18.014 Problem Set Solutions
Sam Elder
November 2, 2015
Problem 1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Prove that there exists
c ∈ (a, b) such that
f (b) − f (a) a + b
+
.
f 0 (c) + c =
b−a
2
Solution. Let g(x) = f (x)+ 21 x2 . This is also continuous on [a, b] and differentiable on (a, b) since polynomials
are continuous and differentiable everywhere. Then g 0 (x) = f 0 (x) + x, so by the Mean Value Theorem for
Derivatives, there is some c ∈ (a, b) such that
g 0 (c)(b − a) = g(b) − g(a)
b2 − a2
2
f
(b)
−
f
(a)
b
+
a
f 0 (c) + c =
+
,
b−a
2
(f 0 (c) + c)(b − a) = f (b) − f (a) +
as desired.
Problem 2. Find the maximum and minimum values of the function
√
√
f (x) = 2 x + 1 − x
on the interval [0, 1].
Answer. 1 ≤ f (x) ≤
√
5 on [0, 1].
Solution. We know that all√extrema occur at critical points (where f 0 (x) = 0), endpoints, or anywhere where
f 0 (x) is not defined. Since x is differentiable for x > 0, f 0 (x) is defined on (0, 1), so we restrict our attention
1
to the endpoints and critical points. For critical points, if f 0 (x) = √1x − 2√1−x
= 0, we have
1
1
4
√ = √
=⇒ x = 4(1 − x) =⇒ x = .
5
x
2 1−x
It suffices to compute f (0), f ( 45 ) and f (1). We have
√
√
f (0) = 2 0 + 1 − 0 = 1
r
r
√
4
4
1
5
f
=2
+
=√ = 5
5
5
5
5
√
√
f (1) = 2 1 + 1 − 1 = 2.
Therefore, √
the minimum value of f (x) on [0, 1] is 1, which occurs at x = 0, and the maximum value of f (x)
on [0, 1] is 5, which occurs at x = 45 .
1
Problem 3. Find the maximum value of the function
f (x) = x +
2
1
+
x2
x+1
on the interval [1, 2].
Answer. f (x) ≤ 3 on [1, 2].
Solution. Notice that f (x) is a sum of rational functions whose denominators are never zero on [1, 2]. Therefore, f (x) is differentiable everywhere, and its maximum must occur either at an endpoint or a critical
point.
Moreover, we compute
f 0 (x) = 1 −
2
2
−
;
3
x
(x + 1)2
f 00 (x) =
4
6
+
.
4
x
(x + 1)3
Therefore, for 1 ≤ x ≤ 2, f 00 (x) > 0 and f is convex. Hence any critical points will be minima and not
2
maxima, so the maximum must occur at an endpoint. Since f (1) = 1 + 112 + 1+1
= 3 and f (2) = 2 + 212 + 32 =
2 + 11
12 < 3, the maximum value of the function on [1, 2] is 3.
Problem 4. Define f : R → R by the integral
x2
Z
f (x) =
x
1
dt.
1 + t2
Compute the derivative f 0 (x).
Solution. Define g : R → R by the integral
Z
x
g(x) =
0
1
dt.
1 + t2
By the first part of the Fundamental Theorem of Calculus, g 0 (x) =
the Chain Rule,
f 0 (x) = 2xg 0 (x2 ) − g 0 (x) =
1
. Since f (x) = g(x2 ) − g(x), by
1 + x2
1
2x
−
.
1 + x4
1 + x2
Problem 5. Compute the integral
Z
0
1
√
x3
dx.
1 + x2
√
Answer. The integral evaluates to
Solution. Let f (x) =
√
2− 2
.
3
1 + x2 , and g(x) = (1 + x2 )3/2 . We compute by the Chain Rule that
1
x
(2x)(1 + x2 )−1/2 = √
2
1 + x2
p
3
3x(1 + x2 )
g 0 (x) = (2x)(1 + x2 )1/2 = 3x 1 + x2 = √
.
2
1 + x2
f 0 (x) =
2
x3
x3
= 13 g 0 (x) − f 0 (x), so 13 g(x) − f (x) is a primitive for √
. By the second part of the
1 + x2
1 + x2
Fundamental Theorem of Calculus, therefore,
Therefore, √
Z
1
0
x3
√
dx =
1 + x2
√
√
1
1
1
1 √
2− 2
.
g(1) − f (1) −
g(0) − f (0) =
2 2− 2 −
·1−1 =
3
3
3
3
3
Problem 6. Let f : R → R be continuous and thrice-differentiable (so f 0 , f 00 , and f 000 all exist everywhere
in R). Let c ∈ R be a point such that f 0 (c) = f 00 (c) = 0 and f 000 (c) > 0.
Part 6.1. Prove that there exists δ > 0 such that f 00 (x) > 0 for c < x < c+δ and f 00 (x) < 0 for c−δ < x < c.
f 00 (x) − f 00 (c)
, Q(c) = f 000 (c). Because f 000 (c) exists and is positive,
x−c
Q(x) is continuous at c, with Q(c) > 0. By the sign-preserving property of continuous functions, there is
some δ > 0 such that Q(x) > 0 for |x − c| < δ. When x > c, this implies that on c < x < c + δ,
Solution. 1 Define the function Q(x) =
f 00 (x) − f 00 (c)
> 0 =⇒ f 00 (x) > f 00 (c) = 0,
x−c
as desired. When x < c, this implies that on c − δ < x < c,
f 00 (x) − f 00 (c)
> 0 =⇒ f 00 (x) < f 00 (c) = 0,
x−c
as desired.
Part 6.2. Prove that f 0 (x) > 0 for 0 < |x − c| < δ, where δ is as in part (a).
Solution. By Theorem 4.7a, since f 00 (x) > 0 on (c, c + δ), f 0 (x) is strictly increasing on [c, c + δ], so f 0 (x) >
f 0 (c) = 0 for c < x < c + δ. By Theorem 4.7b, since f 00 (x) < 0 on (c − δ, c), f 0 (x) is strictly decreasing on
[c − δ, c], so f 0 (x) > f 0 (c) = 0 for c − δ < x < c. Putting these together, if 0 < |x − c| < δ, f 0 (x) > 0.
Part 6.3. Conclude that f does not have a local maximum or minimum at c.
Solution. By Theorem 4.7 again, since f 0 (x) > 0 on (c, c + δ) and (c − δ, c), f is strictly increasing on both
[c − δ, c] and [c, c + δ], so f is strictly increasing on [c − δ, c + δ]. In other words, for c < x < c + δ, f (x) > f (c),
and for c − δ < x < c, f (x) < f (c). Therefore, f does not have a local minimum or maximum at c.
1 This
solution is motivated by the proof of Theorem 4.3.
3