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Functions and Graphs
One of the central principles of economics is that the value
of money is not constant; it is a function of time. Since for-
tunes are made and lost by people attempting to predict the
future value of money, much attention is paid to quantita-
tive measures like the consumer price index, a basic mea-
sure of inflation in various sectors of the economy. See page
159 for a look at how the consumer price index for housing
has behaved over time.
1.1 Modeling and
Equation Solving
1.2 Functions and
Their Properties
1.3 Twelve Basic
Functions
1.4 Building Functions
from Functions
1.5 Parametric
Relations and
Inverses
1.6 Graphical
Transformations
1.7 Modeling with
Functions
C H A P T E R 1
BIBLIOGRAPHY
For students: How to Solve It: A New Aspect of Mathematical Method , Second Edition, George Pólya. Princeton Science Library, 1991. Available through Dale Seymour Publications. Functions and Graphs , I. M. Gelfand, E. G. Glagoleva, E. E. Shnol. Birkhäuser, 1990. For teachers: The Language of Functions and Graphs , Shell Centre for Mathematical Education. Available through Dale Seymour Publications.
Chapter 1 Overview
In this chapter we begin the study of functions that will continue throughout the book.
Your previous courses have introduced you to some basic functions. These functions
can be visualized using a graphing calculator, and their properties can be described
using the notation and terminology that will be introduced in this chapter. A familiarity
with this terminology will serve you well in later chapters when we explore properties
of functions in greater depth.
70 CHAPTER 1 Functions and Graphs
Modeling and Equation Solving
What you’ll learn about
■ (^) Numerical Models ■ (^) Algebraic Models ■ (^) Graphical Models ■ (^) The Zero Factor Property ■ (^) Problem Solving ■ (^) Grapher Failure and Hidden Behavior ■ (^) A Word About Proof
... and why
Numerical, algebraic, and graphical models provide different methods to visualize, analyze, and understand data.
Numerical Models
Scientists and engineers have always used mathematics to model the real world and
thereby to unravel its mysteries. A is a mathematical structure
that approximates phenomena for the purpose of studying or predicting their behav-
ior. Thanks to advances in computer technology, the process of devising mathemati-
cal models is now a rich field of study itself,.
We will be concerned primarily with three types of mathematical models in this book:
numerical models , algebraic models , and graphical models. Each type of model gives
insight into real-world problems, but the best insights are often gained by switching
from one kind of model to another. Developing the ability to do that will be one of the
goals of this course.
Perhaps the most basic kind of mathematical model is the , in which
numbers (or data ) are analyzed to gain insights into phenomena. A numerical model
can be as simple as the major league baseball standings or as complicated as the net-
work of interrelated numbers that measure the global economy.
numerical model
mathematical modeling
mathematical model
EXAMPLE 1 Tracking the Minimum Wage
The numbers in Table 1.1 show the growth of the minimum hourly wage (MHW)
from 1955 to 2005. The table also shows the MHW adjusted to the purchasing power
of 1996 dollars (using the CPI-U, the Consumer Price Index for all Urban
Consumers). Answer the following questions using only the data in the table.
(a) In what five-year period did the actual MHW increase the most?
(b) In what year did a worker earning the MHW enjoy the greatest purchasing power?
(c) A worker on minimum wage in 1980 was earning nearly twice as much as a
worker on minimum wage in 1970, and yet there was great pressure to raise the
minimum wage again. Why?
SOLUTION
(a) In the period 1975 to 1980 it increased by $1.00. Notice that the minimum wage
never goes down, so we can tell that there were no other increases of this magni-
tude even though the table does not give data from every year.
(b) In 1970.
(c) Although the MHW increased from $1.60 to $3.10 in that period, the purchasing
power actually dropped by $0.57 (in 1996 dollars). This is one way inflation can
affect the economy. Now try Exercise 11.
Table 1.1 The Minimum
Hourly Wage
Minimum Purchasing Hourly Wage Power in Year (MHW) 1996 Dollars 1955 0.75 4. 1960 1.00 5. 1965 1.25 6. 1970 1.60 6. 1975 2.10 6. 1980 3.10 5. 1985 3.35 4. 1990 3.80 4. 1995 4.25 4. 2000 5.15 4. 2005 5.15 4. Source: www.infoplease.com
The ability to generate numbers from formulas makes an algebraic model far more
useful as a predictor of behavior than a numerical model. Indeed, one optimistic goal
of scientists and mathematicians when modeling phenomena is to fit an algebraic
model to numerical data and then (even more optimistically) to analyze why it works.
Not all models can be used to make accurate predictions. For example, nobody has
ever devised a successful formula for predicting the ups and downs of the stock mar-
ket as a function of time, although that does not stop investors from trying.
If numerical data do behave reasonably enough to suggest that an algebraic model
might be found, it is often helpful to look at a picture first. That brings us to graphical
models.
Graphical Models
A is a visible representation of a numerical model or an algebraic
model that gives insight into the relationships between variable quantities. Learning to
interpret and use graphs is a major goal of this book.
