50 2 FUNCTIONS, LIMITS, AND THE DERIVATIVE

2.1 Functions and Their Graphs

Input x

.-------.J 1 L--_----;

f

Output

FIGURE 1 A function machine

A B

f(x) x

FIGURE 2 The function f viewed as a mapping

Functions , ., . . to know how his company s proflt lS related to Its pro-

A manufacturer woul~ Ilke d l'k t know how the size of the population of a cer­duction level; a biologlSt woul 1 e 0 r time' a psychologist would like to know the ~

f b . will change ove , tain culture 0 actena . time of an individual and the length of a vocabu_ relationship betwee~ the lealrdnlI~kg t know how the initial speed of a chemical reac-

. d hemist wou 1 eo . lary lIst; an a c f bstrate used. In each lllstance, we are concerned tion is related to th~ amount °d su ne quantity depend upon another? The relation_

. h th arne questIOn' How oes 0 h . WIt e s .. ' . nveniently described in mat ematlcs by using the ship between two quantitIes IS co >

concept of a function.

Function . . ' I that assigns to each element 10 a set A one and only one A functIOn IS a ru e

element in a set B.

The set A is called the do .', of the function. It is cu. tomary to denote a function by a letter of the alphabet, such a the letter f If\' is an elem nt in the domain of a functionf, then the element in B that/ as ociate~ \\ ith .r is written/ex) read "/ of i') }

and is called the value of / at x. The ~et comprising all the values a umed by y=/(x)asxtakesonallpossiblevalue in it<:.domain iscall dthe ng ofthefunc-

tion! We can think of a function/as a machine. The domain i the et of input (raw

material) for the machine, the rule de"cri be" how the input i to b proce ed, and the value(s) of the function are the output~ of the machine Figur", 1). ~

We can also think of a function f a~ a mapping in whi h a.1 lement x in the domain of/is mapped onto a unique element/C\,) in 13 (Figur

Notes

1. The output/ex) associated with an input.r is uniquc. To appr ... t the importance> of this uniqueness property, con"ider a rulc that as ociate l h each Item x in a department store its selling price y. Then, each x ll1L1~t nd to one and only one y. Notice, however, that different \ '~ may he a sO iated ~ ith the arne y. In , the context of the present example, thi.., "ays that dir~ r nt It m mil} ha\e the same price.

2. Alt~ough ~he s~ts A and B that appear in the definition f a function rna} be quite» arbItrary, III thIS book they will denote I.,ch of real nllmb r .

An example of a function may be taken from the familiar relation"hip between the area of a circle and 't d' L' . . ' . . I S fa IUS. ettIng .\ and \' denote the radlll and area ot a CI[-cle, respectively we hay fl ' , e, rom e ementary geometry,

y = 7TX 2 (ll

Equation (1) defines y as a f . f' . . f . " unctIOn 0 x ~Ince lor each admissibl value of.r (that r. or each nonnegative numb . " .

er repreSenllng the radllls 01 a ccrlalll ircle) there carre-

2.1 FUNCTIONS AND THEIR GRAPHS 51

~pond:" I~:'cc iscly on.e number y 7T x2 that gives the area of the circle. The rule defin-

IIlg this area functIOn" may be written as

lex) = 7TX2

(2) \0 C(~l11pute the area of a circle of radius 5 inches, we simply replace x in Equation (-) with the number 5. Thus, the area of the circle is

f(5) = 7T 52 = 257T

or 257T square inches .

In ge~leral, to e~aluate a function at a specific value of x, we replace x with that value, as Illustrated 111 Examples 1 and 2.

EXAMPLE 1 Let the functionfbe defined by the rulef(x) = 2x2 - X + 1. Find:

a. f( I ) b. f( -2) c. f(a) d. f(a + h)

Solution

a. f( I) = 2( ])2 - (1) + 1 = 2 - 1 + 1 = 2 h . . f'( - 2) . 2( - 2)2 - (- 2) + 1 = 8 + 2 + 1 = 11 c. f(([) = 2(a)2 - (a) + 1 = 2a2 - a + 1

d. J( (/ + h) = 2(a + h)2 - (a + h) + 1 = 2a2 + 4ah + 2h2 - a - h + 1

APPLIED EXAMPLE 2 Profit Functions ThermoMaster manufactures

an indoor-outdoor thermometer at its Mexican subsidiary. Management esti­

mate that the profit (in dollars) realizable by ThermoMaster in the manufacture and ale r x thermometer per week is

P(x) = -0.001x2 + 8x - 5000

find Therm 1a ter'., weekiy pr(.fit if its level of production is (a) 1000 ther­III meter p .... r \\' ek and (b) 2000 thClTIlOmetcrs per week.

a. 111 \\ eldy profit when the I~vel of production is 1000 units per week is found

b) e 'aluating the profit function P at x = 1000. Thus,

P(lOOO) - -0.001(1000)2 + 8(1000) - 5000 = 2000

or 2000. b. \Vhen the level of production is 2000 units per week, the weekly profit is

given by

P(2000) = -0.001 (2000? + 8(2000) - 5000 = 7000

or 7000.

