Derivative and properties of functions
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Transcript of 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