EXAMPLE 4 Visualizing Galileo’s Gravity Experiments
Galileo Galilei (1564–1642) spent a good deal of time rolling balls down inclined
planes carefully recording the distance they traveled as a function of elapsed time. His
experiments are commonly repeated in physics classes today, so it is easy to repro-
duce a typical table of Galilean data.
graphical model
72 CHAPTER 1 Functions and Graphs
EXPLORATION EXTENSIONS
Suppose that after the sale, the mer- chandise prices are increased by 25%. If m represents the marked price before the sale, find an algebraic model for the post-sale price, including tax.
EXPLORATION 1 Designing an Algebraic Model
A department store is having a sale in which everything is discounted 25%
off the marked price. The discount is taken at the sales counter, and then a
state sales tax of 6.5% and a local sales tax of 0.5% are added on.
1. The discount price d is related to the marked price m by the formula
d km , where k is a certain constant. What is k? 0.
2. The actual sale price s is related to the discount price d by the formula
s d td , where t is a constant related to the total sales tax. What is t?
7% or 0.
3. Using the answers from steps 1 and 2 you can find a constant p
that relates s directly to m by the formula s pm. What is p? 0.
4. If you only have $30, can you afford to buy a shirt marked $36.99? yes
5. If you have a credit card but are determined to spend no more than $100,
what is the maximum total value of your marked purchases before you
present them at the sales counter? $124.
Elapsed time (seconds) 0 1 2 3 4 5 6 7
Distance traveled (inches) 0 0.75 3 6.75 12 18.75 27 36.
What graphical model fits the data? Can you find an algebraic model that fits?
continued
SECTION 1.1 Modeling and Equation Solving 73
FIGURE 1.1 A scatter plot of the data from
a Galileo gravity experiment. (Example 4)
[–1, 18] by [–8, 56]
Table 1.4 Percentage ( F ) of Females in the Prison Population
t years after 1980
t 0 5 10 15 20 F 3.8 4.4 5.5 5.9 6. Source: U.S. Justice Department.
This insight led Galileo to discover several basic laws of motion that would eventually
be named after Isaac Newton. While Galileo had found the algebraic model to describe
the path of the ball, it would take Newton’s calculus to explain why it worked.
EXAMPLE 5 Fitting a Curve to Data
We showed in Example 2 that the percentage of females in the U.S. prison population
has been steadily growing over the years. Model this growth graphically and use the
graphical model to suggest an algebraic model.
SOLUTION Let t be the number of years after 1980, and let F be the percentage of
females in the prison population from year 0 to year 20. From the data in Table 1.3 we
get the corresponding data in Table 1.4:
SOLUTION A scatter plot of the data is shown in Figure 1.1.
Galileo’s experience with quadratic functions suggested to him that this figure was a
parabola with its vertex at the origin; he therefore modeled the effect of gravity as a
quadratic function:
d kt^2.
Because the ordered pair 1, 0.75 must satisfy the equation, it follows that k 0.75,
yielding the equation
d 0.75 t^2.
You can verify numerically that this algebraic model correctly predicts the rest of the
data points. We will have much more to say about parabolas in Chapter 2.
Now try Exercise 23.
A scatter plot of the data is shown in Figure 1.2.
This pattern looks linear. If we use a line as our graphical model, we can find an
algebraic model by finding the equation of the line. We will describe in Chapter 2
how a statistician would find the best line to fit the data, but we can get a pretty good
fit for now by finding the line through the points (0, 3.8) and (20, 6.7).
The slope is 6.7 3.8 20 0 0.145 and the y -intercept is 3.8. Therefore, the
line has equation y 0.145 x 3.8. You can see from Figure 1.3 that this line does a
very nice job of modeling the data.
Now try Exercises 13 and 15.
FIGURE 1.2 A scatter plot of the data
in Table 1.4. (Example 5)
FIGURE 1.3 The line with equation
y 0.145 x 3.8 is a good model for the data in Table 1.4. (Example 5)
[–5, 25] by [0, 8]
[–5, 25] by [0, 8]
SECTION 1.1 Modeling and Equation Solving 75
x ¬ 0 ¬or 2 x 5 ¬ 0 ¬or 3 x 2 ¬ 0
x ¬ 0 ¬or x ¬
¬or x ¬
Now try Exercise 31.
In Example 6, we used the important Zero Factor Property of real numbers.
The Zero Factor Property
A product of real numbers is zero if and only if at least one of the factors in the
product is zero.
FIGURE 1.4 The graph of
y x^2 4 x 10. (Example 7)
[–8, 6] by [–20, 20]
X=–5.741657 Y=
Zero
It is this property that algebra students use to solve equations in which an expression is
set equal to zero. Modern problem solvers are fortunate to have an alternative way to
find such solutions.
If we graph the expression, then the x -intercepts of the graph of the expression will be
the values for which the expression equals 0.
