Download Lesson 17 Absolute Maximum and Minimum Values

LESSON 17 ABSOLUTE MAXIMUM AND MINIMUM VALUES
Theorem If a function f is continuous on the closed interval [ a , b ] , then f takes
on an absolute maximum value and an absolute minimum value at least once in the
interval [ a , b ] .
Procedure for finding the absolute maximum and minimum values of a
function on a closed interval: Verify that the function is continuous on the closed
interval. Then find the critical number(s) of the function. Evaluate the function at
the critical number(s) which are in the closed interval. Then evaluate the function
at the left-hand and right-hand endpoints of the closed interval. The largest
functional value is the absolute maximum value and the smallest functional value is
the absolute minimum value of the function on the closed interval.
Examples Find the absolute maximum and minimum values of the following
functions on the given interval.
1.
f ( x )  2 x 3  6 x 2  15 ; [  1, 4 ]
Since f is a polynomial function, then it is continuous for all real numbers.
Thus, it is continuous on the closed interval [  1, 4 ] .
f ( x )  6 x 2  12 x  6 x ( x  2 )
Critical Numbers: 0, 2
Both of these critical numbers are in the closed interval [  1, 4 ] .
f ( 0 )   15
f ( 2 )  16  24  15   23
f (  1 )   2  6  15   23
f ( 4 )  128  96  15  17
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Absolute Maximum: 17
Absolute Minimum:  23
2.
f ( x )  2 x 3  6 x 2  15 ; [  4 , 1 ]
Since f is a polynomial function, then it is continuous for all real numbers.
Thus, it is continuous on the closed interval [  4 , 1 ] .
f ( x )  6 x 2  12 x  6 x ( x  2 )
Critical Numbers: 0, 2
Only the critical number 0 is in the closed interval [  4 , 1 ]
f ( 0 )   15
f (  4 )   128  96  15   239
f ( 1 )  2  6  15   19
Absolute Maximum:  15
Absolute Minimum:  239
3.
g ( x )  12  16 x  4 x 2  x 3 ; [  5 , 3 ]
Since g is a polynomial function, then it is continuous for all real numbers.
Thus, it is continuous on the closed interval [  5 , 3 ] .
g ( x )  16  8 x  3x 2  ( 4  x ) (4  3x )
Critical Numbers:  4 ,
4
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Both of these critical numbers are in the closed interval [  5 , 3 ] .
g ( x )  12  16 x  4 x 2  x 3
g (  4 )  12  64  64  64   52
64 64 64
1 
4
1 1
g    12 


12

64



 =
=
3
9
27
3
 3 9 27 

3
1 
 5 
 9
 5
12  64 


 = 12  64 
 = 4  3  16    =
 27  
 27 27 27 
 27 

644
 81 80 
 161 
4

 = 4
 =
27
 27 27 
 27 
g (  5 )  12  80  100  125   43
g ( 3 )  12  48  36  27   3
Absolute Maximum:
644
27
Absolute Minimum:  52
4.
g ( x )  12  16 x  4 x 2  x 3 ; [  6 ,  1 ]
Since g is a polynomial function, then it is continuous for all real numbers.
Thus, it is continuous on the closed interval [  6 ,  1 ] .
g ( x )  16  8 x  3x 2  ( 4  x ) (4  3x )
Critical Numbers:  4 ,
4
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Only the critical number  4 is in the closed interval [  6 ,  1 ] .
g ( x )  12  16 x  4 x 2  x 3
g (  4 )  12  64  64  64   52
g (  6 )  12  16 (  6 )  4 (  6 ) 2  (  6 ) 3  12  16 ( 6 )  4 ( 6 ) 2  63 =
12  6 (  16  24  36 )  12  6 (  4 )  12  24   12
g (  1 )  12  16  4  1   7
Absolute Maximum:  7
Absolute Minimum:  52
5.
y
6
; [  2 , 1]
4t 2  3t  22
The domain of this rational function is the set of all real numbers such that
11
t
and t  2 . Thus, the function is continuous for all t except for
4
11

and 2. Thus, this function is continuous on the closed interval
4
[  2 , 1] .
We differentiated this function in Lesson 9 obtaining that
 6 ( 8t  3 )
 6 ( 8t  3 )
 6 ( 8t  3 )
dy

dt ( 4t 2  3t  22 ) 2 = [ ( t  2 ) ( 4t  11) ]2 = ( t  2 ) 2 ( 4t  11) 2
Critical Number: 
3
8
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
This critical number is in the closed interval [  2 , 1 ] .
Since y 
6
6
=
( t  2 ) ( 4t  11) , then
4t  3t  22
 3
y  
 8
6
= 3
 3

   2     11 
 8
 2


=
6
 19   19 



 8  2 
=
6
361

16
=
96
361
y( 2) 
y (1) 
6.
2
6
2
1


(  4 ) ( 3)  4
2
6
2

(  1 ) ( 15 )
5
Absolute Maximum: 
96
361
Absolute Minimum: 
1
2
h( x )  x  2 cos x ; [  2  , 2  ]
Since h is the difference of the polynomial function y  x , which is
continuous for all real numbers, and the trigonometric function y  cos x ,
which is also continuous for all real numbers, then the function h is
continuous for all real numbers. Thus, it is continuous on the closed interval
[  2 , 2 ] .
h ( x )  1  2 (  sin x )  1  2 sin x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
h ( x )  0  1  2 sin x  0  sin x  
1
2
Since the sine function is negative in the III and IV quadrants, then the angle
solutions for x are in the III and IV quadrants. The reference angle x for
1

sin
x

these angles is obtained by solving the equation
. Thus,
2
1 
x   sin  1  . Thus, the angles in the III quadrant, which are in the
2 6
5
7

[

2

,
2

]
closed interval
, are
and
. The angles in the IV
6
6

11

[

2

,
2

]
quadrant, which are in the closed interval
, are
and
.
6
6
Critical Numbers: 
5
 7  11
,  ,
,
6
6
6
6
h( x )  x  2 cos x

3 
5
5
 5 
 5 




2

h 



2
cos



 =

 =
6
2
6 
6
6 





5

6
3 = 
6 3
6 3  5
5

  0.8859
=
6
6
6
 3 


  
  




2
h    
 2 cos    =

 =
6
2
6
 6 
 6 




6

3 = 

6

6 3
6
= 
  6 3
6

3 
7
7
7
 7 



2

h
 2 cos
 
= 6

2  =
6
6
 6 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
  2.2556
7

6
3 =
6 3
7  6 3
7

 5.3972
=
6
6
6
 3 
11
11
11
 11 



2
h


2
cos

= 6

 =
2
6
6
 6 


11

6
3 =
6 3
11  6 3
11

 4.0275
=
6
6
6
h (  2  )   2   2 cos (  2  )   2   2 ( 1 )   2   2   8.2832
h ( 2  )  2   2 cos 2   2   2 ( 1 )  2   2  4.2832
7  6 3
Absolute Maximum:
6
Absolute Minimum:  2   2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850