HOMEWORK 8 — SOLUTIONS
Question 1. Let f (x) = 14 x4 − 2x2 + 2.
(a) Find all extremal points and the intervals on which f (x) is increasing/decreasing.
(b) Find all inflection points and the intervals on which f (x) is concave up/down.
(c) Draw a graph of f (x) indicating all previous information.
Solution. f (x) is defined everywhere.
(a) Increasing/decreasing intervals and extremal points.
f 0 (x) = x3 − 4x
We solve
x3 − 4x = 0
x(x − 2)(x + 2) = 0
x = −2, x = 0, x = 2
f (x)
sign of f 0 (x)
−
−2
(f 0 (−3)
−
+
0
f 0 (−1)
+
2
f 0 (1)
= −27 + 12 < 0,
= −1 + 4 > 0,
= 1 − 4 < 0, f 0 (3) = 27 − 12 > 0.)
1
(f (−2) = 4 · 16 − 2 · 4 + 2 = −2, f (0) = 0 − 0 + 2 = 2, f (2) = 41 · 16 − 2 · 4 + 2 = −2.)
Answer:
• f (x) is increasing on (−2, 0), (2, ∞).
• f (x) is decreasing on (−∞, −2), (0, 2).
• (−2, −2) and (2, −2) are local minima, (0, 2) is a local maximum.
(b) Concavity and inflection points.
f 00 (x) = 3x2 − 4
We solve:
3x2 − 4 = 0
4
x2 =
3r
r
4
4
x=−
, x=
3
3
sign of f 00 (x)
−
+
−
1
√
4/3
+
√
4/3
2
HOMEWORK 8 — SOLUTIONS
(f 00q
(−2) = 3 · 4 − 4 > 0, f 00 (0) = 3 · 0 − 4 < 0, f 00 (2)q= 3 · 4 − 4 > 0.)
(f (
4
)
3
=
1
4
·
16
9
−2·
4
3
+2=
4
9
−
8
3
+
6
3
= − 29 , f (
4
)
3
= · · · = − 92 .)
Answer:
q q
• f (x) is CD on (− 43 , 43 ).
q
q
• f (x) is CU on (−∞, − 43 ), ( 43 , ∞).
q
q
• (− 43 , − 29 ) and ( 43 , − 92 ) are infection points.
(c) Graph.
f (x)
3
2
1
−3
−2
−1
1
2
3
x
−1
−2
2x
.
x2 + 1
Find all extremal points and the intervals on which f (x) is increasing/decreasing.
Find all inflection points and the intervals on which f (x) is concave up/down.
Find all asymptotes.
Draw a graph of f (x) indicating all previous information.
Question 2. Let f (x) =
(a)
(b)
(c)
(d)
Solution. f (x) is defined for all x.
(a) Increasing/decreasing intervals and extremal points.
f 0 (x) =
2(x2 + 1) − 2x(2x)
−2x2 + 2
=
(x2 + 1)2
(x2 + 1)2
We solve:
−2x2 + 2
=0
(x2 + 1)2
− 2x2 + 2 = 0
x2 − 1 = 0
(x − 1)(x + 1) = 0
x = −1, x = 1
HOMEWORK 8 — SOLUTIONS
3
f (x)
sign of f 0 (x)
−
−
+
−1
(
2
2
+2
+2
= −2·2
< 0, f 0 (0) = −2·0
> 0,
(22 +1)2
(02 +1)2
2(−1)
2·1
= 1, f (−1) = (−1)2 +1 = −1.)
12 +1
f 0 (−2)
(f (1) =
f 0 (2)
1
=
−2·22 +2
(22 +1)2
< 0.)
Answer:
• f (x) is increasing on (−1, 1).
• f (x) is decreasing on (−∞, −1), (1, ∞).
• (−1, −1) is a local minimum, (1, 1) is a local maximum.
(b) Concavity and inflection points.
(−4x)(x2 + 1)2 − (−2x2 + 2)2(x2 + 1)2x
(x2 + 1)4
2
(−4x)(x + 1) − (−2x2 + 2)4x
=
(x2 + 1)3
3
−4x − 4x + 8x3 − 8x
4x3 − 12x
=
=
(x2 + 1)3
(x2 + 1)3
f 00 (x) =
We solve:
4x3 − 12x
=0
(x2 + 1)3
4x3 − 12x = 0
√
√
4x(x + 3)(x − 3) = 0
√
√
x = − 3, x = 0, x = 3
sign of f 00 (x)
−
−
+
√
− 3
(f 00 (−2) =
4(−8)−12(−2)
(4+1)3
< 0, f 00 (−1) =
4·8−12·2
(4+1)3
> 0)
√
√
√
3
(f (− 3) = −2
= − 23 , f (0) =
3+1
2·0
0+1
√
0
4(−1)−12(−1)
(1+1)3
√
= 0, f ( 3) =
√
2 3
3+1
+
3
> 0, f 00 (1) =
4−12
(1+1)3
< 0, f 00 (2) =
√
=
3
)
2
Answer:
√
√
• f (x) is CD on (−∞,
√ − 3),
√ (0, 3).