Determining the Domain of a Function

Suppose we are given the function y = f(x). * Then, the varia~le x is called the

Indent, arjahle . The variable y, whose value depends on x, IS called the

ariable .

* (t i<; cu~tollwry to n: fer to a function I as I(.I) or hy the equation \' = I(.\) defining it.

Ide­

lent

52 2 FUNCTIONS, LIMITS, AND THE DERIVATIVE

. e need to find what restriction ~, if any

FIGURE 3

. of a functIOn, W ...' To determine the domaJO .' ble x. In many practical appllcatlon~, the

. d pendent valla ' 11 to be placed on the 10 e h nature of the problem, as I ustrated in

are . . dictated by t e domain of a functIOn IS

Example 3.

k gl'ng An open box is to be made from a XAMPLE 3 Pac a .' APPLIED E d 16 inches long and 10 Inches wIde by cutting

. rectangular piece of c.ardboabr

'nches) from each corner and foldin g up the . (x lOches Y Xl'

away identIcal squares . d pression that give~ the volume V of the box . F 3) FlO an ex" .

resultmg flaps (Igure : h domain of the functIon? as a function of x. What IS t e

x x

- - -

x

(a) The box is constructed by cutting x-by­x-inch squares from each corner.

(b) The dimensions of the resulting box are (10 - 2x)" by (16 2x)" by x".

Solution The dimensions of the box are (16 - 2x) inch es wide, and x inches high, so it volume (in cuhic inch s) i

V=j(x) = (16 - 2x)(l0 - 2x),r I.en ·th · I

.., = (160 - 52x + 4x-)x

= 4x3 - 52x2 + 160x

ng, ( I () - 2x) inch· 1\ en hy

I 'ht

Since the length of each side of the box Illust h~ great r tha r qua l to zero. we

see that

16 - 2x > 0

simultaneously; that is,

x<8

10 - 2,r > ()

<" -x - ) .\'>0

x

All three inequalities are satisfied ~imultanc()usly provided that 0 <x < 5. Thus. the domain of the function j is the i nten al 10, 51. •

I.n gene.ral, if a function is defined by a rule relating x t j(.\) \vithout 'ipeciti mentIOn of Its domain, it is understood that the dornai n wi 1\ c n ist of all \ alue 01

x for.~h~chj(x) is a real number. In this connection, you sh uld keep in Jl1lOd thJ[

(1) dlvl.sIOn by zero is not permitted and (2) the .... 4uare ro >t of a nelJatl\C number!' not defined. c

2.1 FUNCTIONS AND THEIR GRAPHS

EXAMPLE 4 Find the domain of each function.

3. J(X) :c v'X=- 1 b . . f ·(x) = 2 I 4 c . .r(x) = x 2 + 3 x -

SoLution

53

a. Since the square root of a negative number is undefined, it i, nece"ary that x - I > O. The inequality is satisfied by the set of real numbers x > I . Thu ~ the domain ofJis the interval [1,00). '

b. The only restriction on x is that x 2 - 4 be different from zero since di vi<) ion

by zero is not allowed. But (x2 - 4) = (x + 2)(x - 2) = 0 if x = -2 or

x = 2. Thus, the domain of J in this case consists of the interval s (-x, _ 2), ( - 2, 2), and (2, 00).

c. Here, any real number satisfies the equation, so the domain off is the set of all real numbers.

Graphs of Functions

IfJis a function with domain A, then corresponding to each real number x in A there is precisely one real number J(x). We can also express this fact by using

of real numbers. Write each number x in A as the first member of an ordered pair and each number f(x) corresponding to x as the second member of the ordered pair. This gives exactly one ordered pair (x, f(x» for each x in A. This observation leads to an . f:

Funct . r 'JDve D finition) A functl 'n f with domain A. is the set of all ordered pairs (x, f(x» where x

f long. to A.

Ob erve that the condition that there be one and only one number f (x) correspond­ing to ea\,;h number x in A translates mto the requirement that no two ordered pairs have the same first number.

Since ordered pairs of real numbers correspond to points in the plane, we have found a way to exhibit a function graphically.

Graph of a Function of One VariabLe

The t 0 a f . f is the set of all points (x, y) in the X)'-plane such that x is in the domain off and y = f(x).

Fi ure 4 shows the graph of a function! Observe that the 'y-coordinate of th~ p~int (X gy) on the graph off gives the height of that point (the dIstance above th~ .,-a( ~1 :)'

, . . . . h -f() gives the depth of the pomt .\,-' if f(x) is pOSItIve. If f(x) IS negatIve, t en x . ' f real (the distance below the x-axis). Also, observe that t~e domam ,of! 1 a et 0

numbers lying on the x-axis, whereas the range off lIes on the -' -aXIS.

54 2 FUNCTIONS, LIMITS, AND THE DERIVATIVE

\ .

