Spring 2011 Math 151
Exam III Version B Solutions
1.
E
Since
the
numerator
and
denomina
tor both approach 0, use L’Hospital’s Rule:
lim
x
→
0
e
x

cos
x

2
x
x
2

2
x
= lim
x
→
0
e
x
+ sin
x

2
2
x

2
=
1
2
.
2.
A
Use properties of logarithms:
ln(
x
2
+
x
) =
ln(
x
+4)
, so
x
2
+
x
=
x
+4
, x
2

4 = 0
, which
yields
x
= 2
or
x
=

2
. Since the domain of
the lefthand side of the original equation is
x >
0
, the only solution is
x
=
2
.
3.
A
f
′
positive means
f
is increasing, and
f
′
decreasing means
f
′′
is negative, so
f
is con
cave down.
The only graph increasing and
concave down is graph
A
.
4.
C
The
original
function
f
is
decreasing
when
f
′
is negative, which occurs when
x
∈
(
a
,
c
)
∪
(
e
∞
)
.
5.
E
f
has a critical value when
f
′
= 0
, namely
when
x
=
a, c, e
.
Using the signs of the
derivative
f
′
, we find that
f
is increasing
for
x
∈
(
∞
, a
)
∪
(
c, e
)
and decreasing for
x
∈
(
a, c
)
∪
(
e,
∞
)
.
Therefore,
f
has a local
minimum only when
x
=
c
.
6.
D
Let
y
= log
4
parenleftbigg
1
8
parenrightbigg
. Then
4
y
=
1
8
, or
2
2
y
=
2
−
3
, which means
2
y
=

3
and
y
=

3
2
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 ArtBelmonte
 Calculus, Derivative, Mathematical analysis, Convex function

Click to edit the document details