THE MEAN VALUE THEOREM FOR DERIVATIVES
If y  f  x  is continuous on the interval  a, b , and is differentiable everywhere on the interval
 a, b  , then there is at least one number c between
f b   f  a 
ba
a and b such that:
 f c .
In other words, there’s some point in the interval where the slope of the tangent line equals the
slope of the secant line that connects the endpoints of the interval. (The function has to be
continuous at the endpoints of the interval, but it doesn’t have to be differentiable at the
endpoints.) You can see this graphically below:
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Example 1
Suppose you have the function f  x   x2 , and you’re looking at the interval 1, 3 . The mean
value theorem for derivatives (this is often abbreviated MVTD) states that there is some number
c such that
f c 
32  12
4.
3 1
Because f   x   2 x , plug c in for x and solve. 2c  4  c  2 . Notice that 2 is in the
interval. This is what the MVTD predicted. If you don’t get a value of c that is in the interval,
something went wrong; either the function is not continuous and differentiable in the required
interval, or you made a mistake.
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Example 2
Consider the function f  x   x3 12 x on the interval  2, 2 . The MVTD states that there is a
c in that interval such that:
2
3
f c 


 24   2   24
2   2 
3
  8 .
Then f   c   3c2  12  8 . Solving shows c  
2
(which is approximately 1.155 ).
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Notice that here there are two values of c that satisfy the MVTD. That’s allowed. In fact, there
can be infinitely many values, depending on the function.
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Example 3
Consider the function f  x  
1
on the interval  2, 2 .
x
Follow the MVTD:
1  1
 
2  2 1
f c 

2   2 
4
Then:
f c 
1 1

c2 4
There is no value of c that will satisfy this equation! We expected this. Why? Because f  x  is
not continuous at x  0 , which is in the interval.
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Suppose the interval had been 1, 3 , eliminating the discontinuity. The result would have been:
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 1
1
3
f c 

3  1
3
And
f c 
1
1

2
c
3

c 3
c   3 is not in the interval, but c  3 is in the interval. The answer is c  3 .
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ROLLE’S THEOREM
Rolle’s Theorem is a special case of the MVTD. It says the following:
If y  f  x  is continuous on the interval  a, b , and is differentiable everywhere on the interval
 a, b  , and if f  a   f b   0 , then there is at least one number c between a and b such that
f  c  0 .
Graphically, this means that a continuous, differentiable curve has a horizontal tangent between
any two points where it crosses the x-axis.
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(f-s)(x)
c1
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c2
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Consider the function f  x   x2  x  12 on the interval  3, 4 . Follow the MVTD:
Since f  3  0 and f  4   0 , we can follow the MVTD to see that this is an example of
Rolle’s Theorem:
f   c   2c  1 and f   c  
00
1
 0 . Therefore, 2c  1  0  c  .
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Example 4
Find the values of c that satisfy Rolle’s Theorem for f  x  
x2
 6 x on the interval 0, 12 .
2
First, verify that f  0  0 and f 12  0 .
02
f  0   6  0  0
2
f 12  
122
 6 12   72  72  0 .
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Now find f   x   x  6 .
Therefore, f   c   c  6 .
Set f   c   0 :
c6  0

c  6.
This value of c falls in the interval.
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Homework
1.
Find the values of c that satisfy the MVTD for f  x   3x2  5x  2 on the interval
1, 1 .
2.
Find the values of c that satisfy the MVTD for f  x   x3  24 x  16 on the interval
0, 4 .
3.
Find the values of c that satisfy the MVTD for f  x   x3  12 x2  7 x on the interval
4, 4 .
4.
Find the values of c that satisfy the MVTD for f  x  
6
 3 on the interval 1, 2 .
x
5.
Find the values of c that satisfy the MVTD for f  x  
6
 3 on the interval  1, 2 .
x
6.
Find the values of c that satisfy the Rolle’s theorem for f  x   x2  8x  12 on the
interval  2, 6 .
7.
Find the values of c that satisfy the Rolle’s theorem for f  x   x3  x on the
interval  1, 1 .
8.
Find the values of c that satisfy the Rolle’s theorem for f  x   x 1  x  on the
interval 0, 1 .
9.
Find the values of c that satisfy the Rolle’s theorem for f  x   1 
1
on the
x2
interval  1, 1 .
10.
Find the values of c that satisfy the Rolle’s theorem for f  x   x2/3  x1/3 on the
interval 0, 1 .
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ANSWERS
1.
c0
2.
c
3.
c
4.
c 2
5.
No Solution. The function is not continuous on the interval. The MVTD does not apply.
6.
c4
7.
c
8.
c
9.
No Solution. The function is not continuous on the interval. The MVTD does not apply.
10.
c
4
3
12  8 3
 0.62
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