Review Chapter 8
1. A function is a rule that associates to an allowable
input value one and only one output value.
2. The set of all allowable inputs to a function is
called the domain of the function.
The allowable inputs are those that correspond to
an output.
The set of possible outputs of a function is the
range of the function.
3. Yes, each input has associated with it one and
only 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.
'"
o
Domain
.
35
.
b. To find the maximum amount of sauce for the
range, substitute 35 for diameter in the
formula,
sauce = .2*diamete?
sauce = .2*352
sauce = 245
Range +'"
245
o
c. Enterin the Y= menuYl = X"2 - 3. Onthe
homescreentypeYl(1.46) andpressENTER.
j{1.46)
= ".8684
130
Chapter 8: Functions
7. a. f(x) = x + 1
f(3) =3+1 =4
b. g(x) = x2-x
g(2) = 22- 2 = 4 - 2 = 2
c.
h(t) = -16t2+ 2St+ 6
hC!) = -16Cl) + 2SC!) + 6
= -16*1-25+6
= -16-25+6
= -35
14_ a. We can only take the square root of a
nonnegative number, so x must be greater than
or 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 be
greater than or equal to '2; {x I'2 S;x} ,
range: {y I0 S;y}.
c. The domain excludes any value of x that
causes division by 0; {x 1x * '2}.
The graph has a gap at y = 0 so 0 is excluded
from the range; {y Iy* O}.
d. p(x)=.Jx+5
p(4)=.J4+5 =J9 =3
e.
d. 1 is excluded from the domain because it
causes division by 0; domain: {x 1x * I},
range: {y Iy* O}.
q(x) = 4x3- 2x
qC2) = 4C2)2- 2C2)
=4C8)+ 4
= -32 + 4
= -28
15. a. Linear: I, III;
quadratic: I, IT;
cubic: I, III;
square root: I;
absolute value: I, IT;
reciprocal: I, lIT
9
x-4
=- 9
1-4
9
-3
= -3
f. r(x)=-
b. Reciprocal
16. Local maximum: ('2,4); local minimum: (1, 'I)
17.
Local Maximum
(2, 1.3)
8. The graph represents a function because any
vertical line intersects the graph in at most one
point.
9. The graph does not represent a function because
there is a vertical line that intersects the graph in
more than one point.
10. The graph represents a function because any
vertical line intersects the graph in at most one
point.
11. Domain: {A, B, C, D}; range: {4, 6, 8, 10}
(0,0)
Local
Minimum
.
Local minimum: (0, 0);
Local maximum: (2, 1.3) (answers for the
y-coordinate may vary slightly)
12. Domain: p, 'I, 3, 4}; range: p, '1,1, 2}
13. a. There are no values of x that cause a square
root of a negative number or cause division by
zero so the domain is the set of all real
numbers, see p 316 of the text for review.
Domain
..
..
The graph has a minimumy-value or-I and
continues upward toward 00.
Range
H
b. Domain: {x 1'00< x < oo};range: {y1'1 S;y}
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----
-
Chapter Review
18. a. First we need to fmd the number of printing
cartridges needed to print 4500 pages.
Substitute 4500 for s.
s
p(s)
P
= 1500
(4500) = 4500
1500
pes)
=3
The printer will use 3 printing cartridges.
Now use the fonnula for c(p) to fmd the cost
of printing with 3 cartridges.
c(p) =30p
c(3) = 30 *3
c(3)= 90
It costs $90 to print 4500 pages.
b. Substitute the fonnula for pes) for p in the
fonnula for c(p).
c(p) =30p
c(p(s»
=30p(s)
s
c(p(s»=30*1500
s
c(p(s» =- = .02s
50
It 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 = 0
f(g(2»
b.
g(3) = (3)2- 4 = 9 - 4 = 5
f(g(3»
c. f(g(x»
d.
e.
