SOLUTION OF HW8
MINGFENG ZHAO
April 19, 2013
1. [5 Points] Show that f (x) = O(|x − x0 |2 ) as x → x0 implies f (x) = o(|x − x0 |) as x → x0 , but give
an example to show that the converse is not true.
Proof. Assume that f (x) = O(|x − x0 |2 ) as x → x0 , then there exists some δ > 0 and M > 0 such
that
|f (x)| ≤ M |x − x0 |2 ,
∀|x − x0 | < δ.
So we know that
f (x) |x − x0 | ≤ M |x − x0 |,
∀|x − x0 | < δ.
Hence we get
f (x) ≤ lim M |x − x0 | = 0.
lim
x→x0 |x − x0 | x→x0
That is, f (x) = o(|x − x0 |) as x → x0 .
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Counterexample I: Let x0 = 0, and f (x) = |x| 2 for all x ∈ R, then we know that lim
x→0
1
f (x)
= lim |x| 2 ,
x→0
|x|
that is, f (x) = o(|x|) as x → 0. But we know that
|f (x)|
1
=
1 → ∞,
|x|2
|x| 2
as x → 0.
So we do not have f (x) = O(|x|2 ) as x → 0.
2. [5 Points] Show that if f (x0 ) = 0 and f (x) = o(|x − x0 |) as x → x0 , then f 0 (x0 ) exists. What is
f 0 (x0 )? What does this tell you about x2 sin
1
?
x1000
1
2
MINGFENG ZHAO
Proof. Since f (x) = o(|x − x0 |) as x → x0 , then for any > 0, there exists δ > 0 such that for all
|x − x0 | < δ, we have
f (x) |x − x0 | < .
Since f (x0 ) = 0, then
f (x) − f (x0 ) |x − x0 | < .
By the definition of f 0 (x0 ), we know that f 0 (x0 ) exists and f 0 (x0 ) = 0.
1 1
2
Let g(x) = x sin 1000 for all x ∈ R\{0}. Since sin 1000 ≤ 1, then |g(x)| ≤ |x|2 for all x ∈ R\{0}.
x
x
So if we set g(0) = 0, then g is continuous at 0. Since |g(x)| ≤ |x|2 , then g(x) = O(|x|2 ) as x → 0. By
the result of Problem 1, we know that g(x) = o(|x|) as x → 0. Since g(0) = 0,then g 0 (0) exists and
g 0 (0) = 0.
3. [5 Points] Show that x sin
1
fails to have a derivative at x = 0 and even the one-sided derivative
x
fails to exist.
1
Proof. Let f (x) = x sin for all x ∈ R\{0}. Suppose that f (x) has a derivative at 0, then f is
x
1
continuous at 0. Since x sin ≤ |x| for all x ∈ R\{0}, then f (x) → 0 as x → 0, which implies that
x
f (0) = 0.
4. [5 Points] Given an example of a differentiable function whose tangent line at a point fails to stay
on one sided of the graph even locally.
Proof. Let f (x) = x3 for all x ∈ R. Then f 0 (0) = 0, which implies that the tangent line at 0 is given
by:
y = 0.
But we know that f (x) = x3 does not stay on one sided of y = 0 even locally near 0.
SOLUTION OF HW8
3
5. [5 Points] Let f and g be continuous functions on [a, b] and differentiable at every point in the
interior, with g(a) 6= g(b). Prove that there exists a point x0 in (a, b) such that
f 0 (x0 )
f (b) − f (a)
= 0
.
g(b) − g(a)
g (x0 )
Proof. Consider the function
h(x) =
f (b) − f (a)
g(b) − g(a)
· [g(x) − g(a)] −
· [f (x) − f (a)],
b−a
b−a
∀x ∈ [a, b].
Since f and g are continuous on [a, b] and differentiable at every point in (a, b), then h is continuous
on [a, b] and differentiable at every point in (a, b). It is easy to see that
h(a) = h(b) = 0.
