1. Rolle's Theorem Objectives:
Be able to determine if Rolle’s Theorem can be applied to a
function on a closed interval.
2. Be able to apply Rolle’s Theorem to various functions.
Critical Vocabulary:
Rolle’s Theorem

2. If f is continuous on a closed interval [a, b], then f has both a minimum and maximum on the interval.
However, the minimum and/or maximum may occur at the endpoints
Example 1:
Absolute Max: (3, 9/23)
Absolute Min: (0, 0)

3. Rolle’s Theorem (named after French mathematician Michel Rolle) gives us conditions that guarantees the existence of an extreme value on the interior of the closed interval.
Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one number c in (a, b) such that f’(c) = 0 (horizontal tangent line).
Example 2: Determine if Rolle’s Theorem can be applied
to the function f(x) = x2 + 6x + 13; [-7, 1]
f(-7) = (-7)2 + 6(-7) + 13
f(1) = (1)2 + 6(1) + 13
f(-7) = 20
f(1) = 20
f’(x) = 2x + 6
Rolle’s Theorem can
be applied
2x + 6 = 0
x = -3; (-3,4)

4. Example 3: Determine if Rolle’s Theorem can be applied to the function
Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one number c in (a, b) such that f’(c) = 0 (horizontal tangent line).
Example 3: Determine if Rolle’s Theorem can be applied
to the function
Rolle’s Theorem can’t
be applied to this function.
This function is discontinuous at x = 0

5. Example 4: Find the two x-intercepts of the function f and show
that f’(x) = 0 at some point between the two
intercepts.
1st: Find the x-intercepts (y = 0):
2nd: Find the derivative

6. Example 5 (Problem 10 on page 326):
Determine whether Rolle’s Theorem can be applied to f on the closed interval [a, b]. If Rolle’s Theorem can be applied, find all the values of c in the open interval (a,b) such that f’(c) = 0.
f(x) = (x - 3)(x + 1)2; [-1, 3]
f(-1) = (-1 - 3)(-1 + 1)2
f(3) = (3 - 3)(3 + 1)2
f(-1) = 0
f(3) = 0
Yes, Rolle’s Theorem can be applied!!!!
g(x) = x - 3
g’(x) = 1
h(x) = (x + 1)2
h’(x) = 2(x + 1)
f’(x) = (x + 1)2 + 2(x + 1)(x - 3)
0 = (x + 1)(3x - 5)
x + 1 = 0
3x - 5 = 0
f’(x) = (x + 1)[(x + 1) + 2(x - 3)]
x = -1
x = 5/3
f’(x) = (x + 1)(3x - 5)

7. Page 326 #1-17 odds