1. The Intermediate Value Theorem
Section 5.5
The Intermediate Value Theorem
Rolle’s Theorem
The Mean Value Theorem
3.6

2. 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. b
f(b) N c a
f(a)

3. 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 c
f ‘ (c ) = 0 a b

4. Given the curve:
f(x+h)
x+h x
f(x)

5. 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
Conclusion:
Slope of Secant Line
Equals
Slope of Tangent Line a b
f(a)
f(b) c

6. Show that has exactly one solution on [0, 1].
f(0) = -1 and f(1) = 2
By IVT, there is AT LEAST ONE solution on [0, 1]
Since there is no max/min on (0, 1), there is EXACTLY
One solution on [0, 1].

7. f(0) = f(1) = 2

8. Find the value(s) of c which satisfy Rolle’s Theorem for
on the interval [0, 1].
Verify…..f(0) = 0 – 0 = f(1) = 1 – 1 = 0
which is on [0, 1]

9. Find the value(s) of c that satisfy the Mean Value Theorem for

10. Find the value(s) of c that satisfy the Mean Value Theorem for
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
Since has no real solution, there is no value of c on
[-4, 4] which satisfies the Mean Value Theorem

11. 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.

12. f(x) is continuous and differentiable on [-2, 2]
On the interval [-2, 2], c = 0 satisfies the conclusion of MVT

13. f(x) is continuous and differentiable on [-2, 1]
On the interval [-2, 1], c = 0 satisfies the conclusion of MVT

14. Since f(x) is discontinuous at x = 2, which is part of the interval
[0, 4], the Mean Value Theorem does not apply

15. f(x) is continuous and differentiable on [-1, 2]
c = 1 satisfies the conclusion of MVT

16. CALCULATOR REQUIRED
If , how many numbers on [-2, 3] satisfy
the conclusion of the Mean Value Theorem.
A B C D E. 4
f(3) = f(-2) = 64
For how many value(s) of c is f ‘ (c ) = -5? X X X