Solutions of Homework 8, Mathematics 1
submit by 5.12.
Only problems 1d, 2bcd, 3, 4a, 6, 7abde and 8* will be graded.
Problem 1: Identify the intervals on which the following functions are monotone increasing:
(a)
f (x) = x2 − x,
(b)
f (x) = sin x,
(c)
f (x) = arcsin x,
x
.
f (x) =
2 + x2
(d) [2 points]
Solution: (a): f 0 (x) = 2x − 1 > 0 for x ∈ (1/2, ∞), f 0 (x) = 2x − 1 < 0 for
x ∈ (−∞, 1/2). Moreover, f is continuous on R (in particular at 1/2). Hence,
f is strictly increasing on [1/2,
S ∞) and strictly decreasing on (−∞, 1/2].
(b):
Strictly
increasing
on
k∈Z [2kπ − π/2, 2kπ + π/2], strictly decreasing on
S
[2kπ
+
π/2,
2kπ
+
3π/2].
k∈Z
(c): Strictly increasing on [−1, 1].
(d):
(
√
√
< 0 for x ∈ (−∞, − 2) ∪ ( 2, +∞),
1 · (2 + x2 ) − x · 2x
2 − x2
0
√
√
f (x) =
=
(2 + x2 )2
(2 + x2 )2 > 0 for x ∈ (− 2, 2).
Moreover,
on R and continuous. Hence, f √
is strictly
decreasing on
√ f is defined
√
√
(−∞, − 2] and [ 2, +∞)√and stricty
increasing
on
[−
2,
2].
Note
that f is
√
not decreasing on (−∞, − 2] ∪ [ 2, +∞): for example, f (−2) = −1/3 < 1/3 =
f (2), although −2 < 2.
Problem 2: Find all local maxima and minima (and decide which of the two
each is) of the following functions:
(a) f (x) = x3 + 2x2 ,
(b) [3 points] f (x) = x5 − 2x4 ,
Hint: it might be computationally easier to avoid computing the second
derivative in the critical point – how else can you decide?
(c) [2 points] f (x) = 1 − |x|,
(d) [3 points] f (x) = sin2 x.
Solution: (b): f (x) = x5 − 2x4 is continuous and differentiable everywhere, so
in any local extremum, f 0 must be zero. Since f 0 (x) = 5x4 −8x3 = x3 (5x−8) = 0
1
if and only if x = 0 or x = 8/5, these are the only candidates. Since

3
00

“(−) · (−) > 0 for x < 0,
f 0 (x) = “(+)3 · (−)00 < 0 for 0 < x < 8/5,


“(+)3 · (+)00 > 0 for x > 8/5,
f must have local maximum at 0, with value of the local minimum f (0) = 0,
and a local minimum at 8/5, with value of the local maximum
f
4 4 5
8
8
8
8
8
84 2
−2
=
=
−2 =− 4 · .
5
5
5
5
5
5 5
(c):
(
1 + x for x < 0,
f (x) = 1 − |x| =
1 − x for x ≥ 0.
Obviously f is continuous on R (f (0) = 1 = limx→0− (1 + x) = limx→0+ (1 − x))
and differentiable on R \ {0}. In 0, we have
(1 + x) − 1
= lim 1 = 1,
x→0+
x−0
(1
−
x)
−
1
0
f+
(0) = lim
= lim −1 = −1.
x→0+
x→0+
x−0
0
f−
(0) = lim
x→0−
Hence, f is increasing on (−∞, 0] and decreasing on [0, +∞), making 0 the point
of local (and global) maximum, with value f (0) = 1.
Proof by scatching the graph of 1 − |x| (using the well-known graph of |x|)
is also taken as OK.
(d): f is continuous and differentiable on R, with
f 0 (x) = 2 sin x · cos x = sin(2x) = 0
if and only if x = k π2 , k ∈ Z. Moreover,
(
f 00 (x) = (sin(2x))0 = 2 cos(2x) =
2 > 0 for x = 2k π2 = kπ,
−2 < 0 for x = (2k + 1) π2 .
