Section 8.4 Composite and Inverse Functions
4. a.
()() ()
()
()
()
34
53 4 2
fgx fgx f x
x
==−
=−+
5. a.
( )() ()
()
()
()
2
2
2
2
52
45 2 3
20 8 3
20 11
fgx fgx
fx
x
x
x
=
=−
=−−
=−−
=−
6. a.
()() ()
()
()
()
2
2
2
2
29
72 9 1
14 63 1
14 62
fgx fgx f x
x
x
x
==−
=−+
=−+
=−
b.
( )() ()
()
()
71
gf x gfx g x
==+
7. a.
()() ()
()
()
2
2
fgx fgx
fx
=
=−
()
()
2
2
42
42
22
442
42
gx
x
xx
xx
=+
=+−
=+ +−
=+ +
c.
()() ()
2
4
2242 6
fg =− +
42
213
22
xx
xx
=+ +−
=+ −
c.
()() ()
()
2
4
2262 10
16 6 4 10
16 24 10 2
fg =− +
=− +
=−+=
Chapter 8 Basics of Functions
10. a.
()() ()
()
()
2
2
fgx fgx fx
x
==+
=+
11. a.
( )() ()
()
3
2
fgx fgx
x
f
=
+
=
b.
()() ()
()
()
23
gf x gfx
gx
=
=−
12. a.
()() ()
()
3
6
3
63
6
33
x
fgx fgx f
x
xx
+
==
+
=−
=+−=
13. a.
( )() ()
()
1
fgx fgx
=
1
11
gx
xx
=
==⋅=
14. a.
()() ()
()
2
fgx fgx fx
==
21
222
11
xx
x
x
==÷=⋅=
c.
()()
2
21
2
fg ==
Section 8.4 Composite and Inverse Functions
17.
()
()
8
3
8
38
x
fgx f
x
−
=
−
=+
The functions are inverses.
18.
()
()
99
49
44
99
xx
fgx f
xx
−−
==+
=−+=
19.
()
()
5
9
5
59
9
52581
99
x
fgx f
x
x
+
=
+
=−
+
=−
20.
()
()
()
33
37
77
33
49 3 9 49
xx
fgx f
xx
++
==−
++−
=−=
Since
()
()
fgx x≠
and
()
()
gfx x≠
, we
conclude the functions are not inverses.
21.
()
()
34
fgx f x
=+
()
()
3
4
34
3
4
4
3444
gfx gx
x
xxx
=
−
=+
−
−
=⋅ +=−+=
Chapter 8 Basics of Functions
22.
()
()
22
5255
fgx f x
x
=+=
+−
()
()
22
5
2
5
5
25
2525
52
55
gfx gx
x
x
x
xx
==+
−
−
−
=÷ +=⋅ +
−
=−+=
The functions are inverses.
24.
()
()
() ()
fgx fx x x==−=−
()
()
()
gfx g x x=−=−
The functions are not inverses.
25. a.
()
3
3
fx x
yx
=+
=+
b.
()
()
()
()
1
3
33
33
ff x fx
x
xx
−
=−
=−+
=−+=
26. a.
()
5
5
fx x
yx
=+
=+
()
5fxx
=−
b.
()
()
()()
1
555
55
ff x fx x
xx
−
=−=−+
=−+=
()
()
()()
1
555
55
ffx fx x
xx
−
=+=+−
=+−=
2
()
1
2
x
fx
−
=
b.
()
()
1
2
22
xx
ff x f x
−
===
()
()
()
1
2
22
x
ffx fx x
−
===
4
4
xy
xy
=
=
()
1
x
fx
−
=
Section 8.4 Composite and Inverse Functions
29. a.
()
23
23
fx x
yx
=+
=+
Interchange x and y and solve for y.
b.
()
()
1
3
2
3
23
2
33
x
ff x f
x
xx
−
−
=
−
=+
=−+=
30. a.
()
31
31
fx x
yx
=−
=−
Interchange x and y and solve for y.
b.
()
()
1
11
31
33
11
xx
ff x f
xx
−
++
==−
=+−=
31. a.
()
1
1
fx x
yx
=
=
()
1
1
fxx
−
=
b.
()
()
1
1
11
ff x fx
xx
−
=
11
x
32. a.
()
2
fx x
=
2
yx
=
()
1
2
fxx
−
=
Chapter 8 Basics of Functions
b.
()
()
1
22
2
ff x fx
x
−
==
33. a.