-----/ ---" -- -- ----- /' I

(\ v) /"'f~ _ _ _ _ ____ .:. '.:.. 7j' y == f (x)

I I

Rang' ;_-L_.-_-I--_~I --r---X I ' •

-+ t t ~

7

c. FIGURE 4 The graph of f

I --+­

I I

'i

Domain FIGURE 5

h f a function r i~ ~hn\\,ll in Hgun;; - .

EXAMPLE 5 The grap 0 .

Wh t· the value of 1(3)? The \ aluc nff(: )'!

a. a IS . ' ( .) fn 111 th -a b. What is the height or depth at the pOint _.

. 'I h point

(5,1(5)) from the x-axis? . . c. What is the domain of I? The range nl

Solution \\ h n

c.

Much information about the graph of a un ,ti n 'm

point~ on its graph. Later on we wi II Ievd P 111 r techmques for graphing functions.

EXAMPLE 6 Sketch the graph of th' 1un 'li I I . II

Y = x2 + 1. What is the range of}'!

11 1 . n \ ud that

7,in\ui\e. n \I \ alu in

II 1\ '. {)' ou can r )111 I to

r linat of 71.

pi lllng a fc\\ i phi ti ated

th' quatlOn

Solution The domain of th t' . . in c unctl)n I t '111 r al I lImb'r . B) a ign· g several values to the variable \ '1 I . b . . ,Il 'omputm' th p )J1Jin~ \alu fllf

1 y, we 0 tam the following solution 1 til' 'quatl n ~

(- 3,10)

- 4

fe ~

jan-r;

for

. ." 12 +

f (3. 10) I

; ( I. 2)

-4- ... --t--t---t--+-- or - 4

FIGURE 6 The graph of y xl 1 is a parabola.

r

-4 -2 2

FIGURE 7 The graph of y = f(x) is obtained by graphing y = -x over ( - fJ.-, 0) and y = Yx over [0, %).

2.1 FUNCTIONS AND THEIR GRAPHS 55

x 3 -- 2 - I 0 2 3

Y 10 5 2 2 5 10

By plotting these points and then connectin the ' the graph of y = j'(x) wh' h ' g , m wIth a smooth curve, we obtain , ' IC IS a parabola (FIgure 6) T d '

./, we observe that x 2 > 0 'f ' ' 0 etermlOe the range of - ,) X IS any real number and so x 2 + I >

numbers x, We conclude that the range offis [1 00) - , I fo: all real result visually, ' ' The graph of j confIrm) this

EXPLORING WITH TECHNOLOGY

Letf(x) = x 2,

1. flot the graphs of F(x) = x 2 + c on the same set of axes for c = -2 -1 _1 0 2

1 2 ' , 2' , , , ,

2. PI~t the graphs of G(x) = (x + C)2 on the same set of axes for c = -2 -) _ 1

O 1 2 ' , 2'

.. 2' , .

3. Plot the graphs of H(x) = cx2 on the same set of axes for c = -2 -1 _1 _ 1 o I 1 1 2 ' , 2' of' , 4' 2' , '

4. Study the famil~ of graphs in parts 1-3 and describe the relationship between the graph of a functIOn f and the graphs of the functions defined by (a) y = f(x) + c, (b) y = f(x + c), and (c) y = cf(x) , where c is a constant.

A function that is defined by more than one rule is called a piecewise-defined

function .

• '., ~I"r...'.' 1. .~ j ,', ~~ ~ Y.;; - Sketch the graph of the function f defined by

f(x) = { -x Vx

if x < 0

if x > 0

Solution The function f is defined in a piecewise fashion on the set of all real numbers. In the subdomain (-00, 0), the rule for f is given by f(x) = -.r. The equation y = -xis a linear equation in the slope-intercept form (with slope -1 and intercept 0). Therefore, the graph of f corresponding to the ubdomain (-co, 0) is the half line shown in Figure 7. Next, in the subdomain [0, x ), the rule for fis given by f(x) = Vx. The values off(x) corresponding to x = 0, L 2, 3,4,

9, and 16 are shown in the following table:

x ° j(x) °

1

1

2 3 4

V2 V3 2

9

3

16

4

Using these values, we sketch the graph of the functionfas shown in Figure 7.

56 2 FUNCTIONS , LIMITS, AND THE DERIVATIVE

ank Deposits Ma~i son f'inanee Company APPLIED EXAMPLE 8 fB

f" . , 2 years from now 10 two separate locati ons:

o '0 c o

FIGURE 8

Years

We obtain the graph of the function y = f(x) by graphing y = V2x + 20

over [0, 2] and y = iX2 + 20 over (2, 5],

y

x=9 8

FIGURE 9 Since a vertical line passes through the curve at more than one point, we deduce that it is not the graph of a function,

b-ancho lces ' I ' h ' plans to open tWO I I developed commerCia center In t e City, A~ a ',dustrial complex and a new

