18.014 Exam 2 Solutions
Sam Elder
November 2, 2015
Problem 1. Let f : [a, b] → R be a continuous function. Prove that there exists a positive number M such
that |f (x) − f (y)| < M for all x, y ∈ [a, b].
Solution. Because f is a continuous function on a closed interval, it is bounded. Therefore, there exist
m < n ∈ R such that m < f (x) < n for all x ∈ [a, b]. It follows that for x, y ∈ [a, b],
m − n < f (x) − f (y) < n − m =⇒ |f (x) − f (y)| < n − m.
Therefore, taking M = n − m > 0, we have shown that |f (x) − f (y)| < M for al x, y ∈ [a, b], as desired.
Problem 2. Let f : R≥0 → R be the function on the nonnegative reals given by
f (x) =
x
√ .
1+ x
Compute the derivative f 0 (x) for all x ≥ 0. Deduce from this that f is strictly increasing.
√
2+ x
0
√
Answer. The derivative is f (x) =
.
2(1 + x)2
√
1
Solution. We know that the function g(x) = 1+ x is positive and differentiable for x > 0, with g 0 (x) = √ .
2 x
On this region, therefore, both the numerator and denominator are differentiable and the denominator is
nonzero, so by the quotient rule,
√
√
√
g(x) − xg 0 (x)
1 + x − x/(2 x)
2+ x
x
0
√
√
=⇒ f (x) =
=
=
.
f (x) =
g(x)
g(x)2
(1 + x)2
2(1 + x)2
We still
√ must prove that this formula also holds when x = 0, since g is not differentiable at x = 0. Since
2+ 0
√
= 1, by the definition of the derivative, we must prove that
2(1 + 0)2
lim
h→0
We compute that f (0) =
f (h) − f (0)
= 1.
h
0
h
√ = 0, and that f (h) =
√ . Therefore,
1+ 0
1+ h
f (h) − f (0)
h
1
√ =
√ .
=
h
h(1 + h)
1+ h
√
Taking the limit as h → 0, since 1 + h is continuous, we get
f 0 (0) = lim
h→0
1
1
√ =
√ = 1,
1+ 0
1+ h
1
√
2+ x
√
for all x ≥ 0.
as desired. We have shown that f 0 (x) =
2(1 + x)2
√
Finally, since x ≥ 0, we have f 0 (x) > 0 whenever x ≥ 0, so f is strictly increasing, as desired.
Problem 3. For each of the following statements, f : [a, b] → R is a function on a closed interval.
Part 3.1. If f is differentiable, then f is integrable.
Solution. True. If f is differentiable, then f is continuous, and therefore integrable.
Part 3.2. If there is a differentiable function g : [a, b] → [a, b] such that f (x) = (g(g(x)))2 , then f is
differentiable and
f 0 (x) = 2g(g(x))g 0 (g 0 (x)).
Solution. False. For clarity, call h(x) = g(g(x)), so by the chain rule, h0 (x) = g 0 (g(x))g 0 (x). Then f (x) =
h(x)2 , so again by the chain rule,
f 0 (x) = 2h(x)h0 (x) = 2g(g(x))g 0 (g(x))g 0 (x).
This formula does not match what is written there, so the statement is false.1
Part 3.3. If f is strictly increasing and f is differentiable, then f 0 (x) > 0 for every x ∈ [a, b].
Solution. False. Consider f (x) = x3 . This is strictly increasing and differentiable, but f 0 (0) = 0.
Part 3.4. If f is continuous, f (a) ≥ a2 and f (b) ≤ b2 , then there exists x ∈ [a, b] such that f (x) = x2 .
Solution. True. Define g(x) = f (x) − x2 , so we want to find some x such that g(x) = 0. As the difference of
continuous functions, g is also continuous. Moreover, we are given that g(a) ≥ 0 and g(b) ≤ 0. If g(a) = 0
or g(b) = 0, we are done. Otherwise, g(a) > 0 > g(b), so by the Intermediate Value Theorem, there is some
c ∈ (a, b) such that g(c) = 0, and hence f (c) = c2 , as desired.
Part 3.5. If f is differentiable and |f 0 (x)| ≤ 1 for all x ∈ [a, b], then |f (x)−f (y)| ≤ |x−y| for all x, y ∈ [a, b].
Solution. True. If x = y, |f (x) − f (y)| = 0 = |x − y|. Otherwise, we may assume without loss of generality
x > y. By the Mean Value Theorem for derivatives, since f is differentiable on [y, x], there exists z ∈ (y, x)
f (x) − f (y)
such that f 0 (z) =
. Then
x−y
|f (x) − f (y)| = |(x − y)f 0 (z)| = |x − y||f 0 (z)| ≤ |x − y|,
as desired.
Part 3.6. If f is differentiable and the derivative f 0 is continuous, then
Z b
f (b) = f (a) +
f 0 (x) dx.
a
Solution. True. This is exactly the statement of the second fundamental theorem of calculus, rearranged.
Problem 4. Let f : R → R be a function.
Part 4.1. Suppose that f satisfies
|f (x) − f (y)| ≤ |x − y|
for all x, y ∈ R. Prove that f is continuous.
1 Technically, we should demonstrate that the formulas are unequal for some function g, but almost any function will do.
For one concrete example, choose g(x) = x2 on [0, 1].
2
Solution. Let > 0. Taking δ = , if |x − y| < δ, then |f (x) − f (y)| ≤ |x − y| < δ = , so f (x) is continuous
directly by the definition.
Part 4.2. Suppose that f satisfies
|f (x) − f (y)| ≤ (x − y)2
for all x, y ∈ R. Prove that f is constant.
Solution. We claim that f is differentiable with f 0 (z) = 0. Indeed, for x 6= y, dividing |x − y| on both sides
of the given inequality yields
f (x) − f (y) |f (x) − f (y)|
=
≤ |x − y|.
x−y
|x − y|
To prove the derivative claim at z, let > 0, and define δ = . If 0 < |h| < δ, letting x = z + h and y = z
yields
f (z + h) − f (z) ≤ |(z + h) − z| = |h| < δ = ,
h
f (z + h) − f (z)
= 0, as desired.
h→0
h
0
Now, if f (z) = 0 for all z ∈ R, f is constant, as desired.
so lim
3