Math 414
Professor Lieberman
April 7, 2003
HOMEWORK #10 SOLUTIONS
Section 4.4
2. (e) FALSE. The function f (x) = x is uniformly continuous on D = [0, ∞), but it is not
bounded on D.
7. (e) For n ∈ N, set
xn =
2
,
(2n + 1)π
yn =
2
.
(2n − 1)π
Then
|xn − yn | =
4
,
π(4n2 − 1)
and
2
2
8n
=
|f (xn ) − f (yn )| = +
,
2
(2n + 1)π (2n − 1)π
(4n − 1)π
so |f (xn ) − f (yn )| = 2n|xn − yn |. Therefore f is NOT Lipschitzian.
10. For all examples, we take D = R.
(a) The function f (x) = x + 1 is Lipschitzian with Lipschitz constant L = 1 and has no
fixed points in D.
(b) The function f (x) = x is Lipschitzian with Lipschitz constant L = 1 and has infinitely
many fixed points in D.
(c) The function f (x) = −x is Lipschitzian with Lipschitz constant L = 1 and has exactly
one fixed point in D.
Section 4.6
11. From Theorem 4.6.6, E is closed and bounded so Theorem 4.3.1 implies that f (E) is
bounded. To show that f (E) is closed, let a be an accumulation point of f (E). Then there
is a sequence (yn ) with yn ∈ f (E) \ {a} and yn → a. Because yn ∈ f (E), there is a number
xn ∈ E such that f (xn ) = yn . Now Theorem 2.6.4 implies that there is a convergent
subsequence (xnk ) (because E is bounded and non-empty). Let us set x0 = limk→∞ xnk .
Because E is closed, x0 ∈ E. Now the continuity of f implies that f (xnk ) → f (x0 ). In
addition, Theorem 2.6.5 implies that f (xnk ) → a, so a = f (x0 ), which means that a ∈ f (E),
and therefore f (E) is closed.
Section 5.1
5. (b) It is NOT true. The function f (x) = x3 + x has derivative f 0 (x) = 3x2 + 1, which is
never zero.
1
2
12. First, let’s do some algebra:
f (x + h) − f (x)
f (x)f (h) − f (x)
f (h) − 1
=
= f (x)
,
h
h
h
and f (x) = f (x + 0) = f (x)f (0), so f (0) = 1. Now
lim f (x)
h→0
f (h) − 1
f (h) − 1
= f (x) lim
= f (x)f 0 (0),
h→0
h
h
so
f (x + h) − f (x)
= f 0 (0)f (x).
h→0
h
lim
Section 5.2
15. (a) From exercise 10a, the function f is increasing on (a, c) and decreasing on (c, b). It
follows that f (x) ≤ f (y) for all x and y with a ≤ x ≤ y < c. Because f (c) = limy→c f (y),
it follows that f (x) ≤ f (c) for all x ∈ [a, c]. Similarly, f (x) ≤ f (c) for all x ∈ [c, b] and
therefore f (x) ≤ f (c) for all x ∈ [a, b], so f has a relative maximum at x = c.
20. (h) Consider t ∈ (0, 1) and b as fixed and define f (a) = at b1−t − ta − (1 − t)b. Then
a t−1
f 0 (a) = tat−1 b1−t − t = t
−1 .
b
Because t ∈ (0, 1), it follows that f 0 (a) > 0 if a < b and f 0 (a) < 0 if a > b. Exercise 15(a)
then implies that f has a maximum at a = b. (The argument given here shows that it is a
strict maximum.) Because f (b) = 0, it follows that f (a) < f (b) = 0 if a 6= b.
If t = 0 or t = 1, then the equality follows from simple algebra.
29. Given ε > 0, there is a positive constant M such that |f 0 (y)| < ε if y ≥ M . In addition,
the mean value theorem says that, for any x, there is a y ∈ (x, x + 1) such that g(x) =
f (x + 1) − f (x) = f 0 (y). If x ≥ M , then y ≥ M , so |g(x)| < ε. This says that lim g(x) = 0.
x→∞
40. Suppose first that f+0 (a) > f−0 (b). Because k is between f+0 (a) and f−0 (b), there is a constant
ε1 such that f+0 (a) + ε1 < k < f−0 (b) − ε1 . By the definition of the derivative, there are
δ1 > 0 and δ2 > 0 such that
0
f
(a
+
h)
−
f
(a)
< ε1
f+ (a) −
h
for x ∈ (a, a + δ1 ) and
0
f
(b
−
h)
−
f
(b)
f− (b) −
< ε1
−h
for y ∈ (b − δ2 , b). Now set δ = 21 min{δ1 , δ2 , b − a}. Then
0
f+ (a) − f (a + δ) − f (a) < ε1 ,
δ
and
0
f
(b)
−
f
(b
−
δ)
f− (b) −
< ε1 .
δ
3
Now set g(x) = (f (x+δ)−f (x))/δ. It follows that g(a) < f+0 (a)+ε1 and g(b−δ) > f−0 (b)−ε1 ,
so k is between g(a) and g(b − δ). In addition, g is continuous on [a, b], so the intermediate
value theorem says that there is a point γ ∈ (a, b − δ) such that g(γ) = k. We finish the
proof by applying the mean value theorem to f on the interval [γ, γ + δ]: there is a point
c ∈ (γ, γ + δ) such that
f (γ + δ) − f (γ)
= g(γ) = k.
