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MATH 141, 2nd Examination—Solution Key
Prof. Jonathan Rosenberg
Monday, October 8, 2007
1. (30 points, 10 points per part) Let
f
(
x
) =
x
4

4
x

1
.
(a) What is the biggest interval
I
containing
0
such that the restriction
g
of
f
to
I
is invertible? (Note:
I
is allowed
to extend out to
+
∞
, to
∞
, or both, as appropriate.)
Solution.
The function
f
has derivative
f
′
(
x
) = 4
x
3

4 = 4(
x
3

1)
, so it is increasing when
x
3
>
1
and
decreasing when
x
3
<
1
or
x <
1
. Since
0
<
1
,
f
is decreasing near
x
= 0
and so
I
should be the largest
interval on which
f
is decreasing, i.e.,
I
= (
∞
,
1]
.
(b) What is the domain of the inverse function
g
−
1
?
Solution.
dom(
g
−
1
) = range(
g
) =
f
(
I
)
. Now as
x
→ ∞
,
f
(
x
)
→
+
∞
and
f
is decreasing on
I
, so this is
[
f
(1)
,
∞
) = [

4
,
∞
)
.
(c) Compute
(
g
−
1
)
′
(

1)
.
Solution.
Since
f
(0) =

1
,
g
−
1
(

1) = 0
(
g
−
1
)
′
(

1) =
1
g
′
(
g
−
1
(

1))
=
1
g
′
(0)
=
1

4
.
2. (20 points) Compute
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This note was uploaded on 09/12/2009 for the course MATH 141 taught by Professor Hamilton during the Spring '07 term at Maryland.
 Spring '07
 Hamilton
 Math

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