# Chapter 4. The Derivative in Applications Lecture 1. Mean-Value ...

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8 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)

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Chapter 4.The Derivative in Applications
Lecture 1.Mean-Value Theorems
(Section 4.8 )
We will always concern the functions de¯ned on some intervals.
1.Mean-Value Theorems
Theorem 4.8.1(Rolle's Theorem) Let f be di®erentiable
on (a;b) and continuous on [a;b].If f(a) = f(b),then there is at
least one number c in (a;b) such that f
0
(c) = 0.
Proof By Theorem (M) in Lecture 3 of Chapter 2,there are
u 2 [a;b] and v 2 [a;b],such that f(u) is the minimal value of f
on [a;b] and f(v) is the maximal value.That is,for all x 2 [a;b]
f(u) · f(x) · f(v):
If f(u) = f(v) = f(a) = f(b) then f is a constant function,so
f
0
(x) = 0 for all x.If this is not the case,then at least one of the
1
values f(u) and f(v) must be di®erent than f(a) = f(b).Assume
f(u) < f(a) = f(b).Then a < u < b.Hence we have
lim
u>t!u
f(t) ¡f(u)
t ¡u
= f
0
(u) · 0;
and
lim
u<t!u
f(t) ¡f(u)
t ¡u
= f
0
(u) ¸ 0:
A combination of these inequalities yield f
0
(u) = 0:¤
Theorem 4.8.2(Lagrange's Theorem) Let f be di®eren-
tiable on (a;b) and continuous on [a;b].Then there is at least one
number c in (a;b) such that f(b) ¡f(a) = f
0
(c)(b ¡a).
Proof De¯ne
v(x) = f(x) ¡
f(b) ¡f(a)
b ¡a
(x ¡a):
Then v 2 C[a;b] and v is di®erentiable on (a;b).Moreover,v(a) =
f(a) = v(b).Applying Rolle's Theorem to v we know there is a
number c 2 (a;b) such that v
0
(c) = 0,i.e.f
0
(c) =
f(b) ¡f(a)
b ¡a

2
Theorem 4.1.2 Let f be di®erentiable on (a;b) and contin-
uous on [a;b].
(a) If f
0
(x) > 0 for every x 2 (a;b),then f is strictly increasing
on [a;b];
(b) If f
0
(x) < 0 for every x 2 (a;b),then f is strictly decreas-
ing on [a;b];
(c) If f
0
(x) = 0 for every x 2 (a;b),then f is constant on [a;b].
Proof Let a · u < v · b.In case (a) by Lagrange Theorem
there is c 2 (u;v) such that f(v) ¡f(u) = f
0
(c)(v ¡u) > 0.Hence
f is strictly increasing on [a;b].
The conclusions (b) and (c) can be similarly proved.¤
Remark In (a) if the condition f
0
(x) > 0 is replaced by
f
0
(x) ¸ 0 at every x 2 (a;b),then f is increasing but may not
strictly.In (b) if the condition f
0
(x) < 0 is replaced by f
0
(x) · 0
at every x 2 (a;b),then f is decreasing but may not strictly.
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Example 1 The function sin is strictly increasing on the
interval [¡
¼
2
;
¼
2
];and cos is strictly decreasing on [0;¼].
Example 2( Exercise 38,on Page 451) (a) Showthat f(x) =
x
4
¡2x
3
is not one-to-one on R.
(b) Find the smallest value of k such that f is one-to-one on
the interval [k;+1).
Solution (a) We have f
0
(x) = 4x
3
¡ 6x
2
= 2x
2
(2x ¡ 3);
from which we see f
0
(x) < 0 when 0 6= x <
3
2
and f
0
(x) > 0 when
x >
3
2
.Hence f is strictly decreasing on (¡1;
3
2
] and strictly
increasing on [
3
2
;+1).And it is obvious that
lim
x!¡1
f(x) = lim
x!¡1
x
4
(1 ¡
2
x
) = +1= lim
x!+1
f(x):
We conclude by Theorem (I) in Lecture 3 of Chapter 2 that for
each value y > f(
3
2
) = ¡
27
16
there exist two numbers x
1
<
3
2
and
x
2
>
3
2
such that f(x
1
) = f(x
2
) = y:So,f is not one-to-one on
R.
(b) From the above argument we know k =
3
2
is the desired
4
value.¤
2.Convexity
De¯nition (Convexity) Let f be a function de¯ned on an
open interval I.
If for any three points a;b;c in I whenever a < c < b,the
relation
f(c) ·
b ¡c
b ¡a
f(a) +
c ¡a
b ¡a
f(b)
holds,then f is said to be convex down (or concave up).
If for any three points a;b;c in I whenever a < c < b,the
relation
f(c) ¸
b ¡c
b ¡a
f(a) +
c ¡a
b ¡a
f(b)
holds,then f is said to be convex up (or concave down).
Theorem (Convexity) Let f be a function de¯ned on an
open interval I.
If f has derivative and f
0
is increasing on I,then f is convex
5
down on I.
If f has derivative and f
0
is decreasing on I,then f is convex
up on I.
Proof Let a < c < b be three points in I.Then
b ¡c
b ¡a
f(a) +
c ¡a
b ¡a
f(b) ¡f(c)
=
b ¡c
b ¡a
(f(a) ¡f(c)) +
c ¡a
b ¡a
(f(b) ¡f(c)):
Applying Lagrange Theorem we know there exist u 2 (a;c) and
v 2 (c;b) such that
f(a) ¡f(c) = f
0
(u)(a ¡c);f(b) ¡f(c) = f
0
(v)(b ¡c):
Then we get
b ¡c
b ¡a
f(a) +
c ¡a
b ¡a
f(b) ¡f(c) =
(b ¡c)(c ¡a)
b ¡a
(f
0
(v) ¡f
0
(u)) ¸ 0
since f
0
is increasing.Then we get the convexity of f.
A similar argument shows that if f
0
is decreasing on I then f
is convex up on I.¤
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Corollary Let f be a function de¯ned on an open interval
I.Assume that f has second order derivative on I.If f
00
(x) ¸
0;8x 2 I then f is convex down on I.If f
00
(x) · 0;8x 2 I then f
is convex up on I.
Example 2 sin and cos are all convex up on (0;
¼
2
).
Example 3 The function f(x) = jxj;x 2 R is convex down
on R although it has no derivative at x = 0:
3.Local Extrema
De¯nition 4.2.1 Let f be a function de¯ned on an open
interval I.If x 2 I and there exists a number ± > 0 such that for
all y 2 (x ¡±;x +±),f(y) · f(x) (or f(y) ¸ f(x)),then f(x) is
said to be a local maximum(or local minimum,respectively)
of f.Maximum and minimum are all called extremum.
Theorem 4.2.2 Let f be a function de¯ned on an open
interval I.If f has a local extremumat x 2 I and f has derivative
7
at x then f
0
(x) = 0:
The spirit of the proof is the same as that of the proof of
Rolle's Theorem.You may do it as an exercise.We omit it here.
We call a point x at which f
0
(x) = 0 a stationary point of
f.
Home work:P.249,12,14,16,18,20,22.P.256,24,26,28,
30,32,34.
Prove Theorem 4.2.2.
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