Review Chapter 8

1. A function is a rule that associates to an allowableinput value one and only one output value.

2. The set of all allowable inputs to a function iscalled the domain of the function.The allowable inputs are those that correspond toan output.The set of possible outputs of a function is therange of the function.

3. Yes, each input has associated with it one andonly one output value.

a. Domain: {a, ~, y, E}

b. Range:{l, 2, 3}

4. No, the input x has two outputs, 3 and 8,associated with it.

5. a.Domain .

35.'"o

b. To find the maximum amount of sauce for therange, substitute 35 for diameter in theformula,sauce = .2*diamete?sauce = .2*352sauce = 245

Range'"o

+-245

c. Enterin theY=menuYl = X"2 - 3. OnthehomescreentypeYl(1.46) andpressENTER.j{1.46) =".8684

130 Chapter 8: Functions

7. a. f(x) = x +1f(3) =3+1 =4

b. g(x) = x2-xg(2) = 22- 2= 4 - 2= 2

c. h(t) = -16t2+2St+ 6hC!) = -16Cl) +2SC!)+6

= -16*1-25+6= -16-25+6= -35

d. p(x)=.Jx+5

p(4)=.J4+5 =J9 =3

e. q(x) = 4x3- 2xqC2) = 4C2)2- 2C2)

=4C8)+ 4= -32+ 4= -28

9f. r(x)=-

x-49=-

1-49-3

= -3

8. The graph represents a function because anyvertical line intersects the graph in at most onepoint.

9. The graph does not represent a function becausethere is a vertical line that intersects the graph inmore than one point.

10. The graph represents a function because anyvertical line intersects the graph in at most onepoint.

11. Domain: {A, B, C, D}; range: {4, 6, 8, 10}

12. Domain: p, 'I, 3, 4}; range: p, '1,1, 2}

13. a. There are no values of x that cause a squareroot of a negative number or cause division byzero so the domain is the set of all realnumbers, see p 316 of the text for review.

Domain.. ..The graph has a minimumy-value or-I andcontinues upward toward 00.

RangeH

b. Domain: {x 1'00< x < oo};range: {y1'1 S;y}

14_ a. We can only take the square root of anonnegative number, so x must be greater thanor equal to 2; domain: {x 12 s;x}.The minimumy-value is 0 and the graph of

f (x) =.Jx - 2 continues upward toward 00;

range: {y I0 S;y}.

b. x + 2 must be nonnegative, so x must begreater than or equal to '2; {x I '2 S;x} ,range: {y I0 S;y}.

c. The domain excludes any value of x thatcauses division by 0; {x 1x * '2}.The graph has a gap at y = 0 so 0 is excludedfrom the range; {y Iy* O}.

d. 1 is excluded from the domain because it

causes division by 0; domain: {x 1x * I},range: {y Iy* O}.

15. a. Linear: I, III;quadratic: I, IT;cubic: I, III;square root: I;absolute value: I, IT;reciprocal: I, lIT

b. Reciprocal

16. Local maximum: ('2,4); local minimum: (1, 'I)

17.Local Maximum

(2, 1.3)

(0,0)Local

Minimum .Local minimum: (0, 0);Local maximum: (2, 1.3) (answers for they-coordinate may vary slightly)

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---- -

18. a. First we need to fmd the number of printingcartridges needed to print 4500 pages.Substitute 4500 for s.

sp(s) = 1500

(4500) = 4500P 1500

pes) =3

The printer will use 3 printing cartridges.Now use the fonnula for c(p) to fmd the costof printing with 3 cartridges.c(p) =30pc(3) = 30 *3c(3)= 90

It costs $90 to print 4500 pages.

b. Substitute the fonnula for pes) forp in the

fonnula for c(p).

c(p) =30p

c(p(s» =30p(s)s

c(p(s»=30*-1500

sc(p(s» =- = .02s

50It costs .02s dollars to print s pages.

c. c(p(s» = .02s Yes, we can fmd the cost in

two steps by using the functions pes) and

then c(p), or we can use the composite

function c(p(s» to fmd the cost in one step.

