to (aqt73) – Section 4.1 – isaacson – (55826)
1
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001
10.0points
Find all the critical points of
f
when
f
(
x
) =
x
x
2
+ 25
.
1.
x
= 0
,
5
2.
x
=

25
,
25
3.
x
=

5
,
0
4.
x
=

5
,
25
5.
x
=

25
,
5
6.
x
=

5
,
5
correct
Explanation:
By the Quotient Rule,
f
′
(
x
) =
(
x
2
+ 25)

2
x
2
(
x
2
+ 25)
2
=
25

x
2
(
x
2
+ 25)
2
.
Since
f
is differentiable everywhere, the only
critical points occur at the solutions of
f
′
(
x
) =
0,
i.e.
, at the solutions of
25

x
2
= 0
.
Consequently, the only critical points are
x
=

5
,
5
.
002
10.0points
Find all the critical points of the function
f
(
x
) = 2 sin
x
 
x

on the interval (

π, π
).
1.
x
= 0
2.
x
=

π
3
,
0
,
2
π
3
3.
x
=

2
π
3
,
π
3
4.
x
=

2
π
3
,
0
,
π
3
correct
5.
x
=

π
6
,
0
,
π
6
6.
x
=

5
π
6
,

π
6
,
0
,
π
6
,
5
π
6
7.
x
=

5
π
6
,
0
,
5
π
6
8.
x
=

π
3
,
2
π
3
Explanation:
Since

x

is differentiable everywhere except
at
x
= 0, while sin
x
is differentiable for all
x
, the point
x
= 0 is a critical point of
f
and
all other critical points will be the solutions of
f
′
(
x
) = 0. Now
f
′
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 Spring '11
 Cathy
 Critical Point, Isaacson

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