δ
If f+0 (a) > f−0 (b), a similar argument can be used.
f 0 (c) =
Section 5.3
3. f (x) = x2 |x|.
5. (a) We need to have the derivative equal to 6, and f 0 (x) = x2 + x, so f 0 (x) = 6 for x = 2
and x = −3. The points on the curve are (2, 5/3) and (−3, −11/2).
6. By induction, for each k ∈ N, there is a polynomial pk such that
(
pk (1/x) exp(−1/x2 ) if x > 0,
(k)
f (x) =
0
if x < 0,
so f is infinitely differentiable except possibly at x = 0. In addition,
pk (1/x) exp(−1/x2 )
= 0,
x→0+
x
so f is also infinitely differentiable at 0. Therefore every Taylor polynomial is identically
zero. The same argument works for Example 5.3.3.
f (k+1) (0) = lim
25. Set g(x) = f (x + a) − f (x) for x > 0. Then g 0 (x) = f 0 (x + a) − f 0 (x) ≤ 0 because
f 0 ≤ 0 implies that f 0 is decreasing. It follows that g 0 is decreasing, so g(x) ≤ g(0) = f (a).
Therefore f (x + a) − f (x) ≤ f (a), or f (x + a) ≤ f (x) + f (a). To complete the proof
substitute x = b.
Section 5.5
1. FALSE. We can rewrite f (x) = exp(x ln(4x − 1)), so
0
f (x) = exp(x ln(4x − 1)) ln(4x − 1) + x
4
.
4x − 1
8. TRUE. By differentiation,
2 2 2 x5
x
x
8
x
8
8
0
3
f (x) = 4x exp −
sin 3 −
exp −
sin 3 − 24 exp −
cos 3
4
x
2
4
x
4
x
if x 6= 0 and
2 h
8
0
3
f (0) = lim h exp −
sin 3 = 0.
h→0
4
h
Now comes the big calculation. We use the identity
p
A sin t + B cos t = A2 + B 2 sin(t + δ),
4
√
√
where cos δ = A/ A2 + B 2 and sin δ = B/ A2 + B 2 to see that
2r
x
x5
0
|f (x)| ≤ exp −
(4x3 − )2 + 242 .
4
2
for x ∈ [−1, 1] \ {0}. Now, we set
2r
x
x5
g(x) = exp −
(4x3 − )2 + 242 ,
4
2
and we compute
#
r
2"
3 − x5 /2)(12x2 − 5x4 /2)
5
x
x
x
2(4x
0
p
−
(4x3 − )2 + 242 +
g (x) = exp −
4
2
2
(4x3 − x5 /2)2 + 242
2
x exp − x4
5 4
3
4 2
2
2
4
2
=q
−(4x − x ) − 24 + 2(4x − x )(12x − x ) .
5
2
(4x3 − x2 )2 + 242
Now we estimate:
5
2(4x2 − x4 )(12x2 − x4 ) ≤ 2(4)(12) < 242
2
0
for 0 < x ≤ 1, so g is negative and therefore g is strictly decreasing on [0, 1]. Because g
is even, it follows that g(x) ≤ g(0) = 24 for x ∈ [−1, 1] and g(x) < 24 if x 6= 0. Hence
|f 0 (x)| < 24 if x ∈ [−1, 1], so f 0 is bounded. Now, let’s look at xn = (4/(nπ))1/3 (which
makes 8/x3n = 2nπ) and yn = (8/((2n + 1)π))1/3 (which makes 8/yn3 = (2n + 1)π). Then
lim f 0 (xn ) = −24,
n→∞
and
lim f 0 (yn ) = 24.
n→∞
Therefore sup f 0 = 24 and inf f 0 = −24 and f 0 doesn’t attain either value, so it does not
attain its absolute extrema on [−1, 1].
Section 5.6
1. (a) We have
3f (x + h) − f (x) − 2f (x − h) = 3[f (x) + f 0 (x)h + O(h2 )] − f (x) − 2[f (x) + f 0 (x)(−h) + O(h2 )]
= 5f 0 (x)h + O(h2 ),
so
5
3f (x + h) − f (x) − 2f (x − h)
= f 0 (x) + O(h).
ch
c
If we take c = 5, we conclude that
3f (x + h) − f (x) − 2f (x − h)
+ O(h2 ).
ch
(b) Taylor’s theorem gives
f 0 (x) =
1
f (x + h) = f (x) + f 0 (x)h + f 00 (x)h2 + O(h3 ),
2
and
f (x + 2h) = f (x) + 2f 0 (x)h + 2f 00 (x)h2 + O(h3 ),
5
so
f (x + 2h) − 2f (x + h) + f (x)
1
= [f (x) + 2f 0 (x)h + 2f 00 (x)h2 ] − 2[f (x) + f 0 (x)h + f 00 (x)h2 ] + f (x) + O(h3 )
2
= f 00 (x)h3 + O(h3 ).
Therefore
f 00 (x) =
f (x + 2h) − 2f (x + h) + f (x)
+ O(h).
h2
Section 6.1
4. FALSE. Take f (x) = 0, h(x) = 4, and
(
0 if x is rational
g(x) =
4 if x is irrational.
Then, clearly, f (x) ≤ g(x) ≤ h(x) on the interval [−2, 3], and f and g are Riemann
integrable on [−2, 3] by Exercise 6.1.3, but g is not Riemann integrable [−2, 3] by Example
6.1.8.