()
21
3
21
3
x
fx x
x
yx
+
=−
+
=−
Interchange x and y and solve for y.
()
21
3
321
y
xy
xy y
+
=−
−= +
b.
()
()
131
2
31
21
2
313
2
x
ff x f x
x
x
x
x
−+
=
−
+
+
−
=+
−
−
()
()
()()
()()
11
21
3
3
32 1 1 3
212 3
63 3
2126
7
7
x
ffx f x
x
xx
xx
xx
xx
x
x
−−
+
=
−
−
++ −
=+− −
++−
=+− +
=
=
()
()()
1
123
23
23
23
3
2
xy y
xy x y
xy y x
yx x
x
yx
+= −
+= −
−=−−
−=−+
+
=− −
Section 8.4 Composite and Inverse Functions
b.
()
()
1
3
2
x
ff x f x
−
+
=−
−
3
23
2
31
2
x
x
x
x
+
−−
−
=+
−+
−
()
()
11
23
1
23
3
1
23
2
1
x
ffx f x
x
x
x
x
−−
−
=
+
−+
+
=− −−
+
35. The graph does not satisfy the horizontal line test so
the function does not have an inverse.
36. The graph satisfies the horizontal line test so the
function has an inverse.
39. The graph satisfies the horizontal line test so the
function has an inverse.
40. The graph satisfies the horizontal line test so the
function has an inverse.
41.
43.
46.
()()
(4) 2 1fg f==−
47.
()()( )()
1(1)11gf gf g−= − = =
48.
()()()()
0(0)42gf gf g===
Chapter 8 Basics of Functions
52.
()()()()
1(1)53fg fg f==−=
53.
()()()()
0(0)26gf gf g===−
56.
()()()
()
() ()
0(0)
20 5
545121
gf gf
g
g
=
=⋅−
=−=−−=−
58. Let
()
1
7gx
−
=
. Then
()
7
gx
=
59.
[]
()
()
()
()
()
2
(1) 1 1 2
(4)
24 5
3
gfh gf
gf
g
g
=++
=
=⋅−
=
61. a. f represents the price after a $400 discount; g
represents the price after a 25% discount (75%
of the regular price).
c.
()() ()
()
()
()
400
0.75 400
0.75 300
gf x gfx
gx
x
x
=
=−
=−
=−
gf
represents an additional 25% discount on
Interchange x and y and solve for y.
400
xy
=−
after a $400 discount.
62. a. f is the regular price of the jeans less $5.
g is 60% of the regular price of the jeans.
b.
()() ()
()
()
0.6
0.6 5
fgx fgx f x
x
==
=−
Section 8.4 Composite and Inverse Functions
e.
()
5
5
fx x
yx
=−
=−
63. a. f: {(U.S., 1%), (U.K., 8%), (Italy, 5%),
(France, 5%), (Holland, 30%)}
64. a. f: {(U.S., 84%), (U.K., 62%), (Italy, 42%),
(France, 45%), (Holland, 45%)}
65. a. We know that f has an inverse because no
horizontal line intersects the graph of f in more
than one point.
b.
()
1
0.25f
−
, or approximately 15, represents the
number of people who must be in a room so that
66. a. The graph does not have an inverse because
there are horizontal lines, such as 2,y= that
intersect the graph more than once.
c. No, the graph does not represent a one-to-one
function. The points (12, 3) and (19, 3) have the
same second coordinate but different first
x
=
95
x
=
74.
()
2
1fx x=−
Chapter 8 Basics of Functions
76.
()
3
2
x
fx=
77.
()
4
4
x
fx=
78.
()
2fx x=−
79.
() ( )
3
1fx x=−
80.
()
2
16fx x=− −
81.
()
3
1fx x x=++
()
0.25 1
gx x
=−
()
0.25 1
gx x
=−
Section 8.4 Composite and Inverse Functions
84.
()
() ( )
3
3
2
2
fx x
gx x
=−
=+
85. does not make sense; Explanations will vary.
Sample explanation: The diagram illustrates
()
().gfx
89. false; Changes to make the statement true will vary.
A sample change is: The inverse is
()( )
{
}
4,1 , 7, 2 .
90. false; Changes to make the statement true will vary.
A sample change is: f (x) = 5 does not satisfy the
horizontal line test.
93. To find
()()
1
,fg x
−
first find
()()
.fgx
()() ()
()
()
()
5
35315
fgx fgx fx
xx
==+
=+=+
15
3
xy
−=
()()
1
15
x
fg x
−
−
=
Interchange x and y and solve for y.