= f(O) = 0 + 2 = 2
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 - 3
x
jCl)=~=-1
-1
kUC!»=-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 of
the basic absolute value function up 4 units.
b. 2; subtracting 3 from the input moves the
graph right 3 units.
c. 1; taking the opposite of the output reflects the
graph across the x-axis.
d. 5; multiplying the output by a constant
greater than 1 stretches the graph away from
the x-axis.
e. 3; adding 2 to the input moves the graph left 2
units and subtracting 1 from the output moves
the graph down 1 unit.
22. a. The graph ofj(x) moves left 3 units.
= f(5) = 5 + 2 = 7
b. The graph ofj(x) moves down 3 units.
= f(x2 - 4)
=(x2-4)+2
= x2 - 4 + 2
= x2 - 2
c. The graph ofj(x) is reflected (flipped) across
the x-axis.
f(O) = 0 + 2 = 2
g(f(O» = g(2)
= (2)2- 4
=4-4
=0
fC2)=-2+2=0
g(fC2» =g(O)
= (w - 4
=0-4
= -4
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d. The graph ofj(x) is compressed toward the
x-axis.
23. The graph of the basic square root function moves
left 4 units and moves down 2 units.
24. We are multiplying by 16, taking the opposite of
the output, and subtracting 2 from the input of the
basic quadratic function f(x) = x2. These
modifications stretch the graph of f(x) away
from the x-axis, reflect the graph across the
xaxis, and move the graph right 2 units to give the
function g(x) = -16(x - 2)2.
Chapter 8: Functions
132
25. Subtracting 5 from the output and adding one to
the input of the basic cubic function f(x)
moves the graph of f(x)
=(x + 1)3- 5 .
-
8.
2. a. g(x) =x3 - 3x2
b. g(2)
intersects the x- and
b. No, in the first quadrant the graph of f(x)
continues upward toward 00. In the third
quadrant f(x) continues downward toward
00.
8 Test
1. The graph passes the vertical-line test; any
vertical line intersects the graph in at most one
point.
= x3
y-axes at one point, the origin (0, 0). Thus, the
x-intercept is (0, 0) and the y-intercept is (0,
0).
down 5 units and left 1
unit to give the function g(x)
Chapter
7. a. The graph of f(x)
= x3 ,
y
= 23 - 3 * 22
Jr
=8 -3*4
g(2) =8 -12
g(2) = -4
g(2)
(3. -)
c. Yl = Xl\3 - 3XI\2
YIC3.27) = g("3.27) = -67.044483
3. A value of x = -1 causes division by 0, so -1 is
excluded from the domain.
Domain
11
-\
..
)I
There is a gap in the graph of f(x)
is excluded from the range.
Range
It
$
o
aty = 0, so 0
..
/
I'\.
1'10
t
9. a. c(t) =- and t(b) =90b
2
90b
2
c(t(b)) = 45b
c(t(b))=-
4. a.
Yes, each input is associated with a single
output.
b.
y
14
\2
\0
8
6
4
2
b. C({I~))=C(t(1.75))
= 45*1.75
= 78.75
3
1- poundsof coffeebeansmakes78.75cups
4
of coffee.
1
10. a. f(x) =-x andg(x) = x-I
g(4)=4-1=3
1
f(g( 4)) = f(3) =3
1
b. f(g(x)) = f(x -1) =x-I
0\2345"
Yes, the graph passes the vertical-line test.
5. No, the graph fails the vertical-line test; at least
one vertical line intersects the graph in more than
one point.
6. a. Yes, there is a single output associated with
each input.
d. g(f(X))=g(~)=~-1
b. Domain: {DSL, cable, 56 Kbs, 33.6 Kbs,
28.8 Kbs}
Range: {lOOO,2600, 8600, 15,000, 15,500}
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-
,
jiiIi
-
- -- Chapter Test
11. The graph of the basic square root function
f(x) =E is stretched away from the x-axis and
moved right 3 units to give the graph of
g(x)=S.Jx-3.
12. a. The graph of the basic linear functionj{x)
is moved right 2 units.
=x
b. The graph ofj{x) is moved up 2 units.
c. The graph ofj{x) is reflected across the
x-axis.
d. The graph ofj{x) is stretched away from the
x-axis.
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133