derivative and properties of functions

Derivative and properties of functions

Outline Maximum and minimum Fermat’s theorem The closed interval method

Maximum and minimum Definition. A function f has an absolute maximum (or

global maximum) at c if

where D is the domain of f. The number f(c) is called the maximum value. Similarly, we can define absolute minimum (or global minimum) and minimum value.

The maximum and minimum values are called extreme values of f.

Definition. A function f has a local maximum (or relative maximum) at c if Similarly, we can define local minimum (or relative minimum).

( ) ( ), , f c f x x D

( ) ( ), ( ). f c f x x U c

Remark Both the absolute maximum value and the absolute

minimum value are unique. But the absolute maximum

point or minimum point may not be unique.

When the domain is a closed interval, the endpoint can

NOT be a local maximum point or minimum point.

Fermat’s theorem The extreme value theorem If f is continuous on [a,b], th

en f attains its extreme values.

Fermat’s theorem If f has a local maximum or minimum at c, and if exists, then

Proof. Suppose f has local maximum.

( )f c ( ) 0. f c

0

( ) ( )( ) lim 0

h

f c h f cf c

h

0

( ) ( )( ) lim 0

h

f c h f cf c

h

( ) ( ) 0. f c f c

Remark is only a necessary condition but not sufficient. T

hat is, if c may not be a maximum or minimum point of f.

A typical example is f(x)=x3. At x=0, but f has no maximum or minimum at 0.

If f has maximum or minimum at c, may not exist. For example, f(x)=|x| at x=0.

( ) 0 f c( ) 0, f c

(0) 0, f

( )f c

Critical number Definition A critical number of a function f is a number c i

n the domain of f such that or does not exist. Fermat’s theorem is: If f has a local maximum or minimu

m at c, then c is a critical number of f.

Ex. Find all critical numbers of Sol.

So the critical numbers are 3/2 and 0.

( ) 0 f c ( )f c

3

5( ) (4 ). f x x x2 3

5 52/5

3 12 8 3( ) (4 ) 0 .

5 5 2

x

f x x x x xx

The closed interval methodTo find the absolute maximum and minimum values of a

continuous function f on a closed interval [a,b]: Find the values of f at all critical numbers. Find the values of f at endpoints. The largest of all the above values is the global maximum

value and the smallest is the global minimum value.

Example Ex. Find the extreme values of on [-1,2].

Sol.

So the absolute maximum value is f(2)=4 and the absolute minimum value is f(-1)=–8.

2 3( ) ( 1) f x x x

2 2( ) ( 1) (5 2) 0 0, ,1.

5 f x x x x x

2 108(0) 0, ( ) , (1) 0, ( 1) 8, (2) 4

5 3125 f f f f f

Example Ex. Find the extreme values of on [0,1].

Sol.

No critical numbers!

Absolute maximum value absolute minimum value

1( ) arctan

1

x

f xx

2 2

2( ) 0.

(1 ) (1 )

f xx x

,4

0.

Mean value theorem Outline Rolle’s theorem Lagrange’s mean value theorem

Rolle’s theoremRolle’s Theorem Let f be a function that satisfies the

following hypotheses:

1. f is continuous on the closed interval [a,b]. ( )

2. f is differentiable in the open interval (a,b). ( )

3. f(a)=f(b).

Then there exists a number c in (a,b) such that ( ) 0. f c

[ , ]f C a b

( , )f D a b

Example Ex. Prove that the equation has exactly one root.

Sol. Since f(0)<0, f(1)>0, by the intermediate theorem, there

exists a root. On the other hand, suppose there are two roots,

f(a)=f(b)=0, then by Rolle’s theorem, there is a c such that

But, this is a contraction.2( ) 3 1 1,f x x

3 1 0 x x

( ) 0. f c

ExampleEx. Suppose and

Prove that there is a number such that

Analysis To use Rolle’s theorem, we need to find a function

F, such that since

Does the F exist? No.