EXAMPLE 7 Solving an Equation: Comparing Methods
Solve the equation x^2 10 4 x.
SOLUTION
Solve Algebraically
The given equation is equivalent to x^2 4 x 10 0.
This quadratic equation has irrational solutions that can be found by the quadratic
formula.
x
and
x
While the decimal answers are certainly accurate enough for all practical purposes, it
is important to note that only the expressions found by the quadratic formula give the
exact real number answers. The tidiness of exact answers is a worthy mathematical
goal. Realistically, however, exact answers are often impossible to obtain, even with
the most sophisticated mathematical tools.
Solve Graphically
We first find an equivalent equation with 0 on the right-hand side:
x^2 4 x 10 0. We then graph the equation y x^2 4 x 10, as shown in
Figure 1.4.
We then use the grapher to locate the x -intercepts of the graph:
x 1.7416574 and x 5.741657.
Now try Exercise 35.
We used the graphing utility of the calculator to in Example 7. Most
calculators also have solvers that would enable us to for the same
decimal approximations without considering the graph. Some calculators have com-
puter algebra systems that will solve numerically to produce exact answers in certain
cases. In this book we will distinguish between these two technological methods and
the traditional pencil-and-paper methods used to.
Every method of solving an equation usually comes down to finding where an expres-
sion equals zero. If we use f x to denote an algebraic expression in the variable x , the
connections are as follows:
solve algebraically
solve numerically
solve graphically
76 CHAPTER 1 Functions and Graphs
SOLVING EQUATIONS WITH TECHNOLOGY
Example 7 shows one method of solv- ing an equation with technology. Some graphers could also solve the equation in Example 7 by finding the intersection of the graphs of y x^2 and y 10 4 x. Some graphers have built-in equation solvers. Each method has its advan- tages and disadvantages, but we rec- ommend the “finding the x -intercepts” technique for now because it most closely parallels the classical algebraic techniques for finding roots of equa- tions, and makes the connection between the algebraic and graphical models easier to follow and appreciate.
Fundamental Connection
If a is a real number that solves the equation f x 0, then these three statements
are equivalent:
1. The number a is a (or ) of the f x 0.
2. The number a is a of y f x .
3. The number a is an of the of y f x . (Sometimes the
point a , 0 is referred to as an x -intercept.)
x -intercept graph
zero
root solution equation
Problem Solving
George Pólya (1887–1985) is sometimes called the father of modern problem solving,
not only because he was good at it (as he certainly was) but also because he published the
most famous analysis of the problem-solving process: How to Solve It: A New Aspect of
Mathematical Method. His “four steps” are well known to most mathematicians:
Pólya’s Four Problem-Solving Steps
1. Understand the problem.
2. Devise a plan.
3. Carry out the plan.
4. Look back.
A Problem-Solving Process
Step 1—Understand the problem.
- Read the problem as stated, several times if necessary.
- Be sure you understand the meaning of each term used.
- Restate the problem in your own words. Discuss the problem with others if you
can.
- Identify clearly the information that you need to solve the problem.
- Find the information you need from the given data.
The problem-solving process that we recommend you use throughout this course will
be the following version of Pólya’s four steps.
ALERT
Many students will assume that all solu- tions are contained in their default view- ing window. Remind them that it is fre- quently necessary to zoom-out in order to see the general behavior of the graph, and then zoom-in to find more exact values of x at the intersection points.
ALERT
Students may not recognize the difference between a zero and a root. Functions have zeros, while one-variable equations have roots.
78 CHAPTER 1 Functions and Graphs
Support Graphically
Figure 1.5a shows that the point 440, 60.20 is on the graph of
y 25 0.08 x , supporting our answer to (a). Figure 1.5b shows that the point
850, 93 is on the graph of y 25 0.08 x , supporting our answer to (b). (We could
also have our answer by simply substituting in for each x
and confirming the value of p .)
Interpret
Sally earned $60.20 for driving 440 miles in one day. John drove 850 miles in one day
to earn $93.00. Now try Exercise 47.
supported numerically
It is not really necessary to show written support as part of an algebraic solution, but it
is good practice to support answers wherever possible simply to reduce the chance for
error. We will often show written support of our solutions in this book in order to high-
light the connections among the algebraic, graphical, and numerical models.
Grapher Failure and Hidden Behavior
While the graphs produced by computers and graphing calculators are wonderful tools for
understanding algebraic models and their behavior, it is important to keep in mind that
machines have limitations. Occasionally they can produce graphical models that misrep-
resent the phenomena we wish to study, a problem we call. Sometimes
the viewing window will be too large, obscuring details of the graph which we call
. We will give an example of each just to illustrate what can happen, but
rest assured that these difficulties rarely occur with graphical models that arise from real-
world problems.
EXAMPLE 9 Seeing Grapher Failure
Look at the graph of y 3 2 x 5 on a graphing calculator. Is there an
x -intercept?
