quantitatively the relationship between two variables such as height and time. You will deepen what you have learned in previous courses about functions and the particular relationships that they describe—for example, how height depends on time.

Functions and Mathematical ModelsFunctions and

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• Work with functions that are defi ned graphically, algebraically, numerically, or verbally.

• Make connections among the algebraic equation for a function, its name, and its graph.

• Transform a given pre-image function so that the result is a graph of the image function that has been dilated by given factors and translated by given amounts.

• Given two functions, graph and evaluate the composition of one function with the other.

• Given a function, fi nd its inverse relation, and tell whether the inverse relation is a function. Graph parametric equations both by hand and on a grapher, and use parametric equations to graph the inverse of a function.

• Given a function, transform it by refl ection and by applying absolute value to the function or its argument.

• Start writing a journal to record things learned about precalculus mathematics and questions concerning concepts which are not quite clear.

CHAP TE R O B J EC TIV ES

1A Chapter 1 Interleaf: Functions and Mathematical Models

OverviewIn this chapter students refresh their memories about functions they studied previously in algebra. Th e unifying concept is that familiar functions are built up by transformations of a few basic parent functions. For instance, the point-slope form of the linear function, y 2 y 1 5 m(x 2 x 1 ), is a dilation of the parent function, y 5 x, by a factor of m in the y-direction and translations by x 1 and y 1 in the x- and y-directions, respectively. Students study refl ections as dilations by a factor of 21. Th ey study the formal defi nition of inverses of functions along with the composition of functions. Piecewise functions and their inverses are introduced using a graphing calculator to restrict the domain. Th e Quick Review problems begin in Section 1-3, giving students a time-effi cient review of other concepts and techniques. Students are also introduced to the concept that mathematics can be learned four ways: graphically, algebraically, numerically, and verbally.

Using This ChapterChapter 1 sets the stage for a study of precalculus and provides a solid foundation for this course. Section 1-7, Precalculus Journal, provides a foundation for helping students learn how to express themselves mathematically in oral and written language. Aft er Chapter 1, continue with fi tting functions to real-world data in Chapters 2–4, or branch to the study of periodic functions in Chapters 5–9.

Teaching ResourcesExplorationsExploration 1-1: Paper Cup AnalysisExploration 1-2a: Names of FunctionsExploration 1-2b: Restricted Domains and Boolean VariablesExploration 1-3: Transformations from Graphs Exploration 1-3a: Translations and Dilations, NumericallyExploration 1-3b: Translations and Dilations, AlgebraicallyExploration 1-3c: Transformation ReviewExploration 1-4: Composition of Functions

Exploration 1-5: Parametric Equations GraphExploration 1-5a: Inverses of Functions Exploration 1-5b: Introduction to Parametric EquationsExploration 1-6a: Translation, Dilation, and Refl ection

Blackline MastersSections 1-3 to 1-6, and 1-8

Supplementary ProblemsSections 1-2 and 1-4

Assessment ResourcesTest 1, Sections 1-1 to 1-3, Forms A and BTest 2, Sections 1-4 and 1-5, Forms A and BTest 3, Chapter 1, Forms A and B

Technology ResourcesDynamic Precalculus ExplorationsTranslationDilation

Sketchpad Presentation SketchesTranslation Present.gspDilation Present.gspComposition Present.gspInverse Present.gspRefl ection Present.gspAbsolute Value Present.gsp

ActivitiesSketchpad: Translation of FunctionsSketchpad: Dilation of Functions Fathom: Exploring Translations and Dilations Fathom: Function TransformationsFathom: Reading the NewsCAS Activity 1-2a: Finding Polynomial Equations CAS Activity 1-2b: Functions Defi ned by Two Points CAS Activity 1-3a: Transformed Quadratic Functions

Functions and Mathematical ModelsFunctions and Mathematical ModelsC h a p t e r 1

Chapter 1 Interleaf 1B

Standard Schedule Pacing Guide

Block Schedule Pacing Guide

Day Section Suggested Assignment

1 1-1 Functions: Graphically, Algebraically, Numerically, and Verbally 1–5

2 1-2 Types of Functions 1–39 odd, 40–42

31-3 Dilation and Translation of Function

GraphsRA, Q1–Q10, 1–6

4 7–21

5 1-4 Composition of Functions RA, Q1–Q10, 1, 2, 5, 7, 9, 10, 12–15

61-5 Inverse Functions and Parametric

EquationsRA, Q1–Q10, 1–15 odd, 16

7 17, 19, 21, 25, 31, 35, 37, 38

8 1-6 Refl ections, Absolute Values, and Other Transformations RA, Q1–Q10, 1–5, 7, 9–14

9 1-7 Precalculus Journal 1

101-8 Chapter Review and Test

R1–R6, T1–T28

11 C1, C2, Problem Set 2-1

Day Section Suggested Assignment

11-1 Functions: Graphically, Algebraically,

Numerically, and Verbally 1–5

1-2 Types of Functions 1–39 odd, 40, 41

2 1-3 Dilation and Translation of Function Graphs RA, Q1–Q10, 1–21 odd

31-4 Composition of Functions RA, Q1–Q10, 1–9 odd, 10

1-5 Inverse Functions and Parametric Equations RA, Q1–Q10, 1, 3, 5, 9, 11

41-5 Inverse Functions and Parametric

Equations 17, 19, 21, 29

1-6 Refl ections, Absolute Values, and Other Transformations RA, Q1–Q10, 1–9 odd, 10, 12

51-7 Precalculus Journal 1

1-8 Chapter Review and Test R1–R6, T1–T28

61-8 Chapter Review and Test

2-1 Shapes of Function Graphs 1–4

2 Chapter 1: Functions and Mathematical Models

In previous courses you have studied linear functions, quadratic functions, exponential functions, power functions, and others. In precalculus mathematics you will learn general properties that apply to all types of functions. In particular you will learn how to transform a function so that its graph � ts real-world data. You will gain this knowledge in four ways.

� e graph at right is the graph of a quadratic function. � e y-variable could represent the height of an arrow at various times, x, a� er its release into the air. For larger time values, the quadratic function shows that y is negative. � ese values may or may not be reasonable in the real world.

� e equation of the function isy 4.9 x 2 20x 5

� is table shows corresponding x- and y-values that satisfy the equation of the function.

When the variables in a function stand for things in the real world, the function is being used as a mathematical model. The coef ficients in the equation of the function y 4.9 x 2 20x 5 have a real-world meaning. For example, the coef ficient 4.9 is a constant that is a result of the gravitational acceleration , 20 is the initial velocity, and 5 reflects the initial height of the arrow.

GRAPHICALLY

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20

10

10

20

30

1 2 3 4 5 6x (s)

y (m)

ALGEBRAICALLY

NUMERICALLY

x (s) y (m)

0 5.0

1 20.1

2 25.4

3 20.9

VERBALLY

In previous courses you have studied linear functions, quadratic

Mathematical Overview

3Section 1-1: Functions: Graphically, Algebraically, Numerically, and Verbally

Functions: Graphically, Algebraically, Numerically, and VerballyIf you stack paper cups, the height increases as the number of cups increases. � ere is one and only one height for any given number of cups, so height is called a function of the number of cups. In this course you’ll refresh your memory about some kinds of functions you have studied in previous courses. You’ll also learn some new kinds of functions, and you’ll learn properties of functions so that you will be comfortable with them in later calculus courses. In this section you’ll see that you can study functions in four ways.

Work with functions that are de� ned graphically, algebraically, numerically, or verbally.

In this exploration, you’ll � nd an equation for calculating the height of a stack of paper cups.

Functions: Graphically, Algebraically, Numerically, and Verbally

1-1

Work with functions that are de� ned graphically, algebraically, numerically, or verbally.

Objective

1. Obtain several paper cups of the same kind. Measure the height of stacks containing 5, 4, 3, 2, and just 1 cup. Record the heights to the nearest 0.1 cm in a copy of this table. State what kind of cup you used.

Number Height (cm)

12345

2. Plot the points in the table on graph paper. Show the scale you are using on the vertical axis.

3. On average, by how much did the stack height increase for each cup you added? Show how you got your answer.

4. How tall would you expect a 10-cup stack to be? Show how you get your answer. Would this be twice as tall as a 5-cup stack?

5. Let x be the number of cups in a stack, and let y be the height of the stack, measured in centimeters. Write an equation for y as a function of x.

6. What is the name of the kind of function whose equation you wrote in Problem 5?

7. Show that your equation in Problem 5 gives a height close to the measured height for a stack of 3 cups.

8. Use your equation to predict the height of a stack of 35 cups. Round the answer to 1 decimal place.

9. What are the names of the processes of calculating a value within the range of the data, as in Problem 7, and outside the range of data, as in Problem 8?

10. A cup manufacturer wants to package this kind of cup in boxes that are 45 cm long and hold one stack of cups. What is the maximum number of cups the box could hold? Show how you got your answer.

11. What did you learn as a result of doing this exploration that you did not know before?

1. Obtain several paper cups of the same kind. 5. Let x be the number of cups in a stack, and x be the number of cups in a stack, and xE X P L O R AT I O N 1-1: P a p e r C u p A n a l y s i s


Discuss domain, range, and asymptote, and point them out on the graph, as shown in Figure 1-1b on page 4 . Emphasize domain and range throughout the year so that students become familiar with how to determine both. Real-world problems always have restrictions, and many of the calculus problems students solve next year will require them to consider domain and range restrictions.

Th e description of an asymptote on page 4 suggests that a graph never crosses an asymptote. Th is description is helpful for introducing students to asymptotes, but, when examined more closely, an asymptote describes end behavior. As a function approaches infi nity in the positive or negative direction, it also approaches its horizontal asymptote. Th e graph may cross the asymptote, as in the function f (x) 5 x ______

x 2 1 1 .

S e c t i o n 1-1S e c t i o n 1-1S e c t i o n 1-1S e c t i o n 1-1S e c t i o n 1-1S e c t i o n 1-1PL AN N I N G

Class Time1 day

Homework AssignmentProblems 1–5

Teaching ResourcesExploration 1-1: Paper Cup Analysis

Technology Resources

Refer to pages xv and xxi for a description of Technology Resources and a key to the technology icons.

Exploration 1-1: Paper Cup Analysis

Activity: Reading the News

TE ACH I N G

Important Terms and ConceptsFunctionExpressing mathematical ideas

graphically, algebraically, numerically, and verbally

Mathematical modelDependent variableIndependent variableDomainRangeAsymptoteExtrapolationInterpolation

Section Notes

Th is section reviews functions graphically, numerically, and algebraically. It begins with an exploration that shows that the height of a stack of cups is a function of the number of cups in the stack. Th is is a straightforward introduction to the concept of a function. It then presents a graph and equation for the relationship between coff ee temperature and time.


In previous courses you have studied linear functions, quadratic functions, exponential functions, power functions, and others. In precalculus mathematics you will learn general properties that apply to all types of functions. In particular you will learn how to transform a function so that its graph � ts real-world data. You will gain this knowledge in four ways.

� e graph at right is the graph of a quadratic function. � e y-variable could represent the height of an arrow at various times, x, a� er its release into the air. For larger time values, the quadratic function shows that y is negative. � ese values may or may not be reasonable in the real world.

� e equation of the function isy 4.9 x 2 20x 5

� is table shows corresponding x- and y-values that satisfy the equation of the function.

When the variables in a function stand for things in the real world, the function is being used as a mathematical model. The coef ficients in the equation of the function y 4.9 x 2 20x 5 have a real-world meaning. For example, the coef ficient 4.9 is a constant that is a result of the gravitational acceleration , 20 is the initial velocity, and 5 reflects the initial height of the arrow.

GRAPHICALLY

30

20

10

10

20

30

1 2 3 4 5 6x (s)

y (m)

ALGEBRAICALLY

NUMERICALLY

x (s) y (m)

0 5.0

1 20.1

2 25.4

3 20.9

VERBALLY

In previous courses you have studied linear functions, quadratic

Mathematical Overview


Functions: Graphically, Algebraically, Numerically, and VerballyIf you stack paper cups, the height increases as the number of cups increases. Th ere is one and only one height for any given number of cups, so height is called a function of the number of cups. In this course you’ll refresh your memory about some kinds of functions you have studied in previous courses. You’ll also learn some new kinds of functions, and you’ll learn properties of functions so that you will be comfortable with them in later calculus courses. In this section you’ll see that you can study functions in four ways.

Work with functions that are defi ned graphically, algebraically, numerically, or verbally.

In this exploration, you’ll fi nd an equation for calculating the height of a stack of paper cups.

FunctiNumer

1-1

Workor ve

Objective

1. Obtain several paper cups of the same kind. Measure the height of stacks containing 5, 4, 3, 2, and just 1 cup. Record the heights to the nearest 0.1 cm in a copy of this table. State what kind of cup you used.

Number Height (cm)

12345

2. Plot the points in the table on graph paper. Show the scale you are using on the vertical axis.

3. On average, by how much did the stack height increase for each cup you added? Show how you got your answer.

4. How tall would you expect a 10-cup stack to be? Show how you get your answer. Would this be twice as tall as a 5-cup stack?

5. Let x be the number of cups in a stack, and let y be the height of the stack, measured in centimeters. Write an equation for y as a function of x.

6. What is the name of the kind of function whose equation you wrote in Problem 5?

7. Show that your equation in Problem 5 gives a height close to the measured height for a stack of 3 cups.

8. Use your equation to predict the height of a stack of 35 cups. Round the answer to 1 decimal place.

9. What are the names of the processes of calculating a value within the range of the data, as in Problem 7, and outside the range of data, as in Problem 8?

10. A cup manufacturer wants to package this kind of cup in boxes that are 45 cm long and hold one stack of cups. What is the maximum number of cups the box could hold? Show how you got your answer.


1 Obtain se eral paper c ps of the same kind 5 Let x be the number of cups in a stack andxE X P L O R AT I O N 1-1: P a p e r C u p A n a l y s i s


Exploration Notes

Exploration 1-1 may be used as a time-efficient way to refresh students’ memories about functions and their use as mathematical models. 1. Answers will vary. 2. Answers will vary. 3. Answers will vary. The stack height should increase by the same amount for each additional cup. 4. Answers will vary. No, the 10-cup stack would not be twice as tall as a 5-cup stack. 5. Answers will vary. 6. Linear function 7. Answers will vary. Plug in 3 for x in your equation. 8. Answers will vary. 9. Within: interpolation; Outside: extrapolation 10. Answers will vary. 11. Answers will vary.

0.2

y

xf(x)

2


Example 1 shows you how to describe a function verbally.

� e time it takes you to get home from a football game is related to how fast you drive. Sketch a reasonable graph showing how this time and speed are related. Give the domain and range of the function.

It seems reasonable to assume that the time it takes depends on the speed you drive. So you must plot time on the vertical axis and speed on the horizontal axis.

To see what the graph should look like, consider what happens to the time as the speed varies. Pick a speed and plot a point for the corresponding time (Figure 1-1c). � en pick a faster speed. Because the time will be shorter, plot a point closer to the horizontal axis (Figure 1-1d).

A particularspeed

Correspondingtime

Tim

e

Speed A faster speed

Shorter time

Tim

e

Speed

Figure 1-1c Figure 1-1d

For a slower speed, the time will be longer. Plot a point farther from the horizontal axis (Figure 1-1e). Finally, connect the points with a smooth curve, because it is possible to drive at any speed within the speed limit. � e graph never touches either axis, as Figure 1-1f shows. If the speed were zero, you would never get home. � e length of time would be in� nite. Also, no matter how fast you drive, it will always take you some time to get home. You cannot arrive home instantaneously.

A slowerspeed

Longer time

Tim

e

Speed

Asymptotes

Never toucheseither axisTi

me

Speed

Figure 1-1e Figure 1-1f

Domain: 0 speed speed limit

Range: time minimum time at speed limit

� e problem set will help you see the relationship between variables in the real world and functions in the mathematical world.

� e time it takes you to get home from a football game is related to how fast you drive. Sketch a reasonable

EXAMPLE 1 ➤

It seems reasonable to assume that the time it takes depends on

SOLUTION

➤


If you pour a cup of co�ee, it cools more rapidly at �rst, then less rapidly, �nally approaching room temperature. You can show the relationship between co�ee temperature and time graphically. Figure 1-1a shows the temperature, y, as a function of time, x. At x 0, the co�ee has just been poured. �e graph shows that as time goes on, the temperature levels o�, until it is so close to room temperature, 20°C, that you can’t tell the di�erence.

x (min)Room temperature

302010

y ( C)

20

40

60

80

100

Figure 1-1a

�is graph might have come from numerical data, found by experiment. It actually came from an algebraic equation, y 20 70 (0.8) x .

From the equation, you can �nd numerical information. If you enter the equation into your grapher and then use the table feature, you can �nd these temperatures, rounded to 0.1°C.

Functions that are used to make predictions and interpretations about something in the real world are called mathematical models. Temperature is the dependent variable because the temperature of the co�ee depends on the time it has been cooling. Time is the independent variable. You cannot change time simply by changing co�ee temperature! Always plot the independent variable on the horizontal axis and the dependent variable on the vertical axis.

�e set of values the independent variable of a function can have is called the domain. In the co�ee cup example, the domain is the set of nonnegative numbers, or x 0. �e set of values of the dependent variable corresponding to the domain is called the range of the function. If you don’t drink the co�ee (which would end the domain), the range is the set of temperatures between 20°C and 90°C, including 90°C but not 20°C, or 20 y 90. �e horizontal line at 20°C is called an asymptote. �e word comes from the Greek asymptotos, meaning “not due to coincide.” �e graph gets arbitrarily close to the asymptote but never touches it. Figure 1-1b shows the domain, range, and asymptote.

x (min) y (°C)

0 90.0

5 42.9

10 27.5

15 22.5

20 20.8


302010

y ( C)

20

40

60

80

100

Range, 20 y 90

Asymptote

Domain, x 0Figure 1-1b


Differentiating Instruction• Pass out the list of Chapter 1

vocabulary, available at www.keypress.com/keyonline, for ELL students to look up and translate in their bilingual dictionaries.

• Consider asking pairs or groups of students to show a connection between two representations of the same concept—for example, algebraic and graphic or graphic and verbal.

• Have students do all explorations in pairs or groups.

• Enlarge Figure 1-1b and pass it out so students can draw on it while you discuss it.

• Example 1 may confuse ELL students. Consider specifying a speed and a time before generalizing the example.

• Challenging Vocabulary: asymptote, arbitrarily, and instantaneously. For asymptote, draw examples of vertical, horizontal, and oblique asymptotes. Use the term arbitrarily as you explain your drawings. Instantaneously means “in an instant.”

• Make explicit for students the connection between independent variable and domain, and between dependent variable and range, as well as the difference between integer domains and real number domains.

• A mnemonic such as that domain and range and x and y are both in alphabetical order may help students remember the relationship between them.


Example 1 shows you how to describe a function verbally.

� e time it takes you to get home from a football game is related to how fast you drive. Sketch a reasonable graph showing how this time and speed are related. Give the domain and range of the function.

It seems reasonable to assume that the time it takes depends on the speed you drive. So you must plot time on the vertical axis and speed on the horizontal axis.

To see what the graph should look like, consider what happens to the time as the speed varies. Pick a speed and plot a point for the corresponding time (Figure 1-1c). � en pick a faster speed. Because the time will be shorter, plot a point closer to the horizontal axis (Figure 1-1d).

A particularspeed

Correspondingtime

Tim

e

Speed A faster speed

Shorter time

Tim

eSpeed

Figure 1-1c Figure 1-1d

For a slower speed, the time will be longer. Plot a point farther from the horizontal axis (Figure 1-1e). Finally, connect the points with a smooth curve, because it is possible to drive at any speed within the speed limit. � e graph never touches either axis, as Figure 1-1f shows. If the speed were zero, you would never get home. � e length of time would be in� nite. Also, no matter how fast you drive, it will always take you some time to get home. You cannot arrive home instantaneously.

A slowerspeed

Longer time

Tim

e

Speed

Asymptotes

Never toucheseither axisTi

me

Speed


Domain: 0 speed speed limit

Range: time minimum time at speed limit

� e problem set will help you see the relationship between variables in the real world and functions in the mathematical world.

� e time it takes you to get home from a football game is related to how fast you drive. Sketch a reasonable

EXAMPLE 1 ➤

It seems reasonable to assume that the time it takes depends on

SOLUTION

➤


If you pour a cup of co�ee, it cools more rapidly at �rst, then less rapidly, �nally approaching room temperature. You can show the relationship between co�ee temperature and time graphically. Figure 1-1a shows the temperature, y, as a function of time, x. At x 0, the co�ee has just been poured. �e graph shows that as time goes on, the temperature levels o�, until it is so close to room temperature, 20°C, that you can’t tell the di�erence.


302010

y ( C)

20

40

60

80

100

Figure 1-1a

�is graph might have come from numerical data, found by experiment. It actually came from an algebraic equation, y 20 70 (0.8) x .

From the equation, you can �nd numerical information. If you enter the equation into your grapher and then use the table feature, you can �nd these temperatures, rounded to 0.1°C.

Functions that are used to make predictions and interpretations about something in the real world are called mathematical models. Temperature is the dependent variable because the temperature of the co�ee depends on the time it has been cooling. Time is the independent variable. You cannot change time simply by changing co�ee temperature! Always plot the independent variable on the horizontal axis and the dependent variable on the vertical axis.

�e set of values the independent variable of a function can have is called the domain. In the co�ee cup example, the domain is the set of nonnegative numbers, or x 0. �e set of values of the dependent variable corresponding to the domain is called the range of the function. If you don’t drink the co�ee (which would end the domain), the range is the set of temperatures between 20°C and 90°C, including 90°C but not 20°C, or 20 y 90. �e horizontal line at 20°C is called an asymptote. �e word comes from the Greek asymptotos, meaning “not due to coincide.” �e graph gets arbitrarily close to the asymptote but never touches it. Figure 1-1b shows the domain, range, and asymptote.

x (min) y (°C)

0 90.0

5 42.9

10 27.5

15 22.5

20 20.8


302010

y ( C)

20

40

60

80

100

Range, 20 y 90

Asymptote

Domain, x 0Figure 1-1b


Technology Notes

Exploration 1-1: Paper Cup Analysis has students collect data on the heights of stacks of paper cups and then fi t lines to their data. Students can do this exploration with Fathom. Th ey enter the data into a case table, make a scatter plot, test conjectured functions by plotting them, and use tracing to make a prediction.

Activity: Reading the News in Teaching Mathematics with Fathom has students collect data on how long it takes to read a newspaper, make a scatter plot, and fi t a line to the data in order to make predictions. Th is activity also provides a straightforward introduction to Fathom. Students can review linear equations by experimenting with more activities in Chapter 1 of Teaching Mathematics with Fathom. Allow 25–40 minutes.

CAS Suggestions

Consider presenting some key CAS functionality. In particular, demonstrate how to defi ne and evaluate functions and solve equations.

A key concept when using a CAS is to recognize that once defi ned, a function can be named, manipulated, evaluated, and solved by reference to its name alone. When using a CAS, students need to enter the details of the function expression only once, referring to its name for the rest of the problem. Th is is a critical technology-related aspect to thinking algebraically about problem solving.


3. Mortgage Payment Problem: People who buy houses usually get a loan to pay for most of the house and make payments on the resulting mortgage each month. Suppose you get a $150,000 loan and pay it back at $1,074.64 per month with an interest rate of 6% per year (0.5% per month). Your balance, B, in dollars, a�er n monthly payments is given by the algebraic equation

B 150,000 1.00 5 n 1074.64 _______ 0.005 (1 1.005 n )

a. Make a table of your balances at the end of each 12 months for the �rst 10 years of the mortgage. To save time, use the table feature of your grapher to do this.

b. How many months will it take you to pay o� the entire mortgage? Show how you get your answer.

c. Plot on your grapher the graph of B as a function of n from n 0 until the mortgage is paid o�. Sketch the graph on your paper.

d. True or false: “A�er half the payments have been made, half the original balance remains to be paid.” Show that your conclusion agrees with your graph from part c.

e. Give the domain and range of this function. Explain why the domain contains only integers.

4. Stopping Distance Problem: �e distance your car takes to stop depends on how fast you are going when you apply the brakes. You may recall from driver’s education that it takes more than twice the distance to stop your car if you double your speed.

a. Sketch a reasonable graph showing your stopping distance as a function of speed.

b. What is a reasonable domain for this function?

c. Consult a driver’s manual, the Internet, or another reference source to see what the stopping distance is for the maximum speed you stated for the domain in part b.

d. When police investigate an automobile

accident, they estimate the speed the car was going by measuring the length of the skid marks. Which are they considering to be the independent variable, the speed or the length of the skid marks? Indicate how this would be done by drawing arrows on your graph from part a.

5. Stove Heating Element Problem: When you turn on the heating element of an electric stove, the temperature increases rapidly at �rst, then levels o�. Sketch a reasonable graph showing temperature as a function of time. Show the horizontal asymptote. Indicate on the graph the domain and range.

6. In mathematics you learn things in four ways—algebraically, graphically, numerically, and verbally.

a. In which of Problems 1–5 was the function given algebraically? Graphically? Numerically? Verbally?

b. In which of Problems 1–5 did you go from verbal to graphical? From algebraic to numerical? From numerical to graphical? From graphical to algebraic? From graphical to numerical? From algebraic to graphical?


1. Archery Problem: An archer climbs a tree near the edge of a cli� , then shoots an arrow high into the air. � e arrow goes up, then comes back down, going over the cli� and landing in the valley, 30 m below the top of the cli� . � e arrow’s height, y, in meters above the top of the cli� depends on the time, x, in seconds since the archer released it. Figure 1-1g shows the height as a function of time.

y (m)

1 2 3 4 5 610

10

20

30

20

30

x (s)

Figure 1-1g

a. What was the approximate height of the arrow at 1 s? At 5 s? How do you explain the fact that the height is negative at 5 s?

b. At what two times was the arrow 10 m above the ground? At what time does the arrow land in the valley below the cli� ?

c. How high was the archer above the ground when she released the arrow?

d. Why can you say that height is a function of time? Why is time not a function of height?

e. What is the domain of the function? What is the corresponding range?

2. Gas Temperature and Volume Problem: When you heat a � xed amount of gas, it expands, increasing its volume. In the late 1700s, French chemist Jacques Charles used numerical measurements of the temperature and volume of a gas to � nd a quantitative relationship between these two variables. Suppose that these temperatures and volumes had been recorded for a � xed amount of oxygen.

Jacques Charles invented the hydrogen balloon. He participated in the � rst manned balloon � ight in 1783.

a. On graph paper, plot V as a function of T. Choose scales that go at least from T 300 to T 400, and from V 0 to V 35. You should � nd that the points lie almost in a straight line. With a ruler, construct the best-� tting line you can for these points. Extend the line to the le� until it crosses the T-axis and to the right to T 400.

b. From your graph, read the approximate volumes at T 400 and T 30. Read the approximate temperature at which V 0. How does this temperature compare with absolute zero, the temperature at which molecular motion stops?

c. Finding a value of a variable beyond all given data points is called extrapolation. Extra- means “beyond,” and pol- comes from “pole,” or end. Finding a value between two given data points is called interpolation. Which of the three values in part b did you � nd by extrapolation and which by interpolation?

d. Why can you say that volume is a function of temperature? Is temperature also a function of volume? Explain.

e. Considering volume to be a function of temperature, write the domain and the range of this function.

T (°C) V (L)

0 9.550 11.2

100 12.9150 14.7200 16.4250 18.1300 19.9

1. Archery Problem: An archer climbs a tree near Jacques Charles invented

Exploratory Problem Set 1-1


2b. Answers will vary. V (400) 23, V(30) 11, and V(T) 5 0 when T 2273. Absolute zero is about 2273°C.2c. Extrapolation: V(400) and T such that V(T) 5 0; interpolation: V(30).2d. Th ere is only one volume for a given temperature; yes, because there is only one temperature for a given volume.2e. Domain: x 2273; range: y 0.

Problem 3 provides an equation from which students create both a table and a graph on a grapher. In this problem, the number of months is the independent variable and the balance is the dependent variable. You might ask students to explain why it would be inappropriate to say that the number of months depends on the number of dollars.

PRO B LE M N OTESProblem 1 requires students to analyze and interpret a graph, as well as provide verbal descriptions and explanations of terminology. 1a. 20 m; 217.5 m; it is below the top of the cliff . 1b. 0.3 s; 3.8 s; 5.3 s1c. 5 m 1d. Th ere is only one altitude for any given time; some altitudes correspond to more than one time.1e. Domain: 0 x 5.3; range: 230 y 25.

Problem 2 lists a table of values that students must hand-plot on graph paper. Although a best-fi t line is not formally defi ned, students are asked to draw one in Problem 2a. Students frequently think a best-fi t line needs to go through the fi rst and last points. Th is is not necessarily true. Th is problem introduces the terms extrapolation and interpolation. Help students understand that extrapolation involves using the pattern in the data to estimate values beyond the given values, whereas interpolation involves estimating values between given data values. Caution students that extrapolated results may not be valid because the pattern may not exist beyond the given data.2a. Th is graph also shows the answer for part b.

�300 450

10

�5

20

35V (liters)

T (�C)


3. Mortgage Payment Problem: People who buy houses usually get a loan to pay for most of the house and make payments on the resulting mortgage each month. Suppose you get a $150,000 loan and pay it back at $1,074.64 per month with an interest rate of 6% per year (0.5% per month). Your balance, B, in dollars, a�er n monthly payments is given by the algebraic equation

B 150,000 1.00 5 n 1074.64 _______ 0.005 (1 1.005 n )

a. Make a table of your balances at the end of each 12 months for the �rst 10 years of the mortgage. To save time, use the table feature of your grapher to do this.

b. How many months will it take you to pay o� the entire mortgage? Show how you get your answer.

c. Plot on your grapher the graph of B as a function of n from n 0 until the mortgage is paid o�. Sketch the graph on your paper.

d. True or false: “A�er half the payments have been made, half the original balance remains to be paid.” Show that your conclusion agrees with your graph from part c.

e. Give the domain and range of this function. Explain why the domain contains only integers.

4. Stopping Distance Problem: �e distance your car takes to stop depends on how fast you are going when you apply the brakes. You may recall from driver’s education that it takes more than twice the distance to stop your car if you double your speed.

a. Sketch a reasonable graph showing your stopping distance as a function of speed.

b. What is a reasonable domain for this function?

c. Consult a driver’s manual, the Internet, or another reference source to see what the stopping distance is for the maximum speed you stated for the domain in part b.

d. When police investigate an automobile

accident, they estimate the speed the car was going by measuring the length of the skid marks. Which are they considering to be the independent variable, the speed or the length of the skid marks? Indicate how this would be done by drawing arrows on your graph from part a.

5. Stove Heating Element Problem: When you turn on the heating element of an electric stove, the temperature increases rapidly at �rst, then levels o�. Sketch a reasonable graph showing temperature as a function of time. Show the horizontal asymptote. Indicate on the graph the domain and range.

6. In mathematics you learn things in four ways—algebraically, graphically, numerically, and verbally.

a. In which of Problems 1–5 was the function given algebraically? Graphically? Numerically? Verbally?

b. In which of Problems 1–5 did you go from verbal to graphical? From algebraic to numerical? From numerical to graphical? From graphical to algebraic? From graphical to numerical? From algebraic to graphical?


1. Archery Problem: An archer climbs a tree near the edge of a cli� , then shoots an arrow high into the air. � e arrow goes up, then comes back down, going over the cli� and landing in the valley, 30 m below the top of the cli� . � e arrow’s height, y, in meters above the top of the cli� depends on the time, x, in seconds since the archer released it. Figure 1-1g shows the height as a function of time.

y (m)

1 2 3 4 5 610

10

20

30

20

30

x (s)

Figure 1-1g

a. What was the approximate height of the arrow at 1 s? At 5 s? How do you explain the fact that the height is negative at 5 s?

b. At what two times was the arrow 10 m above the ground? At what time does the arrow land in the valley below the cli� ?

c. How high was the archer above the ground when she released the arrow?

d. Why can you say that height is a function of time? Why is time not a function of height?

e. What is the domain of the function? What is the corresponding range?

2. Gas Temperature and Volume Problem: When you heat a � xed amount of gas, it expands, increasing its volume. In the late 1700s, French chemist Jacques Charles used numerical measurements of the temperature and volume of a gas to � nd a quantitative relationship between these two variables. Suppose that these temperatures and volumes had been recorded for a � xed amount of oxygen.

Jacques Charles invented the hydrogen balloon. He participated in the � rst manned balloon � ight in 1783.

a. On graph paper, plot V as a function of T. Choose scales that go at least from T 300 to T 400, and from V 0 to V 35. You should � nd that the points lie almost in a straight line. With a ruler, construct the best-� tting line you can for these points. Extend the line to the le� until it crosses the T-axis and to the right to T 400.

b. From your graph, read the approximate volumes at T 400 and T 30. Read the approximate temperature at which V 0. How does this temperature compare with absolute zero, the temperature at which molecular motion stops?

c. Finding a value of a variable beyond all given data points is called extrapolation. Extra- means “beyond,” and pol- comes from “pole,” or end. Finding a value between two given data points is called interpolation. Which of the three values in part b did you � nd by extrapolation and which by interpolation?

d. Why can you say that volume is a function of temperature? Is temperature also a function of volume? Explain.

e. Considering volume to be a function of temperature, write the domain and the range of this function.

T (°C) V (L)

0 9.550 11.2

100 12.9150 14.7200 16.4250 18.1300 19.9

1. Archery Problem: An archer climbs a tree near Jacques Charles invented

Exploratory Problem Set 1-1

7

defi ned, it can be graphed by typing the function name and changing the domain variable to x.

Problem 4 gives a description of a situation from which students draw a reasonable graph and then practice using vocabulary.

Problem 5 requires students to draw a reasonable graph to match a given description.

Additional CAS Problems

1. Use Boolean operators in Problem 3 to determine whether the balance at the midpoint in time is higher or lower than half of the original balance.

2. In how many months does the account balance in Problem 3 become half of the original balance?Note: Th is problem is diffi cult to solve by hand, but it is a classic situation in which the CAS enables the student to remain focused on the problem without getting lost in the algebra required to solve it.

3b. Changing Tbl to 1 shows that the balance becomes negative at the end of month 241, so the balance will become 0 during month 241. 3c., 3d. False

3e. Domain: 0 x 241, x is an integer; range: 0 y 150,000.

Problem 3 is eff ective for students who learn to phrase the question correctly using a CAS. Students can use a CAS to evaluate the values requested in Problem 3a directly. Th e key point for Problem 3b is to recognize that the balance is zero. A CAS uses this fact to solve the balance equation for the payment number at which the balance is zero. Because the function has already been 100 200

100,000

x

y

Section 1-1: Functions: Graphically, Algebraically, Numerically, and Verbally

See page 973 for answers to Problems 3a, 4–6 and CAS Problems 1 and 2.

9Section 1-2: Types of Functions

for x, the independent variable. For instance, to substitute 4 for x in the quadratic function f (x) x 2 5x 3, you would write

f (4) 4 2 5(4) 3 7

� e symbol f (4) is pronounced “ f of 4” or sometimes “ f at 4.” You must recognize that the parentheses mean substitution and not multiplication.

� is notation is also useful if you are working with more than one function of the same independent variable. For instance, the height and velocity of a falling object both depend on time, t, so you could write the equations of the two functions this way:

h(t) 4.9t 2 10t 70 (for the height)

v(t) 9.8t 10 (for the velocity)

In f (x), the variable x or any value substituted for x is called the argument of the function. It is important to distinguish between f and f (x). � e symbol f is the name of the function. � e symbol f (x) is the y-value of the function. For instance, if f is the square root function, then f (x)

__ x and f (9)

__ 9 3. Note

that the re� exive axiom, x x, requires that you substitute the same number for x everywhere it appears in an expression or equation. It would be improper format to write f (x)

__ 9 if you have substituted 9 for x.

Names of FunctionsFunctions are named for the operation performed on the independent variable. Here are some types of functions you may recall from previous courses, along with their typical graphs. In these examples, the letters a, b, c, m, and n stand for constants. � e symbols x and f (x) stand for variables, x for the independent variable and f (x) for the dependent variable.

Polynomial function, Figure 1-2c

General equation: f (x) a n x n a n 1 x n 1 . . . a 1 x a 0 , where n is a nonnegative integer

Verbally: f (x) is a polynomial function of x. (If n 3, f is a cubic function. If n 4, f is a quartic function.)

Features: � e graph crosses the x-axis up to n times and has up to n 1 vertices (points where the function changes direction). � e domain is all real numbers.

Quadratic function, Figure 1-2d (a special case of a polynomial function)

General equation: f (x) a x 2 bx c, a 0

Verbally: f (x) varies quadratically with x, or f (x) is a quadratic function of x.

Features: � e graph changes direction at its one vertex. � e domain is all real numbers.

f(x)

x

Figure 1-2c

f(x)

x

Figure 1-2d

De� nition of FunctionIf you plot the function y x 2 5x 3, you get a graph that rises and then falls, as shown in Figure 1-2a. For any x-value you pick, there is only one y-value. � is is not the case for all graphs. For example, in Figure 1-2b there are places where the graph has more than one y-value for the same x-value. Although the two variables are related, the relation is not a function.

y

No two y-valuesfor the same x-value

A function

y-intercept x

x-intercepts

Figure 1-2a Figure 1-2b

Each point on a graph corresponds to an ordered pair of numbers, (x, y). A relation is any set of ordered pairs. A function is a set of ordered pairs for which each value of the independent variable (o� en x) in the domain has only one corresponding value of the dependent variable (o� en y) in the range. So Figures 1-2a and 1-2b are both graphs of relations, but only Figure 1-2a is the graph of a function.

� e y-intercept of a function is the value of y when x 0. It gives the place where the graph crosses the y-axis (Figure 1-2a). An x-intercept is a value of x for which y 0. Functions can have more than one x-intercept.

f (x) TerminologyYou should recall f (x) notation from previous courses. It is used for y, the dependent variable of a function. With it, you show what value you substitute

y

More than oney-value for thesame x-value

Not a function

x

Types of FunctionsIn the previous section you learned that you can describe functions algebraically, numerically, graphically, or verbally. A function de� ned by an algebraic equation o� en has a descriptive name. For instance, the function y x 2 5x 3 is called quadratic, from the latin word quadratum, meaning square, because the function is a polynomial whose highest power of x is x squared and quadrangle is one term for a square. In this section you will refresh your memory about verbal names for algebraically de� ned functions and see what their graphs look like.

Make connections among the algebraic equation for a function, its name, and its graph.

Types of FunctionsIn the previous section you learned that you can describe functions algebraically,

1-2


Objective




Class Time1 day

Homework Assignment

Problems 1–39 odd, 40–42

Teaching ResourcesExploration 1-2a: Names of FunctionsExploration 1-2b: Restricted Domains

and Boolean VariablesSupplementary Problems


Exploration 1-2a: Names of Functions

Exploration 1-2b: Restricted Domains and Boolean Variables

CAS Activity 1-2a: Finding Polynomial Equations

CAS Activity 1-2b: Functions Defi ned by Two Points

TE ACH I N G

Important Terms and ConceptsOrdered pairRelationFunctiony-interceptx-interceptf (x) terminologyArgument of a functionName of a functionConstantVariablePolynomial functionQuadratic functionLinear functionDirect variation functionPower functionExponential functionInverse variation functionRational algebraic functionBoolean variableRestricted domainVertical line test

Section Notes

Begin the section by reviewing the defi nition of function. Make sure students understand that for a relation to be a function, it must be true that for each value in the domain there is only one corresponding value in the range. In other words, the correspondence from the domain to the range must be unique. It is possible for the same range value to correspond to more than one domain

value. To illustrate this, you might discuss the function y 5 x 2 . For this function, every nonzero value in the range corresponds to two values in the domain. For example, the range value 9 corresponds to the domain values 23 and 3.

Students should be familiar with function notation from previous courses. You may want to discuss briefl y this notation, emphasizing that the parentheses in f (x) mean substitution, not multiplication.


for x, the independent variable. For instance, to substitute 4 for x in the quadratic function f (x) x 2 5x 3, you would write

f (4) 4 2 5(4) 3 7

� e symbol f (4) is pronounced “ f of 4” or sometimes “ f at 4.” You must recognize that the parentheses mean substitution and not multiplication.

� is notation is also useful if you are working with more than one function of the same independent variable. For instance, the height and velocity of a falling object both depend on time, t, so you could write the equations of the two functions this way:

h(t) 4.9t 2 10t 70 (for the height)

v(t) 9.8t 10 (for the velocity)

In f (x), the variable x or any value substituted for x is called the argument of the function. It is important to distinguish between f and f (x). � e symbol f is the name of the function. � e symbol f (x) is the y-value of the function. For instance, if f is the square root function, then f (x)

__ x and f (9)

__ 9 3. Note

that the re� exive axiom, x x, requires that you substitute the same number for x everywhere it appears in an expression or equation. It would be improper format to write f (x)

__ 9 if you have substituted 9 for x.

Names of FunctionsFunctions are named for the operation performed on the independent variable. Here are some types of functions you may recall from previous courses, along with their typical graphs. In these examples, the letters a, b, c, m, and n stand for constants. � e symbols x and f (x) stand for variables, x for the independent variable and f (x) for the dependent variable.

Polynomial function, Figure 1-2c

General equation: f (x) a n x n a n 1 x n 1 . . . a 1 x a 0 , where n is a nonnegative integer

Verbally: f (x) is a polynomial function of x. (If n 3, f is a cubic function. If n 4, f is a quartic function.)

Features: � e graph crosses the x-axis up to n times and has up to n 1 vertices (points where the function changes direction). � e domain is all real numbers.

Quadratic function, Figure 1-2d (a special case of a polynomial function)

General equation: f (x) a x 2 bx c, a 0

Verbally: f (x) varies quadratically with x, or f (x) is a quadratic function of x.

Features: � e graph changes direction at its one vertex. � e domain is all real numbers.

f(x)

x

Figure 1-2c

f(x)

x

Figure 1-2d

De� nition of FunctionIf you plot the function y x 2 5x 3, you get a graph that rises and then falls, as shown in Figure 1-2a. For any x-value you pick, there is only one y-value. � is is not the case for all graphs. For example, in Figure 1-2b there are places where the graph has more than one y-value for the same x-value. Although the two variables are related, the relation is not a function.

y

No two y-valuesfor the same x-value

A function

y-intercept x

x-intercepts

Figure 1-2a Figure 1-2b

Each point on a graph corresponds to an ordered pair of numbers, (x, y). A relation is any set of ordered pairs. A function is a set of ordered pairs for which each value of the independent variable (o� en x) in the domain has only one corresponding value of the dependent variable (o� en y) in the range. So Figures 1-2a and 1-2b are both graphs of relations, but only Figure 1-2a is the graph of a function.

� e y-intercept of a function is the value of y when x 0. It gives the place where the graph crosses the y-axis (Figure 1-2a). An x-intercept is a value of x for which y 0. Functions can have more than one x-intercept.

f (x) TerminologyYou should recall f (x) notation from previous courses. It is used for y, the dependent variable of a function. With it, you show what value you substitute

y

More than oney-value for thesame x-value

Not a function

x

Types of FunctionsIn the previous section you learned that you can describe functions algebraically, numerically, graphically, or verbally. A function de� ned by an algebraic equation o� en has a descriptive name. For instance, the function y x 2 5x 3 is called quadratic, from the latin word quadratum, meaning square, because the function is a polynomial whose highest power of x is x squared and quadrangle is one term for a square. In this section you will refresh your memory about verbal names for algebraically de� ned functions and see what their graphs look like.


Types of FunctionsIn the previous section you learned that you can describe functions algebraically,

1-2


Objective


9

a polynomial function with only a linear term. The graph of a direct variation function is a straight line through the origin. An inverse variation function is a power function with a negative exponent. The graph of an inverse variation function has asymptotes at both axes. Students may be familiar with direct and inverse variation functions from their science classes. To help students transfer learning and increase their understanding of both precalculus and science, you might ask them for examples of equations used in science that are direct or inverse variations. For example, Newton’s second law of motion, F 5 ma, is a direct variation if the mass, m, or the acceleration, a, is constant. If the force, F, is constant, the equation is an inverse variation.

Example 1 on page 12 illustrates how to plot a function with a restricted domain using a grapher. This kind of problem prepares students for the piecewise functions (functions composed of two or more functions) they will encounter in future courses. Students may need help understanding the idea of a Boolean variable. A Boolean variable is represented by a condition, not a letter like the variables students are familiar with. A Boolean variable has only two possible values: It is equal to 1 if the condition is true, and it is equal to 0 if the condition is false.

In the given examples, only part of the function is divided by the Boolean variable because it is easier to enter into a grapher. That is, the functions are written in the form f 1 (x) 5 3x 1 26/ (x 3 and x 10) rather than f 1 (x) 5 (3x 1 26)/(x 3 and x 10). You might want to ask students why the two forms yield the same graph. They are both undefined when the Boolean variable is zero.

The Names of Functions subsection presents the graphs and equations of eight types of functions. To help students review these functions, you might assign Exploration 1-2a. Some students have difficulty distinguishing between power functions and exponential functions. The location of the variable is the key to the difference. Power functions have a variable base, whereas exponential functions have a variable exponent. The confusion arises

because some students do not distinguish between exponent and power. Actually, in the expression 5 3 , for instance, the exponent is only the 3. The power is the entire expression, 5 3 . Thus, 3 x has a variable as its exponent, while x 3 is a power of x.

A direct variation function is a special case of a linear function, power function, and polynomial function. Specifically, it is a linear function with y-intercept 0, a power function with an exponent of 1, and

Section 1-2: Types of Functions


Inverse variation function, Figure 1-2i (a special case of a power function)

General equation: f (x) a __ x or f (x) a x 1

or f (x) a __ x n or f (x) a x n , a 0, n 0

Verbally: f (x) varies inversely with x (or with the nth power of x). Alternatively, f (x) is inversely proportional to x (or to the nth power of x).

Features: Both of the axes are asymptotes. � e domain depends on the value of n. For positive integer values of n, the domain is x 0. For most real-world applications, the domain is x 0.

Rational algebraic function, Figure 1-2j

General equation: f (x) n (x) ___ d (x) , where n and d are polynomial functions

Verbally: f (x) is a rational function of x.

Features: A rational function has a discontinuity (asymptote or missing point) where the denominator is zero; it may have horizontal or other asymptotes.

Restricted Domains and Boolean VariablesSuppose that you want to plot a graph using only part of your grapher’s window. For instance, let the height of a growing child between ages 3 and 10 be given by y 3x 26, where x is age in years and y is height in inches. � e domain here is 3 x 10.

Some graphers allow you to enter a restricted domain directly. Other graphers require you to use Boolean variables. A Boolean variable, named for George Boole, an Irish logician and mathematician (1815–1864), equals 1 if a given condition is true and 0 if that condition is false. For instance, the compound statement

(x 3 and x 10)

equals 1 if x 7 (which is between 3 and 10) and equals 0 if x 2 or x 15 (neither of which is between 3 and 10). To plot a graph in a restricted domain using Boolean variables, divide any term of the equation by the appropriate Boolean variable. For the equation above, enter

f 1 (x) 3x 26 / (x 3 and x 10)

If x is between 3 and 10, inclusive, the 26 in 3x 26 is divided by 1, which leaves it unchanged. If x is not between 3 and 10, inclusive, the 26 in 3x 26 is divided by 0 and the grapher plots nothing.

f(x)

x

If n is oddFigure 1-2i

y f(x)

x

Removablediscontinuity

Asymptotes

Figure 1-2j

What if n is even?For example, y = 1 __ x 2

x

y

Linear function, Figure 1-2e (another special case of a polynomial function)

General equation: f (x) ax b (or f (x) mx b)

Verbally: f (x) varies linearly with x, or f (x) is a linear function of x.

Features: � e straight-line graph, f (x), changes at a constant rate as x changes. � e domain is all real numbers.

Direct variation function, Figure 1-2f (a special case of a linear, power, or polynomial function)

General equation: f (x) ax or f (x) mx 0, or f (x) a x 1

Verbally: f (x) varies directly with x, or f (x) is directly proportional to x.

Features: � e straight-line graph goes through the origin. � e domain is all real numbers. However, for most real-world applications, you will use the domain x 0 (as shown).

Power function, Figure 1-2g (a polynomial function if b is a nonnegative integer)

General equation: f (x) a x b (a variable with a constant exponent), a 0, b 0

Verbally: f (x) varies directly with the bth power of x, or f (x) is directly proportional to the bth power of x.

Features: � e domain depends on the value of b. For positive integer values of b, the domain is all real numbers; for negative integer values of b, the domain is x 0. In most real-world applications, the domain is x 0 if b 0 and x 0 if b 0.

f(x)

xIf b is positive

f(x)

xIf b is negative

Figure 1-2g

Exponential function, Figure 1-2h

General equation: f (x) a b x (a constant with a variable exponent), a 0, b 0, b 1

Verbally: f (x) varies exponentially with x, or f (x) is an exponential function of x.

Features: � e graph crosses the y-axis at f (0) a and has the x-axis as an asymptote.

f(x)

x

Figure 1-2e

f(x)

x

Figure 1-2f

f(x)

xa

Figure 1-2h


How about if 0 < b < 1? For example, y = 1 _ 2

x

x

y

1

10.5


Section Notes (continued)

Note: If your students use TI-Nspire

graphers, they can plot piecewise functions by entering them directly into their graphers using a template. See the instruction manual for the TI-Nspire for more information.

Example 1 also asks students to give the range of the function. Students sometimes mistakenly try to determine the range of a function by substituting the endpoints of the domain. Point out that, in this example, the endpoints of the range do not correspond to the endpoints of the domain.

Diff erentiating Instruction• For Romance language speakers, a

mnemonic for quadratic is the word for square (cuadrado, quadrat, etc.).

• Intercept and intersect may be confusing to students because they sound alike and have related meanings. Also, some languages do not make the distinction between exponent and power that is given in the text.

• Make sure students understand how to say f (x) and what it means. Tie this in to the term argument with examples such as f ( x 2 ), f (  

__ x ) and f (4).

• Note that graphers are not allowed in many countries for students this age.

• In Problems 5–18, help ELL students new to the U.S. (or to your school) learn to use their grapher, and consider assigning them a partner who is a strong grapher user.

• Problems 39 and 40 introduce important concepts that are language-heavy. Many ELL students will need help with the vocabulary.

• Problem 42 may be diffi cult for ELL students. Consider letting them answer in their primary language.

Exploration Notes

Exploration 1-2a provides a short summary review of seven of the eight functions covered in the text (the direct variation function is not included). Students working in groups with the aid of graphers can complete the seven questions in 15–20 minutes. It’s probably best to summarize the exploration and answer questions on it before presenting the examples from the textbook.

Exploration 1-2b lets students see a restricted domain in a real-world context. Th e gravitational attraction for an object above Earth’s surface is inversely proportional to the square of the object’s distance from the center of Earth. Below the surface, the attraction is directly proportional to the distance. Students who have graphers using Boolean variables will fi nd the exploration helpful. Th is follow-up exploration takes about 15 minutes.


Inverse variation function, Figure 1-2i (a special case of a power function)

General equation: f (x) a __ x or f (x) a x 1

or f (x) a __ x n or f (x) a x n , a 0, n 0

Verbally: f (x) varies inversely with x (or with the nth power of x). Alternatively, f (x) is inversely proportional to x (or to the nth power of x).

Features: Both of the axes are asymptotes. � e domain depends on the value of n. For positive integer values of n, the domain is x 0. For most real-world applications, the domain is x 0.

Rational algebraic function, Figure 1-2j

General equation: f (x) n (x) ___ d (x) , where n and d are polynomial functions

Verbally: f (x) is a rational function of x.

Features: A rational function has a discontinuity (asymptote or missing point) where the denominator is zero; it may have horizontal or other asymptotes.

Restricted Domains and Boolean VariablesSuppose that you want to plot a graph using only part of your grapher’s window. For instance, let the height of a growing child between ages 3 and 10 be given by y 3x 26, where x is age in years and y is height in inches. � e domain here is 3 x 10.

Some graphers allow you to enter a restricted domain directly. Other graphers require you to use Boolean variables. A Boolean variable, named for George Boole, an Irish logician and mathematician (1815–1864), equals 1 if a given condition is true and 0 if that condition is false. For instance, the compound statement

(x 3 and x 10)

equals 1 if x 7 (which is between 3 and 10) and equals 0 if x 2 or x 15 (neither of which is between 3 and 10). To plot a graph in a restricted domain using Boolean variables, divide any term of the equation by the appropriate Boolean variable. For the equation above, enter

f 1 (x) 3x 26 / (x 3 and x 10)

If x is between 3 and 10, inclusive, the 26 in 3x 26 is divided by 1, which leaves it unchanged. If x is not between 3 and 10, inclusive, the 26 in 3x 26 is divided by 0 and the grapher plots nothing.

f(x)

x

If n is oddFigure 1-2i

y f(x)

x

Removablediscontinuity

Asymptotes

Figure 1-2j

What if n is even?For example, y = 1 __ x 2

x

y

Linear function, Figure 1-2e (another special case of a polynomial function)

General equation: f (x) ax b (or f (x) mx b)

Verbally: f (x) varies linearly with x, or f (x) is a linear function of x.

Features: � e straight-line graph, f (x), changes at a constant rate as x changes. � e domain is all real numbers.

Direct variation function, Figure 1-2f (a special case of a linear, power, or polynomial function)

General equation: f (x) ax or f (x) mx 0, or f (x) a x 1

Verbally: f (x) varies directly with x, or f (x) is directly proportional to x.

Features: � e straight-line graph goes through the origin. � e domain is all real numbers. However, for most real-world applications, you will use the domain x 0 (as shown).

Power function, Figure 1-2g (a polynomial function if b is a nonnegative integer)

General equation: f (x) a x b (a variable with a constant exponent), a 0, b 0

Verbally: f (x) varies directly with the bth power of x, or f (x) is directly proportional to the bth power of x.

Features: � e domain depends on the value of b. For positive integer values of b, the domain is all real numbers; for negative integer values of b, the domain is x 0. In most real-world applications, the domain is x 0 if b 0 and x 0 if b 0.

f(x)

xIf b is positive

f(x)

xIf b is negative

Figure 1-2g

Exponential function, Figure 1-2h

General equation: f (x) a b x (a constant with a variable exponent), a 0, b 0, b 1

Verbally: f (x) varies exponentially with x, or f (x) is an exponential function of x.

Features: � e graph crosses the y-axis at f (0) a and has the x-axis as an asymptote.

f(x)

x

Figure 1-2e

f(x)

x

Figure 1-2f

f(x)

xa

Figure 1-2h


How about if 0 < b < 1? For example, y = 1 _ 2

x

x

y

1

10.5

11

CAS Activity 1-2b: Functions Defi ned by Two Points in the Instructor’s Resource Book has students use Solve and Factor commands on a CAS to fi nd linear, power, and exponential functions determined by two points. Allow 20–25 minutes.

CAS Suggestions

Introduce students to the CAS functionality that solves systems of equations at this time.

Students can fi nd intercepts by setting all other variables equal to zero. Using this idea, x-intercepts can be computed with commands like Solve(f(x) = 0, x) or

Solve(y = f(x), x) y = 0. Th e symbol on the TI-Nspire CAS substitutes any equation that follows the symbol into the preceding command. Th e latter command shows that the student is solving the original function for x when y 5 0.

Th e command can be used to restrict the domain of a function in its equation or on its graph. Th e fi gure shows the same graph as Example 1 with the domain restriction entered alongside the function defi nition.

Th e command can also be used to defi ne functions algebraically. Once defi ned, a CAS evaluates values within the domain values, but returns an undefi ned statement when the input value is outside the domain, even if the input value is defi ned when the domain is unrestricted.

Technology Notes

Exploration 1-2a: Names of Functions in the Instructor’s Resource Book can be done with Sketchpad.

Exploration 1-2b: Restricted Domains and Boolean Variables in the Instructor’s Resource Book can be accomplished using Fathom, but only if students employ nested if-then statements rather than Boolean variables.

CAS Activity 1-2a: Finding Polynomial Equations in the Instructor’s Resource Book, has students use a CAS to fi nd equations for polynomials using systems of equations. Students use a Solve command to explore how many coordinates are needed to fi nd an equation for an nth-degree polynomial. Allow 15–20 minutes.



For Problems 1–4, a. Plot the graph on your grapher using the

given domain. Sketch the result on your paper. b. Give the range of the function. c. Name the kind of function. d. Describe a pair of real-world variables that

could be related by a graph of this shape. 1. f (x) 2x 3 domain: 0 x 10 2. f (x) 0.2 x 3 domain: 0 x 4 3. g (x) 12 ___ x domain: 0 x 10

4. h (x) 5 0. 6 x domain: 5 x 5

For Problems 5–18, a. Plot the graph using a window set to show the

entire graph, when possible. Sketch the result. b. Give the y-intercept and any x-intercepts and

the locations of any asymptotes. c. Give the range.

5. Quadratic (polynomial) function f (x) x 2 4x 12 with the domain 0 x 5


7. Cubic (polynomial) function f (x) x 3 7 x 2 4x 12 with the domain 1 x 7

8. Quartic (polynomial) function f (x) x 4 3 x 3 8 x 2 12x 16 with the domain 3 x 3

9. Power function f (x) 3 x 2/3 with the domain 0 ≤ x ≤ 8

10. Power function f (x) 0.3 x 1.5 with the domain 0 x 9

11. Linear function f (x) 0.7x 4 with the domain 3 x 10

12. Linear function f (x) 3x 6 with the domain 5 x 5

13. Exponential function f (x) 3 1.3 x with the domain 5 x 5


15. Inverse-square variation function f (x) 25 __ x 2

with the domain x 0

16. Direct variation function f (x) 5x with the domain x 0

17. Rational function y x 2 __________ (x 4)(x 1) with the domain 3 x 6, x 4, x 1

18. Rational function y x 2 2x 2 ________ x 3 with the domain 2 x 6, x 3

For Problems 19–28, name the type of function that has the graph shown. 19.

x

y 20.

x

y

21.

x

y 22.

x

y

23.

x

y 24. y

x

25.

x

y 26.

x

y

27.

x

y 28.

x

y

For Problems 1–4, Exponential function f ( ) x with the x with the x

Problem Set 1-2Plot the graph of f (x) x 2 5x 3 in the domain 0 x 4. What kind of function is this? Give the range. Find a pair of real-world variables that could have a relationship described by a graph of this shape.

Enter the equation with restricted domain into your grapher directly. Or, to use Boolean variables, enter

f 1 (x) x 2 5x 3 / (x 0 and x 4)

� e graph in Figure 1-2k shows the restricted domain.

� e function is quadratic because f (x) equals a second-degree polynomial in x.

� e range is 3 f (x) 9.25. You can � nd this interval by tracing to the le� endpoint of the graph where f (0) 3 and to the high point where f (2.5) 9.25. (At the right endpoint, f (4) 7, which is between 3 and 9.25.)

� e function could represent the relationship between something that rises for a while and then falls, such as a punted football’s height as a function of time or (if f (x) is multiplied by 10) the grade you could get on a test as a function of the number of hours you study for it. (� e grade could be lower for longer times if you stay up too late and thus are sleepy during the test.)

DEFINITION: Boolean VariablesA Boolean variable is a variable that has a given condition attached to it. If the condition is true, the variable equals 1. If the condition is false, the variable equals 0.

As children grow older, their height and weight are related. Sketch a reasonable graph to show this relation and then describe it. Identify what kind of function has a graph like the one you drew.

Weight depends on height, so weight is on the vertical axis, as shown on the graph in Figure 1-2l. � e graph curves upward because doubling the height more than doubles the weight. Extending the graph sends it through the origin, but the domainstarts beyond the origin at a value greater than zero, because a person never has zero height or weight. � e graph stops at the person’s adult height and weight. A power function has a graph like this.

Plot the graph of function is this? Give the range. Find a pair of real-world variables that could have

EXAMPLE 1 ➤


SOLUTION

Figure 1-2k

Domain

Range

f (x)

x4

10

➤

As children grow older, their height and weight are related. Sketch a reasonable graph to show this relation and then describe it. Identify what kind of function has

EXAMPLE 2 ➤

Weight depends on height, so weight is on the vertical axis, as shown on the graph

SOLUTION

Birth height Adultheight

Wei

ght

Height

➤ Figure 1-2l



CAS Suggestions (continued)

A CAS is particularly powerful when manipulating functions. Th e student can focus on the mathematics, recognizing the forms of functions and their parameters, while the CAS performs the calculations regardless of how complicated a solution might be. For example, it takes three parameters to defi ne the general equation of a quadratic function, so three points and a Solve command will suffi ce for computing the equation of any quadratic. Th e fi gures show two diff erent approaches to computing a quadratic equation containing the points (1, 5), (2, 7), and (10, 10). Th e fi rst approach shows the mathematics more clearly, while the second approach is more sophisticated, allowing the CAS to do the work with fewer obvious inputs.

PRO B LE M N OTES

Supplementary Problems for this section are available at www.keypress.com/keyonline.

Problems 1–18 are similar to Example 1.

1a.

1b. 3 f (x) 231c. Linear1d. Answers will vary.

2a.

2b. 0 f (x) 12.82c. Power2d. Answers will vary.

4 8

10

20

x

y

2 4

10

y

x


For Problems 1–4, a. Plot the graph on your grapher using the

given domain. Sketch the result on your paper. b. Give the range of the function. c. Name the kind of function. d. Describe a pair of real-world variables that

could be related by a graph of this shape. 1. f (x) 2x 3 domain: 0 x 10 2. f (x) 0.2 x 3 domain: 0 x 4 3. g (x) 12 ___ x domain: 0 x 10

4. h (x) 5 0. 6 x domain: 5 x 5

For Problems 5–18, a. Plot the graph using a window set to show the

entire graph, when possible. Sketch the result. b. Give the y-intercept and any x-intercepts and

the locations of any asymptotes. c. Give the range.



7. Cubic (polynomial) function f (x) x 3 7 x 2 4x 12 with the domain 1 x 7

8. Quartic (polynomial) function f (x) x 4 3 x 3 8 x 2 12x 16 with the domain 3 x 3

9. Power function f (x) 3 x 2/3 with the domain 0 ≤ x ≤ 8

10. Power function f (x) 0.3 x 1.5 with the domain 0 x 9

11. Linear function f (x) 0.7x 4 with the domain 3 x 10

12. Linear function f (x) 3x 6 with the domain 5 x 5



15. Inverse-square variation function f (x) 25 __ x 2

with the domain x 0

16. Direct variation function f (x) 5x with the domain x 0

17. Rational function y x 2 __________ (x 4)(x 1) with the domain 3 x 6, x 4, x 1

18. Rational function y x 2 2x 2 ________ x 3 with the domain 2 x 6, x 3

For Problems 19–28, name the type of function that has the graph shown. 19.

x

y 20.

x

y

21.

x

y 22.

x

y

23.

x

y 24. y

x

25.

x

y 26.

x

y

27.

x

y 28.

x

y

For Problems 1–4, Exponential function f ( ) x with the x with the x

Problem Set 1-2Plot the graph of f (x) x 2 5x 3 in the domain 0 x 4. What kind of function is this? Give the range. Find a pair of real-world variables that could have a relationship described by a graph of this shape.


f 1 (x) x 2 5x 3 / (x 0 and x 4)

� e graph in Figure 1-2k shows the restricted domain.

� e function is quadratic because f (x) equals a second-degree polynomial in x.

� e range is 3 f (x) 9.25. You can � nd this interval by tracing to the le� endpoint of the graph where f (0) 3 and to the high point where f (2.5) 9.25. (At the right endpoint, f (4) 7, which is between 3 and 9.25.)

� e function could represent the relationship between something that rises for a while and then falls, such as a punted football’s height as a function of time or (if f (x) is multiplied by 10) the grade you could get on a test as a function of the number of hours you study for it. (� e grade could be lower for longer times if you stay up too late and thus are sleepy during the test.)

DEFINITION: Boolean VariablesA Boolean variable is a variable that has a given condition attached to it. If the condition is true, the variable equals 1. If the condition is false, the variable equals 0.

As children grow older, their height and weight are related. Sketch a reasonable graph to show this relation and then describe it. Identify what kind of function has a graph like the one you drew.

Weight depends on height, so weight is on the vertical axis, as shown on the graph in Figure 1-2l. � e graph curves upward because doubling the height more than doubles the weight. Extending the graph sends it through the origin, but the domainstarts beyond the origin at a value greater than zero, because a person never has zero height or weight. � e graph stops at the person’s adult height and weight. A power function has a graph like this.

Plot the graph of function is this? Give the range. Find a pair of real-world variables that could have

EXAMPLE 1 ➤


SOLUTION

Figure 1-2k

Domain

Range

f (x)

x4

10

➤

As children grow older, their height and weight are related. Sketch a reasonable graph to show this relation and then describe it. Identify what kind of function has

EXAMPLE 2 ➤

Weight depends on height, so weight is on the vertical axis, as shown on the graph

SOLUTION

Birth height Adultheight

Wei

ght

Height

➤ Figure 1-2l


13

Th e intercepts for Problems 5–18

can be computed quickly using a Solve command on a CAS. 5a.

5b. y-intercept at y 5 12; no x-intercepts; no asymptotes5c. 7 y 16 6a.

6b. y-intercept at y 5 40; no x-intercepts; no asymptotes6c. 31 y 567a.

7b. y-intercept at y 5 12; x-intercepts at x 5 21, x 5 2 and x 5 6; no asymptotes7c. 220.7453… y 40

Problems 19–28 require students to name types of functions from their graphs.19. Exponential 20. Linear21. Linear 22. Exponential 23. Quadratic 24. Cubic25. Power 26. Inverse variation27. Rational 28. Direct variation

4

16y

x

(2, 16)

4

20

40 (3, 31)

y

x

4

y

x

(4.36, �20.75)

20

�20

3a.

3b. g (x) 1.23c. Inverse variation3d. Answers will vary.

4a.

4b. 0.3888 h(x) 64.3 4c. Exponential4d. Answers will vary.

4 8

20

40

y

x

20

40

y

x�4 4


See pages 973–974 for answers to Problems 8–18.

15Section 1-3: Dilation and Translation of Function Graphs

Dilation and Translation of Function GraphsEach of these two graphs shows the unit semicircle and a transformation of it. � e le� graph shows the semicircle dilated (magni� ed) by a factor of 5 in the x-direction and by a factor of 3 in the y-direction. � e right graph shows the unit semicircle translated by 4 units in the x-direction and by 2 units in the y-direction.

y

x

3

1 5Pre-imagefunction f

Imagefunction g

y

x

3


Imagefunction h

Figure 1-3a

� e transformed functions, g and h, in Figure 1-3a are called images of the function f. � e original function, f, is called the pre-image. In this section you will learn how to transform the equation of a function so that its graph will be dilated and translated by given amounts in the x- and y-directions.

Transform a given pre-image function so that the result is a graph of the image function that has been dilated by given factors and translated by given amounts.

DilationsTo get the vertical dilation in the le� graph of Figure 1-3a, multiply each y-coordinate by 3. Figure 1-3b shows the image, y 3f (x).

x

y 3f(x)

y f(x)3

1 4

y

Figure 1-3b

Dilation and Translation of Function Graphs

1-3

Transform a given pre-image function so that the result is a graph of the image function that has been dilated by given factors and translated by

Objective

For Problems 29–32, a. Sketch a reasonable graph showing how the

variables are related. b. Identify the type of function it could be

(quadratic, power, exponential, and so on). 29. � e weight and length of a dog. 30. � e temperature of a cup of co� ee and the time

since the co� ee was poured. 31. � e purchase price of a house in a particular

neighborhood as a function of the number of square feet of � oor space in the house, including a � xed amount for the lot on which the house was built.

32. � e height of a punted football as a function of the number of seconds since it was kicked.

For Problems 33–38, tell whether the relation graphed is a function. Explain how you made your decision. 33. y

x

34. y

x

35.

x

y 36.

x

y

37. y

x

38.

x

y

39. Vertical Line Test Problem: � ere is a graphical way to tell whether a relation is a function. It is called the vertical line test.

PROPERTY: The Vertical Line TestIf any vertical line cuts the graph of a relation in more than one place, then the relation is not a function.

Figure 1-2m illustrates the test.

x

Not a function A vertical linecuts more than once.

y

x

A function No vertical linecuts more than once.

y

Figure 1-2m

a. Based on the de� nition of function, explain how the vertical line test distinguishes between relations that are functions and relations that are not functions.

b. Sketch the graphs in Problems 33 and 35. On your sketch, show how the vertical line test tells you that the relation in Problem 33 is a function but the relation in Problem 35 is not.

40. Explain why a function can have more than one x-intercept but only one y-intercept.

41. What is the argument of the function y f (x 2)?

42. Research Problem: Look up George Boole on the Internet or in another reference source. Describe several of Boole’s accomplishments that you discover. Include your source.



Problem Notes (continued)

Problems 29–32 are similar to Example 2.

Problems 33–38 require students to identify which relations are functions.33. Function; no x-value has more than one corresponding y-value.34. Not a function; some x-values on the left have two corresponding y-values.35. Not a function; there is at least one x-value with more than one corresponding y-value.36. Function; no x-value has more than one corresponding y-value.37. Not a function; there is at least one x-value with more than one corresponding y-value.38. Not a function; the x-value in the middle has infi nitely many corresponding y-values.

Problem 39 presents the vertical line test, which all students should know and use. Be sure to discuss this problem when you review the homework. To reinforce the concept, question students periodically in future assignments about whether graphs “pass the vertical line test.”39a. A vertical line through a given x-value crosses the graph at the y-values that correspond to that x-value. So, if a vertical line crosses the graph more than once, it means that that x-value has more than one y-value.39b. In Problem 33, any vertical line crosses the graph at most once, but in Problem 35, any vertical line between the two endpoints crosses the graph twice.

Problems 40 and 41 require students to demonstrate understanding of some of the terminology associated with functions.

Problem 42 is a research problem that requires students to write.


1. Determine an equation for a quadratic function whose graph contains the points (21, 5), (0.2, 13.64), and (5, 77).

2. What is the y-intercept of the 4th-degree polynomial that contains the points (22, 242), (21, 0), (1, 6), (2, 6), and (4, 90)? Comment on the results.

Note: Th ese points defi ne a cubic polynomial, which the CAS reveals by giving a leading coeffi cient of zero. Th e y-intercept is automatically seen as the constant term in the CAS results. Th e students don’t need to know that this is not a proper quartic polynomial.

3. How many x-intercepts does the graph of the polynomial in CAS Problem 2 have?See page 975 for answers to

Problems 29–32, 40–42 and CAS Problems 1–3.


Dilation and Translation of Function GraphsEach of these two graphs shows the unit semicircle and a transformation of it. Th e left graph shows the semicircle dilated (magnifi ed) by a factor of 5 in the x-direction and by a factor of 3 in the y-direction. Th e right graph shows the unit semicircle translated by 4 units in the x-direction and by 2 units in the y-direction.

y

x

3


Imagefunction g

y

x

3


Imagefunction h

Figure 1-3a

Th e transformed functions, g and h, in Figure 1-3a are called images of the function f. Th e original function, f, is called the pre-image. In this section you will learn how to transform the equation of a function so that its graph will be dilated and translated by given amounts in the x- and y-directions.

Transform a given pre-image function so that the result is a graph of the image function that has been dilated by given factors and translated by given amounts.

DilationsTo get the vertical dilation in the left graph of Figure 1-3a, multiply each y-coordinate by 3. Figure 1-3b shows the image, y � 3f (x).

x

y � 3f(x)

y � f(x)3

1 4

y

Figure 1-3b

Dilatioof Fun

1-3

Transimag

Objective

For Problems 29–32, a. Sketch a reasonable graph showing how the

variables are related. b. Identify the type of function it could be

(quadratic, power, exponential, and so on). 29. � e weight and length of a dog. 30. � e temperature of a cup of co� ee and the time

since the co� ee was poured. 31. � e purchase price of a house in a particular

neighborhood as a function of the number of square feet of � oor space in the house, including a � xed amount for the lot on which the house was built.

32. � e height of a punted football as a function of the number of seconds since it was kicked.

For Problems 33–38, tell whether the relation graphed is a function. Explain how you made your decision. 33. y

x

34. y

x

35.

x

y 36.

x

y

37. y

x

38.

x

y

39. Vertical Line Test Problem: � ere is a graphical way to tell whether a relation is a function. It is called the vertical line test.

PROPERTY: The Vertical Line TestIf any vertical line cuts the graph of a relation in more than one place, then the relation is not a function.

Figure 1-2m illustrates the test.

x

Not a function A vertical linecuts more than once.

y

x

A function No vertical linecuts more than once.

y

Figure 1-2m

a. Based on the de� nition of function, explain how the vertical line test distinguishes between relations that are functions and relations that are not functions.

b. Sketch the graphs in Problems 33 and 35. On your sketch, show how the vertical line test tells you that the relation in Problem 33 is a function but the relation in Problem 35 is not.

40. Explain why a function can have more than one x-intercept but only one y-intercept.

41. What is the argument of the function y f (x 2)?

42. Research Problem: Look up George Boole on the Internet or in another reference source. Describe several of Boole’s accomplishments that you discover. Include your source.




Class Time1–2 days

Homework AssignmentDay 1: Reading Analysis (always assign

these questions), Q1–Q10 (always assign these), Problems 1–6

Day 2: Problems 7–21

Teaching Resources Exploration 1-3: Transformations from

GraphsExploration 1-3a: Translations and

Dilations, NumericallyExploration 1-3b: Translations and

Dilations, AlgebraicallyExploration 1-3c: Transformation ReviewBlackline Masters

Problems 15–20Test 1, Sections 1-1 to 1-3, Forms A and B


Dilation

Translation

Presentation Sketch: Translation Present.gsp

Presentation Sketch: Dilation Present.gsp

Activity: Translation of Functions

Activity: Dilation of Functions

Activity: Exploring Translations and Dilations

Activity: Function Transformations

CAS Activity 1-3a: Transformed Quadratic Functions


Note that the transformation in f 1 _ 5 x is applied to the argument of function f, inside the parentheses. �e transformation in 3f (x) is applied outside the parentheses, to the value of the function. For this reason the transformations are given the names inside transformation and outside transformation, respectively. An inside transformation a�ects the graph in the horizontal direction, and an outside transformation a�ects the graph in the vertical direction.

You may ask, “Why do you multiply by the y-dilation and divide by the x-dilation?” You can see the reason by substituting y for g (x) and dividing both sides of the equation by 3:

y 3f 1 __ 5 x

1 __ 3 y f 1 __ 5 x Divide both sides by 3 or multiply by 1 _ 3 .

You actually divide by both dilation factors, y by the y-dilation and x by the x-dilation.

Translations�e translations in Figure 1-3a that transform f (x) to h(x) are shown again in Figure 1-3e. To �gure out what translation has been done, ask yourself, “To where did the point at the origin move?” As you can see, the center of the semicircle, initially at the origin, has moved to the point (4, 2). So there is a horizontal translation of 4 units and a vertical translation of 2 units.