• f (x) is CU
on (− 3, 0), (√ 3, ∞).
√
√
√
3
• (− 3, − 2 ), (0, 0), ( 3, 23 ) are inflection points.
(c) Asymptotes.
2x
= lim
x→∞ x2 + 1
x→∞
lim f (x) = lim
x→∞
2 xx2
x2
x2
+
1
x2
2 x1
2·0
=
=0
x→∞ 1 + 12
1+0
x
= lim
Likewise, lim f (x) = 0. Thus, y = 0 is a horizontal asymptote.
x→−∞
4
HOMEWORK 8 — SOLUTIONS
The function f (x) is continuous and defined everywhere, hence there are no vertical
asymptotes.
(d) Graph.
f (x)
1
−4
−3
−2
−1
1
2
3
4
x
−1
ex
.
x2
(a) Find all extremal points and the intervals on which f (x) is increasing/decreasing.
(b) Find all inflection points and the intervals on which f (x) is concave up/down.
x
(c) Find all asymptotes. You may use the fact that lim xe 2 = ∞.
Question 3. Let f (x) =
x→∞
(d) Draw a graph of f (x) indicating all previous information.
Solution. f (x) is defined for x 6= 0.
x
Notice that f (x) = xe 2 = x−2 ex . It easier to differentiate the latter expression!
(a) Increasing/decreasing intervals and extremal points.
f 0 (x) = −2x−3 ex + x−2 ex = (x−2 − 2x−3 )ex =
(x − 2)ex
x3
We solve:
(x − 2)ex
=0
x3
x
(x − 2)e = 0
[ex 6= 0]
x−2=0
x=2
f (x)
sign of f 0 (x)
−
+
0
+
2
(Recall that f (x) is not defined at x = 0.)
(f 0 (−1) =
(f (2) =
(−1−2)e−1
−1
> 0, f 0 (1) =
(1−2)e1
1
< 0, f 0 (3) =
e2
)
4
Answer:
• f (x) is increasing on (−∞, 0), (2, ∞).
• f (x) is decreasing on (0, 2).
2
• (2, e4 ) is a local minimum.
(3−2)e3
33
> 0)
HOMEWORK 8 — SOLUTIONS
5
(b) Concavity and inflection points.
f 00 (x) = (−2x−3 + 6x−4 )ex + (x−2 − 2x−3 )ex = (x−2 − 4x−3 + 6x−4 )ex
=
(x2 − 4x + 6)ex
x4
We solve:
(x2 − 4x + 6)ex
=0
x4
2
x
(x − 4x + 6)e = 0
[ex 6= 0]
x2 − 4x + 6 = 0
The discriminant is (−4)2 − 4 · 1 · 6 < 0, so there are no solutions!
sign of f 00 (x)
+
+
0
(Recall that f (x) is not defined at x = 0.)
(f 00 (−1) =
(1−4(−1)+6)e−1
1
> 0, f 00 (1) =
(1−4+6)e1
1
> 0.)
Answer:
• f (x) is CU on (−∞, 0), (0, ∞).
• f (x) has no inflection points.
(c) Asymptotes. Horizontal Asymptotes:
ex given
= ∞
x→∞
x→∞ x2
lim ex
ex
0
0
x→−∞
lim f (x) = lim 2 =
=
=
=0
x→−∞
x→−∞ x
( lim x)2
(−∞)2
∞
lim f (x) = lim
x→−∞
Thus, y = 0 is a horizontal asymptote.
Vertical asymptotes may occur when the denominator of
x = 0.
lim f (x) = lim
x→0+
x→0+
is 0, namely, when
ex
1
= lim ex · ( lim )2 = e0 · (∞)2 = ∞
2
+
+
x
x→0
x→0 x
Likewise, limx→0− f (x) = ∞.
Therefore, x = 0 is a vertical asymptote.
(d) Graph.
ex
x2
6
HOMEWORK 8 — SOLUTIONS
f (x)
5
4
3
2
1
−2
−1
1
2
3
4
x