SOLUTION A graph is shown in Figure 1.6a.
hidden behavior
grapher failure
(a) (b)
FIGURE 1.6 (a) A graph with a mysterious x -intercept. (b) As x approaches 2.5, the value
of 3(2 x 5) approaches. (Example 9)
X
Y 1 = 3/(2X-5)
ERROR 1500 150 15
Y 1
[–3, 6] by [–3, 3]
TECHNOLOGY NOTE One way to get the table in Figure 1.6b is to use the “Ask” feature of your graphing calculator and enter each x value separately.
(a)
(b)
FIGURE 1.5 Graphical support for the
algebraic solutions in Example 8.
[0, 940] by [0, 150]
X=850 Y=
[0, 940] by [0, 150]
X=440 Y=60.
continued
SECTION 1.1 Modeling and Equation Solving 79
The graph seems to show an x -intercept about halfway between 2 and 3. To confirm
this algebraically, we would set y 0 and solve for x :
2 x
0 2 x 5 ¬ 3
The statement 0 3 is false for all x , so there can be no value that makes
y 0, and hence there can be no x -intercept for the graph. What went wrong?
The answer is a simple form of grapher failure: The vertical line should not be there! As
suggested by the table in Figure 1.6b, the actual graph of y 3 2 x 5 approaches ∞ to the left of x 2.5, and comes down from ∞ to the right of x 2.5 (more on this
later). The expression 3 2 x 5 is undefined at x 2.5, but the graph in Figure 1.6a
does not reflect this. The grapher plots points at regular increments from left to right,
connecting the points as it goes. It hits some low point off the screen to the left of 2.5,
followed immediately by some high point off the screen to the right of 2.5, and it con-
nects them with that unwanted line. Now try Exercise 49.
EXAMPLE 10 Not Seeing Hidden Behavior
Solve graphically: x^3 1.1 x^2 65.4 x 229.5 0.
SOLUTION Figure 1.7a shows the graph in the standard 10, 10 by 10, 10 win-
dow, an inadequate choice because too much of the graph is off the screen. Our
horizontal dimensions look fine, so we adjust our vertical dimensions to 500, 500,
yielding the graph in Figure 1.7b.
We use the grapher to locate an x -intercept near 9 (which we find to be 9) and
then an x -intercept near 5 (which we find to be 5). The graph leads us to believe that
we are done. However, if we zoom in closer to observe the behavior near x 5, the
graph tells a new story (Figure 1.8).
In this graph we see that there are actually two x -intercepts near 5 (which we find to
be 5 and 5.1). There are therefore three roots (or zeros) of the equation x^3 1.1 x^2
65.4 x 229.5 0: x 9, x 5, and x 5.1. Now try Exercise 51.
You might wonder if there could be still more hidden x -intercepts in Example 10! We
will learn in Chapter 2 how the Fundamental Theorem of Algebra guarantees that there
are not.
A Word About Proof
While Example 10 is still fresh on our minds, let us point out a subtle, but very impor-
tant, consideration about our solution.
We solved graphically to find two solutions, then eventually three solutions, to the
given equation. Although we did not show the steps, it is easy to confirm numerically
that the three numbers found are actually solutions by substituting them into the equation.
But the problem asked us to find all solutions. While we could explore that equation
FIGURE 1.7 The graph of
y x^3 1.1 x^2 65.4 x 229.5 in two viewing windows. (Example 10)
[–10, 10] by [–500, 500] (b)
[–10, 10] by [–10, 10] (a)
FIGURE 1.8 A closer look at the graph
of y x^3 1.1 x^2 65.4 x 229.5. (Example 10)
[4.95, 5.15] by [–0.1, 0.1]
SECTION 1.1 Modeling and Equation Solving 81
QUICK REVIEW 1.1 (For help, go to Section A.2.)
Factor the following expressions completely over the real numbers.
1. x^2 16 ( x 4)( x 4) 2. x^2 10 x 25 ( x 5)( x 5) 3. 81 y^2 4 (9 y 2)(9 y 2) 4. 3 x^3 15 x^2 18 x 3 x ( x 2)( x 3) 5. 16 h^4 81 6. x^2 2 xh h^2 7. x^2 3 x 4 ( x 4)( x 1) 8. x^2 3 x 4 x^2 3 x 4 9. 2 x^2 11 x 5 10. x^4 x^2 20
SECTION 1.1 EXERCISES
In Exercises 1–10, match the numerical model to the corresponding graphical model ( a–j ) and algebraic model ( k–t ).
