Question

If we apply Rolle's Theorem to the function f(x) = 2x² 20x-5 on the interval [3, 7], how many values of c exist such that f'(c) = 0? What is the value of c? If we try to apply Rolle's Thorem to the function f(x) = 2x² 20x-5 on the interval [0, 14], which of the following conditions is not met? continuty on [0, 14] of(a) f(b) differentiability on [0, 14]

Answer 1

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].