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.

3

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.¤

6

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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