1. (d)(q) 2. (f)(r) 3. (a)(p) 4. (h)(o) 5. (e)(l) 6. (b)(s) 7. (g)(t) 8. (j)(k) 9. (i)(m) 10. (c)(n)
(k) y x^2 x (l) y 40 x^2
(m) y x 1 x 1 (n) y x 3
(o) y 100 2 x (p) y 3 x 2
(q) y 2 x (r) y x^2 2
(s) y 2 x 3 (t) y
x 2
[–5, 40] by [–10, 650] ( j)
[–3, 9] by [–2, 60] (i)
[–5, 30] by [–5, 100] (h)
[–1, 16] by [–1, 9] (g)
[–1, 7] by [–4, 40] (f)
[–1, 7] by [–4, 40] (e)
[–3, 18] by [–2, 32] (d)
[–4, 40] by [–1, 7] (c)
x 3 5 7 9 12 15 y 6 10 14 18 24 30
x 0 1 2 3 4 5 y 2 3 6 11 18 27
x 2 4 6 8 10 12 y 4 10 16 22 28 34
x 5 10 15 20 25 30 y 90 80 70 60 50 40
x 4 7 12 19 28 39 y 1 2 3 4 5 6
x 3 4 5 6 7 8 y 8 15 24 35 48 63
x 4 8 12 14 18 24 y 20 72 156 210 342 600
x 5 7 9 11 13 15 y 1 2 3 4 5 6
x 1 2 3 4 5 6 y 5 7 9 11 13 15
x 1 2 3 4 5 6 y 39 36 31 24 15 4
5. (4 h^2 9)(2 h 3)(2 h 3) 6. ( x h )( x h ) 9. (2 x 1)( x 5) 10. ( x^2 5)( x 2)( x 2)
[–2, 14] by [–4, 36] (a)
[–1, 6] by [–2, 20] (b)
Exercises 11–18 refer to the data in Table 1.6 below showing the percentage of the female and male populations in the United States employed in the civilian work force in selected years from 1954 to 2004.
11. (a) According to the numerical model, what has been the trend in females joining the work force since 1954? (b) In what 5-year interval did the percentage of women who were employed change the most? 1974 to 1979 12. (a) According to the numerical model, what has been the trend in males joining the work force since 1954? (b) In what 5-year interval did the percentage of men who were employed change the most? 1979 to 1984 13. Model the data graphically with two scatter plots on the same graph, one showing the percentage of women employed as a func- tion of time and the other showing the same for men. Measure time in years since 1954. 14. Are the male percentages falling faster than the female percent- ages are rising, or vice versa? vice versa 15. Model the data algebraically with linear equations of the form y mx b. Write one equation for the women’s data and another equation for the men’s data. Use the 1954 and 1999 ordered pairs to compute the slopes. 16. If the percentages continue to follow the linear models you found in Exercise 15, what will the employment percentages for women and men be in the year 2009? Women: 58.5%; men: 74% 17. If the percentages continue to follow the linear models you found in Exercise 15, when will the percentages of women and men in the civilian work force be the same? What percentage will that be? 2018, 69.9% 18. Writing to Learn Explain why the percentages cannot continue indefinitely to follow the linear models that you wrote in Exercise 15. The linear equations will eventually give percentages above 100% (for the women) and below 0% (for the men), neither of which is possible. 19. Doing Arithmetic with Lists Enter the data from the “Total” column of Table 1.2 of Example 2 into list L 1 in your calculator. Enter the data from the “Female” column into list L 2. Check a few computations to see that the procedures in (a) and (b) cause the cal-
Table 1.6 Employment Statistics
Year Female Male 1954 32.3 83. 1959 35.1 82. 1964 36.9 80. 1969 41.1 81. 1974 42.8 77. 1979 47.7 76. 1984 50.1 73. 1989 54.9 74. 1994 56.2 72. 1999 58.5 74. 2004 57.4 71. Source: www.bls.gov
culator to divide each element of L 2 by the corresponding entry in L 1 , multiply it by 100, and store the resulting list of percentages in L 3.
(a) On the home screen, enter the command: 100 L 2 L 1 → L 3.
(b) Go to the top of list L 3 and enter L 3 100 L 2 L 1 .
20. Comparing Cakes A bakery sells a 9 by 13 cake for the same price as an 8 diameter round cake. If the round cake is twice the height of the rectangular cake, which option gives the most cake for the money? rectangular cake 21. Stepping Stones A garden shop sells 12 by 12 square step- ping stones for the same price as 13 round stones. If all of the stepping stones are the same thickness, which option gives the most rock for the money? square stones 22. Free Fall of a Smoke Bomb At the Oshkosh, WI, air show, Jake Trouper drops a smoke bomb to signal the official beginning of the show. Ignoring air resistance, an object in free fall will fall d feet in t seconds, where d and t are related by the algebraic model d 16 t^2. (a) How long will it take the bomb to fall 180 feet? 3.35 sec (b) If the smoke bomb is in free fall for 12.5 seconds after it is dropped, how high was the airplane when the smoke bomb was dropped? 2500 ft 23. Physics Equipment A physics student obtains the following data involving a ball rolling down an inclined plane, where t is the elapsed time in seconds and y is the distance traveled in inches.
Find an algebraic model that fits the data. y 1.2 t^2
24. U.S. Air Travel The number of revenue passengers enplaned in the U.S. over the 14-year period from 1991 to 2004 is shown in Table 1.7.