Hence, f has local minima at x = kπ, k ∈ Z, with values of
f (kπ) = sin2 (kπ) = 0,
and local maxima at x = (2k + 1) π2 , k ∈ Z, with values of
π
f (2k + 1)
= sin2 (kπ + π/2) = (±1)2 = 1.
2
2
Problem 3 [2×3 points]: Find the global maximum and minimum of the
following functions on the specified intervals:
(a) f (x) = x3 − 3x2 , x ∈ [−1, 4];
(b) f (x) = sin x + cos2 x, x ∈ [0, π].
Solution: (a): f is continuous on [−1, 4] and differentiable on (−1, 4). We
need to compare values at endpoints and at any points where f 0 (0) = 0. We
have f 0 (x) = 3x2 − 6x = 3x(x − 2) = 0 if and only if x = 0 or x = 2. Both of
these points lie in the interval [−1, 4], and
f (−1) = (−1)3 − 3(−1)2 = −4,
f (0) = 03 − 3 · 02 = 0,
f (2) = 23 − 3 · 22 = −4,
f (4) = 43 − 3 · 42 = 16.
Hence, within the interval [−1, 4], f has global maximum at 4, with value of 16,
and global minimum both at −1 and 2, with value of −4.
(b): f is continuous on [0, π] and differentiable on (0, π). We need to compare
values at endpoints and at any points where f 0 (0) = 0. We have
f 0 (x) = cos x + 2 cos x · (− sin x) = cos x(1 − 2 sin x) = 0
if and only cos x = 0, i. .e x = π/2, or if sin x = 1/2, i. .e x = π/6 or x = 5π/6
(we only looked for solutions in the interval (0, π)). The values in these points
and endpoints are
f (0) = sin 0 + (cos 0)2 = 0 + 12 = 1,
√ !2
π 2
1
5
3
π = +
= ,
f (π/6) = sin + cos
6
6
2
2
4
π 2
π
f (π/2) = sin + cos
= 1 + 02 = 1,
2
2
√ !2
2
5π
5π
1
3
5
f (5π/6) = sin
+ cos
= + −
= ,
6
6
2
2
4
f (π) = sin π + (cos π)2 = 0 + (−1)2 = 1.
Hence, within the interval [0, π], f has global maximum both at π/6 nad 5π/6,
with value of 5/4, and global minimum at all three points 0, π/2, and π, with
value of 1.
Problem 4: Do the detailed graphing (intersections with the axes, monotonicity, local extrema, convexity/concativy, inflexion points, sketch of graph) for the
3
functions
(a) [6 points] f (x) = sin x −
p
(b) f (x) = x 4 − x2 .
√
3 cos x on the interval [0, 2π],
Solution: See
https://www.math.ucdavis.edu/˜kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html,
problems 9 and 10, and the links there.
Problem 5: Let a ∈ R and x ≥ −1, x 6= 0. Use Lagrange’s Mean Value
Theorem to prove the following inequalities:
(a) if a < 0 or a > 1, then (1 + x)a > 1 + ax;
(b) if 0 < a < 1, then (1 + x)a < 1 + ax.
Hint: Apply Lagrange’s Mean Value Theorem to the function f (z) = (1 + z)a
on the interval [0, x] (if x > 0) or [x, 0] (if x < 0).
Problem 6 [4 points]: Let√a (one-dimensional) displacement of a particle be
given by the formula s(t) = 3t + 4, t ≥ 0. Find time t0 ∈ (0, 4) such that the
instantenous velocity at time t0 is equal to the average velocity over the interval
[0, 4]. Identify what this t0 is corresponding to in the Lagrange’s Mean Value
Theorem.
Solution: The average velocity on the interval [0, 4] is equal to
√
√
s(4) − s(0)
4−2
1
3·4+4− 3·0+4
vavg =
=
=
= .