My

, d' ' on '~ total depo~its during the nex t 5 year< an II ' plans a IS ) anslOn ' , result of these exp " d nce with the rule w 111 accor a are expected to gro v'2x + 20 iro "'= x ::: 2

f(X) == ~X2 + 20 if2 < x ::5 5 2

I t of money (in millions or doll ars) on depoSit ) ' the tota amoun ,

where y == f(X ,gIves == 0 corresponds to the present) , Sketch the graph of with Madison 111 year x (x

the function f , f' d f ned in a piecewise fashi on on the interval

. The functIOn IS e 1 " () ~ r-Solutlon , [0 2] the rule for f is given by I x = v 2x + 20, The [0,5]. In the subdoma1l1

d, ' t ' x == 0 I and 2 may be tabulated as fo ll ows:

values of f(x) correspon 1I1g a "

x o

j(x) 20

----I

21.4

2

22

, h bd mal'n (2 5] the rule for j'i-. given byj{x) = ~X2 t- 20, The val-Next, 111 t e su 0 " , '- , ,

f f( ) ndl'ng to x = 3 4 and .5 are shoWIl 111 th f IloWIng table: ues 0 x correspo ,- . , '-

x 3

j(x) 24,S

4

28

5

32,S

Using the values of f (x) in thi s table, we sketch th graph )f the function f as shown in Figure 8, I

The VerticaL-line Test Although it is true that every fun ction f of a variabk x ha h in the .\),-plane, i! is not true that every curve in the xy-planc is the graph . f a II n. For exampk. consider the curve depicted in Figure 9. This is thc graph f l qUHlion \, 1 = x.ln general, the graph of an l'quatioll i ~ the .... ct of all ordered pair r. y) thdt "at I f) the given equation. Observe that the point~ (9, - 3) and (t) , ) b Hh 'on the cun'e. Thi ~ implies that the number x = 9 is as~oc i ated \\' i th 1H'() n U m b r . - - 3 and r == ~ , But this clearly violates the uniquene~~ propcrty or a fun tion. Thus, we conclude that the curve under consideration cannot bc the !!.raph of a fUll ,ti n.

This example suggests the following ,crtical-lir c t t for t 'nni ning when ~ curve is the graph of a function.

Vertical-Line Test A eurve in the xy-plane I'S the ' h t' ,. . , . '. , grap 0 a unction ), = j(x vertical hne mtersects it in at most one point.

if, IlLI onl y if each a

pan) IOns:

. sa

rval The

'al-

a

In the

FIGURE 10 The vertical-line test can be used to deter­mine which of these curves are graphs of functions.

2.1 FUNCTIONS AND THEIR GRAPHS 57

EXAMPLE 9 Determine which of the curves shown in Figure 10 are the graph, o[ [unctions o[ x .

Solution The curves depicted in Figure lOa, c, and d are graph~ of function~ because each curve satisfies the requirement that each vertical line inter)ects the curve in at most one point. Note that the vertical line shown in Figure lOc does not intersect the graph because the point on the x-axis through which this line passes does not lie in the domain of the function. The Curve depicted in Fig­ure lOb is not the graph of a function because the vertical line shown there inter­sects the graph at three points.

y

--+---------~--~--------~x

(a)

Y I

-------------'t----+--------.~ t

(c)

y

--------------t----,~"'------------~ x

(b)

(d)

y

/ /

/

,-

. __ --I-___ ~/_f_l_ __ ---.. .r

2.1 Self-Check Exercises

1. Let! be the function defined by

Yx + 1 f(x) = x

a. Find the domain off c. Compute!Ca + h). b. Compute!(3). .

2. Statistics show that more and more motorists are pumpmg their own gas. The following function gives self-serve sales as a percent of all U.S. gas sales:

{6t + 17

f(t) = 15.98(t - 6) 1/4 + 53 if 0 -< t -< 6

if 6 < t -< 20

Here t is measured in years, with t = 0 corresponding to the beginning of 1974.

a. Sketch the graph of the function f , b. What percent of all gas sales at the beginning of 1978 were

self-serve? At the beginning of 1994?

•• )(/tl/l.{. l III )L.l, "1' IIUll III

3. Letf(x) = Y2x + 1 + 2. Determine whether the point (-L 6) lies on the graph off

Solutions to Se(f-Check Exercises 2.1 CGn be found 011 page 63.

58 2 FUNCTIONS, LIMITS, AND THE DERIVATIVE

2.1 Concept Questions

1. a. What is a function? h. What is the domain of a function? The range of a func-

tion? c. What is an independent variable? A dependent variable?

2.1 Exercises

1. Let f be the function defined by f(x) = 5x + 6. Find f(3) ,

f(-3),f(a) , f(-a) , andf(a + 3).

2. Let f be the function defined by f(x) = 4x - 3. Find f(4),

f(±),f(O),f(a), andf(a + 1).

3. Let g be the function defined by g(x) = 3x2 - 6x - 3. Find

g(O), g( -1), g(a), g( -a), and g(x + 1).

@. Let h be the function defined by hex) = x3 - x2 + X + 1.

Find h( - 5), h(O), h(a), and h( -a).