Figure 1-3e

To get a vertical translation of 2 units, add 2 to each y-value:

y 2 f (x)

To get a horizontal translation of 4 units, note that what was happening at x 0 in function f has to be happening at x 4 in function h. Again, substituting v for the argument of f gives

h(x) 2 f (v)

x v 4

x 4 v

h(x) 2 f (x 4) Substitute x 4 as the argument of f.

y

x2

1 4Pre-image,function f

Image,function h

y

x515

3

3

f1(x)

f2(x)

� e horizontal dilation is trickier. Each value of the argument must be 5 times what it was in the pre-image to generate the same y-values. Substitute v for the argument of f.

y f (v) Let v represent the original x-values.

x 5v � e x-values of the dilated image must be 5 times the x-values of the pre-image.

1 __ 5 x v Solve for v.

y f 1 __ 5 x Replace v with 1 _ 5 x for the argument to obtain the equation of the dilated image.

Figure 1-3c shows the graph of the image, y f 1 _ 5 x

x1 5

y f ( x)153y f (x)

y

Figure 1-3c

Putting the two transformations together gives the equation for g (x) shown in Figure 1-3a.

g (x) 3f 1 __ 5 x

� e equation of the pre-image function in Figure 1-3a is f (x) ______

1 x 2 . Con� rm on your grapher that g (x) 3f 1 _ 5 x is the transformed image function

a. By direct substitution into the equation

b. By using the grapher’s built-in variables feature

a. g (x) 3 _________

1 (x/5) 2 Substitute x/5 as the argument of f, and multiply the entire expression by 3.

Enter: f 1 (x) ______

1 x 2

f 2 (x) 3 _________

1 (x/5) 2

� e graph in Figure 1-3d shows a dilation by 5 in the x-direction and by 3 in the y-direction. Use the grid-on feature to make the grid points appear. Use equal scales on the two axes so the graphs have the correct proportions.

b. Enter: f 3 (x) 3 f 1 (x/5) f 1 is the function name in this format, not the function value.

� is graph is the same as the graph of f 2 (x) in Figure 1-3d.

� e equation of the pre-image function in Figure 1-3a is on your grapher that

EXAMPLE 1 ➤

a. g (g (gSOLUTION

➤


Figure 1-3d


TE ACH I N G

Important Terms and ConceptsDilationTranslationTransformationInside transformationOutside transformationImagePre-image

Section Notes

It is recommended that you spend two days on this section. Begin the fi rst day of instruction by assigning Exploration 1-3a as a small-group activity. Th en cover the material through Example 2. On the second day, start by assigning Exploration 1-3. Th en cover the remaining material in the section.

Depending on the specifi c grapher

your students use, you may want to discuss the idea of using a “friendly” grapher window. A friendly window is one that gives “nice” coordinate values when a function is traced. When you trace a function, the cursor moves one pixel at a time. Setting the window so that Xmax 2 Xmin is a multiple of the number of pixels in the horizontal direction ensures that the x-coordinates displayed when the function is traced will be nice numbers. For example, the TI-83/84 Plus calculators display 94 pixels in the horizontal direction. If Xmin 5 0 and Xmax 5 94 or if Xmin 5 247 and Xmax 5 47, the x-coordinate will change by 1 with each trace step. If Xmin 5 29.4 and Xmax 5 9.4, the x-coordinate will change by 0.2 with each trace step.

If the coeffi cients in a function are rational, you can also set Ymin and Ymax so that nice y-coordinates are displayed. Th e TI-83/84 Plus displays 62 pixels in the vertical direction. So, for example, using the settings Ymin 5 0 and Ymax 5 62 or Ymin 5 26.2 and Ymax 5 6.2 will give nice y-coordinates when the

function is traced. (Note: Th e y-coordinate is calculated by evaluating the function for the x-coordinate. So, if the coeffi cients in a function are irrational, the y-coordinates will not be nice, even in a friendly window.) Th e documentation for each grapher will explain more about friendly windows.

It is important for students to be familiar with both forms of the dilation rule, 1 _ a g (x) 5 f   1 _ b x and g (x) 5 a f   1 _ b x . Th e form g (x) 5 a f   1 _ b x is the more common,

and it is the form used when a function is entered into a grapher. However, the form 1 _ a g (x) 5 f   1 _ b x allows students to recognize that “the same thing happens to x as to y.” Th at is, both x and y are divided by their respective scale factors. (To make this clearer, you may want to write the rule as 1 _ a y 5 f   1 _ b x rather than 1 _ a g (x) 5 f   1 _ b x ). Similarly, the translation rule can be written as h(x) 2 c 5 f (x 2 d) or h(x) 5 c 1 f (x 2 d). Th e second form


Note that the transformation in f 1 _ 5 x is applied to the argument of function f, inside the parentheses. �e transformation in 3f (x) is applied outside the parentheses, to the value of the function. For this reason the transformations are given the names inside transformation and outside transformation, respectively. An inside transformation a�ects the graph in the horizontal direction, and an outside transformation a�ects the graph in the vertical direction.

You may ask, “Why do you multiply by the y-dilation and divide by the x-dilation?” You can see the reason by substituting y for g (x) and dividing both sides of the equation by 3:

y 3f 1 __ 5 x

1 __ 3 y f 1 __ 5 x Divide both sides by 3 or multiply by 1 _ 3 .

You actually divide by both dilation factors, y by the y-dilation and x by the x-dilation.

Translations�e translations in Figure 1-3a that transform f (x) to h(x) are shown again in Figure 1-3e. To �gure out what translation has been done, ask yourself, “To where did the point at the origin move?” As you can see, the center of the semicircle, initially at the origin, has moved to the point (4, 2). So there is a horizontal translation of 4 units and a vertical translation of 2 units.

Figure 1-3e

To get a vertical translation of 2 units, add 2 to each y-value:

y 2 f (x)

To get a horizontal translation of 4 units, note that what was happening at x 0 in function f has to be happening at x 4 in function h. Again, substituting v for the argument of f gives

h(x) 2 f (v)

x v 4

x 4 v

h(x) 2 f (x 4) Substitute x 4 as the argument of f.

y

x2

1 4Pre-image,function f

Image,function h

y

x515

3

3

f1(x)

f2(x)

� e horizontal dilation is trickier. Each value of the argument must be 5 times what it was in the pre-image to generate the same y-values. Substitute v for the argument of f.

y f (v) Let v represent the original x-values.

x 5v � e x-values of the dilated image must be 5 times the x-values of the pre-image.

1 __ 5 x v Solve for v.

y f 1 __ 5 x Replace v with 1 _ 5 x for the argument to obtain the equation of the dilated image.

Figure 1-3c shows the graph of the image, y f 1 _ 5 x

x1 5

y f ( x)153y f (x)

y

Figure 1-3c

Putting the two transformations together gives the equation for g (x) shown in Figure 1-3a.

g (x) 3f 1 __ 5 x

� e equation of the pre-image function in Figure 1-3a is f (x) ______

1 x 2 . Con� rm on your grapher that g (x) 3f 1 _ 5 x is the transformed image function

a. By direct substitution into the equation

b. By using the grapher’s built-in variables feature

a. g (x) 3 _________

1 (x/5) 2 Substitute x/5 as the argument of f, and multiply the entire expression by 3.

Enter: f 1 (x) ______

1 x 2

f 2 (x) 3 _________

1 (x/5) 2

� e graph in Figure 1-3d shows a dilation by 5 in the x-direction and by 3 in the y-direction. Use the grid-on feature to make the grid points appear. Use equal scales on the two axes so the graphs have the correct proportions.

b. Enter: f 3 (x) 3 f 1 (x/5) f 1 is the function name in this format, not the function value.

� is graph is the same as the graph of f 2 (x) in Figure 1-3d.

� e equation of the pre-image function in Figure 1-3a is on your grapher that

EXAMPLE 1 ➤

a. g (g (gSOLUTION

➤


Figure 1-3d

17

Should you venture into doing more complicated transformations involving combinations of refl ections, dilations, and translations, caution students to be mindful of the order of operations: Th ey must apply refl ections and dilations fi rst and then translations. Students will have signifi cant additional exposure to transformations of sinusoidal functions in Section 6-2.

If your students use TI-Nspire

graphers, consider defi ning a function in a graphical window and using sliders to demonstrate general dilations and translations.

Diff erentiating Instruction• Help ELL students understand dilation

by using expand and shrink.• Have students discuss pages 18 and 19

in pairs. Have them graph y 5 2f (x) and y 5 f (x/2) if f (x) 5 2 x 2 1 3 on white boards, and show you their graphs. Have them describe the vertical and horizontal dilations for each function, and then graph a composition of the two functions.

• Make sure ELL students understand the grapher’s ability to store and use variables. You can do this by circulating, using the overhead, or by having them get help from their partner.

• For Problems 19–28, instruct students to answer in complete sentences.

• When doing Exploration 1-3, oral reporting will help students verify their understanding.illustrates that the vertical translation is

subtracted from y just as the horizontal translation is subtracted from x. You want to avoid students coming to the false conclusion that the two variables are treated in diff erent ways.

Point out that if a function is only dilated, the x-dilation is the number you can substitute for x to make the argument equal to 1. If the function is only translated, the x-translation is the number you can

substitute for x to make the argument equal to zero. Stress that multiplying or dividing variable x can lead to dilations and that adding to or subtracting from variable x leads to translations.

Encouraging students to describe transformations in words (such as “Dilate the function by 2 in the x-direction”) helps them move from the graph to an equation and vice versa.

Section 1-3: Dilation and Translation of Function Graphs


� e three graphs in Figure 1-3g show three di� erent transformations of the pre-image graph to image graphs y g(x). Explain verbally what transformations were done. Write an equation for g(x) in terms of the function f.

Figure 1-3g

Le� graph: Vertical dilation by a factor of 3

Equation: g (x) 3f (x)

Note: Each point on the graph of g is 3 times as far from the x-axis as the corresponding point on the graph of f. Note that the vertical dilation moved points above the x-axis farther up and moved points below the x-axis farther down.

Middle graph: Vertical translation by 6 units

Equation: g (x) 6 f (x)

Note: � e vertical dilation moved all points on the graph of f up by the same amount, 6 units. Also note that the fact that g (1) is three times f (1) is purely coincidental and is not true at other values of x.

Right graph: Horizontal dilation by a factor of 2 and vertical translation by 7 units

Equation: g (x) 7 f 1 __ 2 x

Note: Each point on the graph of g is twice as far from the y-axis as the corresponding point on the graph of f. � e horizontal dilation moved points to the right of the y-axis farther to the right and moved points to the le� of the y-axis farther to the le� .

In this exploration, given a pre-image and an image graph, you’ll identify the transformation.

� e three graphs in Figure 1-3g show three di� erent transformations of the pre-image graph to image graphs

EXAMPLE 3 ➤

f

g

x

10

10

y

f

g

y

x10

10

g

fx

10

10y

Le� graph: Vertical dilation by a factor of 3SOLUTION

➤

� e equation of the pre-image function in Figure 1-3e is f (x) ______

1 x 2 . Con� rm on your grapher that g (x) 2 f (x 4) is the transformed image function by

a. Direct substitution into the equation

b. Using the grapher’s built-in variables feature

a. g (x) 2 ___________

1 (x 4) 2 Substitute x 4 for the argument. Add 2 to the expression.

Enter: f 1 (x) ______

1 x 2

f 2 (x) 2 ___________

1 (x 4) 2

� e graph in Figure 1-3f shows an x-translation of 4 units and a y-translation of 2 units.

b. Enter: f 3 (x) 2 f 1 (x 4)

� e graph is the same as that for f 2 (x) in Figure 1-3f.

Again you may ask, “Why do you subtract an x-translation and add a y-translation?” � e answer again lies in associating the y-translation with the y-variable. You actually subtract both translations:

y 2 f (x 4)

y 2 f (x 4) Subtract 2 from both sides.

� e reason for writing the transformed equation with y by itself is to make it easier to calculate the dependent variable, either by pencil and paper or on your grapher.

� is box summarizes the dilations and translations of a function and its graph.

PROPERTY: Dilations and Translations� e function g given by

g (x) f 1 __ b x or, equivalently, g (x) a f 1 __ b x

represents a dilation by a factor of a in the y-direction and by a factor of b in the x-direction.

� e function h given by

h(x) c f (x d) or, equivalently, h(x) c f (x d)

represents a translation by c units in the y-direction and by d units in the x-direction.

Note: If the function is only dilated, the x-dilation is the number you can substitute for x to make the argument equal 1. If the function is only translated, the x-translation is the number to substitute for x to make the argument equal zero.

� e equation of the pre-image function in Figure 1-3e is Con� rm on your grapher that

EXAMPLE 2 ➤

a. g

Enter:

SOLUTION

x4

2 f1(x)

f2(x)y

Figure 1-3f

➤

1 __ a



Exploration Notes

For notes on Exploration 1-3 see page 20.

Exploration 1-3a introduces vertical and horizontal translations and dilations by having students numerically determine image graphs for a given pre-image function. Th is activity works well as a small-group discovery section at the beginning of the fi rst day of instruction. Allow students 15–20 minutes to complete the activity.

Exploration 1-3b and Exploration 1-3c can be used either as reviews or as quizzes aft er students complete Section 1-3. Exploration 1-3b takes about 20 minutes and Exploration 1-3c takes about 15 minutes.

Technology Notes

Problem 21 asks students to explore two Dynamic Precalculus Explorations at www.keymath.com/precalc. Th e Translation exploration allows students to manipulate sliders to observe how a function plot is translated both algebraically and geometrically; the Dilation exploration allows them to observe how it is dilated.

Presentation Sketch: Translation Present.gsp at www.keypress.com/keyonline uses sliders to translate a function plot both geometrically and algebraically. Th is sketch is related to the activity Translation of Functions.

CAS Activity 1-3a: Transformed Quadratic Functions in the Instructor’s Resource Book has students use a CAS to fi nd equations for transformed quadratic functions in standard, factored, and vertex forms. Allow 20–25 minutes.


� e three graphs in Figure 1-3g show three di� erent transformations of the pre-image graph to image graphs y g(x). Explain verbally what transformations were done. Write an equation for g(x) in terms of the function f.

Figure 1-3g

Le� graph: Vertical dilation by a factor of 3

Equation: g (x) 3f (x)

Note: Each point on the graph of g is 3 times as far from the x-axis as the corresponding point on the graph of f. Note that the vertical dilation moved points above the x-axis farther up and moved points below the x-axis farther down.

Middle graph: Vertical translation by 6 units

Equation: g (x) 6 f (x)

Note: � e vertical dilation moved all points on the graph of f up by the same amount, 6 units. Also note that the fact that g (1) is three times f (1) is purely coincidental and is not true at other values of x.

Right graph: Horizontal dilation by a factor of 2 and vertical translation by 7 units

Equation: g (x) 7 f 1 __ 2 x

Note: Each point on the graph of g is twice as far from the y-axis as the corresponding point on the graph of f. � e horizontal dilation moved points to the right of the y-axis farther to the right and moved points to the le� of the y-axis farther to the le� .

In this exploration, given a pre-image and an image graph, you’ll identify the transformation.

� e three graphs in Figure 1-3g show three di� erent transformations of the pre-image graph to image graphs

EXAMPLE 3 ➤

f

g

x

10

10

y

f

g

y

x10

10

g

fx

10

10y

Le� graph: Vertical dilation by a factor of 3SOLUTION

➤

� e equation of the pre-image function in Figure 1-3e is f (x) ______

1 x 2 . Con� rm on your grapher that g (x) 2 f (x 4) is the transformed image function by

a. Direct substitution into the equation

b. Using the grapher’s built-in variables feature

a. g (x) 2 ___________

1 (x 4) 2 Substitute x 4 for the argument. Add 2 to the expression.

Enter: f 1 (x) ______

1 x 2

f 2 (x) 2 ___________

1 (x 4) 2

� e graph in Figure 1-3f shows an x-translation of 4 units and a y-translation of 2 units.

b. Enter: f 3 (x) 2 f 1 (x 4)

� e graph is the same as that for f 2 (x) in Figure 1-3f.

Again you may ask, “Why do you subtract an x-translation and add a y-translation?” � e answer again lies in associating the y-translation with the y-variable. You actually subtract both translations:

y 2 f (x 4)

y 2 f (x 4) Subtract 2 from both sides.

� e reason for writing the transformed equation with y by itself is to make it easier to calculate the dependent variable, either by pencil and paper or on your grapher.

� is box summarizes the dilations and translations of a function and its graph.

PROPERTY: Dilations and Translations� e function g given by

g (x) f 1 __ b x or, equivalently, g (x) a f 1 __ b x

represents a dilation by a factor of a in the y-direction and by a factor of b in the x-direction.

� e function h given by

h(x) c f (x d) or, equivalently, h(x) c f (x d)

represents a translation by c units in the y-direction and by d units in the x-direction.

Note: If the function is only dilated, the x-dilation is the number you can substitute for x to make the argument equal 1. If the function is only translated, the x-translation is the number to substitute for x to make the argument equal zero.

� e equation of the pre-image function in Figure 1-3e is Con� rm on your grapher that

EXAMPLE 2 ➤

a. g

Enter:

SOLUTION

x4

2 f1(x)

f2(x)y

Figure 1-3f

➤

1 __ a


19

Presentation Sketch: Dilation Present.gsp at www.keypress.com/keyonline uses sliders to translate a function plot both geometrically and algebraically. This sketch is related to the activity Dilation of Functions.

Activity: Translation of Functions in the Instructor’s Resource Book gives students an opportunity to translate points and function plots both geometrically and algebraically. It takes 25–45 minutes, depending on students’ familiarity with Sketchpad.

Activity: Dilation of Functions in the Instructor’s Resource Book asks students to explore stretching and compressing functions both horizontally and vertically, both algebraically and geometrically. Allow 30–40 minutes.

Activity: Exploring Translations and Dilations in the Instructor’s Resource Book provides hands-on experience with translations and dilations of the parent quadratic function. Allow 20–30 minutes.

Activity: Function Transformations in Teaching Mathematics with Fathom leads students who are fairly new to Fathom through a study of translations, dilations, and reflections of the parent quadratic function. It will take 50–100 minutes, depending on student experience. Optionally, you can use a prepared document.



Reading Analysis

From now on most problem sets will begin with an assignment that requires you to spend ten minutes or so reading the section and answering some questions to see how well you understand what you have read. � is will help develop your ability to read a textbook, a very important skill to have in college. It should also make working the problems in the problem set easier.

From what you have read in this section, what do you consider to be the main idea? What is the major di� erence on the image graph between a translation and a dilation, and what operation causes each transformation? How can you tell whether a translation or a dilation will be in the x-direction or the y-direction?

Quick Review

From now on there will be ten short problems at the beginning of most problem sets. Some of the problems are intended for review of skills from previous sections or chapters. Others are intended to test your general knowledge. Speed is the key here, not detailed work. Try to do all ten problems in less than � ve minutes. Q1. y 3 x 2 5x 7 is a particular example of a

? function. Q2. Write the general equation of a power function. Q3. Write the general equation of an exponential

function. Q4. Calculate the product: (x 7)(x 8) Q5. Expand: (3x 5) 2 Q6. Sketch the graph of a relation that is not a

function. Q7. Sketch the graph of y 2 _ 3 x 4. Q8. Sketch an isosceles triangle. Q9. Find 30% of 3000.

Q10. Which one of these is not the equation of a function?

A. y 3x 5 B. f (x) 3 x 2 C. g (x) x D. y

__ x

E. y 5 x 2/3

For Problems 1–6, let f (x) ______

9 x 2 . a. Write the equation for g (x) in terms of x. b. Plot the graphs of f and g on the same screen.

Use a window with integers from about 10 to 10 as grid points. Use the same scale on both axes. Sketch the result.

c. Describe how f (x) was transformed to get g (x), including whether the transformation was an inside or an outside transformation.

1. g (x) 2f (x) 2. g (x) 3 f (x) 3. g (x) f (x 4) 4. g (x) f 1 __ 3 x 5. g (x) 1 f 1 __ 2 x 6. g (x) 1 __ 2 f (x 3)

For Problems 7–12, a. Describe how the pre-image function

f (dashed) was transformed to get the graph of the image function g (solid).

b. Write an equation for g (x) in terms of function f.

7.

f

g

x5–5

5

–5

y

8.

fg x5–5

5

–5

y

5min

Reading Analysis Q10. Which one of these is not the equation of a

Problem Set 1-3For Problems 1–6, identify the transformation of f (dotted) to get g (solid). Describe the transformation verbally, and give an equation for g (x).1.

x

5

10

g

f

1055

5

10

y

2.

x

y

5

10

f g

55

5

10 10

3.

x

y

5

10

f

g

1055

5

10

4.

x

5

10

f g

105

5

10 5

y

5.

x

5

10

1055

5

10

Graphs coincide.Graphs coincide.

fg

y

6.

gf

x

5

10

1055

5

10

y

7. What did you learn as a result of doing this

exploration that you did not know before?

For Problems 1–6, identify the transformation 4.

E X P L O R AT I O N 1-3: Tr a n s f o r m a t i o n s f r o m G r a p h s



CAS Suggestions

Consider the equation y 5 5 2 x 2 . Th e fi ve, the subtraction, and the exponent can all be seen as operations on the variable x. Once this equation is defi ned as a function (e.g., f (x) 5 5 2 x 2 ), operations can be performed on the function, f (x). Th is is particularly helpful when exploring transformations of functions. Th e graph of y 5 3 f (x 2 1) is a transformation of the function originally defi ned by f (x).

Th e function manipulations shown in Example 1 are easily accomplished with a CAS, but sometimes the algebraic results are in unexpected forms. A Boolean operator can be used to verify the algebraic equivalence between the expected result and CAS result. Algebraic equivalence is a key motivation behind mathematical symbol manipulation, and it is central to CAS mathematics.

Additional Exploration Notes

Exploration 1-3 presents six graphs that illustrate various transformations and combinations of transformations. Problem 5 is an absolute value transformation, g (x) 5 | f (x) | , and Problem 6 is a horizontal dilation by a factor of 21. Th ese two problems preview Section 1-6.

You might want to use this exploration to begin the second day of instruction. Allow small groups of students 15–20 minutes to complete the activity.

1. Vertical translation by 26; g(x) 5 f (x) 2 6 2. Horizontal translation by 110; g(x) 5 f (x 2 10) 3. Vertical dilation by 3; g(x) 5 3f (x)4. Horizontal dilation by 2; g(x) 5 f   1 __ 2 x 5. Refl ection across the x-axis of that part of the graph that is below the x-axis; g(x) 5 f (x)

6. Refl ection across the y-axis; g(x) 5 f (2x)7. Answers will vary.


Reading Analysis

From now on most problem sets will begin with an assignment that requires you to spend ten minutes or so reading the section and answering some questions to see how well you understand what you have read. � is will help develop your ability to read a textbook, a very important skill to have in college. It should also make working the problems in the problem set easier.

From what you have read in this section, what do you consider to be the main idea? What is the major di� erence on the image graph between a translation and a dilation, and what operation causes each transformation? How can you tell whether a translation or a dilation will be in the x-direction or the y-direction?

Quick Review

From now on there will be ten short problems at the beginning of most problem sets. Some of the problems are intended for review of skills from previous sections or chapters. Others are intended to test your general knowledge. Speed is the key here, not detailed work. Try to do all ten problems in less than � ve minutes. Q1. y 3 x 2 5x 7 is a particular example of a

? function. Q2. Write the general equation of a power function. Q3. Write the general equation of an exponential

function. Q4. Calculate the product: (x 7)(x 8) Q5. Expand: (3x 5) 2 Q6. Sketch the graph of a relation that is not a

function. Q7. Sketch the graph of y 2 _ 3 x 4. Q8. Sketch an isosceles triangle. Q9. Find 30% of 3000.

Q10. Which one of these is not the equation of a function?

A. y 3x 5 B. f (x) 3 x 2 C. g (x) x D. y

__ x

E. y 5 x 2/3

For Problems 1–6, let f (x) ______

9 x 2 . a. Write the equation for g (x) in terms of x. b. Plot the graphs of f and g on the same screen.

Use a window with integers from about 10 to 10 as grid points. Use the same scale on both axes. Sketch the result.

c. Describe how f (x) was transformed to get g (x), including whether the transformation was an inside or an outside transformation.

1. g (x) 2f (x) 2. g (x) 3 f (x) 3. g (x) f (x 4) 4. g (x) f 1 __ 3 x 5. g (x) 1 f 1 __ 2 x 6. g (x) 1 __ 2 f (x 3)

For Problems 7–12, a. Describe how the pre-image function

f (dashed) was transformed to get the graph of the image function g (solid).

b. Write an equation for g (x) in terms of function f.

7.

f

g

x5–5

5

–5

y

8.

fg x5–5

5

–5

y

5min

Reading Analysis Q10. Which one of these is not the equation of a

Problem Set 1-3For Problems 1–6, identify the transformation of f (dotted) to get g (solid). Describe the transformation verbally, and give an equation for g (x).1.

x

5

10

g

f

1055

5

10

y

2.

x

y

5

10

f g

55

5

10 10

3.

x

y

5

10

f

g

1055

5

10

4.

x

5

10

f g

105

5

10 5

y

5.

x

5

10

1055

5

10

Graphs coincide.Graphs coincide.

fg

y

6.

gf

x

5

10

1055

5

10

y

7. What did you learn as a result of doing this

exploration that you did not know before?

For Problems 1–6, identify the transformation 4.

E X P L O R AT I O N 1-3: Tr a n s f o r m a t i o n s f r o m G r a p h s


21

PRO B LE M N OTESTh e Reading Analysis questions begin in Sec tion 1-3. Th ese questions are designed to help students learn to read a mathematics textbook. Students need to develop the skills to read dense material slowly and carefully. Encourage them to read with a pencil in hand and a calculator nearby.

Problems Q1–Q10 are short problems that do not require detailed work. Some students like the opportunity to solve problems mentally instead of showing all the steps on paper. You may fi nd that students are more willing to show detailed work on other problems if you allow them to show minimal work on the daily Quick Review problems.

Problems 1–6 allow students to see the eff ects of six diff erent transformations on the same function.

In Problems 1–6, the new graphs

are obtained by defi ning f (x) as given. In a calculator screen, algebraic versions of each of the given functions can also be obtained. Encourage students to interpret and predict the outcomes of the transformations before allowing technology to create the image graphs.

Problems 7–12 require students to identify transformations based on graphs of the pre-image and image functions. Because students are not given explicit formulas for the pre-image f, they must write the equation for the image g in terms of f.7a. y-translation by 77b. g (x) 5 7 1 f (x)8a. x-translation by 258b. g (x) 5 f (x 1 5)

Q1. QuadraticQ2. y 5 a x b , a 0, b 0Q3. y 5 a b x , a 0, b 0, b 1Q4. x 2 1 x 2 56Q5. 9 x 2 2 30x 1 25Q6.

Q7.

Q8.

Q9. 900 Q10. D

y

x

4

�6

y

x


See page 975 for answers to Problems 1–6.

23Section 1-4: Composition of Functions

Composition of FunctionsRadiusincreases.

Figure 1-4a

If you drop a pebble into a pond, a circular ripple extends out from the drop point (Figure 1-4a). � e radius of the circle is a function of time. � e area enclosed by the circular ripple is a function of the radius. � us area is a function of time through this chain of functions:

In this case the area is a composite function of time. In this section you will learn some of the mathematics of composite functions.

Given two functions, graph and evaluate the composition of one function with the other.

In this exploration, you’ll � nd the composition of one function with another.

Composition of Functions1- 4


Objective

1. � e � gure shows two linear functions, f and g. Write the domain and range of each function.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

x

f(x)

g(x)

y

2. Read values of g (x) from the graph and write them in a copy of this table. If the value of x is out of the domain, write “none.”

x g (x)

0 1 2 3 456789

1. � e � gure shows two linear functions, f and f and f 2. Read values of ( ) from the graph and write

E X P L O R AT I O N 1- 4 : C o m p o s i t i o n o f Fu n c t i o n s

continued

9.

g xf 5–5

5

–5

y

10.

g

xf

5–5

5

–5

y

11.

g

xf

5–5

5

–5

y

12.

g

xf5–5

5

–5

y

13. � e equation of f in Problem 11 is f (x) 4.5

_____ 1 x 2.5(x 1). Enter this

equation and the equation for g (x) into your grapher and plot the graphs. Does the result agree with the � gure in Problem 11?


_____ 1 x 2.5(x 1). Enter this


Figure 1-3h shows the graph of the pre-image function f. For Problems 15–20,

a. Sketch the graph of the image function g on a copy of Figure 1-3h.

b. Identify the transformation(s) that are done.

xf

5–5

5

–5

y

Figure 1-3h

15. g (x) f (x 6) 16. g (x) f 1 __ 2 x 17. g (x) 5f (x) 18. g (x) 4 f (x) 19. g (x) 5f (x 6) 20. g (x) 4 f 1 __ 2 x

21. Dynamic Transformations Problem: Go to www.keymath.com/precalc and � nd the Dynamic Precalculus Explorations for Chapter 1. Complete the Translation exploration and the Dilation exploration, and explain in writing what you learned.



Problem Notes (continued)9a. x-dilation by 39b. g (x) 5 f   x __ 3 10a. y-dilation by 410b. g (x) 5 4 f (x)11a. x-translation by 6, y-dilation by 311b. g (x) 5 3 f (x 2 6)12a. x-dilation by 3, y-translation by 2412b. g (x) 5 24 1 f   x __ 3 13. No. Th e domain of f (x) is x 1, but the domain of the graph is 23 x 1.14. No. Th e domain of f (x) is x 1, but the domain of the graph is 23 x 1.

Problems 15–20 do not give students an explicit formula for the pre-image function and so require students to perform transformations without plotting points by calculating their coordinates. When an explicit formula for the pre-image function is given, students are more likely to approach the problem by using the formula to plot points and miss the fundamental idea of transforming functions. A blackline master for these problems is available in the Instructor’s Resource Book.15a.

15b. x-translation by 26

Problem 21 requires students to use Dynamic Precalculus Explorations at www.keymath.com/precalc to investigate dilations and translations. Th ese are Java-based sketches that do not require additional soft ware.21. Answers will vary.

�10 �4 4 10�4

2 x

y


1. For some functions, two diff erent transformations on a parent function can be used to obtain the same fi nal image. Let f (x) 5 x. Determine a horizontal translation that produces the same graph as y 5 f (x) 1 5. Compare the magnitude and direction of each translation numerically and graphically.

2. Defi ne f(x) = x and Solve f(x) + k = f(x + h) for h. Interpret the result.

3. What is the eff ect of a on the graph of y 5 a f (x)? What is the eff ect of b in the graph of y 5 f (b x)? Defi ne f(x) = x 2 .

Solve a f(x) = f(b x) for b and explain what the result means.

4. Will there be a solution for b for the equation a f (x) 5 f (b x) for all functions f (x)? Explain.See page 976 for answers to

Problems 16–20 and CAS Problems 1–4.


Composition of FunctionsRadiusincreases.

Figure 1-4a

If you drop a pebble into a pond, a circular ripple extends out from the drop point (Figure 1-4a). � e radius of the circle is a function of time. � e area enclosed by the circular ripple is a function of the radius. � us area is a function of time through this chain of functions:

In this case the area is a composite function of time. In this section you will learn some of the mathematics of composite functions.


In this exploration, you’ll � nd the composition of one function with another.

Composition of Functions1- 4


Objective

1. � e � gure shows two linear functions, f and g. Write the domain and range of each function.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

x

f(x)

g(x)

y

2. Read values of g (x) from the graph and write them in a copy of this table. If the value of x is out of the domain, write “none.”

x g (x)

0 1 2 3 456789

1. � e � gure shows two linear functions, f and f and f 2. Read values of ( ) from the graph and write

E X P L O R AT I O N 1- 4 : C o m p o s i t i o n o f Fu n c t i o n s

continued

9.

g xf 5–5

5

–5

y

10.

g

xf

5–5

5

–5

y

11.

g

xf

5–5

5

–5

y

12.

g

xf5–5

5

–5

y


_____ 1 x 2.5(x 1). Enter this



_____ 1 x 2.5(x 1). Enter this


Figure 1-3h shows the graph of the pre-image function f. For Problems 15–20,

a. Sketch the graph of the image function g on a copy of Figure 1-3h.

b. Identify the transformation(s) that are done.

xf

5–5

5

–5

y

Figure 1-3h

15. g (x) f (x 6) 16. g (x) f 1 __ 2 x 17. g (x) 5f (x) 18. g (x) 4 f (x) 19. g (x) 5f (x 6) 20. g (x) 4 f 1 __ 2 x

21. Dynamic Transformations Problem: Go to www.keymath.com/precalc and � nd the Dynamic Precalculus Explorations for Chapter 1. Complete the Translation exploration and the Dilation exploration, and explain in writing what you learned.



Exploration Notes

Exploration 1-4 can be used to introduce the idea of composition of functions. Blackline masters for Problems 2 and 5 are available in the Instructor’s Resource Book. Allow students 15 minutes to complete the exploration.1. f: Domain: 1 x 5; Range: 2 y 6g: Domain: 2 x 8; Range: 0 y 3

S e c t i o n 1- 4S e c t i o n 1- 4S e c t i o n 1- 4S e c t i o n 1- 4S e c t i o n 1- 4S e c t i o n 1- 4PL AN N I N G

Class Time1 day

Homework AssignmentReading Analysis (RA), Q1–Q10,

Problems 1, 2, 5, 7, 9, 10, 12–15

Teaching Resources Exploration 1-4: Composition of

FunctionsBlackline Masters

Exploration Problems 2 and 5Examples 1 and 5Problems 5 and 6

Supplementary Problems


Presentation Sketch: Composition Present.gsp

TE ACH I N G

Important Terms and ConceptsComposite functionInputOutputInside functionOutside functionNotation for a composite function:

f (g (x)), f + g (x), (f + g)(x)Domain and range of a composite

function

See page 976 for answers to Exploration 1-4, Problem 2.


Mathematicians o� en use f (x) terminology for composite functions. For these radius and area functions, you can write

r (x) 8x x is the input for function r.

a(x) x 2 x is the input for function a.

� e r and a become the names of the functions, and the r(x) and a(x) are the values, or outputs, of the functions. � e x simply stands for the input of the function. You must keep in mind that the input for function r is the time and the input for function a is the radius.

Combining the symbols leads to this way of writing a composite function:

area a(r (x))

� e x is the input for the radius function, and the r(x) is the input for the area function. � e notation a(r(x)) is pronounced “a of r of x.” Function r is called the inside function because it appears inside a pair of parentheses. Function a is called the outside function. Figure 1-4c shows this symbol and its meanings. � e names parallel the terms inside transformation and outside transformation that you learned in the previous section.

area a(r(x))

Outside function Inside functionInput for function r

Input for function a Output of function r Figure 1-4c

� e two function names are sometimes combined this way:

a r(x) or (a r)(x)

� e symbol a r is pronounced “a composition r.” � e parentheses in theexpression (a r) indicate that a r is the name of the function.

Composite Functions from GraphsExample 1 shows you how to � nd a value of the composite function f ( g (x)) from graphs of the two functions f and g.

Functions f and g are graphed in Figure 1-4d. Find f ( g (30)), showing on copies of the graphs how you found this value.

2010 504030

g(x)

21

43

65

x

f(x)

x1 2 3 4 5 6

100200300400

Figure 1-4d

Functions the graphs how you found this value.

EXAMPLE 1 ➤

3. �e symbol f ( g (x)) is read “ f of g of x.” It means �nd the value of g (x) �rst, and then �nd f of the result. For instance, g (5) 1.5. So f ( g (5)) f (1.5) 5.5. Put another column into the table for values of f ( g (x)). Write “none” where appropriate.

4. Show in the table an instance where g (x) is de�ned but f ( g (x)) is not de�ned.

5. Plot the values of f ( g (x)) on a copy of the �gure in Problem 1. If the points do not lie in a straight line, go back and check your work.

6. �e function in Problem 5 is called the composition of f with g, which can be written f g. What are the domain and range of f g?

7. Write equations for functions f and g.

8. Enter into your grapher the equations of f and g as f 1 (x) and f 2 (x), respectively. Use Boolean variables or enter the domain directly, depending on your grapher, to make the functions have the proper domains. �en plot the graphs. Does the result agree with the given �gure?

9. Enter f g into f 3 (x) by entering f 1 f 2 (x) . Plot this graph. Does it agree with the graph you drew in Problem 5?

10. By suitable algebraic operations on the equations in Problem 7, �nd an equation for f ( g (x)). Simplify the equation as much as possible.


Symbols for Composite FunctionsSuppose that the radius of the ripple is increasing at the constant rate of 8 in./s. �en

r 8t

where r is the radius in inches and t is the number of seconds. If t 5, then

r 8 5 40 in.

�e area of the circular region is given by

a r 2

where a is the area in square inches and r is the radius in inches. At time t 5, when the radius is 40, the area is given by

a 40 2 1600 5026.5482… 5027 in 2 or about 35 � 2 .

�e 5 s is the input for the radius function, and the 40 in. is the output. Figure 1-4b shows that the output of the radius function becomes the input for the area function. �e output of the area function is 5026.5....

Radius function Area function

r 8t a r2 5 s 40 in. 5027 in240 in.

Input forradius fn.

Output fromradius fn.

Input forarea fn.

Output fromarea fn.

Figure 1-4b


EXPLORATION, continued


Section Notes

This section explores compositions of functions through application problems and the use of a grapher. It is important that students understand the underlying concept of composition. The first two examples do not give explicit functions, so students cannot depend on their algebraic skills to solve the problems. Make sure it is clear that the inside function is applied first and then the outside function is applied to the result.

A blackline master for Example 1 is available in the Instructor’s Resource Book.

Example 3 on page 27 shows how to evaluate a composite function at various values and then find an explicit formula for f (g (x)).

Example 5 on page 28 shows how to find the domain and range of a composite function. Make sure students understand that for a value x to be in the domain of a composite function f + g, x must be in the domain of the inside function g, and g (x) must be in the domain of the outside function f. A blackline master for Example 5 is available in the Instructor’s Resource Book.