()
()
1
1
53
5
3
5
3
xy xy
xy xy
gxx
x
fx
−
−
=+ =
−=
=
=−
=
Chapter 8 Basics of Functions
94.
()
()
()
3()2
() 5()3
fx
ffx fx
−
=−
32
32
53
32
53
53
x
x
x
x
−
−
−
=−
−
−
95.
()
()
11
22
fx mxb
gx mx b
=+
=+
First find
()()
.fgx
()() ()
()
()
22
fgx fgx fmxb
==+
97.
2
32
32
2
2
916
2723
2
9 2
9 18
xx
xxx x
xx
xx
xx
++
−+−+
−
−
−
16 6 2
25 20
20 4
25 5
xy
x
x
−=
=
==
Instead of substituting 4
5 for x for working with
fractions, go back to the original system and
99. 212 7( 1)
212 7 7
12 7 7 2
19 9
xx
xx
xx
x
−=−
−=−
−−=−−
−=−
Chapter 8 Review Exercises
100.
321
434
321
12 12 12
xx
xx
+−
=+
+−
⋅=⋅+⋅
101.
600 (500,000 400 ) 0
600 500,000 400 0
xx
xx
−+>
−−>
Chapter 8 Review
1. The relation is a function.
Domain {3, 4, 5}
Range {10}
4. a.
()
(0) 7 0 5 0 5 5f=−=−=−
b.
()
(3) 7 3 5 21 5 16f=−=−=
c.
()
( 10) 7 10 5 75f−=−−=−
{
}
5. a.
() ()
2
(0) 3 0 5 0 2 2g=−+=
b.
() ()
2
(5) 3 5 5 5 2
g=−+
e.
() () ()
2
2
434 542
316 20 2
ga a a
aa
=−+
=−+
a function.
8. The vertical line test shows that this is the graph of
a function.
9. The vertical line test shows that this is not the graph
of a function.
12.
{
}
23xx−< ≤
13.
{
}
1.5 2xx−≤≤
Chapter 8 Basics of Functions
17. When 3, ( ) 5.xfx==−
20. a. The eagle’s height is a function of its time in
flight because every time, t, is associated with at
most one height.
e. The eagle began the flight at 45 meters and
remained there for approximately 3 seconds. At
that time, the eagle descended for 9 seconds. It
landed on the ground and stayed there for 5
seconds. The eagle then began to climb back up
to a height of 44 meters.
21. The domain of f is (,).−∞ ∞
22. The domain of f is (,8)or(8,).−∞ − − ∞
25. a.
()
()
()
2
22
()
543
54351
fgx
xx x
xx x x
+
=−++−
= −++−= +
27. The domain of
fg+
is
()
()
()
2
2
2
() 2 5
25
5
fgx x x x
xxx
xx
+=−+−
=−+−
=−−
9351
=−−=
30. 2
() 2, () 5fx x x gx x=− =−
()
()
()
2
2
2
() 2 5
25
35
fgx x x x
xxx
xx
−=−−−
=−−+
=−+
()
( ) () ()
2
2
() 3 5
(1) 1 3 1 5
fgx x x
fg
−=−+
−=−+
16 12 5 9
=−+=
32.
( )() () ()
2
32
(2)(5)
710
fg x f x g x
xxx
xx x
=⋅
=− −
=− +
Chapter 8 Review Exercises
33. 2
() 2, () 5fx x x gx x=− =−
34.
()
2
() 3 5fgx x x−=−+
The domain of is ( , ).fg−−∞∞
36. a.
()() ()
()
()
()
2
41
41 3
fgx fgx
fx
x
=
=−
=−+
c.
( )() () ()
()
2
3163 834
16 9 24 4
144 24 4
124
fg =−+
=−+
=−+
=
37. a.
()() ()
()
fgx fgx
=
38.
() ()
31 5
and g 2
fx x x x=+ =−
53 5 2
61
52
x
=−+
53 1 2
35 2
53 51 2
35 32
x
x
=+−
=+−
()
2
25 5
22 22
x
xxx
−
=−
=− − =−+=
()
()
()
()
25
225
gfx g x
x
=−
−−
Chapter 8 Basics of Functions
40. a.
()
43
43
fx x
yx
=−
=−
b.
()
()
1
3
4
3
43
4
33
x
ff x f
x
xx
−
+
=
+
=−
=+−=
41. a.
