Part II: The Answers
138
Answers
1–100
31.
domain: (–1, 2); range: [–2, 4)
For a one-to-one function (which by definition has an inverse), the domain of f
 
(x)
becomes the range of f
 −1
(x), and the range of f
 
(x) becomes the domain of f
 −1
(x).
Therefore, f
 −1
(x) has domain (–1, 2) and range [–2, 4).
32.
g
 −1
(x) = f
 −1
(x) – c
Replacing x with (x + c) shifts the graph c units to the left, assuming that c > 0. If the point
(a, b) belongs to the graph of f
 
(x), then the point (a – c, b) belongs to the graph of f
 
(x + c).
Now consider the inverse. The point (b, a) belongs to the graph of f
 −1
(x), so the point
(b, a – c) belongs to the graph of g
 −1
(x). Therefore, g
 −1
(x) is the graph of f
 −1
(x) shifted
down c units so that g
 −1
(x) = f
 −1
(x) – c.
The same argument applies if c < 0.
33.
x = 2
Put all terms involving x on one side of the equation and all constants on the other
side, combining all like terms. Finally, divide by the coefficient of x to get the solution:
3
7 13
3
6
2
x
x
x
+ =
=
=
34.
x = –4
Distribute to remove the parentheses. Then put all terms involving x on one side of the
equation and all constants on the other side, combining all like terms. Finally, divide
by the coefficient of x to get the solution:
2
1 3
2
2
2 3
6
4
4
x
x
x
x
x
x
+
(
)=
+
(
)
+ =
+
− =
= −
35.
x = 5
4
Distribute to remove the parentheses. Then put all terms involving x on one side of the
equation and all constants on the other side, combining all like terms. Finally, divide
by the coefficient of x to get the solution:
−
+ −
=
+
−
− − −
=
+
−
− − =
−
=
4
1 2
7
3
8
4
4 2
7
3
24
6
4 10
24
20 16
20
1
(
)
(
)
x
x
x
x
x
x
x
x
x
x
x
6 6
5
4
=
=
x
x