MA 115 Homework Solutions for Week 5
4.2
#6
Absolute maximum value is
f
(8) = 5; absolute minimum value is
f
(2) = 0; local maximum values are
f
(1) = 2,
f
(4) = 4, and
f
(6) = 3; local minimum values are
f
(2) = 0,
f
(5) = 2, and
f
(7) = 1.
4.2
#24
f
(
x
) =
x
3
+
x
2
−
x
⇒
f
′
(
x
) = 3
x
2
+ 2
x
−
1
f
′
(
x
) = 0
⇒
(
x
+ 1)(3
x
−
1) = 0
⇒
x
=
−
1
,
1
3
. These are the only critical numbers.
4.2
#38
f
(
x
) =
x
3
−
3
x
+ 1
,
[0
,
3].
f
′
(
x
) = 3
x
2
−
3 = 0
⇔
x
=
±
1, but
−
1
/
∈
[0
,
3]. So by testing the values
x
= 0
,
1
,
3
we see
f
(0) = 1,
f
(1) =
−
1 and
f
(3) = 19. So
f
(3) = 19 is the absolute maximum value and
f
(1) =
−
1 is
the absolute minimum value.
4.2
#46
f
(
x
) =
x
−
2 cos
x,
[
−
π, π
].
f
′
(
x
) = 1+2 sin
x
= 0
⇔
sin
x
=
−
1
2
⇔
x
=
−
5
π
6
,
−
π
6
. By comparing the function
value at both endpoints and the critical points we see
f
(
−
π
) = 2
−
π
≈ −
1
.
14,
f
(
−
5
π
6
) =
√
3
−
5
π
6
≈ −
0
.
886,
f
(
−
π
6
) =
−
π
6
−
√
3
≈ −
2
.
26,
f
(
π
) =
π
+ 2
≈
5
.
14. So
f
(
π
) =
π
+ 2 is the absolute maximum value and
f
(
−
π
6
) =
−
π
6
−
√
3 is the absolute minimum value.
4.3
#6
(a)
f
is increasing on the intervals where
f
′
(
x
)
>
0, namely, (2
,
4) and (6
,
9).
(b)
f
has a local maximum where it changes from increasing to decreasing, that is, where
f
′
changes from
positive to negative (at
x
= 4). Similarly, where
f
′
changes from negative to positive,
f
has a local
minimum (at
x
= 2 and at
x
= 6).
(c) When
f
′
is increasing, its derivative
f
′′
is positive and hence,
f
is concave upward. this happens on
(1
,
3)
,
(5
,
7) and (8
,
9). Similarly,
f
is concave downward when
f
′
is decreasing–that is, on (0
,
1)
,
(3
,
5)
and (7
,
8).