4
4
4
2
The instanteneous velocity is equal to
0
1
3
v(t) = s0 (t) = (3t + 4)1/2 = (3t + 4)−1/2 · 3 = √
.
2
2 3t + 4
If this is equal to vavg , we must have:
1
3
= √
,
2
2 3t + 4
√
3t + 4 = 3,
3t + 4 = 9,
t = 5/3.
On the other hand, indeed
3
3
1
v(5/3) = p
= √ = = vavg
2
2 9
2 3 · 5/3 + 4
(alternatively, one can check that the manipulations before were equivalent,
namely that we can also get from the bottom to top: in particular, check that
4
in the step where one takes square root, both sides are positive and later we
don’t divide by 0). Hence, the t we were looking for is t0 = 5/3. Since for this
t0 we have s0 (t0 ) = s(4)−s(0)
, this t0 corresponds to the ξ in Lagrange’s Mean
4
Value Theorem applied on interval [0, 4] to the function s (as s satiesfies the
assumptions of the theorem, we are guaranteed the existence of the ξ = t0 ).
Problem 7 [4 × 3 points]: Compute the limits, possibly using l’Hospital rule
(make sure to check it’s assumptions):
ax − xa
for a > 0,
x→a x − a
(a) lim
1
(b) lim x 1−x ,
x→1
1
(c) lim
− cotg x ,
x→0 x
1
2
(d) lim
−
cotg
x
,
x→0 x2
Hint: One saves oneself a lot of differentiation if, after writing the
expression as a fraction, one uses the formula a2 − b2 = (a + b)(a − b)
and splits the limit into a product of two limits.
(e) lim
x→e
ln(ln x)
,
sin(x − e)
(f ) lim (1 − y ln y)
x→0+
1
√
y
.
Solution: (a):
ax · ln a − axa−1
(ax − xa )0
ax − xa “ 00 ”
=
lim
= lim
x→a
x→a (x − a)0
x→a x − a
1
lim
= aa · ln a − aaa−1 = aa (ln a + 1).
Notice we used a > 0 (and hence x > 0 near a) several times. Since the limit
on the right exists, l’Hospital rule is justified.
Solution: (b):
1
ln x
lim x 1−x = lim e 1−x = elimx→1
x→1
x→1
ln x
1−x
,
0
ln x
(ln x)
1/x
= lim
= −1, so
= lim
0
x→1
x→1
1−x
(1 − x)
−1
1
1
lim x 1−x = e−1 = .
x→1
e
lim
“ 00 ”
x→1
Since the limit on the right in the middle equality exists, l’Hospital rule is
justified. Then we used the continuity of y 7→ ey .
5
Solution: (d):
1
sin2 x − x2 cos2 x
2
lim
−
cotg
x
=
lim
x→0 x2
x→0
x2 sin2 x
(sin x − x cos x) (sin x + x cos x)
= lim
x→0
x2 sin x
x
sin x − x cos x
sin x + x cos x
= lim
lim
x→0
x→0
x2 sin x
x
“ 00 ”
sin x
cos x − cos x + x sin x
lim
+ cos x
= lim
x→0
2x sin x + x2 cos x x→0
x
sin x
sin x
= lim
lim
+ lim cos x
x→0 2 sin x + x cos x x→0
x→0
x
sin x/x
= lim
· (1 + 1)
x→0 2 sin x/x + cos x
limx→0 sin x/x
·2
=
limx→0 (2 sin x/x + cos x)
1
2
=
·2=
2+1
3
The calculation is justified backwards: since all the limits in the end existed and
the operations with them were well-defined, one can proceed one-by-one step
upwords.
Solution: (e):
11
1 1
1
ln(ln x) “ 00 ”
ln x x
= 1e = .
