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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert
2
√
7
x
θ
p
x
2
+
7
From this it follows that
y
= sin
θ
=
x
√
x
2
+ 7
.
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 de±ned by
f
(
x
) =
√
6
x

5
,
x
≥
5
6
.
1.
f

1
(
x
) =
1
5
p
x
2
+ 6
,
x
≥
6
5
2.
f

1
(
x
) =
1
6
p
x
2

5
,
x
≥
0
3.
f

1
(
x
) =
1
6
(
x
2
+ 5)
,
x
≥
0
correct
4.
f

1
(
x
) =
1
5
p
x
2

6
,
x
≥
0
5.
f

1
(
x
) =
1
5
(
x
2

6)
,
x
≥
5
6
6.
f

1
(
x
) =
1
6
(
x
2
+ 5)
,
x
≥
6
5
Explanation:
Since
f
has domain [
5
6
,
∞
) and is increasing
on its domain, the inverse of
f
exists and has
range [
5
6
,
∞
); furthermore, since
f
has range
[0
,
∞
), the inverse of
f
has domain [0
,
∞
).
To determine
f

1
we solve for
x
in
y
=
√
6
x

5
and then interchange
x, y
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 Spring '08
 CLARK,C.W./HOY,R.R

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