19. a. f(x)=x+2 and g(x)=x2-4

g(2) = (2)2 - 4 = 4 - 4 = 0f(g(2» = f(O) = 0 + 2 = 2

b. g(3)= (3)2- 4 =9 - 4 =5f(g(3» =f(5) =5 + 2 =7

c. f(g(x» = f(x2 - 4)

=(x2-4)+2=x2 - 4 + 2=x2 - 2

d. f(O) = 0+ 2= 2g(f(O» = g(2)

= (2)2- 4=4-4=0

e. fC2)=-2+2=0g(fC2» =g(O)

= (w - 4=0-4= -4

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Chapter Review 131

f. g(f(x» = g(x + 2)=(x + 2)2- 4=(x+2)(x+2)-4=x2+ 4x +4- 4=x2+ 4x

20. a. j(x) =.!. andk(x) = x - 3x

jCl)=~=-1-1kUC!»=-1-3=-4

b. kU(X»=k(~)=~-3

c. k(2) = 2- 3= -1

j(k(2» =J'Cl) =~ = -1-1

d. j(k(x»=j(x-3)=~ x - 3

21. a. 4; adding 4 to the output moves the graph ofthe basic absolute value function up 4 units.

b. 2; subtracting 3 from the input moves thegraph right 3 units.

c. 1; taking the opposite of the output reflects thegraph across the x-axis.

d. 5; multiplying the output by a constantgreater than 1 stretches the graph away fromthe x-axis.

e. 3; adding 2 to the input moves the graph left 2units and subtracting 1 from the output movesthe graph down 1 unit.

22. a. The graph ofj(x) moves left 3 units.

b. The graph ofj(x) moves down 3 units.

c. The graph ofj(x) is reflected (flipped) acrossthe x-axis.

d. The graph ofj(x) is compressed toward thex-axis.

23. The graph of the basic square root function movesleft 4 units and moves down 2 units.

24. We are multiplying by 16, taking the opposite ofthe output, and subtracting 2 from the input of thebasic quadratic function f(x) =x2. These

modifications stretch the graph of f(x) awayfrom the x-axis, reflect the graph across the x-axis, and move the graph right 2 units to give thefunction g(x) = -16(x - 2)2.

132 Chapter 8: Functions

25. Subtracting 5 from the output and adding one tothe input of the basic cubic function f(x) =x3 ,moves the graph of f(x) down 5 units and left 1

unit to give the function g(x) =(x + 1)3- 5 .

Chapter 8 Test

1. The graph passes the vertical-line test; anyvertical line intersects the graph in at most onepoint.

2. a. g(x)=x3 - 3x2

b. g(2) =23 - 3 * 22

g(2) =8 -3*4g(2) =8 -12g(2) =-4

c. Yl = Xl\3 - 3XI\2YIC3.27) = g("3.27) = -67.044483

3. A value of x = -1 causes division by 0, so -1 isexcluded from the domain.

Domain11 .. ) I

-\

There is a gap in the graph of f(x) aty = 0, so 0is excluded from the range.

RangeIt $ ..

o

4. a.

Yes, each input is associated with a singleoutput.

b.y

14

\2

\08

6

4

2

0\2345"

Yes, the graph passes the vertical-line test.

5. No, the graph fails the vertical-line test; at leastone vertical line intersects the graph in more thanone point.

6. a. Yes, there is a single output associated witheach input.

b. Domain: {DSL, cable, 56 Kbs, 33.6 Kbs,28.8 Kbs}

Range: {lOOO,2600, 8600, 15,000, 15,500}

7. a. The graph of f(x) =x3 intersects the x- and

y-axes at one point, the origin (0, 0). Thus, thex-intercept is (0, 0) and the y-intercept is (0,0).

b. No, in the first quadrant the graph of f(x)

continues upward toward 00. In the thirdquadrant f(x) continues downward toward -00.

8.

tc(t) =- and t(b) =90b

290b

c(t(b))=-2

c(t(b)) = 45b

b. C({I~))=C(t(1.75))= 45*1.75= 78.75

31- poundsof coffeebeansmakes78.75cups4of coffee.

9. a.

110. a. f(x) =- andg(x) = x-Ix

g(4)=4-1=31

f(g( 4)) = f(3) =-3

1b. f(g(x)) = f(x -1) =-

x-I

d. g(f(X))=g(~)=~-1

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-

y

Jr

(3. -)

/I'\.

1'10

, jiiIi - - - - -

11. The graph of the basic square root function

f(x) =E is stretched away from the x-axis andmoved right 3 units to give the graph of

g(x)=S.Jx-3.

12. a. The graph of the basic linear functionj{x) =xis moved right 2 units.

b. The graph ofj{x) is moved up 2 units.

c. The graph ofj{x) is reflected across thex-axis.

d. The graph ofj{x) is stretched away from thex-axis.

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Chapter Test 133