In Example 5d, the range of f + g is found by substituting the endpoints of the domain, 4 x 7, into the equation of the composite function. Emphasize that this technique works because f + g is linear (a linear function does not “change directions,” so the least and greatest y-values occur at the endpoints of the domain). Using the endpoints gives the correct range for any function that is monotone increasing or decreasing on the entire domain. For other types of functions, students should examine the graph to determine the range.

Exploration Notes, continued4. The lines in which x 5 2 and x 5 35.

6. Domain: 4 x 8 Range: 4 y 6

7. f (x) 5 2x 1 7, 1 x 5 g(x) 5 1 __ 2 x 2 1, 2 x 8 8.

Yes

y

x84

4

f � g

g

f

y

x84

4f1(x)

f2(x)

See page 976 for answers to Exploration 1–4, Problem 3.


Mathematicians o� en use f (x) terminology for composite functions. For these radius and area functions, you can write

r (x) 8x x is the input for function r.

a(x) x 2 x is the input for function a.

� e r and a become the names of the functions, and the r(x) and a(x) are the values, or outputs, of the functions. � e x simply stands for the input of the function. You must keep in mind that the input for function r is the time and the input for function a is the radius.

Combining the symbols leads to this way of writing a composite function:

area a(r (x))

� e x is the input for the radius function, and the r(x) is the input for the area function. � e notation a(r(x)) is pronounced “a of r of x.” Function r is called the inside function because it appears inside a pair of parentheses. Function a is called the outside function. Figure 1-4c shows this symbol and its meanings. � e names parallel the terms inside transformation and outside transformation that you learned in the previous section.

area a(r(x))

Outside function Inside functionInput for function r

Input for function a Output of function r Figure 1-4c

� e two function names are sometimes combined this way:

a r(x) or (a r)(x)

� e symbol a r is pronounced “a composition r.” � e parentheses in theexpression (a r) indicate that a r is the name of the function.

Composite Functions from GraphsExample 1 shows you how to � nd a value of the composite function f ( g (x)) from graphs of the two functions f and g.

Functions f and g are graphed in Figure 1-4d. Find f ( g (30)), showing on copies of the graphs how you found this value.

2010 504030

g(x)

21

43

65

x

f(x)

x1 2 3 4 5 6

100200300400

Figure 1-4d

Functions the graphs how you found this value.

EXAMPLE 1 ➤

3. �e symbol f ( g (x)) is read “ f of g of x.” It means �nd the value of g (x) �rst, and then �nd f of the result. For instance, g (5) 1.5. So f ( g (5)) f (1.5) 5.5. Put another column into the table for values of f ( g (x)). Write “none” where appropriate.

4. Show in the table an instance where g (x) is de�ned but f ( g (x)) is not de�ned.

5. Plot the values of f ( g (x)) on a copy of the �gure in Problem 1. If the points do not lie in a straight line, go back and check your work.

6. �e function in Problem 5 is called the composition of f with g, which can be written f g. What are the domain and range of f g?

7. Write equations for functions f and g.

8. Enter into your grapher the equations of f and g as f 1 (x) and f 2 (x), respectively. Use Boolean variables or enter the domain directly, depending on your grapher, to make the functions have the proper domains. �en plot the graphs. Does the result agree with the given �gure?

9. Enter f g into f 3 (x) by entering f 1 f 2 (x) . Plot this graph. Does it agree with the graph you drew in Problem 5?

10. By suitable algebraic operations on the equations in Problem 7, �nd an equation for f ( g (x)). Simplify the equation as much as possible.


Symbols for Composite FunctionsSuppose that the radius of the ripple is increasing at the constant rate of 8 in./s. �en

r 8t

where r is the radius in inches and t is the number of seconds. If t 5, then

r 8 5 40 in.

�e area of the circular region is given by

a r 2

where a is the area in square inches and r is the radius in inches. At time t 5, when the radius is 40, the area is given by

a 40 2 1600 5026.5482… 5027 in 2 or about 35 � 2 .

�e 5 s is the input for the radius function, and the 40 in. is the output. Figure 1-4b shows that the output of the radius function becomes the input for the area function. �e output of the area function is 5026.5....

Radius function Area function

r 8t a r2 5 s 40 in. 5027 in240 in.

Input forradius fn.

Output fromradius fn.

Input forarea fn.

Output fromarea fn.

Figure 1-4b


EXPLORATION, continued

25

Differentiating Instruction• Because the notations f (g (x)) and

f + g (x) are not standard in other countries, check that ELL students understand that the two notations are essentially interchangeable.

• In Example 2, make sure ELL students understand that f (x) and g (x) are different functions than the f (x) and g (x) used before this example. Also, make sure they understand the note about domain.

9.

Yes

10. f (g(x)) 5 2g(x) 1 7 5 2   1 __ 2 x 2 1   1 7

5 2 1 __ 2 x 1 8

11. Answers will vary.

y

x84

4f1(x) f3(x)

f2(x)

Section 1-4: Composition of Functions


Find the other values the same way. Here is a compact way to arrange your work.

f ( g (1)) f (5) 0

f ( g2)) f (3) 6

f ( g (3)) f (2) 4

f ( g (4)) f (1) 3

f ( g (5)) f (7), which does not exist

f ( g (6)) f (4) 2

b. g ( f (2)) g (4) 1, which is not the same as f ( g (2)) 6

Note that in order to � nd a value of a composite function such as f ( g (x)), the value of g (x) must be in the domain of the outside function, f. Because g (5) 7 in Example 2 and there is no value for f (7), the value of f ( g (5)) is unde� ned.

Composite Functions from EquationsExample 3 shows you how to � nd values of a composite function if you know the equations of the two functions.

Let f be the linear function f (x) 3x 5, and let g be the exponential function g (x) 2 x .

a. Find f ( g (4)), f ( g (0)), and f ( g ( 1)).

b. Find g ( f ( 1)) and show that it is not the same as f ( g ( 1)).

c. Find an equation for h(x) f ( g (x)) explicitly in terms of x. Show that h(4) agrees with the value you found for f ( g (4)).

a. g (4) 2 4 16, and f (16) 3 16 5 53, so f ( g (4)) 53

Writing the same steps more compactly for the other two values of x gives

f ( g (0)) f 2 0 f (1) 3 1 5 8

f ( g ( 1)) f 2 1 f (0.5) 3(0.5) 5 6.5

b. g ( f ( 1)) g (3 1 5) g (2) 2 2 4, which does not equal 6.5 from part a

c. h(x) f ( g (x)) f 2 x 3 2 x 5

� e equation is h(x) 3 2 x 5.

So h(4) 3 2 4 5 3 16 5 53, which agrees with part a.

Example 4 shows you that you can compose a function with itself or compose more than two functions.

➤

Let fg (g (g x) x) x

EXAMPLE 3 ➤

a. g (4) g (4) gSOLUTION

➤

First � nd the value of the inside function, g (30). As shown on the le� in Figure 1-4e,

g (30) 2.8

Use this output of function g as the input for function f, as shown on the right in Figure 1-4e. Note that the x in f (x) is simply the input for function f and is not the same number as the x in g (x).

f (2.8) 180

f ( g (30)) 180

2010 504030

g(x)g(30) 2.8

21

43

65

x

f(2.8) 180

x1 2 3 4 5 6

100200300400

f (x)

Figure 1-4e

Composite Functions from TablesExample 2 shows you how to � nd values of a composite function when the two functions are de� ned numerically.

Functions f and g are de� ned only for the integer values of x in the table.

x f(x) g(x)

1 3 52 4 33 6 24 2 15 0 76 1 4

a. Find f ( g (x)) for the six values of x in the table.

b. Find g ( f (2)), and show that it does not equal f ( g (2)).

a. To � nd f ( g (1)), � rst � nd the value of the inside function, g (1), by � nding 1 in the x-column and g (1) in the g (x) column (third column).

g (1) 5

� en use 5 as the input for the outside function f by � nding 5 in the x-column and f (5) in the f (x) column (second column).

f (5) 0

f ( g (1)) 0

First � nd the value of the inside function, Figure 1-4e,

SOLUTION

➤

Functions EXAMPLE 2 ➤

a. To � nd in the

SOLUTION



Technology Notes

Presentation Sketch: Composition Present.gsp at www.keypress.com/keyonline uses dynagraphs to show composition of functions. This sketch also includes a page that illustrates domains and ranges for function composition.


Find the other values the same way. Here is a compact way to arrange your work.

f ( g (1)) f (5) 0

f ( g2)) f (3) 6

f ( g (3)) f (2) 4

f ( g (4)) f (1) 3

f ( g (5)) f (7), which does not exist

f ( g (6)) f (4) 2

b. g ( f (2)) g (4) 1, which is not the same as f ( g (2)) 6

Note that in order to � nd a value of a composite function such as f ( g (x)), the value of g (x) must be in the domain of the outside function, f. Because g (5) 7 in Example 2 and there is no value for f (7), the value of f ( g (5)) is unde� ned.

Composite Functions from EquationsExample 3 shows you how to � nd values of a composite function if you know the equations of the two functions.

Let f be the linear function f (x) 3x 5, and let g be the exponential function g (x) 2 x .

a. Find f ( g (4)), f ( g (0)), and f ( g ( 1)).

b. Find g ( f ( 1)) and show that it is not the same as f ( g ( 1)).

c. Find an equation for h(x) f ( g (x)) explicitly in terms of x. Show that h(4) agrees with the value you found for f ( g (4)).

a. g (4) 2 4 16, and f (16) 3 16 5 53, so f ( g (4)) 53

Writing the same steps more compactly for the other two values of x gives

f ( g (0)) f 2 0 f (1) 3 1 5 8

f ( g ( 1)) f 2 1 f (0.5) 3(0.5) 5 6.5

b. g ( f ( 1)) g (3 1 5) g (2) 2 2 4, which does not equal 6.5 from part a

c. h(x) f ( g (x)) f 2 x 3 2 x 5

� e equation is h(x) 3 2 x 5.

So h(4) 3 2 4 5 3 16 5 53, which agrees with part a.

Example 4 shows you that you can compose a function with itself or compose more than two functions.

➤


EXAMPLE 3 ➤

a. g (4) g (4) gSOLUTION

➤

First � nd the value of the inside function, g (30). As shown on the le� in Figure 1-4e,

g (30) 2.8

Use this output of function g as the input for function f, as shown on the right in Figure 1-4e. Note that the x in f (x) is simply the input for function f and is not the same number as the x in g (x).

f (2.8) 180

f ( g (30)) 180

2010 504030

g(x)g(30) 2.8

21

43

65

x

f(2.8) 180

x1 2 3 4 5 6

100200300400

f (x)

Figure 1-4e

Composite Functions from TablesExample 2 shows you how to � nd values of a composite function when the two functions are de� ned numerically.

Functions f and g are de� ned only for the integer values of x in the table.

x f(x) g(x)

1 3 52 4 33 6 24 2 15 0 76 1 4

a. Find f ( g (x)) for the six values of x in the table.

b. Find g ( f (2)), and show that it does not equal f ( g (2)).

a. To � nd f ( g (1)), � rst � nd the value of the inside function, g (1), by � nding 1 in the x-column and g (1) in the g (x) column (third column).

g (1) 5

� en use 5 as the input for the outside function f by � nding 5 in the x-column and f (5) in the f (x) column (second column).

f (5) 0

f ( g (1)) 0

First � nd the value of the inside function, Figure 1-4e,

SOLUTION

➤

Functions EXAMPLE 2 ➤

a. To � nd in the

SOLUTION


27

CAS Suggestions

You can use the CAS to calculate compositions of functions once the functions have been defi ned algebraically. Th e fi gure below shows the area computation from page 24. Notice the use of composition, the inclusion of units for the area computation, and the presentation of the correct units in the fi nal problem.

Th e fi gure below shows Example 5c. Th e domain restriction of the composite function is clear, even though it is not directly stated in the algebraic form. Students should be expected to produce the algebraic results without using a CAS.

A CAS can provide instant algebraic feedback. Students can do Example 5 on page 28 by hand and get verifi cation of their results from their CAS, gaining effi cient and valuable feedback.



a. � e le� graph in Figure 1-4g shows that g (2) 1 and g (6) 3 but g (8) does not exist, because 8 is outside the domain of function g. � e right graph shows the two output values of function g, 1 and 3, used as inputs for function f. From the graph, f (3) 2 and f ( 1) does not exist, because

1 is outside the domain of function f. Summarize the results:

f (g (6)) f (3) 2

f (g (8)) does not exist because 8 is outside the domain of g.

f (g (2)) f ( 1), which does not exist because 1 is outside the domain of f.

gg(6)

g(2)

No g(8)

x

43

1

26 7 8

y

x2

6

231–1 5

ff(3)

No f(–1)y

Figure 1-4g

b.

� e domain of f g seems to be 4 x 7.

c. Enter: f 1 (x) x 3/(x 2 and x 7) for g (x)

Enter: f 2 (x) 2x 8/(x 1 and x 5) for f (x)

Enter: f 3 (x) f 2 f 1 (x) for f (g (x))

Graph (Figure 1-4h), showing f (g (x)) solid style.� e domain of f g is 4 x 7, in agreement with part b.From the graph, the range of f g is 0 y 6.

a. � e le� graph in Figure 1-4g shows that does not exist, because 8 is outside the domain of function

SOLUTION

x g(x) f (g(x))

1 none none2 1 none3 0 none4 1 65 2 46 3 27 4 08 none none

Use Boolean variables or enter the domain directly, depending on your grapher, to restrict the domain.

f 1 and f 2 become function names in this format.x

4

6

21 21 5 7

f

g

f3(x) f(g(x))y

Figure 1-4h

Let f be the linear function f (x) 3x 5, and let g be the exponential function g (x) 2 x , as in Example 3. Find these values.

a. f ( f (2))

b. f (g( f ( 3)))

a. f (f (2)) f (3 2 5) f (11) 3 11 5 38

b. f (g (f ( 3))) f (g (3 3 5)) f (g ( 4)) f 2 4 f (0.0625) 3 0.0625 5 5.1875

Domain and Range of a Composite FunctionIn Example 2, you saw that the value of the inside function sometimes is not in the domain of the outside function. Example 5 shows you how to � nd the domain of a composite function and the corresponding range under this condition.

� e le� graph in Figure 1-4f shows function g with domain 2 x 7, and the right graph shows function f with domain 1 x 5.

g

x

4

1 2 7

y

f

x

6

21 5

y

Figure 1-4f

a. Show on copies of these graphs what happens when you try to � nd f g (6) , f (g (8)), and f (g (2)).

b. Make a table of values of g (x) and f (g (x)) for integer values of x from 1 through 8. If there is no value, write “none.” From the table, what does the domain of function f g seem to be?

c. � e equations of functions g and f are

g (x) x 3, for 2 x 7

f (x) 2x 8, for 1 x 5

Plot f (x), g (x), and f g (x) on your grapher, with the grapher’s grid showing. Does the domain of f g con� rm what you found numerically in part b? What is the range of f g?

d. Find the domain of f g algebraically and show that it agrees with part c.


EXAMPLE 4 ➤

a. f (SOLUTION

➤

� e le� graph in Figure 1-4f shows function right graph shows function

EXAMPLE 5 ➤




a. � e le� graph in Figure 1-4g shows that g (2) 1 and g (6) 3 but g (8) does not exist, because 8 is outside the domain of function g. � e right graph shows the two output values of function g, 1 and 3, used as inputs for function f. From the graph, f (3) 2 and f ( 1) does not exist, because

1 is outside the domain of function f. Summarize the results:

f (g (6)) f (3) 2

f (g (8)) does not exist because 8 is outside the domain of g.

f (g (2)) f ( 1), which does not exist because 1 is outside the domain of f.

gg(6)

g(2)

No g(8)

x

43

1

26 7 8

y

x2

6

231–1 5

ff(3)

No f(–1)y

Figure 1-4g

b.

� e domain of f g seems to be 4 x 7.

c. Enter: f 1 (x) x 3/(x 2 and x 7) for g (x)

Enter: f 2 (x) 2x 8/(x 1 and x 5) for f (x)

Enter: f 3 (x) f 2 f 1 (x) for f (g (x))

Graph (Figure 1-4h), showing f (g (x)) solid style.� e domain of f g is 4 x 7, in agreement with part b.From the graph, the range of f g is 0 y 6.

a. � e le� graph in Figure 1-4g shows that does not exist, because 8 is outside the domain of function

SOLUTION

x g(x) f (g(x))

1 none none2 1 none3 0 none4 1 65 2 46 3 27 4 08 none none

Use Boolean variables or enter the domain directly, depending on your grapher, to restrict the domain.

f 1 and f 2 become function names in this format.x

4

6

21 21 5 7

f

g

f3(x) f(g(x))y

Figure 1-4h

Let f be the linear function f (x) 3x 5, and let g be the exponential function g (x) 2 x , as in Example 3. Find these values.

a. f ( f (2))

b. f (g( f ( 3)))

a. f (f (2)) f (3 2 5) f (11) 3 11 5 38

b. f (g (f ( 3))) f (g (3 3 5)) f (g ( 4)) f 2 4 f (0.0625) 3 0.0625 5 5.1875

Domain and Range of a Composite FunctionIn Example 2, you saw that the value of the inside function sometimes is not in the domain of the outside function. Example 5 shows you how to � nd the domain of a composite function and the corresponding range under this condition.

� e le� graph in Figure 1-4f shows function g with domain 2 x 7, and the right graph shows function f with domain 1 x 5.

g

x

4

1 2 7

y

f

x

6

21 5

y

Figure 1-4f

a. Show on copies of these graphs what happens when you try to � nd f g (6) , f (g (8)), and f (g (2)).

b. Make a table of values of g (x) and f (g (x)) for integer values of x from 1 through 8. If there is no value, write “none.” From the table, what does the domain of function f g seem to be?

c. � e equations of functions g and f are

g (x) x 3, for 2 x 7

f (x) 2x 8, for 1 x 5

Plot f (x), g (x), and f g (x) on your grapher, with the grapher’s grid showing. Does the domain of f g con� rm what you found numerically in part b? What is the range of f g?

d. Find the domain of f g algebraically and show that it agrees with part c.


EXAMPLE 4 ➤

a. f (SOLUTION

➤

� e le� graph in Figure 1-4f shows function right graph shows function

EXAMPLE 5 ➤




Reading Analysis

From what you have read in this section, what do you consider to be the main idea? � ink of a real-world example, other than the one in the text, in which the value of one variable depends on the value of a second variable and the value of the second variable depends on the value of a third variable. In your example, which is the inside function and which is the outside function? For there to be a value of the composite function, what must be true of the value of the inside function?

Quick Review

Q1. What transformation of f is represented by g (x) 3f (x)?

Q2. What transformation of f is represented by h(x) 5 f (x)?

Q3. If g is a horizontal translation of f by 4 units, then g (x) ? .

Q4. If p is a horizontal dilation of f by a factor of 0.2, then p (x) ? .

Q5. Why is y 3 x 5 not an exponential function, even though it has an exponent?

Q6. Write the general equation of a quadratic function.

Q7. For what value of x will the graph of y x 3 ____ x 5 have a discontinuity?

Q8. Sketch the graph of f (x) x . Q9. Find 40% of 300. Q10. Which of these is a horizontal dilation by a

factor of 2? A. g (x) 2f (x) B. g (x) 0.5f (x) C. g (x) f (0.5x) D. g (x) f (2x)

1. Flashlight Problem: You shine a � ashlight, making a circular spot of light on the wall with radius 5 cm. As you back away from the wall, the radius increases at a rate of 7 cm/s. a. Find the radius at times 4 s and 7 s a� er you

start backing away.

b. Use the radius at times 4 s and 7 s to � nd the area of the spot of light at these times.

c. Why can it be said that the area is a composite function of time?

d. Let t be the number of seconds since you started backing away. Let r(t) be the radius of the spot of light, in centimeters. Let a(r(t)) be the area of the spot, in square centimeters. Write an equation for r(t) as a function of t. Write another equation for a(r(t)) as a function of r(t). Write a third equation for a r(t)) explicitly in terms of t. Show that the last equation gives the correct area for times t 4 s and t 7 s.

2. Bacteria Culture Problem: When you grow a culture of bacteria in a petri dish, the area of the culture is a measure of the number of bacteria present. Suppose that the area of the culture, A(t), in square millimeters, is given by this function of time t, in hours:

A(t) 9(1. 1 t ) a. Find A(0), A(5), and A(10), the area at times

t 0 h, 5 h, and 10 h, respectively. b. Assume that the bacteria culture is circular.

Find the radius of the culture at the three times in part a.

c. Why can it be said that the radius is a composite function of time?

d. Let R be the radius function, with input A(t), the area of the culture. Write an equation for R(A(t)), the radius as a function of area. � en write an equation for R(A(t)) explicitly in terms of t. Show that this equation gives the correct answer for the radius at time t 5 h.

3. Shoe Size Problem: � e size shoe a person wears, S(x), is a function of the length of the person’s foot. � e length of the foot, L(x), is a function of the person’s age. a. Sketch reasonable graphs of functions S

and L. Label the axes of each graph with the name of the variable represented.

5min

Reading Analysis b. Use the radius at times 4 s and 7 s to � nd the

Problem Set 1-4d. To calculate the domain algebraically, � rst observe that g (x) must be within the domain of f.

1 g (x) 5 Write g (x) in the domain of f.

1 x 3 5 Substitute x 3 for g (x).

4 x 8 Add 3 to all three members of the inequality.

Next observe that x must also be in the domain of g, speci� cally, 2 x 7. � e domain of f g is the intersection of these two intervals. Number-line graphs (Figure 1-4i) will help you visualize the intersection.

the domain of f g is 4 x 7.

x 2 x 7

x8

4 x 8

x

20

0

0Intersection

7

4

4 7

Figure 1-4i

DEFINITION AND PROPERTIES: Composite Function� e composite function f g (pronounced “f composition g” or “f of g”) is the function

( f g)(x) f ( g (x))

Function g, the inside function, is evaluated � rst, using x as its input. Function f, the outside function, is evaluated next, using g (x) as its input (the output of function g).

� e domain of f g is the set of all values of x in the domain of g for which g (x) is in the domain of f. � e � gure shows this relationship.

Note: Horizontal dilations and translations are examples of composite functions because they are inside transformations applied to x. For instance, the horizontal translation g (x) 3(x 2) is actually a composite function with the inside function f (x) x 2.

➤



PRO B LE M N OTES

Supplementary Problems for this section are available at www.keypress.com/keyonline.

Q1. y-dilation by 3Q2. y-translation by 5Q3. f (x 1 4)Q4. f (5x)Q5. x is the base, not the exponent.Q6. f (x) 5 a x 2 1 bx 1 c, a 0Q7. x 5 5Q8.

Q9. 120Q10. C

Problem 1 is similar to the example of dropping a pebble into a pond. Students should assume that the radius of the spot of light increases at a constant rate of 7 cm/s. In calculus, students will encounter related-rate problems in which rate changes but not at a constant rate.1a. 33 cm; 54 cm1b. 3421.1943… cm 2 ; 9160.8841… cm 2 1c. Th e area depends on the radius, which in turn depends on the time.1d. r(t) 5 5 1 7t; a  r(t) 5  r(t) 2 ; a  r(t) 5 (5 1 7t ) 2 ; a(4) 5 3421.1943… c m 2 ; a(7) 5 9160.8841… c m 3

x

f(x)

1

1


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? � ink of a real-world example, other than the one in the text, in which the value of one variable depends on the value of a second variable and the value of the second variable depends on the value of a third variable. In your example, which is the inside function and which is the outside function? For there to be a value of the composite function, what must be true of the value of the inside function?

Quick Review

Q1. What transformation of f is represented by g (x) 3f (x)?

Q2. What transformation of f is represented by h(x) 5 f (x)?

Q3. If g is a horizontal translation of f by 4 units, then g (x) ? .

Q4. If p is a horizontal dilation of f by a factor of 0.2, then p (x) ? .

Q5. Why is y 3 x 5 not an exponential function, even though it has an exponent?

Q6. Write the general equation of a quadratic function.

Q7. For what value of x will the graph of y x 3 ____ x 5 have a discontinuity?

Q8. Sketch the graph of f (x) x . Q9. Find 40% of 300. Q10. Which of these is a horizontal dilation by a

factor of 2? A. g (x) 2f (x) B. g (x) 0.5f (x) C. g (x) f (0.5x) D. g (x) f (2x)

1. Flashlight Problem: You shine a � ashlight, making a circular spot of light on the wall with radius 5 cm. As you back away from the wall, the radius increases at a rate of 7 cm/s. a. Find the radius at times 4 s and 7 s a� er you

start backing away.

b. Use the radius at times 4 s and 7 s to � nd the area of the spot of light at these times.

c. Why can it be said that the area is a composite function of time?

d. Let t be the number of seconds since you started backing away. Let r(t) be the radius of the spot of light, in centimeters. Let a(r(t)) be the area of the spot, in square centimeters. Write an equation for r(t) as a function of t. Write another equation for a(r(t)) as a function of r(t). Write a third equation for a r(t)) explicitly in terms of t. Show that the last equation gives the correct area for times t 4 s and t 7 s.

2. Bacteria Culture Problem: When you grow a culture of bacteria in a petri dish, the area of the culture is a measure of the number of bacteria present. Suppose that the area of the culture, A(t), in square millimeters, is given by this function of time t, in hours:

A(t) 9(1. 1 t ) a. Find A(0), A(5), and A(10), the area at times

t 0 h, 5 h, and 10 h, respectively. b. Assume that the bacteria culture is circular.

Find the radius of the culture at the three times in part a.

c. Why can it be said that the radius is a composite function of time?

d. Let R be the radius function, with input A(t), the area of the culture. Write an equation for R(A(t)), the radius as a function of area. � en write an equation for R(A(t)) explicitly in terms of t. Show that this equation gives the correct answer for the radius at time t 5 h.

3. Shoe Size Problem: � e size shoe a person wears, S(x), is a function of the length of the person’s foot. � e length of the foot, L(x), is a function of the person’s age. a. Sketch reasonable graphs of functions S

and L. Label the axes of each graph with the name of the variable represented.

5min

Reading Analysis b. Use the radius at times 4 s and 7 s to � nd the

Problem Set 1-4d. To calculate the domain algebraically, � rst observe that g (x) must be within the domain of f.

1 g (x) 5 Write g (x) in the domain of f.

1 x 3 5 Substitute x 3 for g (x).

4 x 8 Add 3 to all three members of the inequality.

Next observe that x must also be in the domain of g, speci� cally, 2 x 7. � e domain of f g is the intersection of these two intervals. Number-line graphs (Figure 1-4i) will help you visualize the intersection.

the domain of f g is 4 x 7.

x 2 x 7

x8

4 x 8

x

20

0

0Intersection

7

4

4 7

Figure 1-4i

DEFINITION AND PROPERTIES: Composite Function� e composite function f g (pronounced “f composition g” or “f of g”) is the function

( f g)(x) f ( g (x))

Function g, the inside function, is evaluated � rst, using x as its input. Function f, the outside function, is evaluated next, using g (x) as its input (the output of function g).

� e domain of f g is the set of all values of x in the domain of g for which g (x) is in the domain of f. � e � gure shows this relationship.

Note: Horizontal dilations and translations are examples of composite functions because they are inside transformations applied to x. For instance, the horizontal translation g (x) 3(x 2) is actually a composite function with the inside function f (x) x 2.

➤


31

Problem 2 asks students to determine whether the area and radius of a bacterial culture are growing at an increasing rate or at a decreasing rate. This problem is a good preview of the related-rate problems students will solve in calculus. Students taking AP Biology may do experiments similar to the one described in Problem 2. You may want to ask the AP Biology teacher to suggest problems similar to Problem 2 that the students will actually perform in AP Biology.2a. A(0) 5 9 m m 2 ; A(5) 5 14.4945… m m 2 ; A(10) 5 23.3436… m m 2 2b. R(0) 5 1.6925… mm; R(5) 5 2.1479… mm; R(10) 5 2.7258… mm2c. The radius depends on the area (essentially the number of bacteria), which in turn depends on the time.

2d. R  A(t) 5  ____

A(t) ____

.

A(t) 5 9(1.1 ) t , so R  A(t) 5  ______

9(1.1 ) t ______

.

R  A(5) 5  ______

9(1. 1) 5 ______ 5 2.1479… mm

Problem 3 involves a step function.3a. Answers will vary. Note that shoe size is a discrete graph, because shoe sizes come only in half units. Sample answer:

7 8 9

5

10

x (in.)

S(x) (size)

6 10 11

10

10

5

20 30 40 50 60 70 80x (yr)

L(x) (in.)



a. Find g (1) and f ( g (1)). b. Find g (2) and f ( g (2)). c. Find g (3) and f ( g (3)). d. Find f (4) and g ( f (4)). e. Find g ( f (3)). f. Find f ( f (5)). g. Find g ( g (3)). h. Find f ( f ( f (1))).

8. Composite Function Numerically, Problem 2: Functions u and v consist of the discrete points in the table, and only these points. Find the values of the composite functions, or explain why no such value exists. a. Find v(2) and u(v(2)). b. Find v(6) and u(v(6)). c. Find v(4) and u(v(4)). d. Find u(4) and v(u(4)). e. Find v(u(10)). f. Find v(v(10)) g. Find u(u(6)). h. Find v(v(v(8)))

9. Composite Function Algebraically, Problem 1: Let g and f be de�ned by f (x) 9 x 4 x 8 g (x) x 2 1 x 5 a. Make a table showing values of g (x) for each

integer value of x in the domain of g. In another column, show the corresponding values of f ( g (x)). If there is no such value, write “none.”

b. From your table in part a, what does the domain of the composite function f g seem to be? Con�rm (or refute) your conclusion by �nding the domain algebraically.

c. Explain why f ( g (6)) is unde�ned. Explain why f ( g (1)) is unde�ned, even though g (1) is de�ned.

d. Repeat parts a and b for the composite function g f.

e. Figure 1-4l shows functions f and g. Enter the two functions as f 1 (x) and f 2 (x). �en enter f ( g (x)) as f 3 (x) f 1 f 2 (x) , and g ( f (x)) as f 2 f 1 (x) . Plot the graphs using the window

shown, with the grapher’s grid showing and thick style for the two composite function graphs. Sketch the result. Do the domains of the composite functions from the graph agree with your results in parts b and d?

fg

x5

5

y

Figure 1-4l

f. Find f ( f (5)). Explain why g ( g (5)) is unde�ned.

10. Composite Function Algebraically, Problem 2: Let f and g be de�ned by f (x) x 2 8x 4 1 x 6 g (x) 5 x 0 x 7 a. Make a table showing values of g (x) for each



c. Show why f ( g (3)) is de�ned but g ( f (3)) is unde�ned.

d. Figure 1-4m shows the graphs of f and g. Enter these equations as f 1 (x) and f 2 (x). �en enter f ( g (x)) as f 3 (x) f 1 (x) f 2 (x) . Plot the three graphs with the grapher’s grid showing. Sketch the result. Does the domain of the composite function agree with your calculation in part b?

5

5

10

x

y

Figure 1-4m

x u(x) v(x)

2 3 64 8 56 2 48 10 2

10 6 8

b. What does x represent in function S? What does x represent in function L? What does the composite function S(L(x)) represent? What, if any, real-world meaning does L(S(x)) have?

c. Let f (x) (S L)(x). Sketch a reasonable graph of function f. Label the axes of the graph with the name of the variable represented.

4. Tra�c Problem: �e length of time, T(x), it takes you to travel a mile on the freeway depends on the speed at which you travel. �e speed, S(x), depends on the number of other cars on that mile of freeway. a. Sketch reasonable graphs of functions T

and S. Label the axes of each graph with the name of the variable represented.

b. What does x represent in function T? What does x represent in function S? What does the composite function T(S(x)) represent? What, if any, real-world meaning does S(T(x)) have?

c. Let g (x) (T S)(x). Sketch a reasonable graph of function g. Label the axes of the graph with the name of the variable represented.

5. Composite Function Graphically, Problem 1: Functions h and p are de�ned by the graphs in Figure 1-4j, in the domains shown.

1 2 3 4 5 6 7x

7654321

h(x)

p(x)

1 2 3 4 5 6 7x

7654321

Figure 1-4j

a. Find h(3). On a copy of the graphs, draw arrows to show how you found this value.

b. Use the output of h(3) to �nd p(h(3)). Draw arrows to show how you found this value.

c. Find p(h(2)) and p(h(5)) by �rst �nding h(2) and h(5) and then using these values as inputs for function p.

d. Find h(p(2)) by �rst �nding p(2) and then using the result as the input for function h. Draw arrows to show how you found this value. Show that h(p(2)) p(h(2)).

e. Explain why there is no value of h(p(0)), even though there is a value of p(0).

6. Composite Function Graphically, Problem 2: Functions f and g are de�ned by the graphs in Figure 1-4k, in the domains shown.

1 2 3 4 5 6x

605040302010

g(x)

f(x)

20 40 60x

300250200150100

50

Figure 1-4k

a. Find the approximate value of g (4). On a copy of the graphs, show how you found this value.

b. Use the output of g (4) to �nd the approximate value of f ( g (4)). Draw arrows to show how you found this value.

c. Find approximate values of f ( g (3)) and f ( g (2)) by �rst �nding g (3) and g (2) and then using these values as inputs for function f.

d. Explain why there is no value of f ( g (6)). e. Try to �nd f ( g (5)) by �rst �nding g (5)

and then using the result as the input for function f. Draw arrows to illustrate why there is no value of f ( g (5)).

7. Composite Function Numerically, Problem 1: Functions f and g consist of the discrete points in the table, and only these points. Find the values of the composite functions, or explain why no such value exists.

x f(x) g(x)

1 3 22 5 33 4 74 2 55 1 4



Problem Notes (continued)3b. In S(x), x represents foot length (in inches, for the preceding graph). In L(x), x represents age (in years). The composite function S  L(x) gives shoe size as a function of age (x represents age). L  S(x) would be meaningless with the given functions L and S. Because x is substituted into S, x must represent foot length. S then gives shoe size. But this is substituted into L, which expects to have an age, not a shoe size, substituted into it.3c. Answers will vary. Sample Answer:

10 20 30 40 50 60 70 80x (yr)

S(L(x))

5

10

90 Problems 5 and 6 ask students to answer questions about a composite function from graphs without using explicit formulas for the functions. A blackline master for these problems is available in the Instructor’s Resource Book.5a. h(3) 5 5

1 2 3 4 5 6 7

1234567

x

h(x)

5b. p  h(3) 5 p(5) 5 3.5

1 2 3 4 5 6 7

1234567

x

p(x)

5c. p  h(2) 5 4.5; p  h(5) 5 46a. g (4) 48

1 2 3 4 5 6

102030405060

x

g(x)

6b. f  g (4) 51

20 40x � 48

60

50100150200250300

x

f (x)

6c. f  g (3) 75; f  g (2) 150

6d. 6 is not in the domain of g.6e. g (5) is not in the domain of f.

1 2 3 4 5 6

102030405060

x

g(x)

20 40 60

50100150200250300

x

f (x)


a. Find g (1) and f ( g (1)). b. Find g (2) and f ( g (2)). c. Find g (3) and f ( g (3)). d. Find f (4) and g ( f (4)). e. Find g ( f (3)). f. Find f ( f (5)). g. Find g ( g (3)). h. Find f ( f ( f (1))).

8. Composite Function Numerically, Problem 2: Functions u and v consist of the discrete points in the table, and only these points. Find the values of the composite functions, or explain why no such value exists. a. Find v(2) and u(v(2)). b. Find v(6) and u(v(6)). c. Find v(4) and u(v(4)). d. Find u(4) and v(u(4)). e. Find v(u(10)). f. Find v(v(10)) g. Find u(u(6)). h. Find v(v(v(8)))

9. Composite Function Algebraically, Problem 1: Let g and f be de�ned by f (x) 9 x 4 x 8 g (x) x 2 1 x 5 a. Make a table showing values of g (x) for each



c. Explain why f ( g (6)) is unde�ned. Explain why f ( g (1)) is unde�ned, even though g (1) is de�ned.

d. Repeat parts a and b for the composite function g f.

e. Figure 1-4l shows functions f and g. Enter the two functions as f 1 (x) and f 2 (x). �en enter f ( g (x)) as f 3 (x) f 1 f 2 (x) , and g ( f (x)) as f 2 f 1 (x) . Plot the graphs using the window

shown, with the grapher’s grid showing and thick style for the two composite function graphs. Sketch the result. Do the domains of the composite functions from the graph agree with your results in parts b and d?

fg

x5

5

y

Figure 1-4l

f. Find f ( f (5)). Explain why g ( g (5)) is unde�ned.

10. Composite Function Algebraically, Problem 2: Let f and g be de�ned by f (x) x 2 8x 4 1 x 6 g (x) 5 x 0 x 7 a. Make a table showing values of g (x) for each



c. Show why f ( g (3)) is de�ned but g ( f (3)) is unde�ned.

d. Figure 1-4m shows the graphs of f and g. Enter these equations as f 1 (x) and f 2 (x). �en enter f ( g (x)) as f 3 (x) f 1 (x) f 2 (x) . Plot the three graphs with the grapher’s grid showing. Sketch the result. Does the domain of the composite function agree with your calculation in part b?

5

5

10

x

y

Figure 1-4m

x u(x) v(x)

2 3 64 8 56 2 48 10 2

10 6 8

b. What does x represent in function S? What does x represent in function L? What does the composite function S(L(x)) represent? What, if any, real-world meaning does L(S(x)) have?

c. Let f (x) (S L)(x). Sketch a reasonable graph of function f. Label the axes of the graph with the name of the variable represented.

