MATH 100, HOMEWORK 7 SOLUTIONS
Section 4.3, #14: (a) If f (x) = cos2 x−2 sin x, then f 0 (x) = −2 cos x sin x−
2 cos x = −2 cos x(sin x + 1). We first find critical points: if f 0 (x) = 0 then
either sin x = −1, x = 3/pi/2, or cos x = 0, x = π/2 or x = 3π/2. Since
sin x + 1 ≥ 0 for all x, f 0 (x) has the same sign as −2 cos x, i.e. it is negative
on [0, π/2) and (3π/2, 2π], and positive on (π/2, 3π/2). This means that f (x)
is decreasing on [0, π/2] and [3π/2, 2π], and increasing on [π/2, 3π/2).
(b) By the work in (a) and the first derivative test, f (x) has a local minimum
at π/2 with value f (π/2) = 0 − 2 = −2, and a local maximum at 3π/2 with
value f (3π/2) = 0 + 2 = 2.
(c) We have f 00 (x) = 2 sin2 x − 2 cos2 x + 2 sin x = 2 sin2 x − 2(1 − sin2 x) +
2 sin x = 4 sin2 x + 2 sin x − 2 = 2(2 sin2 x + sin x − 1). Thus f 00 (x) = 0 when
√
−1 ± 3
1
−1 ± 1 + 8
=
= , −1,
sin x =
4
4
2
00
so that x = π/6, 5π/6, or 3π/2. We have f (x) > 0 when sin x < −1
(no solutions) or sin x > 1/2, i.e. π/6 < x < 5π/6. Thus f is concave
up on (π/6, 5π/6) and concave down on (0, π/6) and (5π/6, 2π). There are
inflection points at x = π/6, 5π/6, with values f (π/6) = 43 − 1 = − 41 and
f (5π/6) = 34 − 1 = − 14 . (Note that although f 00 (3π/2) = 0, there is no
inflection point there, as f 00 does not change sign at this point.)
4x3
. This is
x4 + 27
0 at x = 0, positive for x > 0, and negative for x < 0. Hence f is decreasing
on (−∞, 0] and increasing on [0, ∞).
Section 4.3, #42: (a) Let f (x) = ln(x4 +27), then f 0 (x) =
(b) By (a) and the first derivative test, there is a local minimum at x = 0,
with f (0) = ln 27.
(c) We have
f 00 (x) =
12x6 + 12 · 27x2 − 16x6
4x2 (81 − x4 )
12x2 (x4 − 27) − 4x3 (4x3 )
=
=
.
(x2 + 27)2
(x2 + 27)2
(x2 + 27)2
This is 0 when x = 0 or x4 = 81, x2 = 9, x = ±3. To determine the sign
of f 00 on the intervals between these points, we try a sample point in each
interval:
400(81 − 10000)
4(81 − 1)
f 00 (±10) =
< 0, f 00 (±1) =
> 0.
2
127
282
1
Hence f 00 < 0 on (−∞, −3) and (3, ∞), and f 00 > 0 on (−3, 0) and (0, 3).
Thus f is concave down on (−∞, 3) and (3, ∞), and concave up on (−3, 3).
There are inflection points at x = ±3, with f (±3) = ln(81 + 27) = ln 108.
ex
is defined and continuous everywhere,
1 + ex
hence there are no vertical asymptotes. To check for horizontal asymptotes,
we find the limits at ±∞:
Section 4.3, #50: (a) f (x) =
0
ex
= 0,
=
x→−∞ 1 + ex
1+0
lim f (x) = lim
x→−∞
ex
1
1
=
= 1,
=
lim
x→∞
x→∞ 1 + ex
x→∞ e−x + 1
0+1
hence there are horizontal asymptotes y = 0 at −∞ and y = 1 at ∞.
ex (1 + ex ) − ex ex
ex
=
> 0 for all x, hence f is increasing
(b) f 0 (x) =
(1 + ex )2
(1 + ex )2
on (−∞, ∞).
lim f (x) = lim
(c) From (b), there are no local minima or maxima.
(d)
ex (1 + ex )2 − ex · 2(1 + ex )ex
ex (1 + 2ex + e2x − 2ex − 2e2x )
f (x) =
=
(1 + ex )4
(1 + ex )4
00
=
ex (1 − e2x )
,
(1 + ex )4
which has the same sign as 1−e2x (since the other factors are always positive).
Hence f 00 (x) = 0 for x = 0, f 00 > 0 for x < 0, and f 00 < 0 for x > 0. Thus f is
concave up on (−∞, 0) and concave down on (0, ∞). There is an inflection
point at x = 0 with f (0) = 21 .
Section 4.5, #10: y is defined for all x 6= 0, 2. Moreover,
x2 − 4
(x + 2)(x − 2)
x+2
2
y= 2
=
=
= 1 + for x 6= 2,
x − 2x
x(x − 2)
x
x
hence the graph of f is exactly the same as the graph of g(x) = 1 + x2 except
that the point with x = 2 is excluded. We have
lim f (x) = lim 1 +
x→±∞
x→±∞
2
2
= 1,
x
hence there is a horizontal asymptote y = 1. There is also a vertical asymptote x = 0, with
lim− f (x) = lim− 1 +
x→0
x→0
2
= −∞,
x
lim+ f (x) = lim+ 1 +
x→0
x→0
2
= +∞.
x
2
2
= 1 + = 2. There is an x-intercept
x
2
at (−2, 0). The function is neither even, nor odd, nor periodic.
Finally, we have lim f (x) = lim 1 +
x→2
x→2
On the domain of f we have f 0 (x) = −2x−2 , which is negative for all x 6= 0.
Hence f is decreasing on each of its intervals of definition: (−∞, 0), (0, 2),
(2, ∞). There are no local extrema.
We also have f 00 (x) = 4x−3 , which is negative for x < 0 and positive for
x > 0. Hence f is concave down on (−∞, 0) and concave up on (0, 2) and
(2, ∞).
Section 4.5, # 32: y = x + cos x is defined for all x. Since −1 ≤ cos x ≤ 1,
limx→−∞ f (x) = −∞ and limx→∞ f (x) = ∞. We have y(0) = 1. There are
no obvious symmetries. There are no asymptotes.
We have y 0 = 1 − sin x. To find critical points, we solve y 0 = 0: sin x = 1,
x = π/2 + 2kπ, k ∈ Z. For all other x we have y 0 > 0. Thus y is increasing
on (−∞, ∞). There are no local minima or maxima.
We have y 00 = − cos x, hence y 00 = 0 when cos x = 0, x = π2 + 2kπ. Also,
y 00 < 0 on each interval (− π2 + 2kπ, π2 + 2kπ), and y 00 > 0 on each interval
( π2 + 2kπ, 3π
+ 2kπ). These are the intervals where f is concave down and
2
concave up, respectively. There are inflection points at x = π2 + 2kπ, with
y( π2 + 2kπ) = π2 + 2kπ.
Section 4.5, #24: By long division, we have 5x4 + x2 + x = (5x + 5)(x3 −
x2 + 2) + (6x2 − 9x − 10). Hence
6x2 − 9x − 10
5x4 + x2 + x
=
(5x
+
5)
+
.
x3 − x2 + 2
x3 − x2 + 2
The second term goes to 0 as x → ±∞. Hence there is a slant asymptote
y = 5x + 5.
3