Vu, Khanh – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels
2
√
5
x
θ
p
x
2
+ 5
From this it follows that
y
= sin
θ
=
x
√
x
2
+ 5
.
Alternatively, we can use the trig identity
csc
2
θ
= 1 + cot
2
θ
to determine sin
θ
.
keywords:
004
(part 1 of 1) 10 points
Find the inverse function,
f

1
, of
f
when
f
is defined by
f
(
x
) =
√
4
x

7
,
x
≥
7
4
.
1.
f

1
(
x
) =
1
7
p
x
2

4
,
x
≥
0
2.
f

1
(
x
) =
1
4
(
x
2
+ 7)
,
x
≥
4
7
3.
f

1
(
x
) =
1
4
p
x
2

7
,
x
≥
0
4.
f

1
(
x
) =
1
7
p
x
2
+ 4
,
x
≥
4
7
5.
f

1
(
x
) =
1
7
(
x
2

4)
,
x
≥
7
4
6.
f

1
(
x
) =
1
4
(
x
2
+ 7)
,
x
≥
0
correct
Explanation:
Since
f
has domain [
7
4
,
∞
) and is increasing
on its domain, the inverse of
f
exists and has
range [
7
4
,
∞
); furthermore, since
f
has range
[0
,
∞
), the inverse of
f
has domain [0
,
∞
).
To determine
f

1
we solve for
x
in
y
=
√
4
x

7
and then interchange
x, y
. Solving first for
x
,
we see that
4
x
=
y
2
+ 7
.
Consequently,
f

1
is defined on [0
,
∞
) by
f

1
(
x
) =
1
4
(
x
2
+ 7)
.
keywords:
005
(part 1 of 1) 10 points
When
g
is the inverse of
f
(
x
) =
x
3
+ 3
x

2
,
find the value of
g
0
(2).
(
Hint
:
f
(1) = 2.)
1.
g
0
(2) =
1
3
2.
g
0
(2) =
2
3
3.
g
0
(2) = 6
4.
g
0
(2) =
3
2
5.
g
0
(2) = 3
6.
g
0
(2) =
1
6
correct
Explanation:
By definition,
g
(
f
(
x
)) =
x ,
so by the Chain Rule,
g
0
(
f
(
x
))
f
0
(
x
) = 1
.
Thus
g
0
(
f
(
x
)) =
1
f
0
(
x
)
.