4. Tra�c Problem: �e length of time, T(x), it takes you to travel a mile on the freeway depends on the speed at which you travel. �e speed, S(x), depends on the number of other cars on that mile of freeway. a. Sketch reasonable graphs of functions T

and S. Label the axes of each graph with the name of the variable represented.

b. What does x represent in function T? What does x represent in function S? What does the composite function T(S(x)) represent? What, if any, real-world meaning does S(T(x)) have?

c. Let g (x) (T S)(x). Sketch a reasonable graph of function g. Label the axes of the graph with the name of the variable represented.

5. Composite Function Graphically, Problem 1: Functions h and p are de�ned by the graphs in Figure 1-4j, in the domains shown.

1 2 3 4 5 6 7x

7654321

h(x)

p(x)

1 2 3 4 5 6 7x

7654321

Figure 1-4j

a. Find h(3). On a copy of the graphs, draw arrows to show how you found this value.

b. Use the output of h(3) to �nd p(h(3)). Draw arrows to show how you found this value.

c. Find p(h(2)) and p(h(5)) by �rst �nding h(2) and h(5) and then using these values as inputs for function p.

d. Find h(p(2)) by �rst �nding p(2) and then using the result as the input for function h. Draw arrows to show how you found this value. Show that h(p(2)) p(h(2)).

e. Explain why there is no value of h(p(0)), even though there is a value of p(0).

6. Composite Function Graphically, Problem 2: Functions f and g are de�ned by the graphs in Figure 1-4k, in the domains shown.

1 2 3 4 5 6x

605040302010

g(x)

f(x)

20 40 60x

300250200150100

50

Figure 1-4k

a. Find the approximate value of g (4). On a copy of the graphs, show how you found this value.

b. Use the output of g (4) to �nd the approximate value of f ( g (4)). Draw arrows to show how you found this value.

c. Find approximate values of f ( g (3)) and f ( g (2)) by �rst �nding g (3) and g (2) and then using these values as inputs for function f.

d. Explain why there is no value of f ( g (6)). e. Try to �nd f ( g (5)) by �rst �nding g (5)

and then using the result as the input for function f. Draw arrows to illustrate why there is no value of f ( g (5)).

7. Composite Function Numerically, Problem 1: Functions f and g consist of the discrete points in the table, and only these points. Find the values of the composite functions, or explain why no such value exists.

x f(x) g(x)

1 3 22 5 33 4 74 2 55 1 4


33

Problems 9 and 10 are similar to Example 5.

Consider supplementing Problems 9–11 with a CAS verifi cation.9b. 2 x 59c. 6 is not in the domain of g, so g (x) is undefi ned. g (1) 5 3, but 3 is not in the domain of f.9e.

Th e domains of the composite functions match the calculations in parts b and d.9f. f   f (5) 5 f (4) 5 5; g (5) 5 7, and 7 is not in the domain of g.1 0a. x g(x) f(g(x))

0 5 111 4 122 3 113 2 84 1 35 0 none6 21 none

7 22 none

10b. 0 x 410c. f  g (3) 5 8; g   f (3) 5 g (11), but 11 is not in the domain of g, so g   f (3) is undefi ned.10d.

4 8

4

y

x

g

f

g f

f g

2 4

5

10f

g

f g

y

xProblems 7 and 8 use numeric information to fi nd values of composite functions.7a. g (1) 5 2; f  g (1) 5 57b. g (2) 5 3; f  g (2) 5 47c. g (3) 5 7; f  g (3) is undefi ned.7d. f (4) 5 2; g   f (4) 5 37e. g   f (3) 5 5 7f. f   f (5) 5 37g. g  g (3) is undefi ned.7h. f   f   f (1) 5 2

8a. v(2) 5 6; u  v(2) 5 28b. v(6) 5 4; u  v(6) 5 88c. v(4) 5 5; u  v(4) 5 is undefi ned because v(4) is not in the domain of u.8d. u(4) 5 8; v  u(4) 5 28e. v  u(10) 5 48f. v  v(10) 5 2 8g. u  u(6) 5 38h. v  v  v(8) 5 v  v(2) 5 v(6) 5 4


See pages 976–977 for answers to Problems 4, 5d, 5e, 9a, and 9d.

35Section 1-5: Inverse Functions and Parametric Equations

Inverse Functions and Parametric Equations� e photograph shows a highway crew painting a center stripe. From records of previous work the crew has done, it is possible to predict how much of the stripe the crew will have painted at any time during a normal eight-hour shi� . It may also be possible to tell how long the crew has been working by how much stripe has been painted. � e input for the distance function is time, and the input for the time function is distance.

If a new relation is formed by interchanging the input and output variables in a given relation, the two relations are called inverses of each other. If both relations turn out to be functions, they are called inverse functions. If not, the relation and its inverse can still be plotted easily using parametric equations in which both x and y are functions of some third variable, such as time.

Given a function, � nd its inverse relation, and tell whether the inverse relation is a function. Graph parametric equations both by hand and on a grapher, and use parametric equations to graph the inverse of a function.

Inverse of a Function Numerically

Suppose that the distance d, in miles, a particular highway crew paints in an eight-hour shi� is given numerically by this function of t, in hours, it has been on the job.

Let d f (t). You can see that f (1) 0.2, f (2) 0.6, . . . , f (8) 3.0. � e input for function f is the number of hours, and the output is the number of miles.

As long as the crew does not stop painting during the eight-hour shi� , the number of hours it has been painting is a function of the distance. Let t g (d). You can see that g (0.2) 1, g (0.6) 2, . . . , g (3) 8. � e input for function g is the number of miles, and the output is the number of hours. � e input and output for functions f and g have been interchanged, and thus the two functions are inverses of each other.

Inverse FunctionParametric Equations

1-5

Given a function, � nd its inverse relation, and tell whether the inverse relation is a function. Graph parametric equations both by hand and on a grapher,

Objective

t (h) d (mi)

1 0.22 0.63 1.04 1.45 1.86 2.27 2.68 3.0

e. Find an equation for f ( g (x)) explicitly in terms of x. Enter this equation as f 4 (x) and plot it on the same screen as the other three functions. What similarities and what di� erences do you see for f 4 (x) and f 3 (x)?

11. Square and Square Root Functions: Let f and g be de� ned by f (x) x 2 , where x is any real number g (x)

__ x , where the values of x make g (x)

a real number a. Find f ( g (3)), f ( g (7)), g ( f (5)), and

g ( f (8)). What do you notice in each case? Make a conjecture: “For all values of x, f ( g (x)) ? and g ( f (x)) ? .”

b. Test your conjecture by � nding f ( g ( 9)) and g( f ( 9)). Does your conjecture hold for negative values of x?

c. Plot f (x), g (x), and f ( g (x)) on the same screen. Use approximately equal scales on both axes, as in Figure 1-4n. Explain why f ( g (x)) x, but only for nonnegative values of x.

5

f

g

5

x5

y

Figure 1-4n

d. Deactivate f ( g (x)), and plot f (x), g (x), and g ( f (x)) on the same screen. Sketch the result.

e. Explain why g ( f (x)) x for nonnegative values of x, but g ( f (x)) x (the opposite of x) for negative values of x. What other familiar function has this property?

12. Horizontal Translation and Dilation Problem: Let f, g, and h be de� ned by

f (x) x 2 2 x 2 g (x) x 3 for all real values of x h(x) 1 _ 2 x for all real values of x

a. f ( g (x)) f (x 3). What transformation is applied to function f by composing it with g?

b. f (h(x)) f 1 _ 2 x . What transformation is applied to function f by composing it with h?

c. Plot the graphs of f, f g, and f h. Sketch the results. Do the graphs con� rm your conclusions in parts a and b?

For Problems 13 and 14, � nd what transformation will turn the dashed graph ( f ) into the solid graph ( g). 13. 14.

g x51

1 f

y

1g

x

f1 5

Both graphscoincide.

y

15. Linear Function and Its Inverse Problem: Let f and g be de� ned by

f (x) 2 _ 3 x 2 g (x) 1.5x 3 a. Find f ( g (6)), f ( g ( 15)), g ( f (10)), and

g ( f ( 8)). What do you notice in each case? b. Plot the graphs of f, g, f g, and g f on the

same screen. How are the graphs of f g and g f related? How are the graphs of f g and g f related to their “parent” graphs, f and g ?

c. Show that f ( g (x)) and g ( f (x)) both equal x. d. Functions f and g in this problem are said to

be inverses of each other. Whatever f does to x, g undoes. Let h(x) 5x 7. Find an equation for the inverse function of h.



Problem Notes (continued)10e. f  g (x) 5 2 x 2 1 2x 1 11, with the domain 0 x 4 found in part b. Th e graph coincides with the graph in part d.

Problem 11 requires students to work with square root functions. Because the square root of a negative number doesn’t exist among real numbers, the square root functions in these problems have restricted domains.11a. f  g (3) 5 3; f  g (7) 5 7; g   f (5) 5 5; g   f (8) 5 8; Conjecture: For all values of x, f  g (x) 5 g  f (x) 5 x.11b. f  g (29) is undefi ned. g   f (29) 5 9 29. No.11c.

g is defi ned only for nonnegative x, so f + g is defi ned only for nonnegative x.

11d.

11e. g  f (x) 5 g ( x 2 ) 5  __

x 2

5 x if x 0

2x if x , 0 5 x

Problem 12 connects Sections 1-3 and 1-4.12a. Translation 3 units to the right12b. Horizontal dilation by a factor of 2

Problems 13–15 prepare students for later sections, so it is important to assign them.13. If the dotted graph is f (x), 1 x 5, then the solid graph is g(x) 5 f (2x), 25 x 21. In terms of

2

4

y

x

f

g

f g

�2 2

2

4

y

xg

f

g fg f

�2 2

 

composition of functions, the solid graph is g(x) 5 f  h(x) , where h(x) 5 2x.

Problem 15 asks students to compose a function and its inverse and to observe that the two functions “undo” one another.


1. Every transformation is mathematically equivalent to a composition of functions. Let f (x) 5 x 1 2. If g (x) is any other

function, explain the meaning of f (g (x)), g ( f (x)), and f ( f (x)).

2. When a linear function is composed with another linear function, what type of function is the result? Prove your claim.

3. What is the slope of the composition of two linear functions? Under what conditions is the composition of two linear functions a decreasing function?See page 977 for answers to Problems 12c,

14–15 and CAS Problems 1–3.


Inverse Functions and Parametric Equations� e photograph shows a highway crew painting a center stripe. From records of previous work the crew has done, it is possible to predict how much of the stripe the crew will have painted at any time during a normal eight-hour shi� . It may also be possible to tell how long the crew has been working by how much stripe has been painted. � e input for the distance function is time, and the input for the time function is distance.

If a new relation is formed by interchanging the input and output variables in a given relation, the two relations are called inverses of each other. If both relations turn out to be functions, they are called inverse functions. If not, the relation and its inverse can still be plotted easily using parametric equations in which both x and y are functions of some third variable, such as time.

Given a function, � nd its inverse relation, and tell whether the inverse relation is a function. Graph parametric equations both by hand and on a grapher, and use parametric equations to graph the inverse of a function.

Inverse of a Function Numerically

Suppose that the distance d, in miles, a particular highway crew paints in an eight-hour shi� is given numerically by this function of t, in hours, it has been on the job.

Let d f (t). You can see that f (1) 0.2, f (2) 0.6, . . . , f (8) 3.0. � e input for function f is the number of hours, and the output is the number of miles.

As long as the crew does not stop painting during the eight-hour shi� , the number of hours it has been painting is a function of the distance. Let t g (d). You can see that g (0.2) 1, g (0.6) 2, . . . , g (3) 8. � e input for function g is the number of miles, and the output is the number of hours. � e input and output for functions f and g have been interchanged, and thus the two functions are inverses of each other.

Inverse FunctionParametric Equations

1-5

Given a function, � nd its inverse relation, and tell whether the inverse relation is a function. Graph parametric equations both by hand and on a grapher,

Objective

t (h) d (mi)

1 0.22 0.63 1.04 1.45 1.86 2.27 2.68 3.0

e. Find an equation for f ( g (x)) explicitly in terms of x. Enter this equation as f 4 (x) and plot it on the same screen as the other three functions. What similarities and what di� erences do you see for f 4 (x) and f 3 (x)?

11. Square and Square Root Functions: Let f and g be de� ned by f (x) x 2 , where x is any real number g (x)

__ x , where the values of x make g (x)

a real number a. Find f ( g (3)), f ( g (7)), g ( f (5)), and

g ( f (8)). What do you notice in each case? Make a conjecture: “For all values of x, f ( g (x)) ? and g ( f (x)) ? .”

b. Test your conjecture by � nding f ( g ( 9)) and g( f ( 9)). Does your conjecture hold for negative values of x?

c. Plot f (x), g (x), and f ( g (x)) on the same screen. Use approximately equal scales on both axes, as in Figure 1-4n. Explain why f ( g (x)) x, but only for nonnegative values of x.

5

f

g

5

x5

y

Figure 1-4n

d. Deactivate f ( g (x)), and plot f (x), g (x), and g ( f (x)) on the same screen. Sketch the result.

e. Explain why g ( f (x)) x for nonnegative values of x, but g ( f (x)) x (the opposite of x) for negative values of x. What other familiar function has this property?

12. Horizontal Translation and Dilation Problem: Let f, g, and h be de� ned by

f (x) x 2 2 x 2 g (x) x 3 for all real values of x h(x) 1 _ 2 x for all real values of x

a. f ( g (x)) f (x 3). What transformation is applied to function f by composing it with g?

b. f (h(x)) f 1 _ 2 x . What transformation is applied to function f by composing it with h?

c. Plot the graphs of f, f g, and f h. Sketch the results. Do the graphs con� rm your conclusions in parts a and b?

For Problems 13 and 14, � nd what transformation will turn the dashed graph ( f ) into the solid graph ( g). 13. 14.

g x51

1 f

y

1g

x

f1 5

Both graphscoincide.

y

15. Linear Function and Its Inverse Problem: Let f and g be de� ned by

f (x) 2 _ 3 x 2 g (x) 1.5x 3 a. Find f ( g (6)), f ( g ( 15)), g ( f (10)), and

g ( f ( 8)). What do you notice in each case? b. Plot the graphs of f, g, f g, and g f on the

same screen. How are the graphs of f g and g f related? How are the graphs of f g and g f related to their “parent” graphs, f and g ?

c. Show that f ( g (x)) and g ( f (x)) both equal x. d. Functions f and g in this problem are said to

be inverses of each other. Whatever f does to x, g undoes. Let h(x) 5x 7. Find an equation for the inverse function of h.




Class Time1–2 days

Homework AssignmentDay 1: RA, Q1–Q10,

Problems 1–15 odd, 16Day 2: Problems 17, 19, 21, 25, 31, 35, 37, 38

Teaching ResourcesExploration 1-5: Parametric Equations

GraphExploration 1-5a: Inverses of FunctionsExploration 1-5b: Introduction to

Parametric EquationsBlackline Masters

Problem 3Problem 5Problem 6Problem 7Problem 8

Test 2, Sections 1-4 to 1-5, Forms A and B


Presentation Sketch: Inverse Present.gsp

TE ACH I N G

Important Terms and ConceptsIncreasingInverseInverse functionInverse function notation: f 21 (x)f and f 21 refl ect across the line y 5 xInvertible functionOne-to-one functionStrictly increasing functionStrictly decreasing functionParametric equationsTh erefore symbol, Q.E.D.


�e linear function that �ts the graph of f 1 is

y 2.5x 0.5 slope 2.5, y-intercept 0.5

If you know the equation of a function, you can transform it algebraically to �nd the equation of the inverse relation by �rst interchanging the variables.

Function: y 0.4x 0.2

Inverse: x 0.4y 0.2

�e equation of the inverse relation can be solved for y in terms of x.

x 0.4y 0.2

y 2.5x 0.5 Solve for y in terms of x.

To distinguish between the function and its inverse, you can write

f (x) 0.4x 0.2 and f 1 (x) 2.5x 0.5

Bear in mind that the x used as the input for function f is not the same as the x used as the input for function f 1 . One is time, and the other is distance.

An interesting thing happens if you take the composition of a function and its inverse. In the highway stripe example,

f (4) 1.4 and f 1 (1.4) 4

f 1 ( f (4)) 4

You get the original input, 4, back again. �is result should not be surprising to you. �e composite function f 1 ( f (4)) means “How many hours does it take the crew to paint the distance it can paint in four hours?” �ere is a similar meaning for f ( f 1 (x)). For instance,

f f 1 (1.4) f (4) 1.4

In this case 1.4 is the original input of the inside function.

Invertibility and the Domain of an Inverse RelationIn the highway stripe example, the length of stripe painted during the �rst half hour was zero because it took some time at the beginning of the shi� for the crew to divert tra�c and prepare the equipment. �e graph of function f in Figure 1-5d includes times at the beginning of the shi�, along with its inverse relation.

Figure 1-5d


Symbols for the Inverse of a FunctionIf function g is the inverse of function f, the symbol f 1 is o�en used for the name of function g. In the highway stripe example, you can write f 1 (0.2) 1, f 1 (0.6) 2, . . . , f 1 (3) 8. Note that f 1 (3) does not mean the reciprocal of f (3).

f 1 (3) 8 and 1 ____ f (3) 1 __ 1 1, not 8

�e 1 used with the name of a function means the function inverse, whereas the 1 used with a number, as in 5 1 , means the multiplicative inverse of that number.

Inverse of a Function GraphicallyFigure 1-5a shows a graph of the data for the highway stripe example. Note that the points seem to lie in a straight line. Connecting the points is reasonable if you assume that the crew paints continuously. �e line meets the t-axis at about t 0.5 h, indicating that it takes the crew about half an hour at the beginning of the shi� to redirect tra�c and set up the equipment before it can start painting.

4 52 31 6 7 8

d f(t)

t

87654321

4 52 31 6 7 8

t f 1(d)

d

87654321

4 52 31 6 7 8

y x

x

87654321

f 1

f

y

Figure 1-5a Figure 1-5b Figure 1-5c

Figure 1-5b shows the inverse function, t f 1 (d). Note that every vertical feature on the graph of f is a horizontal feature on the graph of f 1 , and vice versa. For instance, the graph of f 1 meets the vertical axis at 0.5.

Figure 1-5c shows both graphs on the same set of axes. In this �gure, x is used for the input variable and y for the output variable. Keep in mind that x for function f represents hours and x for function f 1 represents miles. �e graphs are re�ections of each other across the line whose equation is y x.

Inverse of a Function AlgebraicallyIn the highway stripe example, the linear function that �ts the graph of function f in Figure 1-5c is

y 0.4(x 0.5) or, equivalently, y 0.4x 0.2 slope 0.4, x-intercept 0.5


Section Notes

This section discusses the inverse of functions and parametric equations.

You should emphasize that when a function’s variables are inter changed, you get the inverse relation, which may or may not be a function. Discussing Example 1 helps lead to the definition of an invertible function. In Example 2, note that when t 5 23, there is not a point on the (x, y) graph. Example 3 shows how to graph a function and its inverse using parametric equations. Students often think parametric graphing is truly bizarre, but you should reinforce that because x 5 t, x can be replaced by t in the first set of parametric equations. In the second set of parametric equations, simply interchange the equations for x and y in the first set. Example 4 provides an example for an invertible function and shows an important feature of these functions: f 21 ( f (x)) 5 x.

Discussing the concept of a one-to-one function is very important in relation to inverse functions. It will help students’ understanding to give additional examples for invertible functions: f (x) 5 x 3 , or other power functions with odd exponents, and f (x) 5 e x and its transformed forms. Also give additional examples for noninvertible functions, such as f (x) 5 x 4 , or other power functions with even exponents, and f (x) 5 x.

You might want to start your section with an explanation of why the concept of inverse is useful by referring to Problem 37, the Braking Distance Problem. You can use the example of police investigators to show that they would need to find the inverse of a function to find the speed from the skid mark evidence.

Because many students use the word inverse to refer to the reciprocal, or multiplicative inverse, of a number, they may mistakenly think that the inverse of a function y 5 f (x) is y 5 1 ___ f (x) . Explain that f 21 (x) is a functional inverse, whereas y 5 1 ___ f (x) is a multiplicative inverse. Also emphasize that f 21 (x) is read “ f inverse of x.”

Later in the course, students often mistakenly believe trigonometric inverses are equal to recip rocal trigonometric functions (for example, sin 21 x 5 csc x). If you reserve the term inverse to mean functional inverse and refer to the multiplicative inverse as the reciprocal, you can avoid errors later.


�e linear function that �ts the graph of f 1 is

y 2.5x 0.5 slope 2.5, y-intercept 0.5

If you know the equation of a function, you can transform it algebraically to �nd the equation of the inverse relation by �rst interchanging the variables.

Function: y 0.4x 0.2

Inverse: x 0.4y 0.2

�e equation of the inverse relation can be solved for y in terms of x.

x 0.4y 0.2

y 2.5x 0.5 Solve for y in terms of x.

To distinguish between the function and its inverse, you can write

f (x) 0.4x 0.2 and f 1 (x) 2.5x 0.5

Bear in mind that the x used as the input for function f is not the same as the x used as the input for function f 1 . One is time, and the other is distance.

An interesting thing happens if you take the composition of a function and its inverse. In the highway stripe example,

f (4) 1.4 and f 1 (1.4) 4

f 1 ( f (4)) 4

You get the original input, 4, back again. �is result should not be surprising to you. �e composite function f 1 ( f (4)) means “How many hours does it take the crew to paint the distance it can paint in four hours?” �ere is a similar meaning for f ( f 1 (x)). For instance,

f f 1 (1.4) f (4) 1.4

In this case 1.4 is the original input of the inside function.

Invertibility and the Domain of an Inverse RelationIn the highway stripe example, the length of stripe painted during the �rst half hour was zero because it took some time at the beginning of the shi� for the crew to divert tra�c and prepare the equipment. �e graph of function f in Figure 1-5d includes times at the beginning of the shi�, along with its inverse relation.

Figure 1-5d


Symbols for the Inverse of a FunctionIf function g is the inverse of function f, the symbol f 1 is o�en used for the name of function g. In the highway stripe example, you can write f 1 (0.2) 1, f 1 (0.6) 2, . . . , f 1 (3) 8. Note that f 1 (3) does not mean the reciprocal of f (3).

f 1 (3) 8 and 1 ____ f (3) 1 __ 1 1, not 8

�e 1 used with the name of a function means the function inverse, whereas the 1 used with a number, as in 5 1 , means the multiplicative inverse of that number.

Inverse of a Function GraphicallyFigure 1-5a shows a graph of the data for the highway stripe example. Note that the points seem to lie in a straight line. Connecting the points is reasonable if you assume that the crew paints continuously. �e line meets the t-axis at about t 0.5 h, indicating that it takes the crew about half an hour at the beginning of the shi� to redirect tra�c and set up the equipment before it can start painting.

4 52 31 6 7 8

d f(t)

t

87654321

4 52 31 6 7 8

t f 1(d)

d

87654321

4 52 31 6 7 8

y x

x

87654321

f 1

f

y

Figure 1-5a Figure 1-5b Figure 1-5c

Figure 1-5b shows the inverse function, t f 1 (d). Note that every vertical feature on the graph of f is a horizontal feature on the graph of f 1 , and vice versa. For instance, the graph of f 1 meets the vertical axis at 0.5.

Figure 1-5c shows both graphs on the same set of axes. In this �gure, x is used for the input variable and y for the output variable. Keep in mind that x for function f represents hours and x for function f 1 represents miles. �e graphs are re�ections of each other across the line whose equation is y x.

Inverse of a Function AlgebraicallyIn the highway stripe example, the linear function that �ts the graph of function f in Figure 1-5c is

y 0.4(x 0.5) or, equivalently, y 0.4x 0.2 slope 0.4, x-intercept 0.5

37

Differentiating Instruction• Students may be confused that both

the word inverse and the notation 21 as an exponent have more than one mathematical meaning. Point out the two distinct meanings of each, and have students write these in their notes.

• Clarify the language in Example 1 for ELL students.

• Because an equivalent to sketch does not exist in all languages, show ELL students how to sketch a curve, and explain the difference between sketch and plot.

• As ELL students do Exploration 1-5, monitor their work for understanding.

• Because of the difference in topic order from country to country, you may need to show ELL students how to invert a function algebraically and graphically.

• For Problems 33 and 34, clarify the meaning of “show that.”

A memorable way to illustrate that the graph of the inverse of a function is the mirror image of the graph of the function is to have students hold a small mirror so that one edge is on the line y 5 x. The graph of the inverse will appear in the mirror.

Exploration 1-5a can be used to introduce students to inverses of functions.

Section 1-5: Inverse Functions and Parametric Equations


b. Figure 1-5e shows the graphs of function f and its inverse relation. � e inverse relation is not a function because there are two values of y for each value of x 2. � e inverse relation fails the vertical line test.

c. Function: y 0.5 x 2 2 Use y for f (x).

Inverse: x 0.5 y 2 2 Interchange x and y.

y 2 2x 4

y ______

2x 4 Take the square root of both sides.

f 1 (x) 0.5 x 2 2 Enter f (x) as f 1 (x).

f 2 (x) ________

(2x 4)

f 3 (x) ________

(2x 4)

f 4 (x) x Enter y x as f 4 (x).

Figure 1-5f shows the graphs of function f and its inverse relation. � e graphs are re� ections of each other across the line y x.

Parametric Equations� ere is a simple way to plot the graph of the inverse of a function with the help of parametric equations. Here, x and y are both expressed in terms of some third variable, usually t (because time is o� en the independent variable in real-world applications).

In this exploration, you will see how to graph a relation speci� ed by parametric equations, both by hand and on your grapher.

x

y

5 10

10

5

f1(x)y x

f2(x)

f3(x)

Figure 1-5f

Enter the two branches of the inverse relation as f 2 (x) and f 3 (x).

➤

Let x and y be functions of a third variable, t, as speci� ed by these equations:

x t 2 1y t 2

1. Make a table of values of t, x, and y for each integer value of t from 3 to 3.

2. On graph paper, plot the points you found in Problem 1. Connect the points with a smooth curve in the order of increasing values of t.

3. For the relation you plotted in Problem 2, is y a function of x? Explain.

4. Set your grapher to parametric mode. Enter the two equations. Use a window with

3 t 3 and a t-step of 0.1. Set 10 x 10 and 6 y 6. � en plot

the graph. Does your grapher’s graph agree with your pencil-and-paper graph? If not, make changes until the two graphs agree.


Let and be functions of a third variable, t, 3. For the relation you plotted in Problem 2, is

E X P L O R AT I O N 1-5: P a r a m e t r i c E q u a t i o n s G r a p h

5−5

5

−5

f

Inverse of f

y

x

Figure 1-5e


Note that the inverse relation has multiple values of y when x equals zero. � us the inverse relation is not a function. You cannot answer the question “How long has the crew been working when the distance painted is zero?”

If the domain of function f is restricted to times no less than a half hour, the inverse relation is a function. In this case, function f is said to be invertible. If f is invertible, you are allowed to use the symbol f 1 for the inverse function. If the domain of f is 0.5 x 8, then there is exactly one distance for each time and one time for each distance. Function f is said to be a one-to-one function. Any one-to-one function is invertible. A function that is strictly increasing, such as the highway stripe function f, or strictly decreasing is a one-to-one function and thus is invertible.

� e highway stripe problem gives examples of operations with functions from the real world. Example 1 shows you how to operate with a function and its inverse in a strictly mathematical context.

Given f (x) 0.5 x 2 2

a. Make a table of values for f ( 2), f ( 1), f (0), f (1), and f (2). From the numbers in the table, explain why you cannot � nd a unique value of x if f (x) 2.5. How does this result tell you that function f is not invertible?

b. Plot the � ve points in part a on graph paper. Connect the points with a smooth curve. On the same axes, plot the � ve points for the inverse relation and connect them with another smooth curve. How does the graph of the inverse relation con� rm that function f is not invertible?

c. Find an equation for the inverse relation. Plot function f and its inverse on the same screen on your grapher. Show that the two graphs are re� ections of each other across the line y x.

a.

If f (x) 2.5, there are two di� erent values of x, 1 and 1. You cannot uniquely determine the value of x.

Function f is not invertible because there will be two values of y for the same value of x if the variables are interchanged.

Given EXAMPLE 1 ➤

a.SOLUTION x f (x)

2 41 2.50 21 2.52 4


Exploration Notes

Exploration 1-5a introduces students to inversion of functions by having them find the inverses of a linear, a quadratic, and an exponential function. Students do this by a combination of numerical, graphical, and algebraic techniques. Along the way, they demon strate that whether or not the inverse relation is a function, it is a mirror image of the parent function in the line y 5 x. Allow about 20 minutes.

Exploration 1-5b introduces students to parametric equations by having them graph parametric equations both on paper and on their grapher. Allow 20–25 minutes.

Technology Notes

Presentation Sketch: Inverse Present.gsp at www.keypress.com/keyonline demonstrates properties of inverses on a dynagraph. It also includes a page that demonstrates the relationship between the graph of a function and that of its inverse. The final page allows exploration of linear functions that are their own inverses.


b. Figure 1-5e shows the graphs of function f and its inverse relation. � e inverse relation is not a function because there are two values of y for each value of x 2. � e inverse relation fails the vertical line test.

c. Function: y 0.5 x 2 2 Use y for f (x).

Inverse: x 0.5 y 2 2 Interchange x and y.

y 2 2x 4

y ______

2x 4 Take the square root of both sides.

f 1 (x) 0.5 x 2 2 Enter f (x) as f 1 (x).

f 2 (x) ________

(2x 4)

f 3 (x) ________

(2x 4)

f 4 (x) x Enter y x as f 4 (x).

Figure 1-5f shows the graphs of function f and its inverse relation. � e graphs are re� ections of each other across the line y x.

Parametric Equations� ere is a simple way to plot the graph of the inverse of a function with the help of parametric equations. Here, x and y are both expressed in terms of some third variable, usually t (because time is o� en the independent variable in real-world applications).

In this exploration, you will see how to graph a relation speci� ed by parametric equations, both by hand and on your grapher.

x

y

5 10

10

5

f1(x)y x

f2(x)

f3(x)

Figure 1-5f

Enter the two branches of the inverse relation as f 2 (x) and f 3 (x).

➤

Let x and y be functions of a third variable, t, as speci� ed by these equations:

x t 2 1y t 2

1. Make a table of values of t, x, and y for each integer value of t from 3 to 3.

2. On graph paper, plot the points you found in Problem 1. Connect the points with a smooth curve in the order of increasing values of t.

3. For the relation you plotted in Problem 2, is y a function of x? Explain.

4. Set your grapher to parametric mode. Enter the two equations. Use a window with

3 t 3 and a t-step of 0.1. Set 10 x 10 and 6 y 6. � en plot

the graph. Does your grapher’s graph agree with your pencil-and-paper graph? If not, make changes until the two graphs agree.


Let and be functions of a third variable, t, 3. For the relation you plotted in Problem 2, is

E X P L O R AT I O N 1-5: P a r a m e t r i c E q u a t i o n s G r a p h

5−5

5

−5

f

Inverse of f

y

x

Figure 1-5e


Note that the inverse relation has multiple values of y when x equals zero. � us the inverse relation is not a function. You cannot answer the question “How long has the crew been working when the distance painted is zero?”

If the domain of function f is restricted to times no less than a half hour, the inverse relation is a function. In this case, function f is said to be invertible. If f is invertible, you are allowed to use the symbol f 1 for the inverse function. If the domain of f is 0.5 x 8, then there is exactly one distance for each time and one time for each distance. Function f is said to be a one-to-one function. Any one-to-one function is invertible. A function that is strictly increasing, such as the highway stripe function f, or strictly decreasing is a one-to-one function and thus is invertible.

� e highway stripe problem gives examples of operations with functions from the real world. Example 1 shows you how to operate with a function and its inverse in a strictly mathematical context.

Given f (x) 0.5 x 2 2

a. Make a table of values for f ( 2), f ( 1), f (0), f (1), and f (2). From the numbers in the table, explain why you cannot � nd a unique value of x if f (x) 2.5. How does this result tell you that function f is not invertible?

b. Plot the � ve points in part a on graph paper. Connect the points with a smooth curve. On the same axes, plot the � ve points for the inverse relation and connect them with another smooth curve. How does the graph of the inverse relation con� rm that function f is not invertible?

c. Find an equation for the inverse relation. Plot function f and its inverse on the same screen on your grapher. Show that the two graphs are re� ections of each other across the line y x.

a.

If f (x) 2.5, there are two di� erent values of x, 1 and 1. You cannot uniquely determine the value of x.

Function f is not invertible because there will be two values of y for the same value of x if the variables are interchanged.

Given EXAMPLE 1 ➤

a.SOLUTION x f (x)

2 41 2.50 21 2.52 4

39

Additional Exploration Notes

Exploration 1-5 introduces students to graphing parametric equations. Students examine two functions, (t, x(t)) where x is a function of t and (t, y(t)) where y is a function of t but y is not a function of x. If your students have a solid understanding of the definition of functions, you may want to point out that (x, y) is, however, a function of t since a value of t corresponds to exactly one point (x, y).

It may clarify things if you write t 5 23, t 5 22, and so on, next to the discrete points on your graph and draw an arrow to indicate the path as t goes from 23 to 3. Parametric curves have a dynamic quality to them. The discrete points indicate time passing and a direction of motion. When you do Problem 4 in the exploration, spend time explaining how to find reasonable windows for t, x, and y as this is not always obvious to students. Emphasize that a grapher in parametric mode can graph curves that are not functions while a grapher in function mode can only graph functions.3. The relation is not a function because there is more than one value of y for some values of x.4. Grapher graph agrees with graph on paper.5. Answers will vary.

1. 2. t x y

23 10 2122 5 0

21 2 10 1 21 2 32 5 43 10 5

1

y

x1



Put your grapher in parametric mode. � en enter

x 1 (t) t

y 1 (t) 0.5 t 2 2 Because x t, this is equivalent to y 0.5 x 2 2.

x 2 (t) 0.5 t 2 2

y 2 (t) t For the inverse, interchange the equations for x and y.

Use a window with 2 t 4. Use a convenient t-step, such as 0.1. � e result is shown in Figure 1-5h.

� e range of the inverse relation is the same as the domain of the function, and vice versa. � e range and the domain are interchanged.

Example 4 shows you how to demonstrate algebraically that f 1 ( f (x)) x and f f 1 (x) x.

Let f (x) 3x 12.

a. Find an equation for the inverse of f, and explain how that equation con� rms that f is an invertible function.

b. Demonstrate that f 1 ( f (x)) x and f f 1 (x) x.

a. Function: y 3x 12

Inverse: x 3y 12 y 1 __ 3 x 4

Because the equation for the inverse relation has the form y mx b, the inverse is a linear function. Because the inverse relation is a function, f is invertible, so the equation can be written

f 1 (x) 1 __ 3 x 4

b. f 1 ( f (x)) f 1 (3x 12) Substitute 3x 12 for f (x).

1 __ 3 (3x 12) 4 Substitute 3x 12 as the input for function f 1 .

x 4 4 x Show that f 1 ( f (x)) equals x.

Also, f f 1 (x) f 1 __ 3 x 4 Show that f f 1 (x) equals x.

3 1 __ 3 x 4 12

x 12 12 x

f 1 ( f (x)) x and f f 1 (x) x, q.e.d.

Note: � e three-dot mark stands for “therefore.” � e letters q.e.d. stand for the Latin words quod erat demonstrandum, meaning “which was to be demonstrated.”

� e box on the next page summarizes the information of this section regarding inverses of functions.


( )

SOLUTION

x5 10

10Function

Inverse relation5

y

Figure 1-5h

➤

Let fEXAMPLE 4 ➤

a. Function: SOLUTION

➤


Example 2 shows you in general what parametric equations are.

On graph paper, plot the graph of these parametric equations by � rst calculating values of x and y for integer values of t from 3 through 7.

x 1 (t) t 2

y 1 (t) _____

t 2

� e table shows values of t, x, and y.

Figure 1-5g shows the graph of the parametric equations. Note that y is not a function of x because there are two values of y for some values of x.

x 32 541

y

1

2

3

4

5

t 2

t 1

t 7

t 2

� e independent variable t in parametric equations is called the parameter. � e word comes from the Greek para- meaning “alongside,” as in “parallel,” and meter, meaning “measure.” � e values of t do not show up on the graph in Figure 1-5g unless you write them in.

Your grapher is programmed to plot parametric equations. For the equations in Example 2, use parametric mode and enter

x 1 (t) abs(t 2)

y 1 (t) ______

(t 2)

Use a window with 3 t 7, 0 x 5, and 0 y 5 and a convenient t-step such as 0.1. � e graph will be similar to the graph in Figure 1-5g. ➤

Example 3 shows you how to use parametric equations to plot inverse relations on your grapher.

Plot the graph of y 0.5 x 2 2 for x in the domain 2 x 4 and its inverse using parametric equations. What do you observe about the domain and range of the function and its inverse?

On graph paper, plot the graph of these parametric equations by � rst calculating values of

EXAMPLE 2 ➤

� e table shows values of SOLUTION

t x y

–3 5 None

–2 4 0

–1 3 1

0 2 1.4142…

1 1 1.7320…

2 0 2

3 1 2.2360…

4 2 2.4494…

5 3 2.6457…

6 4 2.8284…

7 5 3

Figure 1-5g

Plot the graph of using parametric equations. What do you observe about the domain and range of

EXAMPLE 3 ➤


CAS Suggestions

To solve for the inverse of an equation, defi ne the original function as y 5 f (x). Th en exchange the variables and solve for the new y-variable.

Note: Th is procedure for fi nding the inverse can be used either to solve an application problem or as a tool once students understand what to do. Students need to understand the algebraic steps involved in transforming the original function to its inverse. It is important to use available tools to enhance students’ understanding of the mathematics. Consider having students use a CAS to complete the algebraic steps to fi nd the inverse before teaching them the shortcut described above.

