endpoints) when doing Mean Value Theorem problems. If it is not, then the theorem does not
apply.
- c = 4
Rolle’s Theorem says that if 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 exists at least one number c
on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = x^2 − 8x + 12 and the interval
is [2, 6]. First, we check if the function is equal to zero at both of the endpoints: f(6) = (6)^2 −
8(6) + 12 = 0 and f(2) = (2)^2 − 8(2) + 12 = 0. Next, we take the derivative to find f′(c): f′(x) =
2 x − 8, so f′(c) = 2c − 8. Now, we can solve for c: 2c − 8 = 0 and c = 4. Note that 4 is in the
interval (2, 6), just as we expected.
- c =
Rolle’s Theorem says that if 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 exists at least one number c
on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = x(1 − x) and the interval is
[0, 1]. First, we check if the function is equal to zero at both of the endpoints: f(0) = (0)(1 − 0)
= 0 and f(1) = (1)(1 − 1) = 0. Next, we take the derivative to find f′(c): f′(x) = 1 − 2x, so f′(c) =
1 − 2c. Now, we can solve for c: 1 − 2c = 0 and c = . Note that is in the interval (0, 1), just
as we expected.
- No Solution.
Rolle’s Theorem says that if 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 exists at least one number c
on the interval (a, b) such that f′(c) = 0. Here the function is f(x) = 1 − , and the interval is
[−1, 1]. Note that the function is not continuous on the interval. It has an essential discontinuity
(vertical asymptote) at x = 0. Thus, Rolle’s Theorem does not apply on the interval, and there
is no solution.
Suppose we were to apply the theorem anyway. First, we check if the function is equal to zero