5. Let f be the function defined by f(x) = 2x + 5. Find f(a + h),f( -a),f(a2),f(a - 2h), andf(2a - h).

6. Let g be the function defined by g(x) = - x2 + 2x. Find

g(a + h),g(-a),g(va),a + g(a),and -1-.

g(a)

7. Let s be the function defined by s(t) = 2t . t 2 - 1

Find s(4), s(O), sea), s(2 + a), and set + 1).

8. Let g be the function defined by g(u) = (3u - 2)3/2 F d g(l), g(6), gC31

), and g(u + 1). . m

9. Letfbe the function defined by f(t) = 2t2

. Findf(2) f(a),f(x + 1), andf(x - 1). Yt - 1 '

foJ Letfbe the function defined by f(x) = 2 + 2Y5 - . f( -4),f(l),fC41) , andf(x + 5). x. Fmd

11. Let f be the function defined by

f(x) = {X2 + 1 v'X

Findf( -2),f(0), andf(l).

12. Let g be the function defined by

1

if x < 0

if x > 0

g(x) = -"2x + 1 if x < 2

Yx - 2 if x > 2

Find g( - 2), g(O), g(2), and g( 4).

2. a. What is the graph of a fun ction ? Use a drawing to illu~ d

. d h .)-trate the graph , the oma.ln , an t e range of a function .

b. If you are g i ~en a curve Jl1 th.e xy pl a.ne, how can you tell if the graph IS th at 01 a fun ction.! def Ined by y -- I(x)?

13. Let f be the function defined by

I 1

- x ~ + 3 i f x < I f(x) = 2

if x .> I

Find f( - 1). f(O), Ie I ), and /(2).

L"t et be the function defined by (1)f L f

2 1 I - x if x

f(x) - I if x 1

I - x

Find f(O),f(l). andj(2) .

15. Refer to the graph of the fun ction j in it llov/i ng figure.

1

-1-+---+-----+----+--1- -+ --..... 6 - 1

x

-2

16.

In gr

17.

18.

19.

20.

In

21.

23.

25.

27.

a. Find the value of f(O). 29.

h. F~nd the value of x for which (i ) f(x ) = 3 and (ii) f(x) ::: o. c. Fmd the domain off 31. d. Find the range off

Ill! -n. t 11

)'.

::::: 0,

J6. Refer to the graph of the function f in the following rigure.

\ '

y =f(x)

.3 4 5 6 7 8 9 10

a. Find the value off(7).

b. Find the values of x corresponding to the point(s) on the

graph off located at a height of 5 units from the x-axis.

c. Find the point on the x-axis at which the graph off crosses it. What is the value of f (x) at this point?

d. Find the domain and range off

In Exercises 17-20, determine whether the point Lies on the graph of the function.

~-

17. (2, V3); g(x) = Vx2 - 1

x + 1 18. (3,3);f(x) = ,, /2 +2

vx + 7

It - 11 19. (-2, -3);f(t) = t + 1

( 1 ) It + 11

~O. -3, -13 ; h(t) = t 3 + 1

In Exercises 21-34, find the domain of the function.

21. f(x) = x2 + 3 22. f(x) = 7 - x2

3x + 1 2x + 1 23. f(x) = 2 24. g(x) = x - I

x

25. f(x) = Vx 2 + 1 26. f(x) = Vx - 5

27. f(x) = Vs - x 28. g(x) = V2X2 + 3

x 29·f(x) = x2 - 1

31. f(x) = (x + 3)3/2

I 30. f(x) = x 2 + X - 2

32. g(x) = 2(x - 1)5/2

2.1 FUNCTIONS AND THEIR GRAPHS 59

33. /(x) =- V; x x - 4

:14 j '( ) _ \IX -=--1 u. x---- __ (x t 2)(x - 3)

35. Letfbe a function derined by the rule/(x) =- x 2 - X - 6. a. Find the domain or.r

b. Compute/ex) ror x = - 3, - 2, ' 1, 0, t 1,2,3. c. Use the results obtained in parts (a) and (b) to sketch the

graph of.r

36. Let f be a function defined by the rule j(x) = 2X2 + X - 3. a. Find the domain off

b. Computef(x) for x = - 3, - 2, - I , -~, 0, 1,2,3. c. Use the results obtained in parts (a) and (b) to sketch the

graph off

In Exercises 37-48, sketch the graph of the function with the given ruLe. Find the domain and range of the function.

37. f(x) = 2X2 + I 38. f(x) = 9 - x2

39·f(x) = 2 + Vx 40. g(x) = 4 - vx 41. f(x) = VI -x 42. f(x) = Vx - 1

43. f(x) = Ixl - I 44. f(x) = Ixl + 1

{

X 45. f(x) = 2x + 1

f 4 - x 46, fix) -= . l2x - 2

(-x + I 47 ((.,'\ -- )

• J ,Ai - lx2 - I

-x - 1

48. f(x) =

if x < 0 if x >- 0

if x < 2

if x >- 2

if x < 1

if x > 1

if x < -1

if -1 < x < 1

if x> 1

In Exercises 49-56, use the vertical-line test to determine whether the graph represents y as a function of x.

