Homework 4-Solutions-Introduction to Analysis II
Instructor: Christos Saroglou
Spring 2017
1. Let a1 , . . . , an be real numbers and let f : R → R be defined by
f (x) =
n
X
(ai − x)2 , x ∈ R.
i=1
Find the unique point of relative minimum of f .
Hint: Take the derivative of f (x) and prove that there exists a unique point a∗ ∈ R, such that f 0 (a∗ ) = 0.
Study the sign of f 0 (x) on the intervals (−∞, a∗ ) and (a∗ , ∞).
Solution. We have
0
f (x) =
n
X
2(x − ai ) = 2
i=1
n X
nx − (a1 + · · · + an ) .
i=1
f 0 (x)
Thus, for x < (a1 +· · ·+an )/n, we have
< 0 and for x > (a1 +· · ·+an )/n, we have f 0 (x) > 0, which shows
that f is (strictly) decreasing on (−∞, (a1 + · · · + an )/n] and (strictly) increasing on [(a1 + · · · + an )/n, ∞).
Thus, the point x = (a1 + · · · + an )/n is a point of local minimum for f and f has no other points of local
extremum, since f 0 (x) 6= 0, for x 6= (a1 + · · · + an )/n. 2. Use the Mean Value Theorem to prove that | sin x − sin u| ≤ |x − u|, ∀x, u ∈ R.
Solution. Let x, u ∈ R. The mean value theorem shows ∃c between u and x, such that sin x − sin u =
(sin)0 (c)(x − u) ⇒ | sin x − sin u| = | cos c| · |x − u| ≤ |x − u|. 3. If h(x) := 0, for x < 0 and h(x) := 1, for x ≥ 0, prove that there does not exist a function f : R → R, such
that f 0 (x) = h(x), ∀x ∈ R. Find two functions, that do not differ by a constant, such that their derivatives
equal h(x), ∀x ∈ R \ {0}.
Solution. If there exists such a function f , then by Darboux’s intermediate value theorem for derivatives,
we can find x ∈ R, such that f 0 (x) = 1/2. This is a contradiction since h = f 0 , thus there does not exist such
a function. For the second part, take e.g. f (x) = x if x > 0 and f (x) = 1 if x ≤ 0 and g(x) = x if x > 0 and
g(x) = 2 if x ≤ 0. Then, f 0 (x) = g 0 (x) = h(x), ∀x 6= 0 but f, g do not differ by a constant. 4. Let I be an interval and f : I → R be differentiable on I. If f 0 (x) 6= 0, ∀x ∈ I, prove that f is injective.
Hint: Use the Intermediate Value Theorem for derivatives.
Solution. We will show that either f 0 (x) > 0, for all x ∈ I or f 0 (x) < 0, for all x ∈ I. Then, f will be
strictly increasing or strictly decreasing, hence injective. Assume that ∃x1 , x2 ∈ I, such that f 0 (x1 ) > 0 and
f 0 (x2 ) < 0. Then, by Darboux’s intermediate value theorem for derivatives, ∃c between x1 , x2 (in particular,
c ∈ I), such that f 0 (c) = 0, which is a contradiction, hence f is injective.
1
An easier solution: Let x1 , x2 ∈ I with x1 6= x2 , such that f (x1 ) = f (x2 ). Then, by Rolle’s Theorem, ∃c
between x1 , x2 , such that f 0 (c) = 0. This is a contradiction, so f (x1 ) 6= f (x2 ). Since x1 , x2 are arbitrary points
in I, f is injective. 5. Let I be an interval and f : I → R be differentiable on I. If there exists K > 0, such that |f 0 (x)| ≤ K,
∀x ∈ I, show that
|f (x) − f (u)| ≤ K|x − u| , ∀x, u ∈ I.
This is the inverse of the theorem we proved in class, stating that if a differentiable function is Lipschitz in I,
then it has bounded derivative in I.
Hint: Use the Mean Value Theorem.
Solution. Let x, u ∈ I. As in Problem 2, by the Mean Value Theorem, ∃c between x and u such that
|f (x) − f (u)| = |f 0 (c)| · |x − u| ≤ K|x − u|. 6. Let f : [0, 2] → R be continuous on [0, 2] and differentiable on (0, 2). Assume that f (0) = 0, f (1) = 1 and
f (2) = 1.
a) Show that there exists c1 ∈ (0, 1), such that f 0 (c1 ) = 1.
b) Show that there exists c2 ∈ (1, 2), such that f 0 (c2 ) = 0.
c) Show that there exists c ∈ (0, 2), such that f 0 (c) = 1/3.
Hint: Use the Mean Value Theorem for the first two and the Intermediate Value Theorem for derivatives
together with (a) and (b) for the third.
Solution. (a) By the Mean Value Theorem, ∃c1 ∈ (0, 1), such that f 0 (c1 ) = (f (1) − f (0))/(1 − 0) = 1.
(b) By the Mean Value Theorem, ∃c2 ∈ (1, 2), such that f 0 (c2 ) = (f (2) − f (1))/(2 − 1) = 0.
(c) By Darboux’s intermediate value theorem for derivatives, since f 0 (c2 ) < 1/3 < f 0 (c1 ), ∃c ∈ (c1 , c2 ) ⊆ (0, 1),
such that f 0 (c) = 1/3. 7. (Optional) Let f : [a, b] → R be differentiable on (a, b). If limx→a f 0 (x) = A ∈ R, prove that f 0 (a) exists and equals A.
Hint: Use the definition of f 0 (a) and the Mean Value Theorem.
n→∞
Solution. Let {cn } ⊆ [a, b], such that cn −−−→ a. Then, limn→∞ f 0 (cn ) = A. For any sequence {un } ⊆ [a, b],
n→∞
such that un −−−→ a and for any n ∈ N, by the Mean Value Theorem, ∃cn (a, un ), such that
f 0 (cn ) =
f (un ) − f (a)
.
un − a
n→∞
Notice that {cn } ⊆ [a, b] and cn −−−→ a, thus limn→∞ f 0 (cn ) = A, i.e.
f (un ) − f (a)
= A.
n→∞
un − a
lim
Since {un } is an arbitrary sequence in [a, b] that converges to a, we conclude that
lim
x→a+
f (x) − f (a)
= A,
x−a
which ends our proof. 2