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Create assignment, 57321, Homework 9, Apr 05 at 9:08 pm
1
This
printout
should
have
19
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
CalC7a01b
52:03, calculus3, multiple choice,
<
1 min,
fxed.
001
Use
f
0
(
x
) to determine whether
f
(
x
) =
x
4
4

2
x
2
has an inverse on (
∞
,
∞
).
1.
f
does not have inverse
correct
2.
f
has inverse
Explanation:
The Function will have an inverse iF
f
is in
creasing everywhere on (
∞
,
∞
) or decreas
ing everywhere, so we need to look at the sign
oF
f
0
(
x
). Now
f
0
(
x
) =
x
3

4
x
=
x
(
x
2

4)
,
and so has sign chart
∞

2
0
2
∞

+

+
Thus
f
is decreasing on (
∞
,

2] and on
[0
,
2], but increasing on [

2
,
0] and on [2
,
∞
).
Consequently,
f
does not have an inverse
.
keywords: inverse Functions
CalC7a27s
52:03, calculus3, multiple choice,
>
1 min,
wordingvariable.
002
±ind the inverse Function,
f

1
, oF
f
when
f
is defned by
f
(
x
) =
√
5
x

6
,
x
≥
6
5
.
1.
f

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

1
(
x
) =
1
5
(
x
2
+ 6)
,
x
≥
5
6
3.
f

1
(
x
) =
1
5
p
x
2

6
,
x
≥
0
4.
f

1
(
x
) =
1
6
(
x
2

5)
,
x
≥
6
5
5.
f

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

1
(
x
) =
1
6
p
x
2

5
,
x
≥
0
Explanation:
Since
f
has domain [
6
5
,
∞
) and is increasing
on its domain, the inverse oF
f
exists and has
range [
6
5
,
∞
); Furthermore, since
f
has range
[0
,
∞
), the inverse oF
f
has domain [0
,
∞
).
To determine
f

1
we solve For
x
in
y
=
√
5
x

6
and then interchange
x, y
. Solving frst For
x
,
we see that
5
x
=
y
2
+ 6
.
Consequently,
f

1
is defned on [0
,
∞
) by
f

1
(
x
) =
1
5
(
x
2
+ 6)
.
keywords:
CalC7a41s
52:03, calculus3, multiple choice,
<
1 min,
wordingvariable.
003
On (

1
,
1) the Function
f
(
x
) = 5 +
x
2
+ tan
‡
πx
2
·
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View Full DocumentCreate assignment, 57321, Homework 9, Apr 05 at 9:08 pm
2
has an inverse
g
. Find the value of
g
0
(5).
(Hint:
fnd the value oF
f
(0)).
1.
g
0
(5) =
2
π
correct
2.
g
0
(5) =
π
2
3.
g
0
(5) = 1
4.
g
0
(5) =
5
π
5.
g
0
(5) =
π
5
Explanation:
By de±nition of inverse function:
f
(
g
(
x
)) =
x,
g
(
f
(
x
)) =
x.
Applying the Chain Rule to the ±rst of these
equations, we see that
f
0
(
g
(
x
))
g
0
(
x
) = 1
,
in which case
g
0
(
x
) =
1
f
0
(
g
(
x
))
.
Thus to ±nd the value of
g
0
(5) we need to
know the value of
g
(5) as well as
f
0
(
x
). Now
f
0
(
x
) = 2
x
+
π
2
sec
2
‡
πx
2
·
.
The hint becomes useful in determining the
value of
g
(5).
For
f
(0) = 5, and so by the
second of the equations above,
g
(5) =
g
(
f
(0)) = 0
,.
Consequently,
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 Spring '06
 McAdam

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