SOME PROBLEMS ON THE DERIVATIVE.
TRUE OR FALSE? If true give a proof, if false a counterexample.
1. If f is differentiable on R and f 0 (c) > 0, then there exists a δ > 0
such that f 0 > 0 on the open interval Vδ (c).
Solution. This is false. Consider the function:
f (x) = x + x2 sin(
1
), x 6= 0;
x2
f (0) = 0.
f is differentiable and the derivative function is:
f 0 (x) = 1 + 2x sin(
2
1
1
) − cos( 2 ), x 6= 0;
2
x
x
x
Consider the sequence xn > 0,
1
x2n
f 0 (0) = 1.
= 2πn, n ≥ 1. Then lim xn = 0 and:
f 0 (xn ) = 1 − 2nπ < 0
∀n ≥ 1.
2. The claim in problem 1. is true if the derivative function f 0 is continuous at c. (Use the definition of continuity.)
0
Solution. Let = f 2(c) > 0. Since f 0 is continuous at c we may find
δ > 0 so that:
x ∈ Vδ (0) ⇒ f 0 (x) > f 0 (c) − =
f 0 (c)
> 0.
2
3. If f is differentiable on an interval containing 0 and limx→0 f 0 (x) = L,
then it must be that f 0 (0) = L.
Solution. This is true. To prove this, suppose for example f 0 (0) > L
and let = (1/2)(f 0 (0) − L) > 0. We know there is δ > 0 so that for all
x ∈ Vδ (0), |f 0 (x) − L| < . This implies:
x ∈ Vδ (0) ⇒ f 0 (x) < L + = (1/2)(f 0 (0) + L) < f 0 (0),
in particular f 0 (δ/2) < (1/2)(f 0 (0) + L). Thus f 0 omits a whole interval
of values between f 0 (δ/2) and f 0 (0), and this is not allowed by Darboux’s
theorem.
1
4. If f is differentiable on an interval containing 0, except possibly at
0, and limx→0 f 0 (x) = L, then it must be that f is differentiable at 0 and
f 0 (0) = L.
Solution. This is false, since f may fail to be continuous at 0. To
find an example, take your favorite differentiable function with continuous
derivative and change its value at 0.
5. The claim in problem 4. is true if f is, in addition, assumed to be
continuous at 0.
Solution. Let > 0 be given. We may find δ > 0 so that, for all x with
0 < |x| < δ, |f 0 (x) − L| < . Let x satisfy 0 < |x| < δ. Since f is continuous
in the closed interval [0, x] (or [x, 0]), the Mean Value Theorem implies there
is a c between 0 and x so that:
L−<
f (x) − f (0)
= f 0 (c) < L + .
x
Thus, by definition, f 0 (0) exists and equals L.
6. If f is differentiable on an interval containing 0 and f 0 (0) > 0, then
f must be increasing in a sufficiently small interval around 0.
Solution. This is false. Consider the function:
f (x) =
x
1
+ x2 sin( ), x 6= 0,
2
x
f (0) = 0.
Then f is differentiable on R with f 0 (0) = 1/2, but consider the sequences
xn , yn defined by x1n = 2πn, y1n = π2 + 2πn. Then a short calculation shows
that:
1 + 4(1 − π4 )n
xn > yn , f (xn ) − f (yn ) =
< 0,
4πn(4n + 1)2
(for n large enough) since 4/π > 1. Thus f is not increasing in any neighborhood of 0.
7. If f is differentiable at 0 and f 0 (0) > 0, show that f (x) > f (0) for
x > 0 in a sufficiently small neighborhood around 0.
Solution. This is true. If the claim is not true, we may find a sequence
xn > 0, lim xn = 0, so that f (xn ) ≤ f (0). This implies:
f 0 (0) = lim
f (xn ) − f (0)
≤ 0,
xn
a contradiction.
2
8. The claim in problem 6 is true if, in addition, we assume f 0 is continuous at 0.
Solution. True. By problem 2, f 0 (x) > 0 for x in a neighborhood of
0. Now the Mean Value Theorem implies that, for each x > y in this
neighborhood, we may find c in the interval (y, x) so that:
f (x) − f (y)
= f 0 (c) > 0,
x−y
so f (x) > f (y). Thus f is increasing in this neighborhood of 0.
9. Let f : R → R be differentiable. Suppose f 0 (x) → +∞ as x → +∞.
Recall that this means:
(∀M > 0)(∃L > 0)(∀x)x > L ⇒ f 0 (x) > M.
(i) Show that f can’t be bounded above on R. (ii) Show that f can’t be
Lipschitz on R.
Solution: (i) Let M > 0 be arbitrary, and find L > 0 given by the
definition. Then if x > max{L, 1}, by the MVT we have, for some c in the
interval (L, x):
f (x) − f (0) = xf 0 (c) > M x > M.
Since M is arbitrary, f − f (0) can’t be bounded above, so neither is f .
(ii) Again let M > 0 be arbitrary, and take L > 0 given by the definition.
Then if y > x > L, by the MVT we find c in the interval (x, y) so that:
f (y) − f (x) = f 0 (c)(y − x) > M (y − x) > 0, so |f (y) − f (x)| > M |y − x|.
Since M is arbitrary, f can’t be Lipschitz.
Related problems in text:
1,2: 5.2.8(b) 3: 5.2.8(c), 4,5: 5.2.8. (d), 6,7,8: 5.3.7, 5.3.8
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