husain (aih243) – Exam02Review – Gilbert – (56215)
1
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16
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001
10.0 points
On which interval(s) is
f
(
x
) =
x
4
+ 2
x
2
−
1
increasing?
1.
(
−∞
,
−
1 ]
,
[ 1
,
∞
)
2.
(
−∞
,
0 ]
3.
[
−
1
,
1 ]
4.
[ 0
,
∞
)
correct
5.
[
−
1
,
0 ]
,
[ 1
,
∞
)
6.
(
−∞
,
−
1 ]
,
[ 0
,
1 ]
Explanation:
A function
f
is increasing on an interval
[
a , b
] when
f
′
(
x
)
>
0 on (
a, b
). Now
f
′
(
x
) = 4
x
3
+ 4
x
= 4
x
(
x
2
+ 1)
.
From the sign chart
−
+
0
−∞
∞
for
f
′
it thus follows that
f
is increasing on
[ 0
,
∞
)
.
002
10.0 points
A polynomial function
f
has the properties
(i)
f
(
−
8) = 0 =
f
(
−
2),
(ii)
f
′′
(
x
)
>
0 on (
−
8
,
−
2).
Which of the following statements is a conse
quence of these properties?
1.
f
(
x
)
≥
0 on [
−
8
,
−
2]
2.
f
′
(
x
)
≥
0 on (
−
8
,
−
2)
3.
f
(
x
) = 0 on [
−
8
,
−
2]
4.
f
(
x
)
≤
0 on [
−
8
,
−
2]
correct
5.
f
′
(
x
)
≤
0 on (
−
8
,
−
2)
Explanation:
Condition (ii) ensures that the graph of
f
is concave up on the interval (
−
8
,
−
2).
On
the other hand, by (i),
f
(
−
8) = 0 =
f
(
−
2),
so both endpoints of the graph on [
−
8
,
−
2]
lie on the
x
axis. Hence the graph of
f
must
lie below the
x
axis on the interval (
−
8
,
−
2),
i.e.
,
f
(
x
)
≤
0 on [
−
8
,
−
2]
.
Notice that the inequality
f
(
x
)
<
0 must hold
on the interval (
−
8
,
−
2) because
f
′′
(
x
)
>
0 on
(
−
8
,
−
2).
003
10.0 points
Which one of the following properties does
f
(
x
) =
x
−
1
x
2
+ 24
have?
1.
local max at
x
=
−
6
2.
local max at
x
= 4
3.
local min at
x
= 6
4.
local min at
x
= 4
5.
local min at
x
=
−
6
6.
local max at
x
= 6
correct
Explanation:
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husain (aih243) – Exam02Review – Gilbert – (56215)
2
By the Quotient rule,
f
′
(
x
) =
x
2
+ 24
−
2
x
(
x
−
1)
(
x
2
+ 24)
2
=
24 + 2
x
−
x
2
(
x
2
+ 24)
2
.
The critical points of
f
occur when
f
′
(
x
) = 0,
i.e.
, at the solutions of
f
′
(
x
) =
(4 +
x
)(6
−
x
)
(
x
2
+ 24)
2
= 0
.
Thus the critical points of
f
are
x
=
−
4 and
x
= 6.
To classify these critical points we use the
First Derivative Test. But the sign of
f
′
de
pends only on the numerator, so it is enough,
therefore, to look only at a sign chart for
(4 +
x
)(6
−
x
):
−
4
6
−
−
+
From this it follows that
f
is decreasing on
(
−∞
,
−
4),
increasing on (
−
4
,
6),
and de
creasing on (6
,
∞
). Consequently,
f
has a
local maximum at
x
= 6
.
keywords:
local maximum, local minimum,
critical point, quotient rule, First Derivative
Test, rational function
004
10.0 points
Determine if the limit
lim
x
→ −∞
2 +
x
−
5
x
3
2 + 4
x
3
exists, and if it does, find its value.
1.
limit =
−
1
2.
limit does not exist
3.
limit = 1
4.
limit =
5
4
5.
limit =
−
5
4
correct
Explanation:
Dividing by
x
3
in the numerator and de
nominator we see that
2 +
x
−
5
x
3
2 + 4
x
3
=
2
x
3
+
1
x
2
−
5
2
x
3
+ 4
.
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 Spring '06
 McAdam
 Calculus, Derivative, Husain

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