Graph a scatter plot of the data. Let x be the number of years since 1991. (b) From 1991 to 2000 the data seem to follow a linear model. Use the 1991 and 2000 points to find an equation of the line and superimpose the line on the scatter plot. (c) According to the linear model, in what year did the number of passengers seem destined to reach 900 million? (d) What happened to disrupt the linear model?
Table 1.7 U.S. Air Travel
Passengers Passengers Year (millions) Year (millions)
1991 452.3 1998 612. 1992 475.1 1999 636. 1993 488.5 2000 666. 1994 528.8 2001 622. 1995 547.8 2002 612. 1996 581.2 2003 646. 1997 594.7 2004 697. Source: www.airlines.org
82 CHAPTER 1 Functions and Graphs
t 0 1 2 3 4 5 y 0 1.2 4.8 10.8 19.2 30
Increasing, except for a slight drop from 1999 to 2004.
12. (a) Decreasing, with some minor fluctuations. (^) 15. Women: y 0.582 x 32.3, men: y 0.211 x 83.
(a)
49. Exploring Grapher Failure Let y x^200 ^1 ^200. (a) Explain algebraically why y x for all x 0. (b) Graph the equation y x^200 ^1 ^200 in the window 0, 1 by 0, 1. (c) Is the graph different from the graph of y x? Yes (d) Can you explain why the grapher failed? 50. Connecting Algebra and Geometry Explain how the alge- braic equation x b ^2 x^2 2 bx b^2 models the areas of the regions in the geometric figure shown below on the left:
(Ex. 50) (Ex. 52)
51. Exploring Hidden Behavior Solving graphically, find all real solutions to the following equations. Watch out for hidden behavior. (a) y 10 x^3 7.5 x^2 54.85 x 37.95 3 or 1.1 or 1. (b) y x^3 x^2 4.99 x 3.03 3 52. Connecting Algebra and Geometry The geometric figure shown on the right above is a large square with a small square missing. (a) Find the area of the figure. x^2 bx (b) What area must be added to complete the large square? (c) Explain how the algebraic formula for completing the square models the completing of the square in (b). 53. Proving a Theorem Prove that if n is a positive integer, then n^2 2 n is either odd or a multiple of 4. Compare your proof with those of your classmates. 54. Writing to Learn The graph below shows the distance from home against time for a jogger. Using information from the graph, write a paragraph describing the jogger’s workout. y
x Time
Distance
x
x
b 2
b 2
x
x
b
b
Standardized Test Questions
55. True or False A product of real numbers is zero if and only if every factor in the product is zero. Justify your answer. 56. True or False An algebraic model can always be used to make accurate predictions. False; predictions are not always accurate.
In Exercises 57–60, you may use a graphing calculator to decide which algebraic model corresponds to the given graphical or numeri- cal model.
(A) y 2 x 3 (B) y x^2 5 (C) y 12 3 x (D) y 4 x 3
(E) y 8 x
57. Multiple Choice C 58. Multiple Choice E 59. Multiple Choice B 60. Multiple Choice A
Explorations
61. Analyzing the Market Both Ahmad and LaToya watch the stock market throughout the year for stocks that make significant jumps from one month to another. When they spot one, each buys 100 shares. Ahmad’s rule is to sell the stock if it fails to perform well for three months in a row. LaToya’s rule is to sell in December if the stock has failed to perform well since its purchase.
[0, 9] by [0, 6]
[0, 6] by [–9, 15]
84 CHAPTER 1 Functions and Graphs
x 1 2 3 4 5 6 y 6 9 14 21 30 41
x 0 2 4 6 8 10 y 3 7 11 15 19 23
49. (a) y ( x^200 )1/200^ x 200/200^ x^1 x for all x 0. (d) For values of x close to 0, x^200 is so small that the calculator is unable to distinguish it from zero. It returns a value of 0 1/200^ 0 rather than x. 50. The length of each side of the square is x b , so the area of the whole square is ( x b ) 2. The square is made up of one square with area x x x^2 , one square with area b b b^2 , and two rectangles, each with area b x bx. Using these four figures, the area of the square is x^2 2 bx b^2. 52. (b) ^ b 2
^ b 2
^ b 2
2 (c) x^2 bx ^ b 2
2 x ^ b 2
2 is the algebraic
formula for completing the square, just as the area ^ b 2
2 completes the area x^2 bx to form the area x ^ b 2
2 .
54. One possible story: The jogger travels at an approximately constant speed throughout her workout. She jogs to the far end of the course, turns around and returns to her starting point, then goes out again for a second trip. 55. False; a product is zero if at least one factor is zero.
The graph below shows the monthly performance in dollars (Jan–Dec) of a stock that both Ahmad and LaToya have been watching.
(a) Both Ahmad and LaToya bought the stock early in the year. In which month? March (b) At approximately what price did they buy the stock? $ (c) When did Ahmad sell the stock? (d) How much did Ahmad lose on the stock? About $ (e) Writing to Learn Explain why LaToya’s strategy was better than Ahmad’s for this particular stock in this particular year. (f) Sketch a 12-month graph of a stock’s performance that would favor Ahmad’s strategy over LaToya’s.
