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If f is a continuous function on the closed interval [ a , b ] and differentiable on the open interval ( a , b ), then the exists a number x = c in ( a , b ) , that is, a number x = c where a < c < b , such that f

( c )
 f ( b b
)

 a f ( a )
Note: Sometimes the quantity f ( b )
 f ( a ) b
 a
is referred to as the average or mean slope of the function on the interval [ a , b ].
Geometric Interpretation of the Mean Value Theorem:

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Example 1: Use the graph of the following function to estimate a value of c that satisfies the conclusion of the Mean Value Theorem for the interval [0, 4].
Solution:
█

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Example 2: Determine whether the Mean Value Theorem can be applied to the function f ( x )

2 x
3 
5 x on the closed interval [ a, b ] = [-2, 2]. If so, find a value of c in the open interval ( a, b ) = (-2, 2) where f

( c )
 f ( b ) b

 a f ( a )
.

4
Example 3: Determine whether the Mean Value Theorem can be applied to the function
2 f ( x )

3 x 3 interval ( a, b

2 x on the closed interval [ a, b ] = [0, 1]. If so, find a value of c in the open
) = (0, 1) where f

( c )
 f ( b ) b

 a f ( a )
.
.
Solution: To see whether the function
2 f ( x )

3 x 3

2 x satisfies the criteria for the Mean
Value Theorem, first note that f ( x )
2

3 x
3 
2 x is defined for all values of x in the closed interval [0, 1]. Thus,
2 f ( x )

3 x 3

2 x must be continuous for all values of x in the interval [0, 1]. To test the next criteria, we compute the derivative of the function f , which is given by f

( x )

3 (
2
3
) x

1
3

2

2
1
1 x 3

2

2
3 x

2 ,
For the close interval ( a, b ) = (0, 1), f

( x )

3
2 x

2 is defined for all values in (0, 1)
(note, that the endpoint x = 0 is not included in this interval). Thus, f

( x )

3
2 x

2 must be continuous for all values of x in the interval (0, 1). Since the criteria of the Mean
Value Theorem are satisfied, we look for a value of c where f

( c )
 f ( b )
 b
 a f ( a )
For this example, a = 0 and b = 1. Thus, for the function f ( x )
2

3 x 3 
2 x , f f ( a )

( b )
 f f
(
( 0 )
1 )


3 ( 0 )
2 / 3
3 ( 1 )
2 / 3 

2
2 ( 0 )
( 1 )


0
3


0

2

1
0
.
And for the derivative f

( x )

3
2 x
Formula, we obtain the following:

2 , f

( c )

2
3 c

2 . Thus, using the Mean Value
(Continued on next page)

2
3 c

2
 f

( c )
 f ( b b
)

 a f ( a )
 f ( 1 )

1

0 f ( 0 )
3
2 c

2

1

1
0
(Use fact from above that f ( 0 )

0 and f ( 1 )

1
3
2 c

2

1 (Simplify)
3
2 c

3 (Add 2 to both sides)
3
2 c
3 c

3
3 c (Multiply both sides by
3 c
3
3 c

2 (Rearrange and simplify)
3 c

`
2
3
(Divide both sides by 3)
(
3 c )
3 
`(
2
3
)
3
(Raise both sides to third power to eliminate radical) c

8
27
(Simplify and solve for c )
Thus, the value of x
 c

8
satisfies the Mean Value Theorem. This value is further
27 illustrated with the following graph:
(Continued on next page)
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