Math 414
Professor Lieberman
April 9, 2003
HOMEWORK #9 SOLUTIONS
Section 4.3
1. (c) Take D = [0, 1], and
(
x
f (x) =
0
if 0 ≤ x < 1,
if x = 1.
7. We argue by contradiction. If f is not constant, then there are numbers a1 and b1 in
[a, b] with a1 < b1 and f (a1 ) 6= f (b2 ). By Theorem 1.6.10, there is an irrational number
k between f (a1 ) and f (b1 ), and then the intermediate value theorem implies that there
is a number c ∈ (a1 , b1 ) with f (c) = k. But then c ∈ [a, b] and f (c) is irrational which
contradicts the assumption that the range of f is in Q.
18. (a) Let a ∈ I and ε > 0 be given. Suppose first that a is not an endpoint of I. Then there
are numbers a1 < a and a2 > a in I such that f (a1 ) > f (a) − ε and f (a2 ) < f (a) + ε. Now
set δ = min{a − a1 , a2 − a}. If |x − a| < δ, then f (a1 ) < f (x) < f (a2 ), so |f (x) − f (a)| < ε.
Next, suppose that a is the left endpoint of I. Then there is a number a2 > a in I such
that f (a2 ) < f (a) + ε. Now we set δ = a2 − a. If |x − a| < δ and x ∈ I, then a ≤ x < a2 ,
so |f (x) − f (a)| < ε again. The case that a is the right endpoint of I is completely similar.
(b) Suppose x < y are numbers in f (I), and set a = f −1 (x) and b = f −1 (y). By Definiton
1.6.3, there are three possibilities: (1) a = b, which can’t happen because f (a) = x 6= y =
f (b); (2) a > b, which can’t happen because, in this case, we would have x = f (a) > f (b) =
y; and (3) a < b, which is what we want.
(c) This is just Theorem 4.3.8.
Section 4.4
1. (c) This is uniformly continuous. Given ε > 0, choose δ = 4ε. If |x − t| < δ, then
x
t
4
1
δ
x + 4 − t + 4 = (x + 4)(t + 4) |x − t| ≤ 4 |x − t| < 4 = ε.
2. (c) TRUE. Given ε > 0, there is δ > 0 such that, if x and t are in (a, b) with |x − t| < δ
implies that |f (x) − f (t)| < ε (because f is uniformly continuous). Next, because (xn )
converges, Theorem 2.5.9 implies that the sequence is Cauchy. Hence, there is a natural
number N such that |xn − xm | < δ for all n, m > N . Therefore, if n, m > N , it follows that
|f (xn ) − f (xm )| < ε, so (f (xn )) is a Cauchy sequence.
6. (b) (Here, the limit cannot exist, so you can’t just let f2 go to infinity.) f2 (x) = x sin x. It
is clear that f2 is continuous on [0, ∞). To see that f2 is not uniformly continuous, take
xn = 2πn and tn = xn + 1/n. Then f2 (xn ) = 0 and f2 (tn ) = tn sin(1/n) (by using the
1
2
addition formula for the sine). Because lim (sin x)/x = 1, there is a natural number N such
x→0
that sin(1/n) ≥ 1/(2n) if n ≥ N . For n ≥ N , we therefore have
π
1
1
f2 (tn ) = tn sin
≥ xn
= .
n
2n
2
It follows now from Remark 4.4.4 (with ε = π/2) that f2 is not uniformly continuous.
Finally, Theorem 4.4.8 implies that either lim f2 (x) = ∞ or the limit doesn’t exist. Since
x→∞
xn → ∞ and f2 (xn ) = 0 for the sequence (xn ) described above, the limit can’t be infinite,
so it doesn’t exist.
7. (c) This function is not Lipschitzian. Take t = 0, then
f (x) − f (t) −2/3
,
x−t =x
and this is not bounded, since it goes to ∞ as x goes to zero.
Section 4.5
2. TRUE. If f (a) < f (b), then I claim that f is strictly increasing. First, if x ∈ (a, b), then
there are three cases: (1) f (x) < f (a), (2) f (x) > f (b), (3) f (a) < f (x) < f (b). In the
first case, there would be a point d ∈ (x, b) with f (d) = f (a), which can’t happen and,
similarly, in the second case, there would be a point g ∈ (a, x) with f (g) = f (b), which
can’t happen. This means that case (3) must happen. Now let y be another point in [a, b]
and suppose that x < y. If y = b, we have just shown that f (x) < f (y), so we may assume
that x < y < b. Repeating the previous argument with x in place of a and y in place of
x shows that f (x) < f (y), so f is strictly increasing. Now Exercise 4.3.7 shows that f is
continuous.
9. TRUE. The limit condition implies that f is continuous on the closed bounded set [a, b], so
f is uniformly continuous by theorem 4.4.6.
10. FALSE. We argue by contradiction. Suppose there were such a function, and let a < b be
the two points such that f (a) = f (b) = 1. The extreme value theorem says that there are
numbers c and d in the interval [a, b] such that f takes on its maximum value at c and its
minimum value at d. Since f is equal to 1 only at a and b, at least one of the numbers c
and d must be in the interval (a, b). The arguments for the two cases are virtually identical,
so I will only do the case that c ∈ (a, b). (The idea here is that the graph of f must look
sort of like a downward opening parabola.) By the intermediate value theorem, f takes
on every value between 1 and f (c) somewhere in each of the two intervals (a, c) and (c, b).
Now, write y for the other point such that f (y) = f (c). If y < a, then f takes on every
value between 1 and f (c) somewhere in the interval (y, a), while if y > b, then f takes on
every value between 1 and f (c) somewhere in the interval (b, y). In either case, f would
have to take on some value at least three times, which contradicts our assumption that f
takes on every value only twice.