= lim
x→e sin(x − e)
x→e cos(x − e)
cos 0
e
lim
Problem 8* [4 bonus points]: Consider the function
( 1
e− x2
if x =
6 0,
f (x) =
0
if x = 0.
Compute f (n) (0) for all n ∈ N.
Solution: First, notice that
1
lim e− x2
x→0
y= x12
=
lim e−y = lim
y→+∞
y→+∞
1
= 0,
ey
so f is continuous. Next,
1
0
f+
(0)
e − x2
= lim
x→0+
x
1
0
f−
(0) = lim
x→0−
e − x2
x
1
y= x
y “∞
1
∞”
lim
2 =
2 = 0,
y→+∞ ey
y→+∞ 2yey
1
y= x
y “∞
1
∞”
lim
2 =
2 = 0,
y→−∞ ey
y→−∞ 2yey
=
=
lim
lim
6
(we used l’Hospital rule...) so f 0 (0) = 0.
To compute f 00 (0), we also need to know f 0 (x) for x 6= 0:
f 0 (x) =
2 − 12
e x .
x3
Then
00
f (0) = lim
x→0+
2 − x12 y= 1
x2
x3 e
x
=
2y 2“ ∞
4y“ ∞
4
∞”
∞”
=
lim
=
lim
= 0.
y→+∞ ey
y→+∞ ey
y→+∞ ey
lim
To compute f (n) (0) for general n, we first express f (n) (x) for x 6= 0. Trying
out a few more derivatives suggest that for n ≥ 0 and x 6= 0, f (n) (x) has the
form
1
1
(n)
(1)
f (x) = Pn
e− x2 ,
x
where Pn (y) = an3n y 3n + an3n−1 y 3n−1 + . . . + an1 y + an0 is a polynomial of order 3n
(actually, the lowest nonzero term will be ann+2 y n+2 , but we will not need this).
We prove statement (1) by induction. For n = 0, this is true with P0 (y) = 1
being of order 0. (We even checked the statement for n = 1 with P1 (y) = 2y 3 .)
Now assume (1) holds for n = k (k ∈ N0 ), we will prove it for n = k + 1.
Differentiating (1) for n = k gives
0
1
− x12
(k+1)
(k) 0
e
f
(x) = (f ) (x) = Pk
x
0
0
1
1
1
1
e − x2
e − x2 + P k
= Pk
x
x
0
1
1
1
1
= ak3k 3k + ak3k−1 3k−1 + . . . + ak1 + ak0 e− x2
x
x
x
1
1
2 − 12
1
e x
+ ak3k 3k + ak3k−1 3k−1 + . . . + ak1 + ak0
x
x
x
x3
1
1
1
1
= −3kak3k 3k+1 − (3k − 1)ak3k−1 3k − . . . − ak1 2 e− x2
x
x
x
1
1
1
1
1
+ 2ak3k 3k+3 + 2ak3k−1 3k+2 + . . . + 2ak1 4 + 2ak0 3 e− x2
x
x
x
x
1
1
= Pk+1
e − x2 ,
x
where
Pk+1 (y) =2ak3k y 3k+3 + 2ak3k−1 y 3k+2 + (2ak3k−2 − 3kak3k )y 3k+1 + . . .
+ (2ak0 − 2ak2 )y 3 − ak1 y 2 ,
which is a polynomial of order 3(k + 1), so we proved (1) for n = k + 1 and
hence for any n ∈ N0 .
7
Next, one shows that for any n ∈ N0 ,
lim
x→0
1 − 12
e x = 0.
xn
(2)
For n even, this can be done in the same way as when we computed (2) for
n = 0 and n = 4 above (when checking continuity of f at 0 and computing
f 00 (0)): after the substitution y = x12 , we can use l’Hospital’s rule n/2-times.
For n odd, we need to split the limit into limx→0− and limx→0+ , and proceed
as when we computed (2) for n = 1 (when computing f 0 (0)).
After (1) and (2) are proved, f (n) (0) = 0 follows easily.
8