Math 414
Professor Lieberman
April 17, 2003
HOMEWORK #11 SOLUTIONS
Section 5.5
14. FALSE. f (x) = x3 − x has an inflection point at x = 0 because the second derivative
switches from negative to positive there, but f 0 (0) = −1.
21. TRUE. f 0 (x) = 3x2 + 8x − 1, which is zero when x = (−4 ±
these numbers is in the interval.
√
19)/3, and neither one of
24. (b) FALSE. Take f (x) = x3 − 3x. Then the tangent line to this curve at x = 1 has zero
slope, so it’s y = −2, which intersects the graph also at x = −2.
Section 5.6
11. (a) f (x) = x1/3 on [−1, 1].
(b) f (x) = |x| on [−1, 1].
(c) f (x) = 1/x on (0, 1).
19. (a) Suppose first that f is increasing. Then, for any partition P ,
n
X
|f (xi ) − f (xi−1 )| =
i=1
n
X
f (xi ) − f (xi−1 ) = f (b) − f (a),
i=1
so Vf (a, b) = f (b) − f (a) and f is of bounded variation.
If f is decreasing, then Vf (a, b) = f (a) − f (b) by similar reasoning so f is also of bounded
variation in this case.
Because x1/3 is increasing on any interval, it is also of bounded variation on any bounded
interval.
(h) f (x) = x1/3 on the interval [−1, 1]. (Strictly speaking, this example isn’t correct because
f 0 (0) is infinite. To fix this situation, look at
(
x3/2 sin x1
f (x) =
0
if x 6= 0
if x = 0
on [0, 1]. Then
(
0
f (x) =
3 1/2
sin x1
2x
− x−1/2 cos x1
0
1
if x 6= 0
if x = 0
2
which is unbounded because f 0 (xn ) → ∞ for xn = 1/(nπ). To prove that f is of bounded
variation requires some integration theory. Given a partition P , we have
n Z xk
n
X
X
|f (xk ) − f (xk−1 )| = |f (x1 )| +
f 0 (x) dx
k=1
k=2 xk−1
Z
n
n Z xk
xk
X
X
5 −1/2
0
x
dx
≤ |f (x1 )| +
|f (x)| dx ≤ |f (x1 )| +
2
k=2 xk−1
k=2 xk−1
Z 1
8
5
5 −1/2
x
dx = |f (x1 )| +
1 − (x2 )1/2 ≤ .
= |f (x1 )| +
3
3
x2 2
Because of this technical point, full credit will be awarded for saying f (x) = x1/3 .)
21. (f) Let L be a Lipschitz constant for f and let ε > 0 be given. We then choose δ = ε/L. If
n
P
(bk − ak ) < δ, then
(ak , bk ) (k = 1, . . . , n) are n nonoverlapping intervals with
k=1
n
X
|f (bk ) − f (ak )| ≤
k=1
n
X
L(bk − ak ) = L
k=1
n
X
(bk − ak ) < Lδ = ε.
k=1
24. (b) f (x) = |x|.
(e) Take t ∈ (0, 1) so that b = (1 − t)a + tc. Then t = (b − a)/(c − a) and f (b) ≤
(1 − t)f (a) + tf (c), so
b−a
f (b) − f (a) ≤ t[f (c) − f (a)] =
[f (c) − f (a)],
c−a
and therefore
f (b) − f (a)
f (c) − f (a)
≤
.
b−a
c−a
In addition, 1 − t = (c − b)(c − a), so
c−b
f (b) − f (c) ≤ (1 − t)[f (a) − f (c)] =
[f (a) − f (c)],
c−a
so
f (c) − f (b)
f (c) − f (a)
≤
.
c−a
c−b
(g) From part (f), f+0 (x) is finite, so lim f (x + h) = f (x). (Just as in Theorem 5.1.6.)
h→0+
Similarly, lim f (x + h) = f (x), and therefore lim f (x + h) = f (x), which means that f is
h→0−
h→0
continuous.
Section 6.1
7. Take f to be the function from Example 6.1.8, and set g = −f . Then
Z b
Z b
Z b
Z b
f +g =
0 = 0,
f = 20,
g = 0.
a
a
a
a
3
Section 6.2
7. Since f is continuous on the closed, bounded interval [a, b], it attains its maximum value at
some point c ∈ [a, b]. Suppose f (c) > 0, say f (c) = K. Then there is a number δ > 0 such
that f (x) ≥ K/2 for |x − c| < δ. Let ε > 0 be given and take P to be a partition such that
Z b
ε
|
f − U (P, f )| < ,
2
a
and let Q be the partition P ∪ {max{a, c − δ}, min{b, c + δ}}.