49. )' 50.

----\--H--~.r -----t-- ---~ x

...---

60

51.

53.

55.

LIMITS, AND THE DERIVATIVE

2 FUNCTIONS,

52.

_---t-?L--~ X

y 54.

____ --~----~x

56.

______ ~----+x

y

0--

0--

0- -

_-l-----~x

y

_~---+--t-.-,. x

y

I x --+-+---.-,.

1 i I

6

57. The circumference of a circle is given by C(r) = }"Tr, wh~re r is the radius of the circle. What is the circumlerence a a

circle with a 5-in. radius?

58. The volume of a sphere of radius r is given by V(r) = 11Tr3. Compute V(2.1) and V(2). What does the quantity V(2.1) -

V(2) measure?

5t). GROWTH or A CANCEROUS TUMOR The volume of a spherical cancerous tumor is given by the function

4 3 V(r) = -1Tr

3

where r is the radius of the tumor in centimeters. By what factor is the volume of the tumor increased if its radius is doubled?

6n. GROWTH OF A CANCEROUS TUMOR The surface area of a spheri­cal cancerous tumor is given by the function

S(r) = 41Tr2

where r is the radius of the tumor in centimeters. After exten­sive chemotherapy treatment, the surface area of the tumor is reduced by 75%. What is the radius of the tumor after treatment?

8 7 6

5 4

3 2

' f 11C (,ollowi ng gr:lphs !'.how the USIC

SAl r r; or rl ' r l II . ' • (in hillions of doll'lr"i) hy format . . cordcd tllllSIC .

S'lles y of prel e (. "ll'S) with I () cOITe!'.poncltng to .' . f ' > I In YCl . . as a function 0 lttn l.:

1985.

y (billions of dollars) I C I)"

/

----......... _-- C a eltes

-f-7 8 C) t:::~f----+--.t-- t·_-+-

3 4 5 (1 a

I ' I ol pr recorded cn ette h t e'lrs were t 11: S.I a. In way l , • I '

ter th '1I1 those 01 prerl.:cold d C orea l ' '. d 'd C D b • were the ales 01 pn.:rec r greater

b. In what yc,lrS ') h' those or prerecorded cas He .

t an 'r' the 'Ii ( f prcr orde I cn ctte the c In what year we l:' .

• . I ' I' !')n:n:cllrde I C I ? h tllnale the levelot same as I lose 0 ,

sales in each rormat Ht that lim

(J1I" c. • , II I

'( h' filII wing 'r, ph h )\\ the rallo o~ fr m 19EO thr 1 I 'h _000.

61. THf Gr ,,1'0 GAP

women's earnin!.!s t(l m n'

0.8

0.75

0.7

0.65

0.6

. 10 1ll1.:J1

. 'arnm' ) y (ratio of wom n . ( • 07 )

(0.0.61 )

.. (_0. O. ) ( 10. O.5lJ)

- > t (vear-l-

0 o . 10 _0

a. Write the ruk fur th flln tion I '1\ ing the ratio of , , . ) '1 r I \\ i I h I ~ 0 corre-women S earnll1g tll men 111 t;. ,

sponding to IlJ60. . _ ' ere~cI' Hint: The functilln j'i defined plece .. \ I '.100 I linear ll\ of four suointcrvals. d' ~l

h. In what decade( s) was th g 'nder gap C\pan tn:

Shrinking'! c. Refer to part (h). Ilow 1',

ratio/year ) ex pand i ng or decades?

h I 'r oap {Wi t \\ at' gem I,; = h · k' . , 'h of (h~'~ nn IIlg III e,IL

63.

1.2

1.0

0.8

0.6

64.

65.

.66.

67.

of

the

1.2

1.0

0.8

0.6

I Hr c'r rl' ,., TN IlI(I\TION Th' ,. II . . . , ,.,' ,,' c 0 oWlllg graph sIH)\\S the 1,1110 of bachelor s dcurces C'11'11Cd b

. b " y women to men from 1960 through 1990.

y (ratio of degrees earned)

(30, I. f) -"

(20,0.95)

(0.0.58)

10 20 30 t (years)

3. Write the rule for the function! giving the ratio of bache­lor's degrees earned by women to men in year t, with 1 = 0 corre ponding to 1960. Hin:: The function! is defined piecewise and is linear over each of t\\ a subinten als.

b. How fast was the ratio changing in the period from 1960 to 1980? From 1980 to 1990?

c. In what year (approximately) was the number of bache­lor" s degrees earned by women equal for the first time to that earned by men?

(I [ I I

64. (0 «iU pno Fu enD The consumption function in a cer-tain economy is given by the equation

C{y) = 0.7Sy + 6

where C(y) is the personal cOll'.nmptio!1 expenditure, y is the dispo~able personal income, and both C{y) and yare mea­sured in billions of dolJar'. rind em). C(SO). and C(lOO).

65. SALES TAXES In a certai n state, the ~aje" tax T on the amount of taxable goods is 60/1: of the vulue of the goods purchased (x), where both T and x are measured in dollars. 3. Expre s T as a function of x. b. Find T(200) and T(5.65).