()
1
1
fx x
yx
=−
=−
Interchange x and y and solve for y.
1
xy
=−
42. Since the graph satisfies the horizontal line test, it
has an inverse function.
43. Since the graph does not satisfy the horizontal line
test, it does not have an inverse function.
Range {2, 4, 6}
2. The relation is not a function.
Domain {2, 4, 6}
Range {1, 3, 5, 6}
3.
()()
43 42
3122310
fa a
aa
+= +−
=+−=+
4.
() () ()
2
242 326
f−= − −−+
11. The domain of f is (,10)or(10,).−∞ ∞
Cumulative Review
12.
() ()
2
4and 2fx x x gx x=+ =+
( )() () ()
fgx fx gx
+=+
13.
() ()
2
4and 2fx x x gx x=+ =+
( )() () ()
fgx fx gx
−=−
14.
( )() () ()
2
32
(4)(2)
68
fg x f x g x
xxx
xxx
=⋅
=+ +
=+ +
()()
32
5 ( 5) 6( 5) 8( 5)
fg −=− +− +−
16. Domain of
f
g
is
( , 2) or ( 2, ).−∞ − − ∞
17.
() ()
2
and 3 1fx x x gx x=+ =−
( )() ()
()
()
31
fgx fgx f x
==−
18.
()
57
57
fx x
yx
=−
=−
()
1
7
5
x
fx
−
+
=
c.
()
1
2000f
−
represents the income, $80
thousand, of a family that gives $2000 to charity.
()()
2
25120
23 40
xx
xx
+−=
−+=
230
23
3
2
x
x
x
−=
=
=
or
40
4
x
x
+=
=−
Chapter 8 Basics of Functions
3.
85 4
215 66
xy
xy
−=−
+=−
Eliminate y by multiplying both sides of the first
equation by 3 and adding the two equations.
24 15 12
xy
−=−
4.
15 6
43
xx
−= +
Multiply both sides of the equation by x to eliminate
the fractions.
15 6
43
xx
xx
−= +
5. 378
315
5
x
x
x
−−=
−=
=−
6.
()
2
252fx x x=−+;
()
2
23gx x x=−+
()()
() ()
22
fgx
fx gx
−
=−
4
2
x
x
=− =−
8.
() ( )
834812
812
4
−−− ⋅ =−−−
=− +
=
10.
()()
()()
()()
2
31 2
352
122
31 2
xx
xx
xx
xx
−−
−− +
=+−
Cumulative Review
12.
2
18 77xx−+
We need two factors of 77 whose sum is 18−.
Since the product is positive, the factors have the
13.
()
()()
32
25 25
55
xxxx
xx x
−= −
=− +
15.
()
()
()
()
()()()()
()
() ()
2
2
2
32 2
3
234 69
24 26 29
34 36 39
81218121827
827
xxx
xx xx x
xx
xxxxx
x
−++
=++
−−−
=+ +− −−
=−
16.
2
32
2
2
x
x
xx
−+
+−
17.
22
22
561169
1473
xx x
xxx
−+ −
÷
−++
()()
22
51 1
xx
−−
=
()
1x+
()
1x−
()
43x+
⋅
()
1x+
()()
4343xx−+
51
x
−
=−
24626
25 8
5 18
xyz
xyz
yz
−+ + =
−−=−
−+ =
Using the two reduced equations, we can form the
following system of linear equations in two
variables:
5218
518
yz
yz
+=
−+ =
Multiply the second equation by 5 and add to the
first equation.
5218
52590
yz
yz
+=
−+ =
Chapter 8 Basics of Functions
Back-substitute the values for y and z to solve for x.
35
xyz
+−=
19.
24
24
24
xy
yx
yx
−=
−=− +
=−
()
1, 6−−
must be on the graph as well.
3
y
20.
2
3
yx=−
The slope is
22
33
m−
=− =
and the y-intercept is 0.
One point is
()
0, 0
and using the slope we can get a
21. Since each element from the domain corresponds to
exactly one element of the range, the relation is a
23.
5m=
;
()( )
11
,2,3xy =− −
()
() ()
()
11
35 2
yy mxx
yx
−= −
−− = −−
24.
()()
()
()
82
82
710 310
7 3 10 10
−
−
××
=⋅× ⋅
25.
()
1
15
fx x
=−
This is a rational function so the domain is all real
numbers except where the denominator equals 0.
15 0
15
x
x
−=
=