62. Group Activity Creating Hidden Behavior You can create your own graphs with hidden behavior. Working in groups of two or three, try this exploration. (a) Graph the equation y x 2 x^2 4 x 4 in the window 4, 4 by 10, 10. (b) Confirm algebraically that this function has zeros only at x 2 and x 2. (c) Graph the equation y x 2 x^2 4 x 4.01 in the win- dow 4, 4 by 10, 10 Same visually as the graph in (a). (d) Confirm algebraically that this function has only one zero, at x 2. (Use the discriminant.) (e) Graph the equation x 2 x^2 4 x 3.99 in the
window 4, 4 by [10, 10. Same visually as the graph in (a).
(f) Confirm algebraically that this function has three zeros. (Use the discriminant.)
Stock Index
Jan.
140 130 120 110 100
Feb.Mar.Apr.MayJuneJulyAug.Sept.Oct.Nov.Dec.
Extending the Ideas
63. The Proliferation of Cell Phones Table 1.8 shows the num- ber of cellular phone subscribers in the U.S. and their average monthly bill in the years from 1998 to 2004.
(a) Graph the scatter plots of the number of subscribers and the average local monthly bill as functions of time, letting time t the number of years after 1990. (b) One of the scatter plots clearly suggests a linear model in the form y mx b. Use the points at t 8 and t 14 to find a linear model. (c) Superimpose the graph of the linear model onto the scatter plot. Does the fit appear to be good? (d) Why does a linear model seem inappropriate for the other scatter plot? Can you think of a function that might fit the data better? (e) In 1995 there were 33.8 million subscribers with an average local monthly bill of $51.00. Add these points to the scatter plots. (f) Writing to Learn The 1995 points do not seem to fit the models used to represent the 1998–2004 data. Give a possible explanation for this.
64. Group Activity (Continuation of Exercise 63) Discuss the eco- nomic forces suggested by the two models in Exercise 63 and speculate about the future by analyzing the graphs.
Table 1.8 Cellular Phone Subscribers
Subscribers Average Local Year (millions) Monthly Bill ($) 1998 69.2 39. 1999 86.0 41. 2000 109.5 45. 2001 128.4 47. 2002 140.8 48. 2003 158.7 49. 2004 180.4 50. Source: Cellular Telecommunication & Internet Association.
SECTION 1.1 Modeling and Equation Solving 85
61. (c) June, after three months of poor performance
(e) After reaching a low in June, the stock climbed back to a price near $140 by December. LaToya’s shares had gained $2000 by that point. (f) Any graph that decreases steadily from March to December would favor Ahmad’s strategy over LaToya’s.
62. (b) Factoring, we find y ( x 2)( x 2)( x 2). There is a double zero at x 2, a zero at x 2, and no other zeros (since it is cubic). (d) Since the discriminant of the quadratic x^2 4 x 4.01 is negative, the only real zero of the product y ( x 2)( x^2 4 x 4.01) is at x 2. (f) Since the discriminant of the quadratic x^2 4 x 3.99 is positive, there are two real zeros of the quadratic and three real zeros of the product y ( x 2)( x^2 4 x 3.99). 63. (b) The linear model for subscribers as a function of years after 1990 is y 18.53 x 79.04. (d) The monthly bill scatter plot has an obviously curved shape that could be modeled more effectively by a function with a curved graph. Some possibilities: quadratic (parabola), logarithm, sine, power (e.g., square root), logistic. (We will learn about these curves later in the book.) (f) Cellular phone technology was still emerging in 1995, so the growth rate was not as fast, explaining the lower slope on the subscriber scatter plot. The new technology was also more expensive before competition drove prices down, explaining the anomaly on the monthly bill scatter plot. 64. One possible answer: The number of cell phone users is increasing steadily (as the linear model shows), and the average monthly bill is climbing more slowly as more people share the industry cost. The model shows that the number of users will continue to rise, although the linear model cannot hold up indefinitely.
SECTION 1.2 Functions and Their Properties 87
This uniqueness of the range value is very important to us as we study function behav-
ior. Knowing that f (2) 8 tells us something about f , and that understanding would be
contradicted if we were to discover later that f (2) 4. That is why you will never see
a function defined by an ambiguous formula like f ( x ) 3 x 2.
EXAMPLE 1 Defining a Function
Does the formula y x^2 define y as a function of x?
SOLUTION Yes, y is a function of x. In fact, we can write the formula in func-
tion notation: f ( x ) x^2. When a number x is substituted into the function, the
square of x will be the output, and there is no ambiguity about what the square of
x is. Now try Exercise 3.
Another useful way to look at functions is graphically. The
is the set of all points x , f x , x in the domain of f. We match domain
values along the x -axis with their range values along the y -axis to get the ordered pairs
that yield the graph of y f x .
