Section 5.5
The Intermediate Value Theorem
Rolle’s Theorem
The Mean Value Theorem
3.6
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is a value between f(a) and f(b),
then there is at least one point c between a and b where f takes
on the value N.
f(b)
N
f(a)
a c
b
Rolle’s Theorem
If f is continuous on [a, b], if f(a) = 0, f(b) = 0, then there is at
least one number c on (a, b) where f ‘ (c ) = 0
slope = 0
f ‘ (c ) = 0
a
c
b
Given the curve:
mtan  lim
h0
f  x  h  f  x 
h
f(x+h)
x
x+h
f(x)
msec 
f  x  h  f  x 
h
The Mean Value Theorem (MVT)
aka the ‘crooked’ Rolle’s Theorem
If f is continuous on [a, b] and differentiable on (a, b)
There is at least one number c on (a, b) at which
f b   f  a 
 f ' c 
ba
Conclusion:
Slope of Secant Line
Equals
Slope of Tangent Line
f(b)
a
c
b
f(a)
If f  x   x  2x  1, a  0, b  1, and
2
f(0) = -1
f b   f  a 
ba
f b   f  a 
ba
f(1) = 2

2   1
1 0
f '  x   2x  2
3  2x  2
1
x
2
3
 f ' c  , find c.
Find the value(s) of c which satisfy Rolle’s Theorem for
f  x   x 4  x on the interval [0, 1].
Verify…..f(0) = 0 – 0 = 0
f(1) = 1 – 1 = 0
f '  x   4x3  1
0  4x3  1
c
3
1
4
which is on [0, 1]
Find the value(s) of c that satisfy the Mean Value Theorem for
1
f  x   x  on  4, 4 
x
1 17
f  4  4  
4 4
1 17
f  4   4 

4
4
17  17 

f b   f  a 
4  4  17


ba
4   4 
16
17
1
f ' c  
 1 2
16
x
Find the value(s) of c that satisfy the Mean Value Theorem for
1
f  x   x  on  4, 4
x
Note: The Mean Value Theorem requires the function to be
continuous on [-4, 4] and differentiable on (-4, 4). Therefore, since
f(x) is discontinuous at x = 0 which is on [-4, 4], there may be no
value of c which satisfies the Mean Value Theorem
1 1
 2 has no real solution, there is no value of c on
Since
16 x
[-4, 4] which satisfies the Mean Value Theorem
Given the graph of f(x) below, use the graph of f to estimate the
numbers on [0, 3.5] which satisfy the conclusion of the Mean Value
Theorem.
Determine whether f  x   x 2  2x  2 satisfies the hypothesis of
the Mean Value Theorem on  -2, 2. If it does, find all numbers
c in (a, b) such that f '  c  
f b   f  a 
ba
f(x) is continuous and differentiable on [-2, 2]
f  2  f  2 
2   2

6   2 
4
2
f '  x   2x  2
2x  2  2  c  0
On the interval [-2, 2], c = 0 satisfies the conclusion of MVT
x2  1
Determine whether f  x  
satisfies the hypothesis of
x2
the Mean Value Theorem on -2, 1. If it does, find all numbers
c in (a, b) such that f '  c  
f b   f  a 
ba
f(x) is continuous and differentiable on [-2, 1]
 3 
0
 1
f 1  f  2 
4



1   2 
3
4
f 'x 


2x  x  2   1 x 2  1
 x  2
2
x 2  4x  1 1
2
2
2


3x
 12x  0

4x

16x

4

x

4x

4
2
x  4x  4 4
3x  x  4   0
On the interval [-2, 1], c = 0 satisfies the conclusion of MVT
x2  1
Determine whether f  x  
satisfies the hypothesis of
x2
the Mean Value Theorem on 0, 4 . If it does, find all numbers
c in (a, b) such that f '  c  
f b   f  a 
ba
Since f(x) is discontinuous at x = 2, which is part of the interval
[0, 4], the Mean Value Theorem does not apply
Determine whether f  x   x 3  3x  1 satisfies the hypothesis of
the Mean Value Theorem on -1, 2. If it does, find all numbers
c in (a, b) such that f '  c  
f b   f  a 
ba
f(x) is continuous and differentiable on [-1, 2]
f  2   f  1
2   1
33

0
3
f '  x   3x 2  3
3x2  3  0
3  x  1 x  1  0
c = 1 satisfies the conclusion of MVT


CALCULATOR REQUIRED

If f  x   x 2  12 x 2  4 , how many numbers on [-2, 3] satisfy
the conclusion of the Mean Value Theorem.
A. 0
B. 1
C. 2
D. 3
E. 4
f(3) = 39
f b   f  a 
f(-2) = 64
64  39
 5
ba
2  3
For how many value(s) of c is f ‘ (c ) = -5?

X
X
X