66. SURFACE AREA OF A SI~Gl[-(fL lfD Opr, The surface area S of a single-celled organism may be found by multiplying 47T times the square of the radius r of the cell. Express S as a function of r.

67. FRIEND'S Ru F Friend's rule, a method for calculating pedi­atric drug dosages, is based on a child's age. If a denotes the adult dosage (in milligrams) and jf t is the age of the child (in years), then the child's dosage is given by

2 D(t) = 25 ta

If the adult dose of a substance is 500 mg, how much should a 4-yr-old child receive?

2.1 FUNCTIONS AND THEIR GRAPHS 61

COLAs Social S . . . . . . ecunty reCIpients receive an automatic cost-

of-Ilv~n~ a~Justment (COLA) once each year. Their monthly ?en.eflt IS m~reased by the amount that consumer prices 1I1~leas.ed dunng the preceding year. Suppose that consumer pnces lllcreased b~ 5.3% during the preceding year. 3. ~x~r~ss the adJuste? monthly benefit of a Social Security

Ie~Ipient as a functIon of his or her current monthly ben­efIt.

b. If Harrington's monthly Social Security benefit is now $620, what will be his adjusted monthly benefit?

69. BROADBAND INTERNET HOUSEHOLDS The number of U.S. broad­band Internet households stood at 20 million at the beginning of 2002 and is projected to grow at the rate of 7.5 mil/ion households per year for the next 6 yr.

3. Find a function!(t) giving the projected U.S. broadband Internet households (in millions) in year t, where t = 0 corresponds to the beginning of 2002. Hi nt: The graph off is a straight line.

h. What is the projected size of U.S. broadband Internet households at the beginning of 2008?

SOlin t. Strategy Analyt lls Inc.

70. (OSI OF RENTING A TRUCK Ace Truck leases its 10-ft box truck at $30/day and $.45/mi, whereas Acme Truck leases a simi­lar truck at $25/day and $.50/mi. 3. Find the daily cost of leasing from each company as a

function of the number of miles driven. h. Sketch the graphs of the two functions on the same set of

axes.

c. Which company should a customer rent a truck from for 1 day if she plans to drive at most 70 mi and wishes to minimize her cost?

~ l, t 1'0. ... ·1( '" . r i ' A new machine was purchased by

72.

National Textile for $120,000. For income tax purposes. the machine is depreciated linearly over 10 yr; that is, the book value of the machine decreases at a constant rate, so that at the end of 10 yr the book value is zero. 3. Express the book value of the machine (V) as a function

of the age, in years, of the machine (n).

b. Sketch the graph of the function in part (a). c. Find the book value of the machine at the end of the sixth

year. d. Find the rate at which the machine is being depreciated

each year.

LINEAR DEPRECIATION Refer to Exercise 71. An office build­ing worth $1 million when completed in 1986 was dep~·eci­ated linearly over 50 yr. What was the book value of the building in 2001? What will be the book valu~ i~1 200~'? In 2009? (Assume that the book value of the buddtng wrll be zero at the end of the 50th year.)

62 2 FUNCTIONS, LIMITS, AND THE DERIVATIVE

f' B I" , l'lW the pressure 7.'. BOYlE'S LAW As a conscqucnce 0 oy e s' , .

," I f' I Id t· '()11St'111t temperature IS P of a flxcd samp c 0 gas lC a ,\ C . <

relatcd to the volume V of the gas by the rule k

p = f(V) = V

where k is a constant. What is the domain of the function f?

Sketch thc graph of the function f 7.t. POISEUILlE'S LAW According to a law discovered by the 19th­

century physician Poiseuille, the velocity (in centimet.ers/.sec

-ond) of blood,. cm from the central axis of an artery IS gIven

by vCr) = k(R 2

- r2)

where k is a constant and R is the radius of the artery. Suppo e that for a celtain artery, k = 1000 and R = 0.2 so

that vCr) = 1000(0.04 - r2). a. What is the domain of the function vCr)? h. Compute v(O), v(O, 1), and v(0.2) and interpret your

results. c. Sketch the graph of the function von the interval [0, 0.2]. d. What can you say about the velocity of blood as we movc

away from the central axis toward the artery wall?

75. CANCER SURVIVORS The number of living Americans who havc had a cancer diagnosis has increased drastically since 1971, In part, this is due to more testing for cancer and better treat­ment for some cancers. In part, it is because the population is older, and cancer is largely a disease of the elderly. The num­ber of cancer survi vors (in millions) between 1975 (t = 0) and 2000 (t = 25) is approximately

N(t) = 0.003lt 2 + O.l6t + 3.6 (0 < t < 25)

a. How many living Americans had a cancer diagno..,i.., in 1975? In 2000?

h. Assuming the trend continued, how many canccr sur­vivors were there in 2005?