By mean value theorem, we know that there exists some x0 ∈ (a, b) such that
h0 (x0 ) = 0.
That is,
g(b) − g(a) 0
f (b) − f (a) 0
· g (x0 ) −
· f (x0 ) = 0.
b−a
b−a
Since g(b) 6= g(a), then we have
f (b) − f (a)
f 0 (x0 )
= 0
.
g(b) − g(a)
g (x0 )
6. [5 Points] Is the converse of the mean value theorem true, in the sense that if f is continuous on
[a, b] and differentiable on (a, b), given a point x0 in (a, b) must there exist points x1 , x2 in [a, b] such
that
f (x2 ) − f (x1 )
= f 0 (x0 )?
x2 − x1
4
MINGFENG ZHAO
Proof. The converse of the mean value theorem true is not true. For example, let f (x) = x3 for all
x ∈ [−1, 1], then f 0 (0) = 0, but we know that f is strictly increasing, which implies that for any
x1 6= x2 ∈ [−1, 1], we have
f (x2 ) − f (x1 )
6= 0.
x2 − x1
7. [5 Points] Show that if f is differentiable and f 0 (x) ≥ 0 on (a, b), then f is strictly increasing
provided there is no subinterval (c, d) with c < d on which f 0 is identically zero.
Proof. For any x, y ∈ (a, b) such that x < y, by mean value theorem, we know that there exists some
z ∈ [x, y] such that
f (x) − f (y) = f 0 (z)(x − y).
By the assumption we know that f 0 (z) ≥ 0, which implies that f (x) ≥ f (y), that is, f is increasing
on (a, b). Now assume that f is not strictly increasing, then there exists some x0 , y0 ∈ (a, b) with
x0 < y0 such that f (x0 ) = f (y0 ). Since f is increasing on [x0 , y0 ], then f (x) = f (x0 ) for all x ∈ (x0 , y0 ),
which implies that f 0 (x) ≡ 0 for all x ∈ (x0 , y0 ), contradiction. Therefore, we know that f is strictly
increasin.
8. [5 Points] If f assumes a local maximum or minimum at an endpoint of its domain [a, b], what can
you say about the one-sided derivative?
Proof. Case I: f takes a local maximum at a, then we have f (x) ≤ f (a) for all x ∈ (a, b], then
f (x) − f (a)
≤ 0,
x−a
Hence we get f 0 (a+) ≤ 0.
∀x ∈ (a, b).
SOLUTION OF HW8
5
Case II: f takes a local minimum at a, then we have f (x) ≥ f (a) for all x ∈ (a, b], then
f (x) − f (a)
≥ 0,
x−a
∀x ∈ (a, b).
Hence we get f 0 (a+) ≥ 0.
Case III: f takes a local maximum at b, then we have f (x) ≤ f (b) for all x ∈ [a, b), then
f (x) − f (b)
≥ 0,
x−b
∀x ∈ (a, b).
Hence we get f 0 (b−) ≥ 0.
Case IV: f takes a local minimum at b, then we have f (x) ≥ f (b) for all x ∈ [a, b), then
f (x) − f (b)
≤ 0,
x−b
∀x ∈ (a, b).
Hence we get f 0 (b−) ≤ 0.
9. [5 Points] Given an example of a function that is differentiable on a bounded interval (a, b) but
cannot be made continuous on [a, b] be any definition of f (a) or f (b). Can you given an example
where f is bounded?
Proof. Let f (x) = tan x on (− π2 , π2 ), then we know that
lim
x→− π
2
f (x) = −∞,
and
lim f (x) = ∞.
x→ π
2
So f can not be made continuous on [− π2 , π2 ].
Another example, we can take
1
f (x) = sin ,
x
∀0 < x < 1.
Since lim f (x) does not exist, so we can not extend f to a continuous function on [0, 1], even
x→0+
though f is bounded by 1.
6
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu