Rolle's theorem cannot be applied to the function f(x) = x1/3 on the interval [–1, 1] because
Answer Choices:
f is not differentiable on the interval [–1, 1]
f(–1) ≠ f(1)
f is not differentiable on the interval [–1, 1] and f(–1) ≠ f(1)
Rolle's theorem can be applied to f(x) = x1/3 on the interval [–1, 1]
Check the graph. Is f'(x)=0 anywhere on the interval?
Clearly not, so now you have to figure out why.
Is f(-1) = f(1)?
Is df/dx defined everywhere on the interval? (check out x=0)
Rolle's theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and the function values at the endpoints are equal (f(a) = f(b)), then there exists at least one point c in the open interval (a, b) where the derivative of the function is zero (f'(c) = 0).
To determine if Rolle's theorem can be applied to the function f(x) = x^(1/3) on the interval [–1, 1], we need to check if all the conditions of Rolle's theorem are satisfied.
1. First, we need to check if the function is continuous on the closed interval [-1, 1] as specified in Rolle's theorem. The function f(x) = x^(1/3) is indeed continuous on the interval [-1, 1] since it is defined for all values in that interval.
2. Next, we need to check if the function is differentiable on the open interval (-1, 1) as specified in Rolle's theorem. The function f(x) = x^(1/3) is differentiable on the open interval (-1, 1) since its derivative exists for all values in that interval.
3. Finally, we need to check if the function values at the endpoints are equal (f(-1) = f(1)). For the function f(x) = x^(1/3), we have f(-1) = (-1)^(1/3) = -1 and f(1) = 1^(1/3) = 1. Therefore, f(-1) is not equal to f(1).
Since the condition f(-1) = f(1) is not satisfied, we can conclude that Rolle's theorem cannot be applied to the function f(x) = x^(1/3) on the interval [-1, 1].
Therefore, the correct answer choice is: f is not differentiable on the interval [-1, 1] and f(-1) ≠ f(1).