Th e algebraic parallel to the non-invertibility of f (x) 5 0.5 x 2 1 2 in Example 1 is that there is no single algebraic form for the inverse. Notice two things in the fi gure. First, two diff erent equations are given for the inverse. Th is shows that a single inverse equation does not exist. Second, each inverse is given a domain (which is the range of the original function). Students are oft en careless in noting domain restrictions when fi nding inverses, but a CAS notes them meticulously, encouraging students to be careful in their computations.



x 1 (t) t

y 1 (t) 0.5 t 2 2 Because x t, this is equivalent to y 0.5 x 2 2.

x 2 (t) 0.5 t 2 2

y 2 (t) t For the inverse, interchange the equations for x and y.

Use a window with 2 t 4. Use a convenient t-step, such as 0.1. � e result is shown in Figure 1-5h.

� e range of the inverse relation is the same as the domain of the function, and vice versa. � e range and the domain are interchanged.

Example 4 shows you how to demonstrate algebraically that f 1 ( f (x)) x and f f 1 (x) x.

Let f (x) 3x 12.

a. Find an equation for the inverse of f, and explain how that equation con� rms that f is an invertible function.

b. Demonstrate that f 1 ( f (x)) x and f f 1 (x) x.

a. Function: y 3x 12

Inverse: x 3y 12 y 1 __ 3 x 4

Because the equation for the inverse relation has the form y mx b, the inverse is a linear function. Because the inverse relation is a function, f is invertible, so the equation can be written

f 1 (x) 1 __ 3 x 4

b. f 1 ( f (x)) f 1 (3x 12) Substitute 3x 12 for f (x).

1 __ 3 (3x 12) 4 Substitute 3x 12 as the input for function f 1 .

x 4 4 x Show that f 1 ( f (x)) equals x.

Also, f f 1 (x) f 1 __ 3 x 4 Show that f f 1 (x) equals x.

3 1 __ 3 x 4 12

x 12 12 x

f 1 ( f (x)) x and f f 1 (x) x, q.e.d.

Note: � e three-dot mark stands for “therefore.” � e letters q.e.d. stand for the Latin words quod erat demonstrandum, meaning “which was to be demonstrated.”

� e box on the next page summarizes the information of this section regarding inverses of functions.


( )

SOLUTION

x5 10

10Function

Inverse relation5

y

Figure 1-5h

➤

Let fEXAMPLE 4 ➤

a. Function: SOLUTION

➤


Example 2 shows you in general what parametric equations are.

On graph paper, plot the graph of these parametric equations by � rst calculating values of x and y for integer values of t from 3 through 7.

x 1 (t) t 2

y 1 (t) _____

t 2

� e table shows values of t, x, and y.

Figure 1-5g shows the graph of the parametric equations. Note that y is not a function of x because there are two values of y for some values of x.

x 32 541

y

1

2

3

4

5

t 2

t 1

t 7

t 2

� e independent variable t in parametric equations is called the parameter. � e word comes from the Greek para- meaning “alongside,” as in “parallel,” and meter, meaning “measure.” � e values of t do not show up on the graph in Figure 1-5g unless you write them in.

Your grapher is programmed to plot parametric equations. For the equations in Example 2, use parametric mode and enter

x 1 (t) abs(t 2)

y 1 (t) ______

(t 2)

Use a window with 3 t 7, 0 x 5, and 0 y 5 and a convenient t-step such as 0.1. � e graph will be similar to the graph in Figure 1-5g. ➤

Example 3 shows you how to use parametric equations to plot inverse relations on your grapher.

Plot the graph of y 0.5 x 2 2 for x in the domain 2 x 4 and its inverse using parametric equations. What do you observe about the domain and range of the function and its inverse?

On graph paper, plot the graph of these parametric equations by � rst calculating values of

EXAMPLE 2 ➤

� e table shows values of SOLUTION

t x y

–3 5 None

–2 4 0

–1 3 1

0 2 1.4142…

1 1 1.7320…

2 0 2

3 1 2.2360…

4 2 2.4494…

5 3 2.6457…

6 4 2.8284…

7 5 3

Figure 1-5g

Plot the graph of using parametric equations. What do you observe about the domain and range of

EXAMPLE 3 ➤

41

Consider defining the inverse function using a name like finv instead of f 21 to avoid confusion with the reciprocal notation. A CAS allows the student to compute f + f 21 and f 21 + f and show the results, or test the inverse function using Boolean operators to show that the composition actually provides the expected results.

Remember that it is easy to switch between function and parametric modes without leaving a graphing window. A significant advantage of the TI-Nspire over previous models is that different graph types can be graphed on the same screen.

The parts of parametric functions can also be defined on a calculator screen and then graphed by name in parametric mode. While it is unnecessary here, using this method allows students to manipulate the function by name.



x (min) y (psi)

0 365 24

10 1615 10.720 7.125 4.730 035 0

a. Let y f (x). Find f (5), f (10), and f (15). b. Why is it reasonable to assume that f is an

invertible function if x is in the domain 0 x 25? Find f 1 (24) and f 1 (16), and give their real-world meanings.

c. Why is function f not invertible on the whole interval 0 x 35? What do you suppose happens between 25 min and 30 min that causes f to not be invertible?

d. Plot the eight given points for function f and the eight corresponding points for the inverse relation. Connect each set of points with a smooth curve. Draw y x and explain how the two graphs are related to this line.

e. Suppose f is restricted to the domain 0 x 25. What is the di�erence in the meaning of x as an input for function f and x as an input for function f 1 ?

2. Cricket Chirping Problem: �e rate at which crickets chirp is a function of the temperature of the air around them. Suppose that the following data have been measured for chirps, c, per minute, y, at temperatures in degrees Fahrenheit, x.

x (°F) y (c/min)

20 030 040 550 3060 5570 8080 105

a. Let y c(x). Find c(40), c(50), and c(60). b. For temperatures of 40°F and above, the

chirping rate seems to be a one-to-one function of time. How does this fact imply that function c is invertible for x 40? Find the values of c 1 (30) and c 1 (80). How do these values di�er in meaning from c(30) and c(80)?

c. Why is function c not invertible for x in the interval 20 x 80? What is true in this real-world situation that makes c not invertible?

d. On graph paper, plot the seven given points for function c and the corresponding points for the inverse relation. Connect each set of points with a line or a smooth curve. Draw the line y x and explain how the two graphs are related to this line.

e. Suppose c is restricted to the domain 40 x 80. What is the di�erence in the meaning of x as an input for function c and x as an input for function c 1 ?

3. Punted Football Problem: Figure 1-5i shows the height of a punted football y, in meters, as a function of time x, in tenths of a second since it was punted. On a copy of the �gure, sketch the graph of the inverse relation and show that the two graphs are re�ections across the line y x. How does the graph of the inverse relation reveal that the height function is not invertible?

y

x

10

20

10 20 Figure 1-5i

DEFINITIONS AND PROPERTIES: Function Inversesinverse of a relation in two variables is formed by interchanging the

two variables.

re� ections of each other across the line y x.

f is also a function, then f is invertible.

f is invertible and y f (x), then you can write the inverse of f as y f 1 (x).

Interchange the variables, solve for y, and plot the resulting equation(s), or

use parametric mode, as in Example 3.

f is invertible, then the compositions of f and f 1 are

f 1 ( f (x)) x, provided x is in the domain of f and f (x) is in the domain of f 1

f f 1 (x) x, provided x is in the domain of f 1 and f 1 (x) is in the domain of f

functions are one-to-one functions.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? Why is it possible to � nd the inverse of a function even if the function is not invertible? Under what conditions are you permitted to use the symbol f 1 for the inverse of function f ? How does the meaning of 1 in the function name f 1 di� er from the meaning of 1 in a numerical expression such as 7 1 ?

Quick Review

Q1. In the composite function m(d(x)), function d is called the ? function.

Q2. In the composite function m(d(x)), function m is called the ? function.

Q3. Give another symbol for m(d(x)). Q4. If f (x) 2x and g (x) x 3, � nd f ( g (1)).

Q5. Find g ( f (1)) for the functions in Problem Q4. Q6. Find f ( f (1)) for the functions in Problem Q4. Q7. 3 5 ? Q8. Identify the function graphed.

y

x

Q9. If f (x) 2x, � nd f (0). Q10. If f (x) 2x, � nd an equation for g (x), a

horizontal translation of f (x) by 3 units.

1. Punctured Tire Problem: Suppose that your car runs over a nail. � e table shows the pressure y, in pounds per square inch (psi), of the air inside the tire as a function of x, the number of minutes that have elapsed since the nail punctured the tire.

5min

Reading Analysis Q5. Find g (g (g f (1)) for the functions in Problem Q4.

Problem Set 1-5


PRO B LE M N OTES

Q1. Inside Q2. OutsideQ3. (m + d)(x) Q4. 8Q5. 5 Q6. 4Q7. 2 Q8. y 5 xQ9. 1 Q10. g (x) 5 2 x13

Problems 1 and 2 are applications of inverses to real-world situations. Th ey emphasize inverse function notation and the refl ection of a function and its inverse across the line y 5 x.1a. f (5) 5 24 psi; f (10) 5 16 psi; f (15) 5 10.7 psi1b. Th e air leaks out of the tire as time passes, so the pressure is constantly getting lower. Th us, f is a decreasing function and hence is invertible. f 21 (24) 5 5 min, which answers the question “At what time was the pressure 24 psi?” f 21 (16) 5 10 min, which answers the question “At what time was the pressure 16 psi?”1c. Somewhere between x 5 25 and x 5 30 min , all the air goes out of the tire, and the pressure remains zero. So it is not possible to give a unique time corresponding to a pressure of 0 psi; f 21 (0) cannot be defi ned.1d. Th e graph of the inverse relation is dotted. Th e two graphs are refl ections of each other over the line y 5 x. (Th ey coincidentally happen to be very close over most of their length.)

10 20 30 40

10

20

30

40

x

y

y � f(x)y � x

1e. As an input for f, x represents time in minutes. As an input for f 21 , it represents pressure in psi.


x (min) y (psi)

0 365 24

10 1615 10.720 7.125 4.730 035 0

a. Let y f (x). Find f (5), f (10), and f (15). b. Why is it reasonable to assume that f is an

invertible function if x is in the domain 0 x 25? Find f 1 (24) and f 1 (16), and give their real-world meanings.

c. Why is function f not invertible on the whole interval 0 x 35? What do you suppose happens between 25 min and 30 min that causes f to not be invertible?

d. Plot the eight given points for function f and the eight corresponding points for the inverse relation. Connect each set of points with a smooth curve. Draw y x and explain how the two graphs are related to this line.

e. Suppose f is restricted to the domain 0 x 25. What is the di�erence in the meaning of x as an input for function f and x as an input for function f 1 ?

2. Cricket Chirping Problem: �e rate at which crickets chirp is a function of the temperature of the air around them. Suppose that the following data have been measured for chirps, c, per minute, y, at temperatures in degrees Fahrenheit, x.

x (°F) y (c/min)

20 030 040 550 3060 5570 8080 105

a. Let y c(x). Find c(40), c(50), and c(60). b. For temperatures of 40°F and above, the

chirping rate seems to be a one-to-one function of time. How does this fact imply that function c is invertible for x 40? Find the values of c 1 (30) and c 1 (80). How do these values di�er in meaning from c(30) and c(80)?

c. Why is function c not invertible for x in the interval 20 x 80? What is true in this real-world situation that makes c not invertible?

d. On graph paper, plot the seven given points for function c and the corresponding points for the inverse relation. Connect each set of points with a line or a smooth curve. Draw the line y x and explain how the two graphs are related to this line.

e. Suppose c is restricted to the domain 40 x 80. What is the di�erence in the meaning of x as an input for function c and x as an input for function c 1 ?

3. Punted Football Problem: Figure 1-5i shows the height of a punted football y, in meters, as a function of time x, in tenths of a second since it was punted. On a copy of the �gure, sketch the graph of the inverse relation and show that the two graphs are re�ections across the line y x. How does the graph of the inverse relation reveal that the height function is not invertible?

y

x

10

20

10 20 Figure 1-5i

DEFINITIONS AND PROPERTIES: Function Inversesinverse of a relation in two variables is formed by interchanging the

two variables.

re� ections of each other across the line y x.

f is also a function, then f is invertible.

f is invertible and y f (x), then you can write the inverse of f as y f 1 (x).

Interchange the variables, solve for y, and plot the resulting equation(s), or

use parametric mode, as in Example 3.

f is invertible, then the compositions of f and f 1 are

f 1 ( f (x)) x, provided x is in the domain of f and f (x) is in the domain of f 1

f f 1 (x) x, provided x is in the domain of f 1 and f 1 (x) is in the domain of f

functions are one-to-one functions.


Reading Analysis

From what you have read in this section, what do you consider to be the main idea? Why is it possible to � nd the inverse of a function even if the function is not invertible? Under what conditions are you permitted to use the symbol f 1 for the inverse of function f ? How does the meaning of 1 in the function name f 1 di� er from the meaning of 1 in a numerical expression such as 7 1 ?

Quick Review

Q1. In the composite function m(d(x)), function d is called the ? function.

Q2. In the composite function m(d(x)), function m is called the ? function.

Q3. Give another symbol for m(d(x)). Q4. If f (x) 2x and g (x) x 3, � nd f ( g (1)).

Q5. Find g ( f (1)) for the functions in Problem Q4. Q6. Find f ( f (1)) for the functions in Problem Q4. Q7. 3 5 ? Q8. Identify the function graphed.

y

x

Q9. If f (x) 2x, � nd f (0). Q10. If f (x) 2x, � nd an equation for g (x), a

horizontal translation of f (x) by 3 units.

1. Punctured Tire Problem: Suppose that your car runs over a nail. � e table shows the pressure y, in pounds per square inch (psi), of the air inside the tire as a function of x, the number of minutes that have elapsed since the nail punctured the tire.

5min

Reading Analysis Q5. Find g (g (g f (1)) for the functions in Problem Q4.

Problem Set 1-5

43

2a. c(40) 5 5 c/min; c(50) 5 30 c/min; c(60) 5 55 c/min2b. Any one-to-one function is invertible. c 21 (30) 5 50°F; c 21 (80) 5 70°F; these give the temperature corresponding to 30 c/min and 80 c/min. By contrast, c(30) and c(80) give the number of chirps/min corresponding to 30°F and 80°F.2c. The cricket does not begin chirping until the temperature is at least 30°F. For 20 x 30, the number of chirps/min remains zero, so c 21 (0) cannot be defined.2d.

The graphs are reflections of each other across the line y 5 x.2e. As the input to c, x represents temperature in °F. As the input to c 21 , it represents the number of chirps/min.

Problems 3 and 5–8 provide practice in sketching the inverse relation for a given graph. A blackline master for these problems is available in the Instructor’s Resource Book.3.

Throughout most of its domain, the inverse relation has two y-values for every x-value.

50 100

50

x

y

y � c(x) y � x

10 20

10

20

x

y



16. Two Ships Problem: At time t 0 h, a freighter is at the point (90, 10) to the east-northeast of a lighthouse located at the origin of a Cartesian coordinate system, where x and y are distance in miles. At time t 2 h, a Coast Guard cutter starts from the lighthouse to intercept the freighter. Figure 1-5l shows the graph of these parametric equations representing the ships’ paths:

Freighter: x 90 10t Cutter: x 8(t 2) y 10 5t y 10(t 2)

100

Cutter

x

50

50

Freighter

Lighthouse

y

Figure 1-5l

a. Find the value of t at which the y-values of the two paths are equal. At this value of t, are the two x-values equal?

b. Do the two ships arrive at the intersection point at the same time? If so, how can you tell? If not, which ship arrives at the intersection point �rst?

For Problems 17–28, plot the function in the given domain using parametric mode. On the same screen, plot the inverse relation. Tell whether the inverse relation is a function. Sketch the graphs. 17. f (x) 2x 6 1 x 5 18. f (x) 0.4x 4 7 x 10 19. f (x) x 2 4x 1 0 x 5 20. f (x) x 2 2x 4 2 x 4 21. f (x) 2 x x is any real number. 22. f (x) 0. 5 x x is any real number. 23. f (x)

_____ 3 x 6 x 3

24. f (x) 3 __

x 1 x 8

25. f (x) 1 _____ x 3 2 x 8

26. f (x) x _____ x 1 6 x 4

27. f (x) x 3 2 x 1 28. f (x) 0.016 x 4 4 x 5

For Problems 29–32, write an equation for the inverse relation by interchanging the variables and solving for y in terms of x. �en plot the function and its inverse on the same screen, using function mode. Sketch the result, showing that the function and its inverse are re�ections across the line y x. Tell whether the inverse relation is a function. 29. y 2x 6 30. y 0.4x 4 31. y 0.5 x 2 2 32. y 0.4 x 2 3 33. Show that f (x) 1 _ x is its own inverse function. 34. Show that f (x) x is its own inverse function. 35. Cost of Owning a Car Problem: Suppose that you

have �xed costs (car payments, insurance, and so on) of $500 per month and operating costs of $0.40 per mile you drive. �e monthly cost of owning the car is given by the linear function

c(x) 0.40x 500 where x is the number of miles you drive the

car in a given month and c(x) is the number of dollars per month you spend.

a. Find c(1000). Explain the real-world meaning of the answer.

b. Find an equation for c 1 (x), where x now stands for the number of dollars you spend instead of the number of miles you drive. Explain why you can use the symbol c 1 for the inverse relation. Use the equation of c 1 (x) to �nd c 1 (758), and explain its real-world meaning.

c. Plot f 1 (x) c(x) and f 2 (x) c 1 (x) on the same screen, using function mode. Use a window with 0 x 1000 and use equal scales on the two axes. Sketch the two graphs, showing how they are related to the line y x.


4. Discrete Function Problem: Figure 1-5j shows a function that consists of a discrete set of points. Show that the function is one-to-one and thus is invertible, even

though the function is increasing in some parts of the domain and decreasing in other parts.

For Problems 5–8, sketch the line y x and the inverse relation on a copy of the given �gure. Be sure that the inverse relation is a re�ection of the function graph across the line y x. Tell whether the inverse relation is a function. 5. 6.

x

5

–5

5–5

y

x

5

–5

5–5

y

7. 8.

x

5

–5

5–5

y

x

5

–5

5–5

y

For Problems 9–14, a. Plot the parametric equations on graph paper

using the given domain for t. Connect the points with lines or smooth curves.

b. Tell whether y is a function of x. c. Con�rm your results by grapher, using the

given domain for t.

9. x t 3 5 t 5 y 2 t

10. x 5 t 7 t 4 y t 1 11. x 7 t 2 3 t 4 y t 2 12. x t 3 1 t 4 y (t 2) 2 13. x t 3 1 t 3 y t 3 2t 2 14. x t 2 2t 2 1 t 2 y t 3 t 2 t 1 15. Two Paths Problem: Two particles (small objects)

move along the paths shown in Figure 1-5k. �e paths are given by these parametric equations, where x and y are distance in meters and t is time in seconds.

Particle 1: x t 1 Particle 2: x 1.5t 2 y 7 t 2 y 1.5t 6

Particle 1

5x

5

Particle 2

10

3

y

Figure 1-5k

a. �e paths intersect at two points. For each point, determine whether the particles reach that point at the same time or at di�erent times. Give numbers to support your conclusion.

b. Con�rm your answer to part a graphically by plotting the two sets of parametric equations dynamically on the same screen, setting your grapher to simultaneous mode. Write a sentence or two explaining how your graph con�rms your answer to part a.

54321 109876

21

43

65

x

y

Figure 1-5j



Problem 4 requires students to apply the idea of invertibility to a new situation. Th e function presented in this problem is made up of a discrete set of points (a set of points with “gaps” between them). Although the function is not strictly increasing or strictly decreasing, it is one-to-one and is therefore invertible.4. No y-value comes from more than one x-value. Also, no horizontal line passes through more than one point of the function.5. Function

y

x

y � x

6. Not a function

y

x

y � x

7. Not a function

y

x

y � x

8. Not a function

y

x

y � x

Problems 9–14 ask the students to graph the parametric equations “by hand,” decide if y is a function of x, and then use a grapher to check. Once students learn how to use the grapher for parametric

equations, graphing by hand may seem tedious and unnecessary, but students develop a stronger understanding of parametric equations when they fi nd the (t, x(t)), (t, y(t)), and (t, x(t), y(t)) points.

Problems 9–14 parts c can be solved using a CAS by checking whether the x-equation for each parametric equation can be solved for t in a single equation. If so, y is a function of x.

Problems 15 and 16 are application problems involving parametric equations.15a. Paths intersect simultaneously at point (21, 3) when t 5 22 s. Paths intersect at point (2, 6) but not simultaneously.15b. Grapher graph confi rms that the paths intersect simultaneously only at point (21, 3) when t 5 22 s.16a. Th e y-values are equal at t 5 6 h. Freighter: x 5 30 mi; Cutter: x 5 32 mi. Th e x-values are not equal at t 5 6 h .


16. Two Ships Problem: At time t 0 h, a freighter is at the point (90, 10) to the east-northeast of a lighthouse located at the origin of a Cartesian coordinate system, where x and y are distance in miles. At time t 2 h, a Coast Guard cutter starts from the lighthouse to intercept the freighter. Figure 1-5l shows the graph of these parametric equations representing the ships’ paths:

Freighter: x 90 10t Cutter: x 8(t 2) y 10 5t y 10(t 2)

100

Cutter

x

50

50

Freighter

Lighthouse

y

Figure 1-5l

a. Find the value of t at which the y-values of the two paths are equal. At this value of t, are the two x-values equal?

b. Do the two ships arrive at the intersection point at the same time? If so, how can you tell? If not, which ship arrives at the intersection point �rst?

For Problems 17–28, plot the function in the given domain using parametric mode. On the same screen, plot the inverse relation. Tell whether the inverse relation is a function. Sketch the graphs. 17. f (x) 2x 6 1 x 5 18. f (x) 0.4x 4 7 x 10 19. f (x) x 2 4x 1 0 x 5 20. f (x) x 2 2x 4 2 x 4 21. f (x) 2 x x is any real number. 22. f (x) 0. 5 x x is any real number. 23. f (x)

_____ 3 x 6 x 3

24. f (x) 3 __

x 1 x 8

25. f (x) 1 _____ x 3 2 x 8

26. f (x) x _____ x 1 6 x 4

27. f (x) x 3 2 x 1 28. f (x) 0.016 x 4 4 x 5

For Problems 29–32, write an equation for the inverse relation by interchanging the variables and solving for y in terms of x. �en plot the function and its inverse on the same screen, using function mode. Sketch the result, showing that the function and its inverse are re�ections across the line y x. Tell whether the inverse relation is a function. 29. y 2x 6 30. y 0.4x 4 31. y 0.5 x 2 2 32. y 0.4 x 2 3 33. Show that f (x) 1 _ x is its own inverse function. 34. Show that f (x) x is its own inverse function. 35. Cost of Owning a Car Problem: Suppose that you

have �xed costs (car payments, insurance, and so on) of $500 per month and operating costs of $0.40 per mile you drive. �e monthly cost of owning the car is given by the linear function

c(x) 0.40x 500 where x is the number of miles you drive the

car in a given month and c(x) is the number of dollars per month you spend.

a. Find c(1000). Explain the real-world meaning of the answer.

b. Find an equation for c 1 (x), where x now stands for the number of dollars you spend instead of the number of miles you drive. Explain why you can use the symbol c 1 for the inverse relation. Use the equation of c 1 (x) to �nd c 1 (758), and explain its real-world meaning.

c. Plot f 1 (x) c(x) and f 2 (x) c 1 (x) on the same screen, using function mode. Use a window with 0 x 1000 and use equal scales on the two axes. Sketch the two graphs, showing how they are related to the line y x.


4. Discrete Function Problem: Figure 1-5j shows a function that consists of a discrete set of points. Show that the function is one-to-one and thus is invertible, even

though the function is increasing in some parts of the domain and decreasing in other parts.

For Problems 5–8, sketch the line y x and the inverse relation on a copy of the given �gure. Be sure that the inverse relation is a re�ection of the function graph across the line y x. Tell whether the inverse relation is a function. 5. 6.

x

5

–5

5–5

y

x

5

–5

5–5

y

7. 8.

x

5

–5

5–5

y

x

5

–5

5–5

y

For Problems 9–14, a. Plot the parametric equations on graph paper

using the given domain for t. Connect the points with lines or smooth curves.

b. Tell whether y is a function of x. c. Con�rm your results by grapher, using the

given domain for t.

9. x t 3 5 t 5 y 2 t

10. x 5 t 7 t 4 y t 1 11. x 7 t 2 3 t 4 y t 2 12. x t 3 1 t 4 y (t 2) 2 13. x t 3 1 t 3 y t 3 2t 2 14. x t 2 2t 2 1 t 2 y t 3 t 2 t 1 15. Two Paths Problem: Two particles (small objects)

move along the paths shown in Figure 1-5k. �e paths are given by these parametric equations, where x and y are distance in meters and t is time in seconds.

Particle 1: x t 1 Particle 2: x 1.5t 2 y 7 t 2 y 1.5t 6

Particle 1

5x

5

Particle 2

10

3

y

Figure 1-5k

a. �e paths intersect at two points. For each point, determine whether the particles reach that point at the same time or at di�erent times. Give numbers to support your conclusion.

b. Con�rm your answer to part a graphically by plotting the two sets of parametric equations dynamically on the same screen, setting your grapher to simultaneous mode. Write a sentence or two explaining how your graph con�rms your answer to part a.

54321 109876

21

43

65

x

y

Figure 1-5j

45

Problem 27b could be solved

without fi nding the equation of the inverse function by recognizing that 200 is an input of the inverse function and therefore an output value of the original function.

Problems 29–32 require students to write equations for, and then graph the inverses of, given functions and decide whether the inverses are functions.

Problems 33 and 34 present two functions that are their own inverses.

Problems 33 and 34 can be

solved on a CAS using compositions of functions.33. f (f (x)) 5 1 ____ f (x) 5 1 _____ (1/x) 5 x, x 034. f (f (x)) 5 2f (x) 5 2(2x) 5 x for all x.35a. c(1000) 5 900. If you drive 1000 mi in a month, your monthly cost is $900.35b. c 21 (x) 5 2.5x 2 1250. c 21 (x) is a function because no input produces more than one output. c 21 (758) 5 645. You would have a monthly cost of $758 if you drove 645 mi in a month.35c.

Problems 35–37 are application problems with multiple questions and require students to use several skills. Problems structured like this appear in the free-response section of the AP Calculus test. If most of your students will take AP Calculus next year, try creating similar problems on your chapter tests in addition to assigning these kinds of problems for homework.

y

x

c(x)

c�1(x)200

200 600 1000

400600800

1000

16b. Th e ships do not arrive at the intersection point at the same time because the two x-values are not equal when the two y-values are equal. Freighter arrives at x 5 31.4285… mi when t 5 5.8571… h. Cutter arrives at x 5 31.4285… mi when t 5 5.9285… h. Freighter arrives at the intersection point 0.0714… h, or about 4 minutes, before cutter.

Problems 17–28 require students to write equations for, and then graph inverses of, given functions.

Problems 17–28 can be solved by algebraically reversing the variables and solving for y. (See CAS Suggestions for additional information.)


See pages 977–979 for answers to Problems 9–14, and 17–32.

47Section 1-6: Re� ections, Absolute Values, and Other Transformations

Re� ections, Absolute Values, and Other TransformationsIn Section 1-3, you learned that if y f (x), then multiplying x by a nonzero constant causes a horizontal dilation. Suppose that the constant is 1. Each x-value will be 1/( 1) or 1 times what it was in the pre-image. Figure 1-6a shows that the resulting image is a horizontal re� ection of the graph across the y-axis. � e new graph is the same size and shape, simply a mirror image of the original. Similarly, a vertical dilation by a factor of 1 re� ects the graph vertically across the x-axis.

In this section you will learn special transformations of functions that re� ect their graphs in various ways. You will also learn what happens when you take the absolute value of a function or of the independent variable x. Finally, you will learn about odd and even functions.

Given a function, transform it by re� ection and by applying absolute value to the function or its argument.

Re� ections Across the x-axis and y-axisExample 1 shows you how to plot the graphs in Figure 1-6a.

� e pre-image function y f (x) in Figure 1-6a is f (x) x 2 8x 17, where 2 x 5.

a. Write an equation for the re� ection of this function across the y-axis.

b. Write an equation for the re� ection of this function across the x-axis.

c. Plot the pre-image and the two re� ections on the same screen.

a. A re� ection across the y-axis is a horizontal dilation by a factor of 1. So

y f ( x) ( x) 2 8( x) 17 Substitute x for x.

y x 2 8x 17

Domain: 2 x 5

2 x 5 or 5 x 2 Multiply all three sides of the inequality by 1. � e inequalities reverse.

b. y f (x) For a re� ection across the x-axis, � nd the opposite of f (x).

y x 2 8x 17 � e domain remains 2 x 5.

Re� ections, Absolute Values, and Other Transformations

1- 6

y f x

y f x y f x

x

y

Figure 1-6a

Given a function, transform it by re� ection and by applyingthe function or its argument.

Objective

� e pre-image function 2

EXAMPLE 1 ➤

a. A re� ection across the SOLUTION


36. Deer Problem: � e surface area of a deer’s body is approximately proportional to the 2 _ 3 power of the deer’s weight. (� is is true because the area is proportional to the square of the length and the weight is proportional to the cube of the length.) Suppose that the particular equation for area as a function of weight is given by the power function

A(x) 0.4 x 2/3 where x is the weight in pounds and A(x) is the

surface area measured in square feet.

a. Find A(50), A(100), and A(150). Explain the real-world meaning of the answers.

b. True or false: “A deer twice the weight of another deer has a surface area twice that of the other deer.” Give numerical evidence to support your answer.

c. Find an equation for A 1 (x), where x now stands for area instead of weight.

d. Plot A and A 1 on the same screen using function mode. Use a window with 0 x 250. How are the two graphs related to the line y x?

37. Braking Distance Problem: � e length of skid marks, d(x) feet, le� by a car braking to a stop is a direct square power function of x, the speed in miles per hour when the brakes were applied. Based on information in the Texas Drivers Handbook (2002), d(x) is given approximately by

d(x) 0.057 x 2 for x 0

� e graph of this function is shown in Figure 1-5m.

x (mi/h)50 100

50

d(x) (�)

Figure 1-5m

a. When police o� cers investigate automobile accidents, they use the length of the skid marks to calculate the speed of the car at the time it started to brake. Write an equation for the inverse function, d 1 (x), where x is now the length of the skid marks. Explain why you need to take only the positive square root.

b. Find d 1 (200). What does this number represent in the context of this problem?

c. Suppose that the domain of function d started at 20 instead of zero. With your grapher in parametric mode, plot the graphs of function d and its inverse relation. Use the window shown in Figure 1-5m with 20 t 70. Sketch the result.

d. Explain why the inverse of function d in part c is not a function. What relationship do you notice between the domain and range of d and its inverse?

38. Horizontal Line Test Problem: � e vertical line test of Section 1-2, Problem 39, helps you see graphically that a relation is a function if no vertical line crosses the graph more than once. � e horizontal line test allows you to tell whether a function is invertible. Sketch two graphs, one for an invertible function and one for a non-invertible function, that illustrate this test.

PROPERTY: The Horizontal Line Test

If a horizontal line cuts the graph of a function in more than one place, then the function is not invertible because it is not one-to-one.


Problem Notes (continued)36a. A(50) 5 5.4288… ; A(100) 5 8.6177… ; A(150) 5 11.2924…; Deer that weigh 50, 100, and 150 lb have hides of areas approximately 5.43, 8.62, and 11.29 ft 2 , respectively.36b. False. A(100) 2 A(50)36c. A 21 (x) 5 (2.5x ) 1.5 36d.

Th e two curves are refl ections of each other across the line y 5 x.37a. d 21 (x) 5  

_____ x _____ 0.057 . Because the

domain of d is x 0 , the range of d 21 is d 21 (x) 0.37b. d 21 (200) 5 59.234… Th is means that a 200-ft skid mark is caused by a car moving at a speed of about 59 mi/h.37d. Because the domain of d now contains negative numbers, the range of the inverse relation contains negative numbers. Now, because the range of the inverse relation contains negative numbers, y 5  

_____ x _____ 0.057 , which is not a

function.

Problem 38 introduces the horizontal line test to determine invertibility. It is important to familiarize students with this visual clue to determining the invertibility of functions.


1. Find the equations of the inverse functions (if they exist) for functions of the form y 5 x a , where a can be any nonzero integer.

100 200

100

200

y

x

A�1(x)

A(x)

y = x

a. For what values of a do a function in this form and its inverse have the same equation?

b. For what values of a are functions in this form invertible? From your knowledge of exponents, explain why your answer is reasonable.

2. A student once claimed that f (x) 5 x 2 and g (x) 5  

__ x were inverse functions.

a. Find f (g (x)) and g (f (x)).

b. Based on the results to part a, explain why the student’s claim is not completely correct.

c. Under what domain restrictions are f and g inverses?

d. Graph f and g under the conditions you determined in part c. How does the graph confi rm your answer to part c?

See page 979 for answers to Problems 37c and 38, and CAS Problems 1 and 2.

47Section 1-6: Refl ections, Absolute Values, and Other Transformations

Refl ections, Absolute Values, and Other TransformationsIn Section 1-3, you learned that if y � f (x), then multiplying x by a nonzero constant causes a horizontal dilation. Suppose that the constant is �1. Each x-value will be 1/(�1) or �1 times what it was in the pre-image. Figure 1-6a shows that the resulting image is a horizontal refl ection of the graph across the y-axis. Th e new graph is the same size and shape, simply a mirror image of the original. Similarly, a vertical dilation by a factor of �1 refl ects the graph vertically across the x-axis.

In this section you will learn special transformations of functions that refl ect their graphs in various ways. You will also learn what happens when you take the absolute value of a function or of the independent variable x. Finally, you will learn about odd and even functions.

Given a function, transform it by refl ection and by applying absolute value to the function or its argument.

Refl ections Across the x-axis and y-axisExample 1 shows you how to plot the graphs in Figure 1-6a.

Th e pre-image function y � f (x) in Figure 1-6a is f (x) � x 2 � 8x � 17, where 2 � x � 5.

a. Write an equation for the refl ection of this function across the y-axis.

b. Write an equation for the refl ection of this function across the x-axis.

c. Plot the pre-image and the two refl ections on the same screen.

a. A refl ection across the y-axis is a horizontal dilation by a factor of �1. So

y � f (�x) � (� x) 2 � 8(�x) � 17 Substitute �x for x.

y � x 2 � 8x � 17

Domain: 2 � �x � 5

�2 � x � �5 or �5 � x � �2 Multiply all three sides of the inequality by �1. Th e inequalities reverse.

b. y � �f (x) For a refl ection across the x-axis, fi nd the opposite of f (x).

y � � x 2 � 8x � 17 Th e domain remains 2 � x � 5.

Refl ectand Ot

1- 6

52

Verticalreflectiony � �f (x)

Horizontalreflectiony � f (�x)

y � f (x)

�2�51

5

�5

x

y

Figure 1-6a

Givethe f

Objective

Th e2 �

EXAMPLE 1

a. ASOLUTION


36. Deer Problem: � e surface area of a deer’s body is approximately proportional to the 2 _ 3 power of the deer’s weight. (� is is true because the area is proportional to the square of the length and the weight is proportional to the cube of the length.) Suppose that the particular equation for area as a function of weight is given by the power function

A(x) 0.4 x 2/3 where x is the weight in pounds and A(x) is the

surface area measured in square feet.

a. Find A(50), A(100), and A(150). Explain the real-world meaning of the answers.

b. True or false: “A deer twice the weight of another deer has a surface area twice that of the other deer.” Give numerical evidence to support your answer.

c. Find an equation for A 1 (x), where x now stands for area instead of weight.

d. Plot A and A 1 on the same screen using function mode. Use a window with 0 x 250. How are the two graphs related to the line y x?

37. Braking Distance Problem: � e length of skid marks, d(x) feet, le� by a car braking to a stop is a direct square power function of x, the speed in miles per hour when the brakes were applied. Based on information in the Texas Drivers Handbook (2002), d(x) is given approximately by

d(x) 0.057 x 2 for x 0

� e graph of this function is shown in Figure 1-5m.

x (mi/h)50 100

50

d(x) (�)

Figure 1-5m

a. When police o� cers investigate automobile accidents, they use the length of the skid marks to calculate the speed of the car at the time it started to brake. Write an equation for the inverse function, d 1 (x), where x is now the length of the skid marks. Explain why you need to take only the positive square root.

b. Find d 1 (200). What does this number represent in the context of this problem?

c. Suppose that the domain of function d started at 20 instead of zero. With your grapher in parametric mode, plot the graphs of function d and its inverse relation. Use the window shown in Figure 1-5m with 20 t 70. Sketch the result.

d. Explain why the inverse of function d in part c is not a function. What relationship do you notice between the domain and range of d and its inverse?

38. Horizontal Line Test Problem: � e vertical line test of Section 1-2, Problem 39, helps you see graphically that a relation is a function if no vertical line crosses the graph more than once. � e horizontal line test allows you to tell whether a function is invertible. Sketch two graphs, one for an invertible function and one for a non-invertible function, that illustrate this test.

PROPERTY: The Horizontal Line Test

If a horizontal line cuts the graph of a function in more than one place, then the function is not invertible because it is not one-to-one.

