Download Sol.

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 f (c)  f ( x), x  D,
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 f (c)  f ( x), x U (c). Similarly, we
can define local minimum (or relative minimum).
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],
then f attains its extreme values.
Fermat’s theorem If f has a local maximum or minimum
at c, and if f (c ) exists, then f (c)  0.
Proof. Suppose f has local maximum.
f ( c  h )  f (c )
f (c)  lim
0
h 0 
h
f (c  h )  f (c )
f (c)  lim
0
h 0 
h
 f (c)  f (c)  0.
Remark

f (c)  0 is only a necessary condition but not sufficient.
That is, if f (c)  0, c may not be a maximum or minimum
point of f.


A typical example is f(x)=x3. At x=0, f (0)  0, but f has no
maximum or minimum at 0.
If f has maximum or minimum at c, f (c ) may not exist.
For example, f(x)=|x| at x=0.
Critical number




Definition A critical number of a function f is a number c
in the domain of f such that f (c)  0 or f (c ) does not exist.
Fermat’s theorem is: If f has a local maximum or
minimum at c, then c is a critical number of f.
3
5
Ex. Find all critical numbers of f ( x)  x (4  x).
2
3

3
12  8 x
3
Sol.
5
5
f ( x)  x (4  x)  x  2/5  0  x  .
5
5x
2
So the critical numbers are 3/2 and 0.
The closed interval method
To 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 f ( x)  x ( x  1) on [-1,2].
2
3
2
Sol. f ( x)  x( x  1) (5 x  2)  0  x  0, ,1.
5
2
108
f (0)  0, f ( )  
, f (1)  0, f (1)  8, f (2)  4
5
3125
2
So the absolute maximum value is f(2)=4 and the absolute
minimum value is f(-1)=–8.
Example


1 x
Ex. Find the extreme values of f ( x)  arctan
on [0,1].
1 x
2
Sol. f ( x) 
 0.
2
2
(1  x)  (1  x)
No critical numbers!

Absolute maximum value  , absolute minimum value  0.
4
Mean value theorem



Outline
Rolle’s theorem
Lagrange’s mean value theorem
Rolle’s theorem
Rolle’s Theorem Let f be a function that satisfies the
following hypotheses:
1. f is continuous on the closed interval [a,b]. ( f  C[a, b] )
2. f is differentiable in the open interval (a,b). ( f  D(a, b) )
3. f(a)=f(b).
Then there exists a number c in (a,b) such that f (c)  0.
Example
Ex. Prove that the equation x3  x  1  0 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
f (c)  0. But, f ( x)  3x 2  1  1, this is a contraction.
Example
Ex. Suppose f  C[a, b], f  D(a, b) and f (a)  f (b)  0.
Prove that there is a number  ( a, b) such that f ( )  f ( ).
Analysis To use Rolle’s theorem, we need to find a function
F, such that F ( x)  f ( x)  f ( x) sincef ( )  f ( )  F ( )  0.
Does the F exist? No.
Can we change f ( )  f ( ) into [ f ( )  f ( )]e  0
Sol. Let F ( x)  e x f ( x). By the given condition, we have
F  C[a, b], F  D(a, b), F (a)  F (b)  0. Then by Rolle’s
theorem,   (a, b) such that F ( )  0 and hence f ( )  f ( ).
Question
2

(
x
)

(
x

1)
f ( x).

Suppose f exists on [1,2], f(1)=f(2)=0,
Prove that there is a number   (1, 2) such that  ( )  0.
Sol.
 (1)   (2)  0   ( )  0
 ( x)  2( x  1) f ( x)  ( x  1)2 f ( x)   (1)  0
 (1)  0,  ( )  0   ( )  0.
Question
Q1. Suppose f  C[a, b], f  D(a, b) and f (a)  f (b)  0.
Prove that for any   R,   (a, b) such that f ( )   f ( ).
Sol. F ( x)  e x f ( x).
Q2. Suppose f  C[0,1], f  D(0,1), f (0)  0. Let k be a
positive integer. Prove that   (0,1) such that
 f ( )  kf ( )  f ( ).
k
Sol. F ( x)  ( x  1) f ( x)
Question
Q1 Suppose f has second derivative on [0,1] and f(0)=f(1)=0.
Prove that   (0,1) such that 2 f ( )   f ( )  0.
Sol. F ( x)  xf ( x)  F  ( x)  xf ( x)  f ( x)
F (0)  F (1)  0  F  ( )  0
F  ( x)  xf  ( x)  2 f ( x), F (0)  F ()  0  F  ( )  0
Lagrange’s mean value theorem
Theorem Let f  C[a, b] and f  D(a, b). Then there is a
number c in (a,b) such that
f (b)  f (a )
f (c) 
.
ba
Proof. Let F ( x)  f ( x) 
f (b)  f (a)
( x  a).
ba
Then F  C[a, b], F  D(a, b) and F(a)=F(b)=f(a). By Rolle’s
theorem, there is a number c in (a,b) such that F (c)  0, or,
f (b)  f (a )
f (c) 
.
ba
Applications


Corollary If f ( x)  0 for all x in an interval (a,b), then f is
constant in (a,b).
Proof. x1 , x2  (a, b), f ( x1 )  f ( x2 )  f ( )( x1  x2 )  0.
Corollary If f ( x)  g ( x) for all x in (a,b), then f  g is
constant, that is, f(x)=g(x)+c where c is a constant.
Example

Ex. Prove the identity arcsin x  arccos x  .
2
Sol. Let f ( x)  arcsin x  arccos x, then
1
1

f ( x) 

 0  f ( x)  c  f (0)  .
2
1  x2
1  x2
Ex. Suppose f ( x)  f ( x), prove that f ( x)  ce x where c is a
constant.
x
Sol. F ( x)  e f ( x)  F ( x)  0
Example
1 x
1 x 1
Ex. Prove the inequality (1  )  e  (1  )
( x  0).
x
x
1
1 1
 ln(1  )  , or
Sol. The inequality is equivalent to
1 x
x x
1
1
 ln( x  1)  ln x  .
1 x
x
Let f(x)=lnx. By Lagrange’s mean value theorem,
1
ln( x  1)  ln x  f ( x  1)  f ( x)  f ( )( x  1  x)  ,

1 1
where x    x  1, hence 1
1 x
Therefore the inequality follows.



x
.
Question
Prove (1) e  1  x ( x  0),
x
1
(2) 2 x  3  ( x  1).
x
Sol. (1) f ( x)  e x  1  x, f ( x)  f (0)  f ( )( x  0)  e x
1
1
(2) f ( x)  2 x  3  , f ( x)  f (1)  f ( )( x  1) 
( x  1)
x

1 1
  (1, x)      x 

 x
Question
Prove when b>a>e,
ab  ba .
ln b ln a
a  b  b ln a  a ln b 

Sol.
b
a
ln x
1  ln 
f ( x) 
, f (b)  f (a)  f ( )(b  a)  2 (b  a)
x

b
a
Homework 8

Section 4.1: 53, 54, 55, 63, 74, 75

Section 4.2: 5, 18, 20, 25, 27, 28, 29, 30, 36