Can we change into

Sol. Let By the given condition, we have

Then by Rolle’s

theorem, such that and hence

[ , ], ( , ) f C a b f D a b ( ) ( ) 0. f a f b( , ) a b ( ) ( ). f f

( ) ( ) ( ) F x f x f x ( ) ( ) ( ) 0. f f F

( ) ( ). xF x e f x( ) ( ) f f [ ( ) ( )] 0 f f e

[ , ], ( , ), ( ) ( ) 0. F C a b F D a b F a F b( , ) a b ( ) 0 F ( ) ( ). f f

Question Suppose exists on [1,2], f(1)=f(2)=0,

Prove that there is a number such that

Sol.

f

(1) (2) 0 ( ) 0 2( ) 2( 1) ( ) ( 1) ( ) (1) 0 x x f x x f x

2( ) ( 1) ( ). x x f x(1,2) ( ) 0.

(1) 0, ( ) 0 ( ) 0.

QuestionQ1. Suppose and

Prove that for any such that

Sol.

Q2. Suppose Let k be a

positive integer. Prove that such that

Sol.

[ , ], ( , ) f C a b f D a b ( ) ( ) 0. f a f b( , ) a b ( ) ( ). f f , R

( ) ( ). xF x e f x

[0,1], (0,1), (0) 0. f C f D f

( ) ( 1) ( ) kF x x f x( ) ( ) ( ). f kf f

(0,1)

Question Q1 Suppose f has second derivative on [0,1] and f(0)=f(1)=0.

Prove that such that

Sol. (0,1) 2 ( ) ( ) 0. f f

( ) ( ) ( ) ( ) ( ) F x xf x F x xf x f x(0) (1) 0 ( ) 0 F F F ( ) ( ) 2 ( ), (0) ( ) 0 ( ) 0 F x xf x f x F F F

Lagrange’s mean value theoremTheorem Let and Then there is a

number c in (a,b) such that

Proof. Let

Then and F(a)=F(b)=f(a). By Rolle’s

theorem, there is a number c in (a,b) such that or,

[ , ]f C a b ( , ).f D a b

( ) ( )( ) .

f b f af c

b a

( ) ( )( ) ( ) ( ).

f b f a

F x f x x ab a

[ , ], ( , ) F C a b F D a b( ) 0, F c

( ) ( )( ) .

f b f af c

b a

Applications Corollary If for all x in an interval (a,b), then f is

constant in (a,b).

Proof.

Corollary If for all x in (a,b), then is

constant, that is, f(x)=g(x)+c where c is a constant.

( ) 0 f x

1 2 1 2 1 2, ( , ), ( ) ( ) ( )( ) 0. x x a b f x f x f x x

( ) ( ) f x g x f g

ExampleEx. Prove the identity

Sol. Let then

Ex. Suppose prove that where c is a

constant.

Sol.

( ) arcsin arccos , f x x xarcsin arccos .

2 x x

2 2

1 1( ) 0 ( ) (0) .

21 1

f x f x c f

x x

( ) ( ), f x f x ( ) xf x ce

( ) ( ) ( ) 0 xF x e f x F x

ExampleEx. Prove the inequality

Sol. The inequality is equivalent to or

Let f(x)=lnx. By Lagrange’s mean value theorem,

where hence

Therefore the inequality follows.

11 1(1 ) (1 ) ( 0). x xe x

x x1 1 1

ln(1 ) ,1

x x x1 1

ln( 1) ln .1

x xx x

1ln( 1) ln ( 1) ( ) ( )( 1 ) , x x f x f x f x x

1, x x 1 1 1.

1

x x

Question Prove (1) (2)

Sol. (1)

(2)

1 ( 0), xe x x

( ) 1 , ( ) (0) ( )( 0)xf x e x f x f f x e x

1 1( ) 2 3 , ( ) (1) ( )( 1) ( 1)f x x f x f f x x

x

12 3 ( 1). x x

x

1 1(1, ) x x

x

Question Prove when b>a>e,

Sol. ln ln

ln ln b a b aa b b a a b

b a

2

ln 1 ln( ) , ( ) ( ) ( )( ) ( )

x

f x f b f a f b a b ax

.b aa b

Homework 8 Section 4.1: 53, 54, 55, 63, 74, 75

Section 4.2: 5, 18, 20, 25, 27, 28, 29, 30, 36