Question

If we apply rolles theorem to the function f(x) = 2x^3 - 12x - 3 on the interval (1,5) how many values of c exist such that f'(c)=0

a) 1
b) 2
c) 0
d) 3

Answer 1

Upon checking the conditions for Roll's Theorem, it's found that the given function does not satisfy the criteria because the function values at the endpoints are not equal. Therefore, Roll's Theorem cannot be applied, and there are zero values of c that satisfy f'(c) = 0.

Step-by-step explanation:

To apply Roll’s Theorem to the function f(x) = 2x3 - 12x - 3 on the interval (1,5), we first need to check whether the function meets the criteria of the theorem:

1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
3. The function must have the same value at the endpoints of the interval [a, b].

Since polynomial functions are continuous and differentiable everywhere, the first two criteria are satisfied. We must now check if f(1) = f(5):

f(1) = 2(1)3 - 12(1) - 3 = -13
f(5) = 2(5)3 - 12(5) - 3 = 247

Since f(1) ≠ f(5), Roll’s Theorem cannot be applied as the third criteria is not met. Therefore, there are zero values of c in the interval (1,5) that satisfy the condition f’(c) = 0.