The graph of
(
)
y
F
x
h
=
+
is not the same as the graph of
(
)
.
y
F
x
h
=
+
The graph of
(
)
y
F
x
h
=
+
represents a vertical shift of the graph of
(
)
.
y
F
x
=
67.
The graph of
(
)
2
6
F
x
x
=
+
is translated 6
units up from the graph of
(
)
2
.
f
x
x
=
68.
(
)
6
G x
x
=
+
69.
The graph of
(
)
6
G x
x
=
+
is translated 6
units up from the graph of
(
)
.
g x
x
=
70.
The graph of
(
)
(
)
2
6
F
x
x
=
−
is translated 6
units to the right from the graph of
(
)
2
.
f
x
x
=
71.
(
)
6
G x
x
=
−
72.
The graph of
(
)
6
G x
x
=
−
is translated 6
units to the right from the graph of
(
)
.
g x
x
=
73. (a)
Choose
any
value
x
.
Find
the
corresponding value of
y
on both graphs
and
compare
these
values.
For
example, choose
2.
x
=
From the
graph,
( )
( )
2
1 and
2
3.
f
g
=
=
For any
value
x
, the
y
-value for
(
)
g x
is 2
greater than the
y
-value for
(
)
,
f
x
so
the
graph
of
(
)
g x
is
a
vertical
translation of the graph of
(
)
f
x
up 2
units.
Therefore
(
)
(
)
2,
g x
f
x
=
+
that
is
2.
c
=
(b)
Choose
any
value
y
.
Find
the
corresponding value of
x
on both graphs
and
compare
these
values.
For
example, choose
3.
y
=
From the
graph,
( )
( )
6
3 and
2
3.
f
g
=
=
For any
value
y
, the
x
-value for
(
)
g x
is 4 less
than the
x
-value for
(
)
,
f
x
so the graph
of
(
)
g x
is a horizontal translation of
the graph of
(
)
f
x
to the left 4 units.
Therefore
(
)
(
)
4 ,
g x
f
x
=
+
that
is
4.
c
=

Section 2.7: Function Operations and Composition
273
Section 2.7: Function Operations and Composition
In Exercises 1–8,
2
( )
5
2
f x
x
x
=
−
and
g
(
x
) = 6
x
+ 4.
1.
[
]
(
)
(
)
2
(
)(3)
(3)
(3)
5(3)
2(3)
6(3)
4
45
6
18
4
39
22
61
f
g
f
g
+
=
+
=
−
+
+
=
−
+
+
=
+
=
2.
(
)
(
)
2
(
)( 5)
( 5)
( 5)
[5( 5)
2( 5)]
[6( 5)
4]
125
10
30
4
135
( 26)
161
f
g
f
g
−
−
=
−
−
−
=
−
−
−
−
−
+
=
+
− −
+
=
− −
=
3.
(
) (
)
2
(
)(4)
(4)
(4)
[5(4)
2(4)] [6(4)
4]
[5(16)
2(4)] [24
4]
80
8
28
72(28)
2016
fg
f
g
=
⋅
=
−
⋅
+
=
−
⋅
+
=
−
⋅
=
=
4.
(
) (
)
2
(
)( 3)
( 3)
( 3)
[5( 3)
2( 3)] [6( 3)
4]
[5(9)
2( 3)] [ 18
4]
45
6
14
51( 14)
714
fg
f
g
−
=
−
⋅
−
=
−
−
−
⋅
−
+
=
−
−
⋅ −
+
=
+
⋅ −
=
−
= −
5.
2
( 1)
5( 1)
2( 1)
( 1)
( 1)
6( 1)
4
5
2
7
6
4
2
f
f
g
g
−
−
−
−
−
=
=
−
−
+
+
=
= −
−
+
6.
2
(4)
5(4)
2(4)
(4)
(4)
6(4)
4
80
8
72
18
24
4
28
7
f
f
g
g
−
=
=
+
−
=
=
=
+
7.
2
2
2
(
)(
)
(
)
(
)
(5
2
)
(6
4)
5
2
6
4
5
8
4
f
g
m
f m
g m
m
m
m
m
m
m
m
m
−
=
−
=
−
−
+
=
−
−
−
=
−
−
8.
2
2
2
2
(
)(2 )
(2 )
(2 )
[5(2 )
2(2 )]
[6(2 )
4]
[5(4)
2(2 )]
[12
4]
(20
4 )
(12
4)
20
8
4
f
g
k
f
k
g
k
k
k
k
k
k
k
k
k
k
k
k
+
=
+
=
−
+
+
=
−
+
+
=
−
+
+
=
+
+
9.
( )
3
4, ( )
2
5
f
x
x
g x
x
=
+
=
−
i)
(
)( )
( )
( )
(3
4)
(2
5)
5
1
f
g
x
f
x
g x
x
x
x
+
=
+
=
+
+
−
=
−
ii)
(
)( )
( )
( )
(3
4)
(2
5)
9
f
g
x
f
x
g x
x
x
x
−
=
−
=
+
−
−
=
+
iii)
2
2
(
)( )
( )
( )
(3
4)(2
5)
6
15
8
20
6
7
20
fg
x
f
x
g x
x
x
x
x
x
x
x
=
⋅
=
+
−
=
−
+
−
=
−
−
iv)
( )
3
4
( )
( )
2
5
f
f
x
x
x
g
g x
x
+
=
=
−
The domains of both
f
and
g
are the set of all real numbers, so the domains of
f
+
g
,
f
–
g
, and
fg
are all
(
)
,
.
−∞ ∞
The domain of
f
g
is the set of all real numbers for which
(
)
0.
g x
≠
This is the set of all
real numbers except
5
2
, which is written in interval notation as
(
)
(
)
5
5
2
2
,
,
.
−∞
∪
∞
10.
f
(
x
) = 6 – 3
x
,
g
(
x
) = –4
x
+ 1
i
)
(
)( )
( )
( )
(6
3 )
(
4
1)
7
7
f
g
x
f
x
g x
x
x
x
+
=
+
=
−
+ −
+
= −
+
ii)
(
)( )
( )
( )
(6
3 )
(
4
1)
5
f
g
x
f
x
g x
x
x
x
−
=
−
=
−
− −
+
=
+
iii)
2
2
(
)( )
( )
( )
(6
3 )(
4
1)
24
6
12
3
12
27
6
fg
x
f
x
g x
x
x
x
x
x
x
x
=
⋅
=
−
−
+
= −
+
+
−
=
−
+
iv)
( )
6
3
( )
( )
4
1
f
f x
x
x
g
g x
x
−
=
=
−
+
The domains of both
f
and
g
are the set of all real numbers, so the domains of
f
+
g
,
f
–
g
, and
fg
are all
(
)
,
.
−∞ ∞

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