47Section 1-6: Re� ections, Absolute Values, and Other Transformations

S e c t i o n 1- 6S e c t i o n 1- 6S e c t i o n 1- 6S e c t i o n 1- 6S e c t i o n 1- 6S e c t i o n 1- 6PL AN N I N G

Class Time1 day

Homework AssignmentRA, Q1–Q10, Problems 1–5, 7, 9–14

Teaching Resources Exploration 1-6a: Translation, Dilation,

and Re� ectionBlackline Masters

Problem 1Problem 2Problem 3Problem 4Problem 7


Dilation

Presentation Sketch: Re� ection Present.gsp

Presentation Sketch: Absolute Value Present.gsp

Activity: Exploring Translations and Dilations

TE ACH I N G

Important Terms and ConceptsRe� ectionRe� ection across the y-axis Re� ection across the x-axisDisplacementPiecewise functionAbsolute value transformationsEven functionOdd functionStep discontinuity Greatest integer function, x

49Section 1-6: Re�ections, Absolute Values, and Other Transformations

Because there are two di�erent rules for g (x) in di�erent “pieces” of the domain, g is called a piecewise function of x.

PROPERTY: Absolute Value Transformations

�e transformation g (x) f (x)f across the x-axis if f (x) is negative

f unchanged if f (x) is nonnegative

�e transformation g (x) f x f unchanged for nonnegative values of x

x to the corresponding negative values of x

f for negative values of x

Even Functions and Odd FunctionsFigure 1-6e shows the graph of f (x) x 4 5 x 2 1, a polynomial function with only even exponents. (�e number 1 equals 1 x 0 , which has an even exponent.) Figure 1-6f shows the graph of f (x) x 3 6x, a polynomial function with only odd exponents. What symmetries do you observe?

4aa

8

x

f ( a) f (a)

f (x) x 4 + 5x2 1

Evenfunction

y

4aa

8f(a)

x

f ( a) f (a)

f(a)

f (x) x3 + 6x

Oddfunction

y


Re�ecting the graph of the even function f (x) x 4 5 x 2 1 horizontally across the y-axis leaves the graph unchanged. You can see this algebraically given the property of powers with even exponents.

f ( x) ( x) 4 5( x) 2 1 Substitute x for x.

f ( x) x 4 5x 2 1 Negative number raised to an even power.

f ( x) f (x)

Figure 1-6g shows that re�ecting the graph of the odd function f (x) x 3 6x horizontally across the y-axis has the same e�ect as re�ecting it vertically across the x-axis. Algebraically,

f ( x) ( x) 3 6( x) Substitute x for x.

f ( x) x 3 6x Negative number raised to an odd power.

f ( x) f (x)

4x

x- or y-re�ection

Oddfunction

y

Figure 1-6g

5

g

f f and gcoincide.

x5

6y

Figure 1-6d


c. f 1 (x) x 2 8x 17 / (x 2 and x 5)

f 2 (x) x 2 8x 17 / (x 5 and x 2)

f 3 (x) x 2 8x 17 / (x 2 and x 5)

� e graphs are shown in Figure 1-6a.

You can check the algebraic solutions by plotting f 4 (x) f 1 ( x) and f 5 (x) f 1 (x) using thick style. � e graphs should overlay f 2 (x) and f 3 (x).

PROPERTY: Reflections Across the Coordinate Axesg (x) f (x) is a vertical re� ection of function f across the x-axis.

g (x) f ( x) is a horizontal re� ection of function f across the y-axis.

Absolute Value TransformationsSuppose you shoot a basketball. While in the air, it is above the basket level sometimes and below it at other times. Figure 1-6b shows y f (x), the displacement from the level of the basket as a function of time. If the ball is above the basket, its displacement is positive; if the ball is below the basket, its displacement is negative.

Distance, however, is the magnitude (or size) of the displacement, which is never negative.

Distance equals the absolute value of the displacement. � e solid graph in Figure 1-6c is the graph of y g (x) f (x) . Taking the absolute value of f (x) retains the non-negative values of y and re� ects the negative values vertically across the x-axis.

Figure 1-6d shows what happens for g (x) f x ,for which you take the absolute value of the argument (this is a di� erent function f than in the last example). For positive values of x, x x, so g (x) f (x) and the graphs coincide. For negative values of x,

x x, so g (x) f ( x), across the y-axis of the part of function f where x 0. Notice that the graph of f for the negative values of x is not a part of the graph of f x .

� e equation for g (x) can be written this way:

g (x) f (x)

f ( x)

if x 0 if x 0

Divide by a Boolean variable or enter the domain directly, depending on your grapher, to restrict the domain.

➤

2

y (�)10

10

5

5

x (s)Basket level

Floor levelFigure 1-6b

2

y (�)10

10

5

5

x (s)

Graphs coincide.

y g(x)

Basket level

Floor levelFigure 1-6c


Section Notes

This section discusses reflections across the x- and y-axes, absolute value transformations, and even and odd functions.

You can teach this section in one day. Start by discussing Example 1, and consider challenging students to sketch the graphs of functions with which they are familiar and then to graph y 5 f (2x) and y 5 2f (x). Because y 5 f (2x) is a horizontal dilation by a factor of 21, students may mistakenly think it is a reflection across the horizontal axis. Make sure students understand that y 5 f (2x) represents a reflection across the vertical axis, or y-axis. Similarly, make sure they understand that although y 5 2f (x) is a vertical dilation, it is a reflection across the horizontal axis, or x-axis.

Before discussing absolute value transformations, you may need to review the definition of absolute value. The fact that x 5 2x for x , 0 is troubling to many students because they mistakenly think that 2x always represents a negative number. Using the words “the opposite of” instead of “negative” can help clarify the definition for students. Try saying, “If the number inside the absolute value sign is negative, then its absolute value is the opposite of the negative number.” You might illustrate with an example:

If x 5 23, then x 5 (23) 5 2(23). The value of 2(23) is the opposite of 23, which is 3.

Spend some time in class explaining the difference between the y 5 f (x) and y 5 f  x   transformations. Use the graphs of familiar functions to demonstrate how the y 5 f (x) transformation affects the range and the f  x   transformation affects the domain of f.

Finally, you can introduce the discussion of even and odd functions by raising

these questions: For which functions will the y 5 f (2x) transformation result in the original function, and for which does it give the opposite of the original function? In other words, for which functions is f (2x) 5 f (x), and for which functions is f (2x) 5 2f (x)?

Ask students to give examples (such as y 5 kx, y 5 x 2 , y 5 x 3 , y 5 x), and perhaps have them discuss the question in small groups. Once students have recognized the

properties of functions that satisfy these equations (functions symmetrical to the y-axis or to the origin), you can introduce the terms even and odd. If you run out of time, you could assign this question as homework.

Trigonometric functions are either odd or even functions. For example, sin(2x) 5 2sin x and cos(2x) 5 cos x. Studying odd and even polynomial functions now will prepare students for


Because there are two di�erent rules for g (x) in di�erent “pieces” of the domain, g is called a piecewise function of x.

PROPERTY: Absolute Value Transformations

�e transformation g (x) f (x)f across the x-axis if f (x) is negative

f unchanged if f (x) is nonnegative

�e transformation g (x) f x f unchanged for nonnegative values of x

x to the corresponding negative values of x

f for negative values of x

Even Functions and Odd FunctionsFigure 1-6e shows the graph of f (x) x 4 5 x 2 1, a polynomial function with only even exponents. (�e number 1 equals 1 x 0 , which has an even exponent.) Figure 1-6f shows the graph of f (x) x 3 6x, a polynomial function with only odd exponents. What symmetries do you observe?

4aa

8

x

f ( a) f (a)

f (x) x 4 + 5x2 1

Evenfunction

y

4aa

8f(a)

x

f ( a) f (a)

f(a)

f (x) x3 + 6x

Oddfunction

y


Re�ecting the graph of the even function f (x) x 4 5 x 2 1 horizontally across the y-axis leaves the graph unchanged. You can see this algebraically given the property of powers with even exponents.

f ( x) ( x) 4 5( x) 2 1 Substitute x for x.

f ( x) x 4 5x 2 1 Negative number raised to an even power.

f ( x) f (x)

Figure 1-6g shows that re�ecting the graph of the odd function f (x) x 3 6x horizontally across the y-axis has the same e�ect as re�ecting it vertically across the x-axis. Algebraically,

f ( x) ( x) 3 6( x) Substitute x for x.

f ( x) x 3 6x Negative number raised to an odd power.

f ( x) f (x)

4x

x- or y-re�ection

Oddfunction

y

Figure 1-6g

5

g

f f and gcoincide.

x5

6y

Figure 1-6d


c. f 1 (x) x 2 8x 17 / (x 2 and x 5)

f 2 (x) x 2 8x 17 / (x 5 and x 2)

f 3 (x) x 2 8x 17 / (x 2 and x 5)

� e graphs are shown in Figure 1-6a.

You can check the algebraic solutions by plotting f 4 (x) f 1 ( x) and f 5 (x) f 1 (x) using thick style. � e graphs should overlay f 2 (x) and f 3 (x).

PROPERTY: Reflections Across the Coordinate Axesg (x) f (x) is a vertical re� ection of function f across the x-axis.

g (x) f ( x) is a horizontal re� ection of function f across the y-axis.

Absolute Value TransformationsSuppose you shoot a basketball. While in the air, it is above the basket level sometimes and below it at other times. Figure 1-6b shows y f (x), the displacement from the level of the basket as a function of time. If the ball is above the basket, its displacement is positive; if the ball is below the basket, its displacement is negative.

Distance, however, is the magnitude (or size) of the displacement, which is never negative.

Distance equals the absolute value of the displacement. � e solid graph in Figure 1-6c is the graph of y g (x) f (x) . Taking the absolute value of f (x) retains the non-negative values of y and re� ects the negative values vertically across the x-axis.

Figure 1-6d shows what happens for g (x) f x ,for which you take the absolute value of the argument (this is a di� erent function f than in the last example). For positive values of x, x x, so g (x) f (x) and the graphs coincide. For negative values of x,

x x, so g (x) f ( x), across the y-axis of the part of function f where x 0. Notice that the graph of f for the negative values of x is not a part of the graph of f x .

� e equation for g (x) can be written this way:

g (x) f (x)

f ( x)

if x 0 if x 0

Divide by a Boolean variable or enter the domain directly, depending on your grapher, to restrict the domain.

➤

2

y (�)10

10

5

5

x (s)Basket level

Floor levelFigure 1-6b

2

y (�)10

10

5

5

x (s)

Graphs coincide.

y g(x)

Basket level

Floor levelFigure 1-6c

49

• Have ELL students write a summary of the section rather than doing the Reading Analysis questions. If they write in their primary language, have them translate a couple of sentences for you.

Exploration Notes

Exploration 1-6a presents students with eight transformations of the same function. (The function is cubic with a restricted domain, but this fact is not mentioned in the exploration.) In each case, students are presented with an equation for y in terms of function f, and they are asked to name the transformation and sketch the transformed graph. Allow about 20 minutes.

Technology Notes

Problem 14 asks students to use a Dynamic Precalculus Exploration at www.keymath.com/precalc. The Dilation exploration allows students to explore reflections across the x- and y-axes by changing one or both of the dilation sliders to 21.

Presentation Sketch: Reflection Present.gsp at www.keypress.com/keyonline demonstrates the effects of reflecting the graph of a square root function across coordinate axes. The presentation demonstrates the effects on the coordinates as well as on the algebraic description of the function.

Presentation Sketch: Absolute Value Present.gsp at www.keypress.com/keyonline demonstrates the graphical effect of composing a quadratic function with absolute value.

studying trigonometric functions and their properties.

Exploration 1-6a may be used as a follow-up assignment or quiz to see if students have learned about the transformations of this section.

Differentiating Instruction• Discussing displacement and distance

using an enlargement of Figures 1-6c and 1-6d will help clarify the difference between these two concepts.

• If you assign Problem 13, help ELL students understand the vocabulary.

Section 1-6: Reflections, Absolute Values, and Other Transformations


5. �e equation for the function in Problem 3 is f (x) x 3 6 x 2 13x 42 for 6 x 4. Plot the function as f 1 (x) on your grapher. Plot f 2 (x) f 1 x using thick style. Does the result con�rm your answer to Problem 3, part d?

6. �e equation for the function in Problem 4 is f (x) 3

____________ 25 (x 2) 2 for 3 x 7.

Plot the function as f 1 (x) on your grapher. Plot f 2 (x) f 1 x using thick style. Does the result con�rm your answer to Problem 4, part d?

7. Absolute Value Transformations Problem: Figure 1-6h shows the graph of f (x) 0.5(x 2) 2 4.5 in the domain

2 x 6.

f

y

x5 5

5

5

Figure 1-6h

a. Plot the graph of f 1 (x) f (x). On the same screen, plot f 2 (x) f (x) using thick style. Sketch the result and describe how this transformation changes the graph of f.

b. Deactivate f 2 (x). On the same screen as f 1 (x), plot the graph of f 3 (x) f x using thick style. Sketch the result and describe how this transformation changes the graph of f.

c. Use the equation for function f to �nd the value of f (3) and the value of f 3 . Show that both results agree with your graphs in parts a and b. Explain why 3 is in the domain of f x even though it is not in the domain of f itself.

d. Figure 1-6i shows the graph of a function g, but you don’t know the equation for the function. On a copy of this �gure, sketch the graph of y g (x) , using the conclusion you reached in part a. On another copy of this

�gure, sketch the graph of y g x , using the conclusion you reached in part b.

x5 5

5

5g

y

Figure 1-6i

8. Displacement vs. Distance Absolute Value Problem: Calvin’s car runs out of gas as he is going uphill. He continues to coast uphill for a while, stops, then starts rolling backward without applying the brakes. His displacement, y, in meters, from a gas station on the hill as a function of time, x, in seconds, is given by

y 0.1 x 2 12x 250 a. Plot the graph of this function. Sketch

the result. b. Find Calvin’s displacement at 10 s and at

40 s. What is the real-world meaning of his negative displacement at 10 s?

c. What is Calvin’s distance from the gas station at times x 10 s and x 40 s? Explain why both values are positive.

d. De�ne Calvin’s distance from the gas station. Sketch the graph of distance versus time.

e. If Calvin keeps moving as indicated in this problem, when will he pass the gas station as he rolls back down the hill?

9. Even Function and Odd Function Problem: Figure 1-6j shows the graph of the even function f (x) x 4 3 x 2 4. Figure 1-6k on the next page shows the graph of the odd function g (x) x 5 6 x 3 6x.

f

y

x4 4

8

8

4

4

Figure 1-6j


Any function having the property f ( x) f (x) is called an even function. Any function having the property f ( x) f (x) is called an odd function. � ese names apply even if the equation for the function does not have exponents.

DEFINITION: Even Function and Odd Function� e function f is an even function if and only if f ( x) f (x) for all x in the domain.

� e function f is an odd function if and only if f ( x) f (x) for all x in the domain.

Note: For odd functions, re� ection across the y-axis gives the same image as re� ection across the x-axis. For even functions, re� ection across the y-axis is the same as the pre-image. So odd functions are symmetric about the origin, and even functions are symmetric across the y-axis. Most functions do not possess the property of oddness or evenness.

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? Reread the paragraph on page 48 that discusses Figure 1-6d. Use numerical values from the graph to guide yourself through this paragraph. Explain in your own words what the sentence about negative values of x means. Why is part of the graph of f “lost” in the graph of f x ? Write down speci� c questions about what you may not understand, and � nd someone who can answer them.

Quick Review

Q1. If f (x) 2x, then f 1 (x) ? . Q2. If f (x) x 3, then f 1 (x) ? . Q3. If f (x) 2x 3, then f 1 (x) ? . Q4. If f (x) x 2 , write the equation for the

inverse relation. Q5. Explain why the inverse relation in Problem Q4

is not a function. Q6. If f (x) 2 x , then f 1 (8) ? . Q7. If the inverse relation for function f is also a

function, then f is called ? .

Q8. Write the de� nition of a one-to-one function.

Q9. Give a number x for which x x.


For Problems 1–4, sketch the graphs of a. g (x) f (x) b. h(x) f ( x)

c. a(x) f (x) d. v(x) f x

1. 2.

x5 5

5

5 y f (x)y

x5 5

5

5

y f (x)

y

3. 4.

x46

50

50 y f (x)

y

x5 5

5

5

y f (x)

y

5min

Reading Analysis Q8. Write the de� nition of a one-to-one function.

Problem Set 1-6


Technology Notes (continued)

Activity: Exploring Translations and Dilations in the Instructor’s Resource Book focuses on translations and dilations, but you could extend it to include investigations of refl ections.

CAS Suggestions

Functions can be defi ned and transformed directly. For example, to graph the refl ection image of f (x) 5 x 2 2 2x 1 7 across the y-axis, defi ne f (x) on a calculator screen and graph y 5 f (2x).

To verify whether functions are even or odd using a CAS, defi ne the function and use Boolean operators to determine if f (x) 5 f (2x) or f (x) 5 2f (x). A Boolean “true” result will be displayed only if the equation is universally true for the given function. If the equation is not universally true, then the function is not even or odd, and the CAS “thinks” an equation is being defi ned and displays the result.

Note: Performing the algebraic manipulations required to determine if a function is even or odd is arguably a diff erent skill than understanding that f (2x) 5 f (x) for all even functions and f (2x) 5 2f (x) for all odd functions. A CAS allows you to assess students’ understanding independent of their ability to perform the algebraic manipulations.

PRO B LE M N OTES

Q1. 1 __ 2 xQ2. x 1 3Q3. x 1 3 _____ 2 Q4. y 5  

__ x

Q5. Th ere are two y-values for every positive x-value.

Q6. 3Q7. InvertibleQ8. A function for which each y-value in the range corresponds to only one x-value Q9. Sample Answer: 5Q10. Sample Answer: 25

Problems 1–4 provide practice in applying the transformations introduced in this section to the graphs of functions. Blackline masters for these problems are available in the Instructor’s Resource Book.1a. y

x�5 5

�5

5


5. �e equation for the function in Problem 3 is f (x) x 3 6 x 2 13x 42 for 6 x 4. Plot the function as f 1 (x) on your grapher. Plot f 2 (x) f 1 x using thick style. Does the result con�rm your answer to Problem 3, part d?

6. �e equation for the function in Problem 4 is f (x) 3

____________ 25 (x 2) 2 for 3 x 7.

Plot the function as f 1 (x) on your grapher. Plot f 2 (x) f 1 x using thick style. Does the result con�rm your answer to Problem 4, part d?

7. Absolute Value Transformations Problem: Figure 1-6h shows the graph of f (x) 0.5(x 2) 2 4.5 in the domain

2 x 6.

f

y

x5 5

5

5

Figure 1-6h

a. Plot the graph of f 1 (x) f (x). On the same screen, plot f 2 (x) f (x) using thick style. Sketch the result and describe how this transformation changes the graph of f.

b. Deactivate f 2 (x). On the same screen as f 1 (x), plot the graph of f 3 (x) f x using thick style. Sketch the result and describe how this transformation changes the graph of f.

c. Use the equation for function f to �nd the value of f (3) and the value of f 3 . Show that both results agree with your graphs in parts a and b. Explain why 3 is in the domain of f x even though it is not in the domain of f itself.

d. Figure 1-6i shows the graph of a function g, but you don’t know the equation for the function. On a copy of this �gure, sketch the graph of y g (x) , using the conclusion you reached in part a. On another copy of this

�gure, sketch the graph of y g x , using the conclusion you reached in part b.

x5 5

5

5g

y

Figure 1-6i

8. Displacement vs. Distance Absolute Value Problem: Calvin’s car runs out of gas as he is going uphill. He continues to coast uphill for a while, stops, then starts rolling backward without applying the brakes. His displacement, y, in meters, from a gas station on the hill as a function of time, x, in seconds, is given by

y 0.1 x 2 12x 250 a. Plot the graph of this function. Sketch

the result. b. Find Calvin’s displacement at 10 s and at

40 s. What is the real-world meaning of his negative displacement at 10 s?

c. What is Calvin’s distance from the gas station at times x 10 s and x 40 s? Explain why both values are positive.

d. De�ne Calvin’s distance from the gas station. Sketch the graph of distance versus time.

e. If Calvin keeps moving as indicated in this problem, when will he pass the gas station as he rolls back down the hill?

9. Even Function and Odd Function Problem: Figure 1-6j shows the graph of the even function f (x) x 4 3 x 2 4. Figure 1-6k on the next page shows the graph of the odd function g (x) x 5 6 x 3 6x.

f

y

x4 4

8

8

4

4

Figure 1-6j


Any function having the property f ( x) f (x) is called an even function. Any function having the property f ( x) f (x) is called an odd function. � ese names apply even if the equation for the function does not have exponents.

DEFINITION: Even Function and Odd Function� e function f is an even function if and only if f ( x) f (x) for all x in the domain.

� e function f is an odd function if and only if f ( x) f (x) for all x in the domain.

Note: For odd functions, re� ection across the y-axis gives the same image as re� ection across the x-axis. For even functions, re� ection across the y-axis is the same as the pre-image. So odd functions are symmetric about the origin, and even functions are symmetric across the y-axis. Most functions do not possess the property of oddness or evenness.

Reading Analysis

From what you have read in this section, what do you consider to be the main idea? Reread the paragraph on page 48 that discusses Figure 1-6d. Use numerical values from the graph to guide yourself through this paragraph. Explain in your own words what the sentence about negative values of x means. Why is part of the graph of f “lost” in the graph of f x ? Write down speci� c questions about what you may not understand, and � nd someone who can answer them.

Quick Review

Q1. If f (x) 2x, then f 1 (x) ? . Q2. If f (x) x 3, then f 1 (x) ? . Q3. If f (x) 2x 3, then f 1 (x) ? . Q4. If f (x) x 2 , write the equation for the

inverse relation. Q5. Explain why the inverse relation in Problem Q4

is not a function. Q6. If f (x) 2 x , then f 1 (8) ? . Q7. If the inverse relation for function f is also a

function, then f is called ? .

Q8. Write the de� nition of a one-to-one function.



For Problems 1–4, sketch the graphs of a. g (x) f (x) b. h(x) f ( x)

c. a(x) f (x) d. v(x) f x

1. 2.

x5 5

5

5 y f (x)y

x5 5

5

5

y f (x)

y

3. 4.

x46

50

50 y f (x)

y

x5 5

5

5

y f (x)

y

5min

Reading Analysis Q8. Write the de� nition of a one-to-one function.

Problem Set 1-6

51

1d.

2a.

2b.

2c.

Problems 5 and 6 ask students to use their graphers to plot the functions in Problems 3 and 4 and to verify the results for parts d.5. Th e graphs match.6. Th e graphs match.

Problems 6 and 7 can be handled using any grapher, but a CAS allows the functions to be defi ned and transformed directly.

A blackline master for Problem 7d is available in the Instructor’s Resource Book.

Problems 7–10 provide practice for and promote understanding of the concepts of absolute value transformations and even and odd functions. If you have time aft er discussing the concepts with them, have students work in small groups to solve these problems in class.

y

x–5 5

–5

5

x�5 5

�5

5y

y

x�5 5

�5

5

x�5 5

�5

5y

1b. 1c. y

x�5 5

�5

5

y

x�5 5

�5

5

Section 1-6: Refl ections, Absolute Values, and Other Transformations

See pages 979–980 for answers to Problems 2d–4, 7 and 8.


12. Step Functions—�e Postage Stamp Problem: Figure 1-6p shows the graph of the greatest integer function, f (x) x . In this function, x is the greatest integer less than or equal to x. For instance, 3.9 3, 5 5, and 2.1 3.

x1

1

y

Figure 1-6p

a. Plot the greatest integer function using dot style so that points will not be connected. Most graphers use the symbol int(x) for x . Trace to x 2.9, x 3, and x 3.1. What do you �nd for the three y-values?

b. In the year 2005, the postage for a �rst-class letter was 37 cents for weights up to 1 oz and 23 cents more for each additional ounce or fraction of an ounce. Sketch the graph of this function.

+ + + 23 23 23

c. Using a transformation of the greatest integer function, write an equation for the 2005 postage as a function of the weight. Plot it on your grapher. Does the graph agree with the one you sketched in part b?

d. In 2005, �rst-class postage rates applied only until the letter reached the weight at which the postage would exceed $3.13. What is the domain of the function in part c?

e. Check the Internet or another source to �nd the �rst-class postage rates this year. What di�erences do you �nd from the 2005 rates? Cite the source you used.

13. Piecewise Functions—Weight Above and Below Earth’s Surface Problem: When you are above the surface of Earth, your weight is inversely proportional to the square of your distance from the center of Earth. �is is because the farther

away you are, the weaker the gravitational force between Earth and you. When you are below the surface of Earth, your weight is directly proportional to your distance from the center. At the center you would be “weightless” because Earth’s gravity would pull you equally in all directions.

Figure 1-6q shows the graph of the weight function for a 150-lb person. �e radius of Earth is about 4000 mi. �e weight is called a piecewise function of the distance because it is given by di�erent equations in di�erent “pieces” of the domain. Each piece is called a branch of the function. �e equation of the function can be written

y ax b __ x 2

if 0 x 4000

x (mi)

4000

y (lb)

50

100

150

8000Figure 1-6q

a. Find the values of a and b that make y 150 when x 4000 for each branch.

b. Plot the graph of f. Use piecewise functions or Boolean variables to restrict the domain of the graph.

c. Find y if x 3000 and if x 5000. d. Find the two distances from the center at

which the weight would be 50 lb.

14. Dynamic Re�ection Problem: Go to www.keymath.com/precalc and open the Dilation exploration. Set slider c equal to 1 and slider d equal to 1 and describe what you observe. �en set slider c equal to 1 and slider d equal to 1 and describe what you observe. Finally, set both sliders equal to 1 and describe what you observe. Explain how re�ections are related to dilations.

if x 4000


g

y

x4 4

8

8

4

4

Figure 1-6k

a. On the same screen, plot f 1 (x) f (x) and f 2 (x) f ( x). Use thick style for f 2 (x). Based on the properties of negative numbers raised to even powers, explain why the two graphs are identical.

b. Deactivate f 1 (x) and f 2 (x). On the same screen, plot f 3 (x) g (x), f 4 (x) g ( x), and f 5 (x) g (x). Use thick style for f 5 (x). Based on the properties of negative numbers raised to odd powers, explain why the graphs of f 4 (x) and f 5 (x) are identical.

c. Even functions have the property f ( x) f (x). Odd functions have the property f ( x) f (x). Figure 1-6l shows two functions, h and j, but you don’t know the equation of either function. Tell which function is an even function and which is an odd function.

x

j

hy

Figure 1-6l

d. Let e(x) 2 x . Sketch the graph. Based on the graph, is function e an odd function, an even function, or neither? Con�rm your answer algebraically by �nding e( x).

10. Absolute Value Function—Odd or Even? Plot the graph of f (x) x . Sketch the result. Based on the graph, is function f an odd function, an even function, or neither? Con�rm your answer algebraically by �nding f ( x).

11. Step Discontinuity Problem: Figure 1-6m shows the graph of

f (x) xx

�e graph has a step discontinuity at x 0, where f (x) jumps instantaneously from 1 to 1.

a. Plot the graph of f 1 (x) f (x). Use a window that includes x 0 as a grid point. Does your graph agree with the �gure?

b. Figure 1-6n is a vertical dilation of function f with vertical and horizontal translations. Enter an equation for this function as f 2 (x), using operations on the variable f 1 (x). Use a window that includes x 4 as a grid point. When you have duplicated the graph in Figure 1-6n, write an equation for the transformed function in terms of function f.

x4

4

8y

5

10

5

x

y

Figure 1-6n Figure 1-6o

c. Figure 1-6o shows the graph of the quadratic function y (x 3) 2 to which something has been added or subtracted to give it a step discontinuity of 4 units at x 5. Find an equation of the function. Verify that your equation is correct by plotting it on your grapher.

x

y

Figure 1-6m



Problems 9c, 9d, and 10 can be solved using direct defi nitions.

9a.

Th e polynomial function f (x) is the sum of even powers of x. A negative number raised to an even power is equal to the absolute value of that number raised to the same power. So, for x, the same corresponding y-value occurs, and therefore f (x) 5 f (2x) . 9b.

A negative number raised to an odd power is equal to the opposite of the absolute value of that number raised to the same power. Because each term in g (x) is a monomial in x raised to an odd power, g (2x) has the same eff ect on g (x) as 2g (x).9c. Function h is odd; function j is even.9d.

Th e function e(x) is neither odd nor even. e(2x) 2e(x)

�4 4

�8

�4

4

8

x

y

�4 4

�8

�4

4

8

x

y

y

x�5

�5

5

5

10.

Th e function f is an even function.f (2x) 5 2x 5 x 5 f (x)

Problems 11–13 are important to assign and discuss with students because of the signifi cance of the concepts of discontinuity, piecewise functions, the greatest integer function, and some interesting applications.

y

x


12. Step Functions—�e Postage Stamp Problem: Figure 1-6p shows the graph of the greatest integer function, f (x) x . In this function, x is the greatest integer less than or equal to x. For instance, 3.9 3, 5 5, and 2.1 3.

x1

1

y

Figure 1-6p

a. Plot the greatest integer function using dot style so that points will not be connected. Most graphers use the symbol int(x) for x . Trace to x 2.9, x 3, and x 3.1. What do you �nd for the three y-values?

b. In the year 2005, the postage for a �rst-class letter was 37 cents for weights up to 1 oz and 23 cents more for each additional ounce or fraction of an ounce. Sketch the graph of this function.

+ + + 23 23 23

c. Using a transformation of the greatest integer function, write an equation for the 2005 postage as a function of the weight. Plot it on your grapher. Does the graph agree with the one you sketched in part b?

d. In 2005, �rst-class postage rates applied only until the letter reached the weight at which the postage would exceed $3.13. What is the domain of the function in part c?

e. Check the Internet or another source to �nd the �rst-class postage rates this year. What di�erences do you �nd from the 2005 rates? Cite the source you used.

13. Piecewise Functions—Weight Above and Below Earth’s Surface Problem: When you are above the surface of Earth, your weight is inversely proportional to the square of your distance from the center of Earth. �is is because the farther

away you are, the weaker the gravitational force between Earth and you. When you are below the surface of Earth, your weight is directly proportional to your distance from the center. At the center you would be “weightless” because Earth’s gravity would pull you equally in all directions.

Figure 1-6q shows the graph of the weight function for a 150-lb person. �e radius of Earth is about 4000 mi. �e weight is called a piecewise function of the distance because it is given by di�erent equations in di�erent “pieces” of the domain. Each piece is called a branch of the function. �e equation of the function can be written

y ax b __ x 2

if 0 x 4000

x (mi)

4000

y (lb)

50

100

150

8000Figure 1-6q

a. Find the values of a and b that make y 150 when x 4000 for each branch.

b. Plot the graph of f. Use piecewise functions or Boolean variables to restrict the domain of the graph.

c. Find y if x 3000 and if x 5000. d. Find the two distances from the center at

which the weight would be 50 lb.

14. Dynamic Re�ection Problem: Go to www.keymath.com/precalc and open the Dilation exploration. Set slider c equal to 1 and slider d equal to 1 and describe what you observe. �en set slider c equal to 1 and slider d equal to 1 and describe what you observe. Finally, set both sliders equal to 1 and describe what you observe. Explain how re�ections are related to dilations.

if x 4000


g

y

x4 4

8

8

4

4

Figure 1-6k

a. On the same screen, plot f 1 (x) f (x) and f 2 (x) f ( x). Use thick style for f 2 (x). Based on the properties of negative numbers raised to even powers, explain why the two graphs are identical.

b. Deactivate f 1 (x) and f 2 (x). On the same screen, plot f 3 (x) g (x), f 4 (x) g ( x), and f 5 (x) g (x). Use thick style for f 5 (x). Based on the properties of negative numbers raised to odd powers, explain why the graphs of f 4 (x) and f 5 (x) are identical.

c. Even functions have the property f ( x) f (x). Odd functions have the property f ( x) f (x). Figure 1-6l shows two functions, h and j, but you don’t know the equation of either function. Tell which function is an even function and which is an odd function.

x

j

hy

Figure 1-6l

d. Let e(x) 2 x . Sketch the graph. Based on the graph, is function e an odd function, an even function, or neither? Con�rm your answer algebraically by �nding e( x).

10. Absolute Value Function—Odd or Even? Plot the graph of f (x) x . Sketch the result. Based on the graph, is function f an odd function, an even function, or neither? Con�rm your answer algebraically by �nding f ( x).

11. Step Discontinuity Problem: Figure 1-6m shows the graph of

f (x) xx

�e graph has a step discontinuity at x 0, where f (x) jumps instantaneously from 1 to 1.

a. Plot the graph of f 1 (x) f (x). Use a window that includes x 0 as a grid point. Does your graph agree with the �gure?

b. Figure 1-6n is a vertical dilation of function f with vertical and horizontal translations. Enter an equation for this function as f 2 (x), using operations on the variable f 1 (x). Use a window that includes x 4 as a grid point. When you have duplicated the graph in Figure 1-6n, write an equation for the transformed function in terms of function f.

x4

4

8y

5

10

5

x

y

Figure 1-6n Figure 1-6o

c. Figure 1-6o shows the graph of the quadratic function y (x 3) 2 to which something has been added or subtracted to give it a step discontinuity of 4 units at x 5. Find an equation of the function. Verify that your equation is correct by plotting it on your grapher.

x

y

Figure 1-6m

53

12b.

12c. Dilated by a factor of 23; translated up 37 cents;

y 5 0,

2232x 1 1 1 37, x 0

Th e graphs match.12d. 0 x 1312e. Answers will vary.13a. a 5 0.0375; b 5 2,400,000,00013b.

13c. y(3000) 5 112.5 lb; y(5000) 5 96 lb13d. f 1 (x) 5 50 ⇒ x 5 1.333.

_ 3 mi;

f 2 (x) 5 50 ⇒ x 5 6928.2032… mi

Problem 14 refers students to www.keymath.com/precalc and gives them opportunity to work with refl ections dynamically.14. Answers will vary.


1. For any function f (x), is g (x) 5 f (x) even, odd, or neither? Explain your response both algebraically and with respect to the transformation involved.

2. For any function f (x), is h(x) 5 f (x)

even, odd, or neither? Explain your response both algebraically and with respect to the transformation involved.

1 2 3 4 5

40

80

120

Weight (oz)

Price (cents)

4000 8000

50

100

150y

x

11a. Th e graphs match.

11b. g (x) 5 3 x 2 4 ________ x 2 4 1 5; g (x) 5 3f (x 2 4) 1 511c. f (x) 5 (x 2 3 ) 2 2 2 x 2 5 ________ x 2 5 Th e graphs match.

12a.

f (2.9) 5 2, f (3) 5 3, f (3.1) 5 3

3

3

y

x

Section 1-6: Refl ections, Absolute Values, and Other Transformations

x 5 0

See page 980 for answers to CAS Problems 1 and 2.

55Section 1-8: Chapter Review and Test

Chapter Review and TestIn this chapter you saw how you can use functions as mathematical models algebraically, graphically, numerically, and verbally. Functions describe a relationship between two variable quantities, such as distance and time for a moving object. Functions de� ned algebraically are named according to the way the independent variable appears in the equation. If x is an exponent, the function is an exponential function, and so forth. You can transform the graphs of functions by dilating and translating them in the x- and y-directions. Some of these transformations re� ect the graph across the x- or y-axis or the line y x. A good understanding of functions will prepare you for later courses in calculus, in which you will learn how to � nd the rate of change of y as x varies.

You may continue your study of precalculus mathematics either with periodic functions in Chapters 5 through 9, which will probably be quite new to you, or with the � tting of other functions to real-world data in Chapters 2 through 4, which may be more familiar to you from previous courses.

� e Review Problems are numbered according to the sections of this chapter. Answers are provided at the back of the book. � e Concept Problems allow you to apply your knowledge to new situations. Answers are not provided, and, in some chapters, you may be required to do research to � nd answers to open-ended problems. � e Chapter Test is more like a typical classroom test your instructor might give you. It has a calculator part and a noncalculator part, and the answers are not provided.

Chapter Review and TestIn this chapter you saw how you can use functions as mathematical models

1- 8

R1. Punctured Tire Problem: For parts a–d, suppose that your car runs over a nail. � e tire’s air pressure, y, in pounds per square inch (psi), decreases with time, x, in minutes, as the air leaks out. A graph of pressure versus time is shown in Figure 1-8a.

42 1086

35y (psi)

x (min)

Figure 1-8a

a. Find graphically the pressure a� er 2 min. Approximately how many minutes can you drive before the pressure reaches 5 psi?

b. � e algebraic equation for the function in Figure 1-8a is

y 35 0. 7 x Make a table of numerical values of pressure

for times of 0, 1, 2, 3, and 4 min. c. Suppose the equation in part b gives

reasonable answers until the pressure drops to 5 psi. At that pressure, the tire comes loose from the rim and the pressure drops to zero. What is the domain of the function described by this equation? What is the corresponding range?

d. � e graph in Figure 1-8a gets closer and closer to the x-axis but never quite touches it. What special name is given to the x-axis in this case?

R1. Punctured Tire Problem: For parts a–d, suppose b. � e algebraic equation for the function in

Review Problems


Precalculus JournalIn this chapter you have been learning mathematics graphically, numerically, and algebraically. An important ability to develop for any subject you study is to verbalize what you have learned, both orally and in writing. To gain verbal practice, you should start a journal. In it you will record topics you have studied and topics about which you are still unsure. � e word journal comes from the same word as the French jour, meaning “day.” Journey has the same root and means “a day’s travel.” Your journal will give you a written record of your travel through mathematics.

Start writing a journal in which you can record things you have learned about precalculus mathematics and questions you have concerning concepts about which you are not quite clear.

You should use a bound notebook or a spiral notebook with large index cards for pages so that your journal will hold up well under daily use. Researchers use such notebooks to record their � ndings in the laboratory or in the � eld. Each entry should start with the date and a title for the topic. A typical entry might look like this sample.