8. First notice that L(P, f ) = 0 for any partition P . Therefore, we only have to show that
we can make U (P, f ) small by choosing the partition P appropriately. Specifically, given
ε > 0, chose q ∈ N so that 1/q < ε/2 and then n ∈ N so that q(q + 1)/n < ε/2. If P is
the partition {x0 , . . . , xn } with xi = i/n, then there are at most q(q + 1)/2 subintervals in
the partition on which the maximum of f is greater than 1/q, and the maximum on each
of these subintervals is at most 1/2. Therefore
q(q + 1) 1 1 1
U (P, f ) ≤
+
2
n 2 q
because the sum of the lengths of the subintervals on which the maximum of f is less than
or equal to 1/q is less than or equal to 1. Therefore U (P, f ) < ε. It follows from Theorem
R1
6.2.1 that f is integrable and, because the lower integral is zero, 0 f = 0. This does not
contradict Exercise 7 because f is discontinuous.
Section 6.3
1 (a) NO. Take f (x) = 0 and g(x) = x on the interval [−1, 1].
4. (a) First, notice that, for any y and z in [a, b], we have
f (y)g(y) − f (z)g(z) = f (y)[g(y) − g(z)] + g(z)[f (y) − f (z)]
≤ sup |f ||g(y) − g(z)| + sup |g||f (y) − f (z)|.
Now fix a partition P = {x0 , . . . , xn }. If y and z are both in [xk−1 , xk ], then |g(y) − g(z)| ≤
Mk (g) − mk (g) and |f (y) − f (z)| ≤ Mk (f ) − mk (f ). Now fix k and let (yn ) be a sequence
in [xk−1 , xk ] such that f (yn )g(yn ) → Mk (f g) and let (zn ) be a sequence in [xk−1 , xk ] such
that f (zn )g(zn ) → mk (f g). Then
Mk (f g) − mk (f g) ≤ sup |f |[Mk (g) − mk (g)] + sup |g|[Mk (f ) − mk (f )].
After multiplying this inequality by xk − xk−1 and summing on k, it follows that
U (P, f g) − L(P, f g) ≤ sup |f |[U (P, g) − L(p, g)] + sup |g|[U (P, f ) − L(P, f )].
To finish, we suppose that ε > 0 is given and choose a partition Q such that U (Q, f ) −
L(Q, f ) < ε/(2 sup |g|) and a partition R such that U (R, g) − L(R, g) < ε/(2 sup |f |). If we
then take P = Q ∪ R, then we see that U (P, f g) − L(P, f g) < ε, so f g is integrable.
5. (b) Define f on [0, 1] by
(
1
f (x) =
−1
if x is rational,
if x is irrational.
4
then f is not Riemann integrable for the same reason that the function in Example 6.1.8
isn’t. But |f |(x) = 1 for any x ∈ [0, 1], so |f | is continuous and therefore integrable.
13. For any real numbers α and β, we have
Z b
0≤
(αf + βg)2 ,
a
so
Z
β2 b 2
f +
g .
−αβ
2 a
a
a
Rb
Rb
Suppose first that a g 2 = 0. Then take β = − a f g and suppose α > 0. Then we can
divide by α:
Z b 2
Z
α b 2
fg ≤
f
2 a
a
for any α > 0. Taking the limit as α → 0 then gives
Z b 2
Z b Z b fg ≤ 0 =
f2
g2 .
Z
b
α2
fg ≤
2
Z
a
If
Rb
a
b
2
a
a
g 2 isn’t zero, then take
b
Z
g2,
α=
fg
a
to see that
Z b
g
a
2
Z
2
b
fg
a
a
1
≤
2
Z
Z
b
b
g
b
g
2
2 Z
a
b
f
2
Z
≤
Rb
+
fg
b
g
2
2 Z
f
a
a
a
a
a
b
2
b
g
a
Dividing both sides of this equation by a g 2 gives
Z b 2 Z b Z b fg ≤
f2
g2 .
a
2 Z
b
Z
a
2
fg
a
2
a
Therefore
Z
b
Z
β=−
.
2
.