14. TRUE. If f is contractive, then f is Lipschitzian, so it’s uniformly continuous by Theorem
4.4.10, and then continuous.
3
21. FALSE. Take D = [0, ∞), f (x) = x, g(x) = sin x. Then f and g are uniformly continuous
by Theorem 4.4.10, and g is bounded, but f g is not uniformly continuous as shown in
Exercise 4.3.6b.
23. TRUE. Given ε > 0, choose δ = min{1, ε}. If |x − t| < ε, then
|f (x) − f (t)| ≤ (x − t)2 ≤ |x − t| < δ ≤ ε.
Section 4.6
5. (a) This is not compact. (Follow example 4.6.2). The collection of open intervals {Gn : n ∈
N} with Gn = (−n, 1) covers the interval (−∞, 0] but it has no finite subcover.
(d) This is not compact. The collection of intervals {Gn } with n ∈ N defined by
1
Gn =
, 2
n
is an open cover of the set, but it has no finite subcover.
(e) This is compact. If {Gα } is an open cover of the set, there is some index α0 such that
0 ∈ Gα0 . Because Gα0 is open, there is a δ > 0 such that (−δ, δ) ⊂ Gα0 . Now take N ≥ 1/δ,
and, for each n ≤ N , there is an index αn such that 1/n ∈ Gαn . The collection
{Gα0 , . . . , GαN }
is then a finite subcover.
Section 5.1
3. (b) This function is differentiable at x = 1 because
lim
x→1−
f (x) − f (1)
x2 + 1 − 2
= lim
= lim (x + 1) = 2,
x−1
x−1
x→1−
x→1−
and
f (x) − f (1)
2x − 2
= lim
= lim 2 = 2.
+
x−1
x→1 x − 1
x→1+
Since the one-sided limits exist and agree, the full limit exists and the derivative is 2.
(c) This function is not differentiable at x = 1 because
lim
x→1+
lim
x→1−
f (x) − f (1)
x2 + 1 − 2
= lim
= lim (x + 1) = 2,
x−1
x−1
x→1−
x→1−
but
f (x) − f (1)
3x − 2
= lim
= lim 3 = 3.
x−1
x→1+ x − 1
x→1+
x→1+
Since the one-sided limits exist but disagree, the limit doesn’t exist.
lim
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 are the curve are (2, 5/3) and (−3, −11/2).
10. (b) To show that this limit exists and equals f 0 (x), we note that
3f (x + h) − f (x) − 2f (x − h)
3[f (x + h) − f (x)] 2[f (x) − f (x − h)]
=
+
.
5h
5h
5h
4
Then
lim
h→0
3[f (x + h) − f (x)]
3
= f 0 (x),
5h
5
and
2
2[f (x) − f (x − h)]
= f 0 (x),
h→0
5h
5
lim
so
3f (x + h) − f (x) − 2f (x − h)
= f 0 (x).
5h
Despite the claim in the book, if the limit exists, then f is differentiable at x. To prove
this statement, we first note that
lim
h→0
3f (x − h) − f (x) − 2f (x + h)
3f (x + h) − f (x) − 2f (x − h)
= lim
,
h→0
h→0
5h
5(−h)
lim
and also
5[f (x + h) − f (x)] = 3[3f (x + h) − f (x) − 2f (x − h)] − 2[f (x + h) + f (x) − 3f (x − h)].
(To see this more easily, set f1 (x) = f (x + h) − f (x) and f2 (x) = f (x) − f (x − h), so the
issue to solve the system of equations 5f1 (x) = α[3f1 (x)2f2 (x)] + β[2f1 (x) − 3f2 (x)] for α
and β.) Therefore
f (x + h) − f (x)
3f (x + h) − f (x) − 2f (x − h)
2f (x + h) + f (x) − 3f (x − h)
=3
−2
.
h
5h
5h
It follows that lim
h→0
f (x+h)−f (x)
h
exists and is equal to
3f (x + h) − f (x) − 2f (x − h)
2f (x + h) + f (x) − 3f (x − h)
− 2 lim
h→0
h→0
5h
5h
3f (x + h) − f (x) − 2f (x − h)
= lim
.
h→0
5h
3 lim
11. If f is even, then
f 0 (−x) = lim
h→0
f (−x + h) − f (−x)
f (x − h) − f (x)
= lim
= −f 0 (x),
h→0
h
h
so f 0 is odd. A similar argument shows that, if f is odd, then f 0 is even
Section 5.2
1. (f) By implicit differentiation, we have 2x + y + xy 0 − 2yy 0 = 0, so, at the point (2, 3), we
have 4 + 3 + 2y 0 − 6y 0 = 0, or y 0 = 7/4. The normal line has slope −4/7, so the normal line
has the equation
4
y − 3 = − (x − 2).
7
10. (a) ⇒) If f 0 (x) ≥ 0 for all x, let c and d be numbers in the interval [a, b] with c < d. Then
the mean value theorem implies there is a number k ∈ (c, d) such that
f 0 (k) =
f (d) − f (c)
.
d−c
5
Because f 0 (k) ≥ 0, it follows that f (d) ≥ f (c), so f is increasing.
⇐) If f is increasing, then the difference quotient
f (x + h) − f (x)
h
is nonnegative: If h > 0, then the numerator is nonnegative (because x + h ≥ x) and the
denominator is positive, so this fraction is nonnegative in this case. On the other hand,
if h < 0, then the numerator is nonpositive (because x + h ≤ x) and the denominator is
negative, so the fraction is nonnegative in this case as well. Theorem 3.2.5 (j) then implies
that the derivative is nonnegative.