Topic: Inverse of a Function 9/15I’ve learned that you invert a function by interchanging the variables. Sometimes an inverse is a relation that is n� a function. If it is a function, the inverse � y = f (x) isy = f –1 (x). At first, I thought this meant 1 ___ f (x) but after losing 5 points on a quiz , I realized that wasn’t correct. The graphs of f and f –1 are reflections of each other across the line y = x, like this:

y

x

Precalculus JournalIn this chapter you have been learning mathematics graphically, numerically,

1-7

Start writing a journal in which you can record things you have learned about precalculus mathematics and questions you have concerning concepts about

Objective

1. Start a journal for recording your thoughts about precalculus mathematics. � e � rst entry should include things such as these:

Sketches of graphs from real-world information

Familiar kinds of functions from previous courses

invert the graph of a function

as its potential usefulness to you

overcame

Any topics about which you are still unsure

1. Start a journal for recording your thoughts about

Problem Set 1-7



Class Time 1 __ 2 day

TE ACH I N G

Section Notes

Th is section introduces writing in a journal. In recent years, college and high school mathematics curricula and standardized tests have placed increased emphasis on having students verbalize mathematical concepts and ideas. Students will improve their writing skills if you require them to write periodically. In addition to assigning journal entry problems for homework, you may want to include writing questions on chapter tests.

Establish guidelines for how journal entries will be graded. Here is a sample.

• Answer the writing prompt. Write at least four complete sentences.

• Be specifi c. Use at least three new vocabulary words. Explain their meanings, and give examples.

• Give real-world examples for the applications of the new concepts and procedures.

• Exemplary journal entries receive extra credit. Vague, rambling statements receive no credit.

Have students read this section, including the sample journal entry. Discuss with students whether the sample meets all of your guidelines.

Assign Problem 1. Allow 15–20 minutes of class time for students to write. Th en call on volunteers to read their entries. Point out something in each volunteer’s entry that meets the guidelines, is unusually well written, or sheds new insight. Avoid negative comments.

You may want to assign the Section 1-8 Chapter Review problems the same day you explain Section 1-7 so students who fi nish their journal entries early can start working on those problems. Alternatively, you may want to assign the journal as homework following your test on Chapter 1.

Be sure your students realize that the journal is not the place to write their class notes. Rather, it is the place to record briefl y some things they can distill from their

notes, especially when they have fi nally mastered diffi cult topics.

Consider allowing students to use their journals on tests. For instance, the test on trigonometric identities in Chapter 7 could have a part where students use the properties and techniques for proving identities that they have recorded in their journals.

Periodically ask students to read from their journals. Th e last fi ve minutes of class are a


Chapter Review and TestIn this chapter you saw how you can use functions as mathematical models algebraically, graphically, numerically, and verbally. Functions describe a relationship between two variable quantities, such as distance and time for a moving object. Functions de� ned algebraically are named according to the way the independent variable appears in the equation. If x is an exponent, the function is an exponential function, and so forth. You can transform the graphs of functions by dilating and translating them in the x- and y-directions. Some of these transformations re� ect the graph across the x- or y-axis or the line y x. A good understanding of functions will prepare you for later courses in calculus, in which you will learn how to � nd the rate of change of y as x varies.

You may continue your study of precalculus mathematics either with periodic functions in Chapters 5 through 9, which will probably be quite new to you, or with the � tting of other functions to real-world data in Chapters 2 through 4, which may be more familiar to you from previous courses.

� e Review Problems are numbered according to the sections of this chapter. Answers are provided at the back of the book. � e Concept Problems allow you to apply your knowledge to new situations. Answers are not provided, and, in some chapters, you may be required to do research to � nd answers to open-ended problems. � e Chapter Test is more like a typical classroom test your instructor might give you. It has a calculator part and a noncalculator part, and the answers are not provided.

Chapter Review and TestIn this chapter you saw how you can use functions as mathematical models

1- 8

R1. Punctured Tire Problem: For parts a–d, suppose that your car runs over a nail. � e tire’s air pressure, y, in pounds per square inch (psi), decreases with time, x, in minutes, as the air leaks out. A graph of pressure versus time is shown in Figure 1-8a.

42 1086

35y (psi)

x (min)

Figure 1-8a

a. Find graphically the pressure a� er 2 min. Approximately how many minutes can you drive before the pressure reaches 5 psi?

b. � e algebraic equation for the function in Figure 1-8a is

y 35 0. 7 x Make a table of numerical values of pressure

for times of 0, 1, 2, 3, and 4 min. c. Suppose the equation in part b gives

reasonable answers until the pressure drops to 5 psi. At that pressure, the tire comes loose from the rim and the pressure drops to zero. What is the domain of the function described by this equation? What is the corresponding range?

d. � e graph in Figure 1-8a gets closer and closer to the x-axis but never quite touches it. What special name is given to the x-axis in this case?

R1. Punctured Tire Problem: For parts a–d, suppose b. � e algebraic equation for the function in

Review Problems


Precalculus JournalIn this chapter you have been learning mathematics graphically, numerically, and algebraically. An important ability to develop for any subject you study is to verbalize what you have learned, both orally and in writing. To gain verbal practice, you should start a journal. In it you will record topics you have studied and topics about which you are still unsure. � e word journal comes from the same word as the French jour, meaning “day.” Journey has the same root and means “a day’s travel.” Your journal will give you a written record of your travel through mathematics.

Start writing a journal in which you can record things you have learned about precalculus mathematics and questions you have concerning concepts about which you are not quite clear.

You should use a bound notebook or a spiral notebook with large index cards for pages so that your journal will hold up well under daily use. Researchers use such notebooks to record their � ndings in the laboratory or in the � eld. Each entry should start with the date and a title for the topic. A typical entry might look like this sample.

Topic: Inverse of a Function 9/15I’ve learned that you invert a function by interchanging the variables. Sometimes an inverse is a relation that is n� a function. If it is a function, the inverse � y = f (x) isy = f –1 (x). At first, I thought this meant 1 ___ f (x) but after losing 5 points on a quiz , I realized that wasn’t correct. The graphs of f and f –1 are reflections of each other across the line y = x, like this:

y

x

Precalculus JournalIn this chapter you have been learning mathematics graphically, numerically,

1-7

Start writing a journal in which you can record things you have learned about precalculus mathematics and questions you have concerning concepts about

Objective

1. Start a journal for recording your thoughts about precalculus mathematics. � e � rst entry should include things such as these:

Sketches of graphs from real-world information

Familiar kinds of functions from previous courses

invert the graph of a function

as its potential usefulness to you

overcame

Any topics about which you are still unsure

1. Start a journal for recording your thoughts about

Problem Set 1-7


S e c t i o n 1- 8PL AN N I N G

Class Time2 days (including 1 day for testing)

Homework AssignmentDay 1: R1–R6, T1–T28 Day 2 (aft er Chapter 1 Test):

Problems C1–C2 and Problem Set 2-1

Teaching ResourcesBlackline Masters

Problems R3, R5, R6Problem C2Problems T13–T16

Test 3, Chapter 1, Forms A and B

TE ACH I N G

Important Terms and ConceptsSine functionPeriodic functionPeriod of a periodic function

Section Notes

Th e last section of each chapter includes a set of review problems, numbered according to the sections in the chapter, and a sample chapter test, which can also be used for review. Most chapters also have a set of concept problems that extend the concepts in the chapter or introduce concepts for the next chapter. Concept problems make excellent group activities or projects. Answers for the review problems are provided in the back of the book so that students can monitor their own progress. Answers are not provided for the chapter test or concept problems.

You may want to assign Section 1-7 and Section 1-8 on the same day. Th is would also be a good time to assign any problems you will have due the day aft er a chapter test.

Section 1-7 (continued)

good time for this activity. It’s good to keep track of which students have participated so that all students have a turn.

Diff erentiating Instruction• Th is section will be challenging for

some students, but it will also help them tremendously to refl ect on and explain

what they’ve learned and what they do and do not understand fully.

• Because students may never have written journals in any course, they may need help getting started. Let ELL students write in their primary language. You can check their work by asking them to translate pieces.

1. Answer will vary. See page 980 for answers to Problem R1a–d.


d. Assuming that the height increases at the constant rate of 3 in. per year, does the weight also increase at a constant rate? Explain how you arrived at your answer.

e. What is a reasonable domain for t for the composite function W h?

f. Composite Functions Numerically Problem: Functions f and g are de�ned only for the values of x in the table.

x f(x) g(x)

3 2 4

4 6 5

5 3 8

6 5 3

Find these values, or explain why they are unde�ned: f ( g (3)), f ( g (4)), f ( g (5)), f ( g (6)), g ( f (6)), f ( f (3)), and g ( g (3)).

Two Linear Functions Problem: For parts g–j, let functions f and g be de�ned by

f (x) x 2 4 x 8

g (x) 2x 3 2 x 6 g. Plot the graphs of f, g, and f ( g (x)) on the

same screen. Sketch the results. h. Find f ( g (4)). i. Show that f ( g (3)) is unde�ned, even though

g (3) is de�ned. j. Calculate the domain of the composite

function f g and show that it agrees with the graph you plotted in part g.

R5. Figure 1-8i shows the graph of f (x) x 2 1 in the domain 1 x 2.

x5 5

5

5 y f(x)y

Figure 1-8i

a. On a copy of the �gure, sketch the graph of the inverse relation. Explain why the inverse is not a function.

b. Plot the graphs of f and its inverse relation on the same screen using parametric equations. Also plot the line y x. How are the graphs of f and its inverse relation related to the line y x? How are the domain and range of the inverse relation related to the domain and range of function f ?

c. Write an equation for the inverse of the function y x 2 1 by interchanging the variables. Solve the new equation for y in terms of x. How does this solution reveal that there are two di�erent y-values for some x-values?

d. On a copy of Figure 1-8j, sketch the graph of the inverse relation. What property does the function graph have that allows you to conclude that the function is invertible? What are the vertical lines at x 3 and at x 3 called?

x

5

5

55

y

Figure 1-8j


e. Earthquake Problem: Earthquakes happen when rock plates slide past each other. �e stress between plates that builds up over a number of years is relieved by the quake in a few seconds. �en the stress starts building up again. Sketch a reasonable graph showing stress as a function of time.

In 1989, a magnitude 7.1 earthquake struck Northern California, destroying houses in San Francisco's Marina district.

R2. For parts a–e, name the kind of function for each equation given. a. f (x) 3x 7 b. f (x) x 3 7 x 2 12x 5 c. f (x) 1. 3 x d. f (x) x 1.3 e. f (x) x 5 __________

x 2 2x 3

f. Name a pair of real-world variables that could be related by the function in part a.

g. If the domain of the function in part a is 2 x 10, what is the range?

h. In a �u epidemic, the number of people currently infected depends on time. Sketch a reasonable graph of the number of people infected as a function of time. What kind of function has a graph that most closely resembles the one you drew?

i. For Figures 1-8b through 1-8d, what kind of function has the graph shown?

x

y

x

y

x

y

Figure 1-8b Figure 1-8c Figure 1-8d

j. Explain how you know that the relation graphed in Figure 1-8e is a function but the relation graphed in Figure 1-8f is not a function.

x

y

x

y


R3. a. For functions f and g in Figure 1-8g, identify how the pre-image function f (dashed) was transformed to get the image function g (solid). Write an equation for g (x) in terms of x given that the equation of f is

f (x) ______

4 x 2 Con�rm the result by plotting the image

and the pre-image on the same screen on your grapher.

g

xf

5

–5

5–5

y

xf

y

Figure 1-8g Figure 1-8h

b. If g (x) 3f (x 4), explain how function f was transformed to get function g. Using the pre-image in Figure 1-8h, sketch the graph of g on a copy of this �gure.

R4. Height and Weight Problem: For parts a–e, the weight of a growing child depends on his or her height, and the height depends on age. Assume that the child is 20 in. when born and grows 3 in. per year. a. Write an equation for h(t) (in inches) as a

function of t (in years). b. Assume that the weight function W is given

by the power function W(h(t)) 0.004h(t ) 2.5 . Find h(5), and use the result to calculate the predicted weight of the child at age 5.

c. Plot the graph of y W(h(t)). Sketch the result.


Diff erentiating Instruction• Go over the review problems in class,

perhaps by having students present their solutions. You might assign students to write up their solutions before class starts.

• Because many cultures’ norms highly value helping peers, ELL students oft en help each other on tests. You can limit this tendency by making multiple versions of the test.

• Consider giving a group test the day before the individual test, so that students can learn from each other as they review, and they can identify what they don’t know prior to the individual test. Give a copy of the test to each group member, have them work together, then randomly choose one paper from the group to grade. Grade the test on the spot, so students know what they need to review further. Make this test worth 1 _ 3 the value of the individual test, or less.

• ELL students may not be used to the type of exam given in this course. Doing the chapter test in the book will help them get used to the format and type of questions they will be expected to answer.

• ELL students may need more time to take the test.

• ELL students will benefi t from having access to their bilingual dictionaries while taking the test.

PRO B LE M N OTES

R2a. LinearR2b. Polynomial (cubic)R2c. ExponentialR2d. PowerR2e. RationalR2f. Answers will vary; e.g., number of items manufactured and total manufacturing cost.R2g. 13 f (x) 37

R2h.

A quadratic function (with a negative x 2 -coeffi cient) fi ts this pattern.R2i. 1-8b: exponential; 1-8c: polynomial (probably quadratic); 1-8d: power

R2j. Figure 1-8e passes the vertical line test, so no x-value corresponds to more than one y-value. Figure 1-8f fails the vertical line test, so more than one y-value corresponds to the same x-value.

Problems R3 and R5 both require students to answer questions about functions based on their graphs. A blackline master for these problems is available in the Instructor’s Resource Book.

y

x


d. Assuming that the height increases at the constant rate of 3 in. per year, does the weight also increase at a constant rate? Explain how you arrived at your answer.

e. What is a reasonable domain for t for the composite function W h?

f. Composite Functions Numerically Problem: Functions f and g are de�ned only for the values of x in the table.

x f(x) g(x)

3 2 4

4 6 5

5 3 8

6 5 3

Find these values, or explain why they are unde�ned: f ( g (3)), f ( g (4)), f ( g (5)), f ( g (6)), g ( f (6)), f ( f (3)), and g ( g (3)).

Two Linear Functions Problem: For parts g–j, let functions f and g be de�ned by

f (x) x 2 4 x 8

g (x) 2x 3 2 x 6 g. Plot the graphs of f, g, and f ( g (x)) on the

same screen. Sketch the results. h. Find f ( g (4)). i. Show that f ( g (3)) is unde�ned, even though

g (3) is de�ned. j. Calculate the domain of the composite

function f g and show that it agrees with the graph you plotted in part g.

R5. Figure 1-8i shows the graph of f (x) x 2 1 in the domain 1 x 2.

x5 5

5

5 y f(x)y

Figure 1-8i

a. On a copy of the �gure, sketch the graph of the inverse relation. Explain why the inverse is not a function.

b. Plot the graphs of f and its inverse relation on the same screen using parametric equations. Also plot the line y x. How are the graphs of f and its inverse relation related to the line y x? How are the domain and range of the inverse relation related to the domain and range of function f ?

c. Write an equation for the inverse of the function y x 2 1 by interchanging the variables. Solve the new equation for y in terms of x. How does this solution reveal that there are two di�erent y-values for some x-values?

d. On a copy of Figure 1-8j, sketch the graph of the inverse relation. What property does the function graph have that allows you to conclude that the function is invertible? What are the vertical lines at x 3 and at x 3 called?

x

5

5

55

y

Figure 1-8j


e. Earthquake Problem: Earthquakes happen when rock plates slide past each other. �e stress between plates that builds up over a number of years is relieved by the quake in a few seconds. �en the stress starts building up again. Sketch a reasonable graph showing stress as a function of time.

In 1989, a magnitude 7.1 earthquake struck Northern California, destroying houses in San Francisco's Marina district.

R2. For parts a–e, name the kind of function for each equation given. a. f (x) 3x 7 b. f (x) x 3 7 x 2 12x 5 c. f (x) 1. 3 x d. f (x) x 1.3 e. f (x) x 5 __________

x 2 2x 3

f. Name a pair of real-world variables that could be related by the function in part a.

g. If the domain of the function in part a is 2 x 10, what is the range?

h. In a �u epidemic, the number of people currently infected depends on time. Sketch a reasonable graph of the number of people infected as a function of time. What kind of function has a graph that most closely resembles the one you drew?

i. For Figures 1-8b through 1-8d, what kind of function has the graph shown?

x

y

x

y

x

y

Figure 1-8b Figure 1-8c Figure 1-8d

j. Explain how you know that the relation graphed in Figure 1-8e is a function but the relation graphed in Figure 1-8f is not a function.

x

y

x

y


R3. a. For functions f and g in Figure 1-8g, identify how the pre-image function f (dashed) was transformed to get the image function g (solid). Write an equation for g (x) in terms of x given that the equation of f is

f (x) ______

4 x 2 Con�rm the result by plotting the image

and the pre-image on the same screen on your grapher.

g

xf

5

–5

5–5

y

xf

y

Figure 1-8g Figure 1-8h

b. If g (x) 3f (x 4), explain how function f was transformed to get function g. Using the pre-image in Figure 1-8h, sketch the graph of g on a copy of this �gure.

R4. Height and Weight Problem: For parts a–e, the weight of a growing child depends on his or her height, and the height depends on age. Assume that the child is 20 in. when born and grows 3 in. per year. a. Write an equation for h(t) (in inches) as a

function of t (in years). b. Assume that the weight function W is given

by the power function W(h(t)) 0.004h(t ) 2.5 . Find h(5), and use the result to calculate the predicted weight of the child at age 5.

c. Plot the graph of y W(h(t)). Sketch the result.

57

R4c.

R4d. No; the graph is curved.R4e. Sample answer: 0 t 13R4f. f  g (3) 5 6; f  g (4) 5 3; f  g (5) 5 f (8), which is undefi ned; f  g (6) 5 2; g   f (6) 5 8; f   f (3) 5 f (2) which is undefi ned; g  g (3) 5 5R4g.

R4h. f  g (4) 5 3R4i. f  g (3) is undefi ned because g(3) is not in the domain of f.R4j. Th e domain is 7 _ 2 x 11 __ 2 , which agrees with the graph.R5a. Th e inverse does not pass the vertical line test.

R5b.

Th e graphs are each other’s refl ections across the line. Th e domain of f corresponds to the range of the inverse relation. Th e range of f corresponds to the domain of the inverse relation.

Problem R5c can be solved directly on a CAS.

8

40

t

y

5

5

y

x

f

g

f g

y

x

5

�5

�5 5

�5 5

�5

5

x

y

R3a. Horizontal dilation by a factor of 3, vertical translation by 25;

g (x) 5  __________

4 2   x __ 3 2 2 5 R3b. Horizontal translation by 14, vertical dilation by a factor of 3

5

5

x

y

fg

Problem R4j can be defi ned algebraically with domain restrictions and the fi nal domain read from the graph of the composition. Th is is not a proof, but it does off er very strong evidence for the fi nal result.R4a. h(t) 5 3t 1 20R4b. h(5) 5 35 in.; W  h(5) 29 lb

Section 1-8: Chapter Review and Test

See page 981 for answers to Problems R1e, R5c and R5d.


C1. Four Transformations Problem: Figure 1-8l shows a pre-image function f (dashed) and a transformed image function g (solid). Dilations and translations were performed in both directions to get the graph for g. Figure out what the transformations were. Write an equation for g (x) in terms of f. Let f (x) x 2 with domain 2 x 2. Plot the graph of g on your grapher. Does your grapher agree with the � gure?

f g

x

5

5

55

y

Figure 1-8l

C2. Sine Function Problem: If you enter f 1 (x) sin(x) into your grapher and plot the graph, the result resembles Figure 1-8m. (Your grapher should be in radian mode.) � e function is called the sine function (pronounced “sign”), which you will study starting in Chapter 5.

1010

5

x

5

f1(x)

y

Figure 1-8m

a. � e sine function is an example of a periodic function. Why do you think this name is given to the sine function?

b. � e period of a periodic function is the di� erence in x-values from a point on the graph to the point where the graph � rst starts repeating itself. Approximately what does the period of the sine function seem to be?

c. Is the sine function an odd function, an even function, or neither? How can you tell?

d. On a copy of Figure 1-8m, sketch a vertical dilation of the sine function graph by a factor of 5. What is the equation of this transformed function? Check your answer by plotting the sine graph and the transformed image graph on the same screen.

e. Figure 1-8n shows a two-step transformation of the sine graph in Figure 1-8m. Name the two transformations. Write an equation for the transformed function, and check your answer by plotting both functions on your grapher.

1010

5

x

5

f1(x)

f2(x)

y

Figure 1-8n

f. Let f (x) sin x. What transformation would g (x) sin 1 _ 2 x be? Check your answer by plotting both functions on your grapher.

C1. Four Transformations Problem: Figure 1-8l a. � e sine function is an example of a periodic

Concept Problems


e. Plot these parametric equations on graph paper, using each integer value of t from 3 to 3. Con�rm the results by plotting on your grapher. Is y a function of x? Explain.

x 3 4 t 2

y t 4 f. Spherical Balloon Problem: �e table shows

the volume of a spherical balloon, v(x), in cubic meters, as a function of its radius, x, in meters.

x (m) v (x) ( m 3 )0.2 0.30.4 0.270.6 0.900.8 2.141.0 4.19

Plot function v on graph paper by plotting y v(x) for these points and connecting the points with a smooth curve. What evidence do you have that function v is invertible?

Plot the graph of y v 1 (x) on the same axes. What is the di�erence in the meaning of x as the input for function v and x as the input for function v 1 ? Explain why v 1 (v(x)) equals x.

Echo 1, the �rst communication satellite developed by NASA, was a giant metal balloon that �oated in orbit. It was used to bounce sound signals from one place on Earth to another.

g. Sketch the graph of a one-to-one function. Explain why it is invertible.

R6. a. On four copies of y f (x) in Figure 1-8k, sketch the graphs of these four functions:

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

y f x .

x5 5

5

5y f(x)

y

Figure 1-8k

b. Function f in part a is de�ned piecewise by

f (x) 3 _____

x 2 3 3

_____ 2 x 3

2 x 6 7 x 2

Plot the two branches of this function as f 1 (x) and f 2 (x) on your grapher. Does the graph agree with Figure 1-8k? Plot y f x by plotting f 3 (x) f 1 (x) and f 4 (x) f 2 (x). Does the graph agree with your result in the corresponding portion of part a?

c. Explain why functions with the property f ( x) f (x) are called odd functions and functions with the property f ( x) f (x) are called even functions.

d. Plot the graph of

f (x) 0.2 x 2 x 3________ x 3

Use a window that includes x 3 as a grid point. Sketch the result. Name the feature that appears at x 3.

R7. In Section 1-7 you started a precalculus journal. In what ways do you think keeping this journal will help you? How could you use the completed journal at the end of the course? What is your responsibility throughout the year to ensure that writing the journal has been a worthwhile project?


Problem Notes (continued)R5e.

Grapher graph agrees with graph on paper; Not a function because every x in the domain has multiple values of y.R5f.

Th e curve is invertible because it is increasing. As the input to v, x represents radius in meters. As the input to v 21 , it represents volume in cubic meters. If x 0 is a particular input to v, then   x 0 , v( x 0 ) is a point on the graph of v(x) . Plugging the output, v( x 0 ), into v 21 gives the point  v( x 0 ), v 21  v( x 0 ) on the graph of v 21 (x) . But the graph of v 21 (x) is just the graph of v(x) with all the x- and y-values exchanged, so this point is actually  v( x 0 ), x 0 . Th us, v 21  v( x 0 ) 5 x 0 .R5g. Since no y corresponds to more than one x in the original function, no x corresponds to more than one y in the inverse relation, so the inverse relation is a function. Sample graph:

y

x

A blackline master for Problem R6a is available in the Instructor’s Resource Book.

10

5

y

x

10

5

4

3

2

1

4321

Problem R6b can be defi ned in piecewise mode on a TI-Nspire or TI-Nspire CAS. Using separate f 1 (x) and f 2 (x) defi nitions is unnecessary.R6b. Th e graph agrees with Figure 1-8k; each of the graphs agrees with those in part a.R6c. Because power functions with odd powers satisfy the property f (2x) 5 2f (x) and power functions with even powers satisfy the property f (2x) 5 f (x)

R6d.

DiscontinuityR7. Answers will vary.

6

�4 4

y

x

See page 981 for answers to Problem R6a.


C1. Four Transformations Problem: Figure 1-8l shows a pre-image function f (dashed) and a transformed image function g (solid). Dilations and translations were performed in both directions to get the graph for g. Figure out what the transformations were. Write an equation for g (x) in terms of f. Let f (x) x 2 with domain 2 x 2. Plot the graph of g on your grapher. Does your grapher agree with the � gure?

f g

x

5

5

55

y

Figure 1-8l

C2. Sine Function Problem: If you enter f 1 (x) sin(x) into your grapher and plot the graph, the result resembles Figure 1-8m. (Your grapher should be in radian mode.) � e function is called the sine function (pronounced “sign”), which you will study starting in Chapter 5.

1010

5

x

5

f1(x)

y

Figure 1-8m

a. � e sine function is an example of a periodic function. Why do you think this name is given to the sine function?

b. � e period of a periodic function is the di� erence in x-values from a point on the graph to the point where the graph � rst starts repeating itself. Approximately what does the period of the sine function seem to be?

c. Is the sine function an odd function, an even function, or neither? How can you tell?

d. On a copy of Figure 1-8m, sketch a vertical dilation of the sine function graph by a factor of 5. What is the equation of this transformed function? Check your answer by plotting the sine graph and the transformed image graph on the same screen.

e. Figure 1-8n shows a two-step transformation of the sine graph in Figure 1-8m. Name the two transformations. Write an equation for the transformed function, and check your answer by plotting both functions on your grapher.

1010

5

x

5

f1(x)

f2(x)

y

Figure 1-8n

f. Let f (x) sin x. What transformation would g (x) sin 1 _ 2 x be? Check your answer by plotting both functions on your grapher.

C1. Four Transformations Problem: Figure 1-8l a. � e sine function is an example of a periodic

Concept Problems


e. Plot these parametric equations on graph paper, using each integer value of t from 3 to 3. Con�rm the results by plotting on your grapher. Is y a function of x? Explain.

x 3 4 t 2

y t 4 f. Spherical Balloon Problem: �e table shows

the volume of a spherical balloon, v(x), in cubic meters, as a function of its radius, x, in meters.

x (m) v (x) ( m 3 )0.2 0.30.4 0.270.6 0.900.8 2.141.0 4.19

Plot function v on graph paper by plotting y v(x) for these points and connecting the points with a smooth curve. What evidence do you have that function v is invertible?

Plot the graph of y v 1 (x) on the same axes. What is the di�erence in the meaning of x as the input for function v and x as the input for function v 1 ? Explain why v 1 (v(x)) equals x.

Echo 1, the �rst communication satellite developed by NASA, was a giant metal balloon that �oated in orbit. It was used to bounce sound signals from one place on Earth to another.

g. Sketch the graph of a one-to-one function. Explain why it is invertible.

R6. a. On four copies of y f (x) in Figure 1-8k, sketch the graphs of these four functions:

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

y f x .

x5 5

5

5y f(x)

y

Figure 1-8k

b. Function f in part a is de�ned piecewise by

f (x) 3 _____

x 2 3 3

_____ 2 x 3

2 x 6 7 x 2

Plot the two branches of this function as f 1 (x) and f 2 (x) on your grapher. Does the graph agree with Figure 1-8k? Plot y f x by plotting f 3 (x) f 1 (x) and f 4 (x) f 2 (x). Does the graph agree with your result in the corresponding portion of part a?

c. Explain why functions with the property f ( x) f (x) are called odd functions and functions with the property f ( x) f (x) are called even functions.

d. Plot the graph of

f (x) 0.2 x 2 x 3________ x 3

Use a window that includes x 3 as a grid point. Sketch the result. Name the feature that appears at x 3.

R7. In Section 1-7 you started a precalculus journal. In what ways do you think keeping this journal will help you? How could you use the completed journal at the end of the course? What is your responsibility throughout the year to ensure that writing the journal has been a worthwhile project?

59

C1. Horizontal dilation by 3, vertical dilation by 2, horizontal translation by 13, vertical translation by 25; g (x) 5 2   x 2 3 _____ 3

2 2 5

Problem C2 introduces the sine function, which students will study in Chapter 5. In the problem, students look at the periodic behavior of the function, determine whether it is odd or even, and explore transformations of the function. A blackline master for this problem is available in the Instructor’s Resource Book.C2a. Answers will vary. The function repeats itself periodically.C2b. About 6.3, or 2

C2c. Odd. It is its own reflection through the origin, so f (2x) 5 2f (x).C2d.

10�10

5

�5

y

x

y 5 5 sin(x)C2e. Horizontal translation 12, vertical translation 13; y 5 sin(x 2 2) 1 3 C2f. Horizontal dilation by 2

y

x

f g

�

1



For Problems T13–T16, sketch the indicated transformations on copies of Figure 1-8o. Describe the transformations.

T13. y 1 __ 2 f (x)

T14. y f 2 __ 3 x

T15. y f (x 3) 4 T16. �e inverse relation of f (x) T17. Explain why the inverse relation in

Problem T16 is not a function. T18. Let f (x)

__ x . Let g (x) x 2 4. Find f ( g (3)).

Find g f (3) . Explain why f g (1) is not a real number, even though g (1) is a real number.

T19. Use the absolute value function to write a single equation for the discontinuous function graphed in Figure 1-8p. Check your answer by plotting it on your grapher.

x4

4

8y

Figure 1-8p

T20. Plot these parametric equations on graph paper, using each integer value of t from

3 to 3. Con�rm the results by plotting them on your grapher. Is y a function of x? Explain.

x t 4

y 1 4 t 2

Wild Oats Problem: Problems T21–T28 refer to the competition of wild oats, a kind of weed, with the wheat crop. Based on data in A. C. Madgett’s book Applications of Mathematics: A Nationwide Survey, the percent loss in wheat crop, L(x), is approximately

L(x) 3.2 x 0.52

where x is the number of wild oat plants per square meter of land.

T21. Describe how L(x) varies with x. What kind of function is L?

T22. Find L(150). Explain verbally what this number means.

T23. Suppose the wheat crop is reduced to 60% of what it would be without the wild oats. How many wild oats per square meter are there?

T24. Let y L(x). Find an equation for y L 1 (x). For what kind of calculations would the equation y L 1 (x) be more useful than y L(x)?

T25. Find L 1 (100). Explain its real-world meaning. T26. Based on your answer to Problem T25, what

would be a reasonable domain and range for L? T27. Plot f 1 (x) L(x) and f 2 (x) L 1 (x) on the

same screen. Use equal scales for the two axes. Use the domain and range from Problem T25. Sketch the results along with the line y x.

T28. How can you tell that the inverse relation is a function?

T29. What did you learn as a result of taking this test that you didn’t know before?


Part 1: No calculators allowed (T1–T11)

For Problems T1–T4, name the type of function that each graph shows. T1. f(x)

x

T2. f(x)

x

T3. f(x)

x

T4. f(x)

x

T5. Which of the functions in Problems T1–T4 are one-to-one functions? What conclusion can you make about a function that is not one-to-one?

T6. When you turn on the hot water faucet, the time the water has been running and the temperature of the water are related. Sketch a reasonable graph of this function.

For Problems T7 and T8, tell whether the function is odd, even, or neither.

T7. f (x)

x

T8. f (x)

x

For Problems T9–T11, describe how the graph of f (dashed) was transformed to get the graph of g (solid). Write an equation for g (x) in terms of f.

T9.

x

f

g

5

5

55

y

T10.

f

g

x

5

5

55

y

T11.

f

g

x

5

5

55

y

Part 2: Graphing calculators allowed (T12–T29) T12. Figure 1-8o shows the graph of a function,

y f (x). Give the domain and the range of f.

x

f

55

y

5

Figure 1-8o

Part 1: No calculators allowed (T1–T11) T10. y

Chapter Test


Problem Notes (continued)T1. ExponentialT2. LinearT3. Polynomial (quadratic)T4. PowerT5. All except T3; Functions that are not one-to-one are not invertible; that is, their inverses are not functions.T6. Answers will vary. Sample Answer:

Time

Temperature

T7. OddT8. NeitherT9. Horizontal dilation by 2; g (x) 5 f   x __ 2 T10. Horizontal translation by 21, vertical translation by 15; g (x) 5 f (x 1 1) 1 5T11. Horizontal translation by 16, vertical dilation by 2; g (x) 5 2 f (x 2 6) T12. Domain: 22 x 7; range: 4 y 9

Blackline masters for Problems T13–T16 are available in the Instructor’s Resource Book.

Problems T13–T16 can be solved directly if students take the time to fi nd a domain-restricted algebraic representation of the given graph.T13. Vertical dilation by 1 __ 2

5

5

y

x

T14. Horizontal dilation by 3 __ 2

5

5

y

x

T15. Horizontal translation by 23, vertical translation by 24

5

5

y

x


For Problems T13–T16, sketch the indicated transformations on copies of Figure 1-8o. Describe the transformations.

T13. y 1 __ 2 f (x)

T14. y f 2 __ 3 x

T15. y f (x 3) 4 T16. �e inverse relation of f (x) T17. Explain why the inverse relation in

Problem T16 is not a function. T18. Let f (x)

__ x . Let g (x) x 2 4. Find f ( g (3)).

Find g f (3) . Explain why f g (1) is not a real number, even though g (1) is a real number.

T19. Use the absolute value function to write a single equation for the discontinuous function graphed in Figure 1-8p. Check your answer by plotting it on your grapher.

x4

4

8y

Figure 1-8p

T20. Plot these parametric equations on graph paper, using each integer value of t from

3 to 3. Con�rm the results by plotting them on your grapher. Is y a function of x? Explain.

x t 4

y 1 4 t 2

Wild Oats Problem: Problems T21–T28 refer to the competition of wild oats, a kind of weed, with the wheat crop. Based on data in A. C. Madgett’s book Applications of Mathematics: A Nationwide Survey, the percent loss in wheat crop, L(x), is approximately

L(x) 3.2 x 0.52

where x is the number of wild oat plants per square meter of land.

T21. Describe how L(x) varies with x. What kind of function is L?

T22. Find L(150). Explain verbally what this number means.

T23. Suppose the wheat crop is reduced to 60% of what it would be without the wild oats. How many wild oats per square meter are there?

T24. Let y L(x). Find an equation for y L 1 (x). For what kind of calculations would the equation y L 1 (x) be more useful than y L(x)?

T25. Find L 1 (100). Explain its real-world meaning. T26. Based on your answer to Problem T25, what

would be a reasonable domain and range for L? T27. Plot f 1 (x) L(x) and f 2 (x) L 1 (x) on the

same screen. Use equal scales for the two axes. Use the domain and range from Problem T25. Sketch the results along with the line y x.

T28. How can you tell that the inverse relation is a function?

T29. What did you learn as a result of taking this test that you didn’t know before?


Part 1: No calculators allowed (T1–T11)

For Problems T1–T4, name the type of function that each graph shows. T1. f(x)

x

T2. f(x)

x

T3. f(x)

x

T4. f(x)

x

T5. Which of the functions in Problems T1–T4 are one-to-one functions? What conclusion can you make about a function that is not one-to-one?

T6. When you turn on the hot water faucet, the time the water has been running and the temperature of the water are related. Sketch a reasonable graph of this function.

For Problems T7 and T8, tell whether the function is odd, even, or neither.

T7. f (x)

x

T8. f (x)

x

For Problems T9–T11, describe how the graph of f (dashed) was transformed to get the graph of g (solid). Write an equation for g (x) in terms of f.

T9.

x

f

g

5

5

55

y

T10.

f

g

x

5

5

55

y

T11.

f

g

x

5

5

55

y

Part 2: Graphing calculators allowed (T12–T29) T12. Figure 1-8o shows the graph of a function,

y f (x). Give the domain and the range of f.

x

f

55

y

5

Figure 1-8o

Part 1: No calculators allowed (T1–T11) T10. y

Chapter Test

61

T19. Horizontal translation by 14, vertical translation by 15, and vertical dilation by 3 of x ____

x   ;

y 5 3 x 2 4 ________ x 2 4

   1 5

T20.

Grapher graph agrees with graph on paper. Function, because it passes the vertical line test.

Problem T21, T24, and T25 can be found using substitution or a Solve command on a CAS.T21. L(x) varies proportionately to the 0.52 power of x. Power function.T22. L(150) 5 3.2(150 ) 0.52 5 43.3228… If there are 150 wild oat plants per square meter of land, the percentage loss to the wheat crop will be about 43%.T23. About 129 plants per square meterT24. x 5 3.2 y 0.52 ⇒ y 5   x ___ 3.2

1 ___ 0.52

If you know the percentage loss and want to fi nd the number of wild oat plants per square meter.T25. L 21 (100) 5   100 ___ 3.2

1 ___ 0.52 5 749.3963…

If the crop loss is 100% (i.e., the total crop is lost), there must have been about 750 wild oat plants (or more) per square meter. T26. Domain: 0 x 750; range: 0 y 100 T27.

T28. It passes the vertical line test. (Th e original function passes the horizontal line test—it is one-to-one.) T29. Answers will vary.

10

5

y

x

10

5

500

500

x

y

L�1(x)

L(x)

T16. Refl ection through the line y 5 x

5

5

y

x

Problem T17 can be solved directly if students fi nd the algebraic representation for Problems T13–T16.T17. Th e graph fails the vertical line test. (Th e pre-image graph fails the horizontal line test—it is not one-to-one.)T18. f  g (3) 5 f (5) 5  

__ 5 ;

g   f (3) 5 g    __

3 5 2 1; f  g (1) 5 f (23), which is not defi ned, because 23 is not in the domain of f.


Documents

Algebraic, Exponential, and Logarithmic Functions · ˜ is unit focuses on a study of functions, establishing the algebraic basis for precalculus. Linear, exponential, power, logarithmic,