SUl/rct: ,'illH1nJI Clncer IIl"lltlile

76. WORKER EFFICIENCY An efficiency study conducted for Elcktra Electronics showed that the number of "Space Commander" walkie-talkies assembled by the average worker t hr after starting work at 8:00 a,m. is given by

77.

N(t) = -t3 + 6t 2 + 1St (0 < t < 4)

How many walkie-talkies can an average worker be expected to assemble between 8:00 and 9:00 a.m.? Between 9'00' d 10'00 ? . an . a.m ..

POLI~I~S Political scientists have discovered the followin u

empmcal rule, known as the "cube rule" h' h' e I' . ' W IC gl ves the re atlOnshlp between the proportion of seats in the H " f Representati ves won b D '. Ouse 0 . y emocratlc candIdates s(x) and the proportion of popular votes x received by th D . presidential candidate: e emocratlc

x 3

s(x) =~---x3 + (I _ x) 3 (0 < x < I)

Compute s(0.6) and interpret your result.

7~. "

I R' PAll NT PflrV1l1 ENCE III !HE

d. I(H)7 the I) Ie Ilt 01 th

ducLC In '/ . . ' afflIcted WIth Al/h~itl1 I' S ell a I

pc\) _ 0.0726.\ 2 r O.7!)()l,\

where \ is t11ca'illlcd in y al , with age 65. What pcru!llt ~)r tho . P expected LO have /\1/11 'ItllMr

~(l/m ( 1/11 1111 r

79. REGISTERED VEHI lES I A

tered vehicle (ill milli on III

(I - ()) und 2(}() I I .. ) i

N(t) -= - 0.00 I ~I \ U.

a. Ilow achll

IJ. How achu

n. POSTAL

wa ral

in cnt. fun til n" :

h.

Ir atm nl

I) Ih

(/ I

I

d

... I

I 1 I

I on a tUdy con ·.S. p pul<ltion by \::

en hy th functl n

o

o

to

t 12

th r In

th r tn

t

t

I I

2

L

2. a

:991

1"s-

{

1.31 + 22.9

'/(/) - 0.712 + 7.21 + 1/.5

2.61 + 9.4

ira =::; 1 s 3

irJ < 1 =::; 7

if7 < 1 =::; 10 where f is measured in decades with 1 = 0 " .

". ' cOllespondln cr to the begll1l1l11g 01 1900. b

a. What was the median acre of the US po 1 t' h ". b '. pll a lOll at t e

beg~nn~ng O! 1900'1 At the beginning of J 950? At the begll1l1l11g of 1990?

b. Sketch the graph off I'

2.1 Solutions to Self-Check Exercises

1. a. The expression under the radical sign must be nonnegative, so x + 1 :> ° or x :> -1. Also, x*-o because division by zero is not permitted. Therefore, the domain of fis [-1,0) U (0, (0).

b. f(3) = V3 + 1 = V4 = 2 3 3 3

V(a + h) + 1 Va + h + 1 c. f(a + h) = = --__

a+h a+h

2. a. For t in the sllbdomain [0, 6], the rule for f is given by f(t) = 6t + 17. The equation y = 6t + 17 is a linear equation, so that portion of the graph off is the line seg­ment joining the points (0, 17) and (6, 53). Next, in the subdomain (6, 20], the rule for f is given by f(t) =

15 .98(t - 6) 114 + 53. Using a calculator, we construct the following table of values of f(t) for selected values of t.

t 6 8 10 12 14 16 18 20 --- ---- ---J(t) 53 72 75.6 78 79.9 81.4 82.7 83.9

We have included t = 6 in the table, although it does not lie in the subdomain of the function under considera­tion, in order to help us obtain a better sketch of that por­tion of the graph off in the subdomain (6, 20]. The graph of f follows:

2.1 FUNCTIONS AND THEIR GRAPHS 63

tIn Exerfcisles 83-86, determine whether the statement is rue or a se If it is tru l . . . . . e, exp am why 1t 1S true. If it is false

9lVe an example to show why it is false. '

83. If a = b, then/(a) = feb).

84. If f(a) = feb), then a = b.

85. [ffis a function, thenf(a + b) = f(a) + feb).

86. A verti.calline must intersect the graph of y = f(x) at exactly one pomt.

y

100

80

~

t:: 8 60 1

! .... Q)

0...

40

20 j ! ,

2 4 6 8 10 12 14 16 18 ::0

Years

b. The percent of all self-serve gas sales at the beginning of 1978 is found by evaluatingfat t = 4. Since thi , point lies in the interval [0, 6], we use the rulef(t) = 6t + 17 and find

f(4) = 6(4) + 17 = 41

giving 41 % as the required figure. The percent of all self­serve gas sales at the beginning of 199-+ is gi \ en b)

f(20) = 15.98(20 - 6)1/ .. + 53 = 3.9

or approximately 83.9%.

3. A point (x, y) lies on the graph of the fUllction f if nnu only if the coordinates satisfy the equation y = f(x). 0\\.

f( 4) = V2( 4) + 1 + 2 = V9 + 2 = 5 =1:. 6

and we conclude that the given point does flot lit: on th~ graph

off