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```Maximum and Minimum Values
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MATH 1310
Lecture 23
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Definitions:
A function f has an absolute maximum at c if f ( c )  f ( x ) for all x in D,
where D is the domain of f . The number f (c ) is called the maximum value of f
on D. (Also called the global maximum.)
A function f has an absolute minimum at c if f ( c )  f ( x ) for all x in D.
The number f (c ) is called the minimum value of f on D. (Also called the
global minimum.)
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Lecture 23
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A function f has an relative maximum at c if f ( c )  f ( x ) for all x near c.
The number f (c ) is called the maximum value of f on D. (Also called the local
maximum.)
A function f has an relative minimum at c if f ( c )  f ( x ) for all x near c.
The number f (c ) is called the minimum value of f on D. (Also called the local
minimum.)
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MATH 1310
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Lecture
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Terminology:
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Points vs. Values: c is the point where something happens, the input;
f (c ) is the value, or the output.
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Extrema: Max’s and min’s are also called extrema.
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Local vs. Global: Local extrema also called relative extrema. This is
because they are extreme relative to points nearby. Global extrema are
also called absolute extrema, and are so named because they are the
value that bound the function over its entire domain.
MATH 1310
Lecture 23
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Extreme Value Theorem
If f (x ) is continuous on the closed interval [a, b],
then f (x ) achieves both a global (absolute)
maximum and global minimum at some numbers c
and d in [a, b].
This theorem guarantees solutions to many problems you will encounter.
Fermat’s Theorem
If f (x ) has a local maximum or minimum value at
c, and if f (c ) exists, then f ( c )  0
MATH 1310
Lecture 23
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Definition
A critical number (point) of a function f is a number c in the
domain of f such that either f ( c )  0 or f (c ) does not exist.
Example:
Find all critical numbers for the function f ( x )  x 3  3x 2  24 x .
Solution:
First find the critical points which is where f ( c )  0, or D.N.E.
f ( x )  3x 2  6 x  24  0 , so we need to solve x 2  2 x  8  0 .
Factoring is easiest so rewrite this as ( x  4)( x  2)  0 . Critical
points are x  2, and x  4 .
Example:
Solution:
MATH 1310
x2
Find all critical numbers for the function f ( x ) 
.
x 1
2
2
2 x ( x  1)  x (1) x  2 x

f
x
(
)


.
Compute
2
( x  1)
( x  1) 2
x ( x  2)
Rewrite f ( x ) 
. So critical numbers are x  0, 2 . Since
2
( x  1)
x  1 is not in the domain of f, it is not s critical point.
Lecture 23
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The Closed Interval Method
To find the absolute maximum and minimum values of a continuous
function f on a closed interval [a, b]:
1) Find the values of f at the critical numbers of f in (a, b)
2) Find the values of f at the endpoints of the interval.
3) The largest value from steps 1 and 2 is the absolute maximum
value, and the smallest of these values is the absolute minimum value.
Example:
Find the absolute maximum and minimum values for the
function f ( x )  x 3  3x 2  24 x on [1, 5].
We already found the critical numbers to be x  2, and
x  4 . We consider only x  2, since x  4 is not in the
interval. Now check the values:
f ( 1)  26 ,
MATH 1310
f ( 2)  28 ,
Lecture 23
and f (5)  80 .
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f ( x )  x 3  3x 2  24 x
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Example:
Find the absolute maximum and minimum values
x2
on [, 5].
for the function f ( x ) 
x 1
Solution:
First note that f is not continuous in this interval. It may or
may not take on any maximum or minimum value. Critical
numbers were x = 0 and 2. Since f (1) , is not defined one
need examine the function closer.
MATH 1310
Lecture 23
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Example:
Find the absolute maximum and minimum values
for the function f ( x )  x  2 sin x on [0,  ].
Solution:
f ( x )  1  2 cos x , and so critical numbers are
1
determined by 1  2 cos x  0 , or cos x   . This
2
2
occurs for x 
in the given interval. Now
3
checking all the values:
2
2
2
 2 sin 
f   
 3  3
 3 
 3  2
2
 2  
 3  3.83
3
 2  3
Checking the endpoints f (0)  0  2 sin(0)  0 , and
f ( )    2 sin( )   so the absolute min is 0,

and max is
MATH 1310
2
 3.
3
Lecture 23
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MATH 1310
Lecture 23
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Example:
Determine the absolution maximum and minimum values of the
function f ( x )  x 2 e  x on the interval [0, 10].
Solution:
First we calculate f ( x )  ( 2 x ) e  x  x 2 (  e  x ) using the product rule.
Factoring we get f ( x )  x ( 2  x ) e  x , so that critical numbers are
solutions to
f ( x )  x ( 2  x ) e  x  0 which happens when x  0, 2 .
Comparing values
f ( 0)  0 ,
f ( 2)  4e  2 
4
100
10
,
and
f
e

0
.
54
(
10
)

100

 0.0045
2
10
e
e
So, the absolute maximum is f ( 2) 
f ( 0)  0 .
MATH 1310
Lecture 23
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, and the absolute minimum is
2
e
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MATH 1310