To apply Rolle's Theorem to the function f(x) = 2x² + 20x - 5 on the interval [3, 7], we need to check the following conditions:
1. Continuity: The function f(x) is a polynomial, which is continuous for all real numbers.
2. Differentiability: The function f(x) is a polynomial, which is differentiable for all real numbers.
3. Endpoint values: f(3) = 2(3)² + 20(3) - 5 = 48 and f(7) = 2(7)² + 20(7) - 5 = 138.
If the conditions of Rolle's Theorem are satisfied, then there exists at least one value c in the interval (3, 7) such that f'(c) = 0. To find this value, we need to find the derivative of f(x):
f'(x) = 4x + 20.
Setting f'(x) = 0 and solving for x:
4x + 20 = 0,
4x = -20,
x = -5.
Therefore, there is one value of c, namely c = -5, in the interval (3, 7) such that f'(c) = 0.
Now, if we try to apply Rolle's Theorem to the function f(x) = 2x² + 20x - 5 on the interval [0, 14], we need to check the conditions of the theorem. In this case, the condition that is not met is the differentiability of f(x) on the interval [0, 14]. The function f(x) is a polynomial and is differentiable for all real numbers, so the differentiability condition is met on any closed interval, including [0, 14].