Homework 8 Solutions
Math 132 Section 01
Chapter 5.3, problem 24. Solution:
ln (1 + 2x)
.
x2
The top and the bottom each approach 0 so we may apply L’Hospital’s rule. So
lim
x→0−
ln (1 + 2x)
lim
= lim
x2
x→0−
x→0−
Chapter 5.3, problem 42. Solution:
2
(1+2x)
2x
= lim
x→0−
1
= −∞.
(1 + 2x)x
e2x − 1 − 2x − 2x2
.
x→0
x3
The top and the bottom each approach 0 so we can apply L’Hospital’s rule.
lim
2e2x − 2 − 4x
4e2x − 4
8e2x
8
4
e2x − 1 − 2x − 2x2
=
lim
=
lim
=
lim
= = .
3
2
x→0
x→0
x→0 6
x→0
x
3x
6x
6
3
We used L’Hospital’s rule several times. Each time we used L’Hospital’s rule, the numerator and
denominator each were 0.
Chapter 5.4, problem 26. Solution:
lim
x2
.
1 + x2
First, f (x) is defined for all x, f (x) = 0 only for x = 0, and f (x) is positive for all x. Also,
f (x) =
x2
x→∞ 1 + x2
lim
is an
∞
∞
form, so we use L’Hospital’s rule:
x2
2x
= lim
= 1.
2
x→∞ 1 + x
x→∞ 2x
lim
Likewise,
x2
= 1.
x→−∞ 1 + x2
So f (x) has horizontal asymptotes along y = 1 for x → ±∞.
Now we look at local extrema, where f (x) is increasing and decreasing.
lim
f 0 (x) =
2x · (1 + x2 ) − x2 · (2x)
2x
=
.
2
2
(1 + x )
(1 + x2 )2
f 0 (x) = 0 only at x = 0 and f 0 (x) is defined everywhere. Furthermore, f 0 (x) < 0 for x < 0 and
f 0 (x) > 0 for x > 0, so f (x) is increasing on (0, ∞) and f (x) is decreasing on (−∞, 0), and x = 0
gives a local minimum, and the value of the local minimum is f (0) = 0.
Now we look at inflection points and concavity of f (x).
0
2x
2 · (1 + x2 )2 − 8x2 · (1 + x2 )
2 · (1 − 3x2 )
f 00 (x) =
=
=
.
2
2
2
4
(1 + x )
(1 + x )
(1 + x2 )3
1
2
−1 √1
−1
So f 00 (x) = 0 for 1−3x2 = 0 so x = ± √13 . Also, f 00 (x) > 0 on ( √
, 3 ) and f 00 (x) < 0 on (−∞, √
)∪
3
3
−1 √1
−1
( √13 , ∞), so f (x) is concave up on ( √
, 3 ) and f (x) is concave down on (−∞, √
) ∪ ( √13 , ∞). The
3
3
concavity changes at x = ± √13 so f (x) has inflection points at x = ± √13 .
Chapter 5.4, problem 46. Solution:
f (x) = lnxx . f (x) is defined only for x > 0. f (x) = 0 at x = 1, f (x) < 0 on (0, 1) and f (x) > 0 on
(1, ∞).
lim
x→0+
ln x
= −∞
x
and
ln x
1
= lim
= 0,
x→∞
x
x
where we used L’Hospital’s rule for the second limit. So f (x) has a vertical asymptote at x = 0
and a horizontal asymptote along y = 0.
Now we look at local extrema, where f (x) is increasing and decreasing.
lim
x→∞
1
x
· x − ln x
1 − ln x
=
x2
x2
0
0
0
so f (x) = 0 for x = e, f (x) > 0 on (0, e) and f (x) < 0 on (e, ∞). So f (x) is increasing on (0, e)
and f (x) is decreasing on (e, ∞). Furthermore, x = e gives a local max since f (x) changes from
increasing to decreasing. The value of the local max is f (e) = lnee = 1e .
Now for concavity,
− 1 · x2 − 2x · (1 − ln x)
2 ln x − 3
f 00 (x) = x
=
,
4
x
x3
so f 00 (x) = 0 implies 2 ln x − 3 = 0 so ln x = 32 so x = e3/2 . Also, f 00 (x) < 0 on (0, e3/2 ) and
f 00 (x) > 0 on (e3/2 , ∞) so f (x) is concave down on (0, e3/2 ) and f (x) is concave up on (e3/2 , ∞),
and f (x) has an inflection point at x = e3/2 .
f 0 (x) =
3
Chapter 5.5, problem 26. Solution:
2
f (x) = x9−x
2 +4 . The absolute extrema, if they exist, must occur at the critical values.
f 0 (x) =
−26x
−2x · (x2 + 4) − 2x · (9 − x2 )
= 2
.
2
2
(x + 4)
(x + 1)2
The only critical value is x = 0. f (0) = 49 . Also,
9 − x2
= −1.
x→±∞ x2 + 4
and there is no absolute minimum.
lim
So the absolute maximum is f (0) =
9
4