EXAMPLE 2 Seeing a Function Graphically
Of the three graphs shown in Figure 1.11, which is not the graph of a function? How
can you tell?
SOLUTION The graph in (c) is not the graph of a function. There are three points
on the graph with x -coordinate 0, so the graph does not assign a unique value to 0.
(Indeed, we can see that there are plenty of numbers between –2 and 2 to which
the graph assigns multiple values.) The other two graphs do not have a compara-
ble problem because no vertical line intersects either of the other graphs in more
than one point. Graphs that pass this vertical line test are the graphs of functions.
Now try Exercise 5.
function y f ( x )
graph of the
FIGURE 1.11 One of these is not the graph of a function. (Example 2)
[–4.7, 4.7] by [–3.3, 3.3] (c)
[–4.7, 4.7] by [–3.3, 3.3] (b)
[–4.7, 4.7] by [–3.3, 3.3] (a)
TEACHING NOTE
Many students will find it helpful to remember that the value of y depends on the value of x , so x is the independent variable and y is the dependent variable.
TEACHING NOTE
In our definition of Function, Domain, and Range, the statement that R is the set of all output values (or range) means that the function is "onto". That is, we are describing a map from D , the domain, onto R , the range. We typically also con- sider a function defined from D into Y where R , the range, is a subset of Y. This idea is a bit subtle and we include it here only for your information.
OBJECTIVE
Students will be able to represent func- tions numerically, algebraically, and graphically, determine the domain and range for functions, and analyze function characteristics such as extreme values, symmetry, asymptotes, and end behavior.
MOTIVATE
Ask students why it is important to be able to tell whether a graph on a grapher is reasonable.
LESSON GUIDE
Day 1: Function Definition and Notation; Domain and Range; Continuity Day 2: Increasing and Decreasing Functions; Boundedness; Local and Absolute Extrema Day 3: Symmetry; Asymptotes; End Behavior.
Students who have not learned many func- tion concepts in previous courses might need to take more time in this section.
Vertical Line Test
A graph set of points x , y in the xy -plane defines y as a function of x if and only
if no vertical line intersects the graph in more than one point.
TEACHING NOTE
A familiarity with the concept of func- tions is very important preparation for calculus. The graphing calculator enables students to understand functions in a way that never existed before.
Domain and Range
We will usually define functions algebraically, giving the rule explicitly in terms of the
domain variable. The rule, however, does not tell the complete story without some con-
sideration of what the domain actually is.
For example, we can define the volume of a sphere as a function of its radius by the
formula
V r
r^3 (Note that this is “ V of r ”—not “ V • r ”).
This formula is defined for all real numbers, but the volume function is not defined for
negative r values. So, if our intention were to study the volume function, we would
restrict the domain to be all r 0.
88 CHAPTER 1 Functions and Graphs
Agreement
Unless we are dealing with a model (like volume) that necessitates a restricted
domain, we will assume that the domain of a function defined by an algebraic
expression is the same as the domain of the algebraic expression, the.
For models, we will use a domain that fits the situation, the relevant domain.
implied domain
EXAMPLE 3 Finding the Domain of a Function
Find the domain of each of these functions:
(a) f x x 3
(b) g x
x
x
(c) A s 3 4 s^2 , where A s is the area of an equilateral triangle with sides of
length s.
SOLUTION
Solve Algebraically
(a) The expression under a radical may not be negative. We set x 3 0 and solve
to find x 3. The domain of f is the interval –3, .
(b) The expression under a radical may not be negative; therefore x 0. Also, the
denominator of a fraction may not be zero; therefore x 5. The domain of g is
the interval 0, with the number 5 removed, which we can write as the union
of two intervals: 0, 5 5, .
(c) The algebraic expression has domain all real numbers, but the behavior being
modeled restricts s from being negative. The domain of A is the interval 0, .
Support Graphically
We can support our answers in (a) and (b) graphically, as the calculator should not
plot points where the function is undefined.
(a) Notice that the graph of y x 3 (Figure 1.13a) shows points only for x 3,
as expected.
TEACHING NOTE
Many students have difficulty with the concepts of domain and range. Provide opportunities to discuss and write down the domains and ranges of many func- tions. The examples and exercises in this section lend themselves to using coopera- tive groups in the classroom. Working in groups allows students to help each other and to take advantage of a variety of exploratory activities.
TEACHING NOTE
It may be necessary to review the notation for the union of two sets in the context of intervals.
WHAT ABOUT DATA? When moving from a numerical model to an algebraic model we will often use a function to approximate data pairs that by themselves violate our defini- tion. In Figure 1.12 we can see that several pairs of data points fail the ver- tical line test, and yet the linear func- tion approximates the data quite well.
NOTE The symbol “” is read “union.” It means that the elements of the two sets are combined to form one set.
FIGURE 1.12 The data points fail the
vertical line test but are nicely approximated by a linear function.
[–1, 10] by [–1, 11]
continued
