garcia (jjg2564) – Hw 09 – gualdani – (56410)
3
can be the graph of
f
.
keywords: absolute minimum, absolute maxi-
mum, local maximum, local minimum
004
10.0 points
Find all the critical points of
f
when
f
(
x
) =
x
x
2
+ 16
.
1.
x
=
−
4
,
16
2.
x
=
−
16
,
16
3.
x
=
−
16
,
4
4.
x
=
−
4
,
4
correct
5.
x
= 0
,
4
6.
x
=
−
4
,
0
Explanation:
By the Quotient Rule,
f
′
(
x
) =
(
x
2
+ 16)
−
2
x
2
(
x
2
+ 16)
2
=
16
−
x
2
(
x
2
+ 16)
2
.
Since
f
is differentiable everywhere, the only
critical points occur at the solutions of
f
′
(
x
) =
0,
i.e.
, at the solutions of
16
−
x
2
= 0
.
Consequently, the only critical points are
x
=
−
4
,
4
.
005
10.0 points
Find all the critical points of
f
when
f
(
x
) =
x
4
/
5
(
x
−
5)
2
.
1.
x
= 0
,
10
7
2.
x
= 0
,
5
7
,
5
3.
x
= 0
,
10
7
,
5
correct
4.
x
=
10
7
,
5
5.
x
=
5
7
,
5
6.
x
= 0
,
5
7
Explanation:
Since
x
1
/
5
is defined for all
x
, the function
f
is defined for all
x
, but it will not be differ-
entiable at
x
= 0. Thus
x
= 0 is one critical
point. Now by the Chain and Product Rules
f
′
(
x
) =
4
5
x
−
1
/
5
(
x
−
5)
2
+ 2
x
4
/
5
(
x
−
5)
=
2(
x
−
5)
5
braceleftBig
2(
x
−
5)
x
1
/
5
+ 5
x
4
/
5