Rolle's theorem cannot be applied to the function f(x) = x^1/3 on the interval [–1, 1] because...
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]
Rolle's theorem states that if a function is continuous on a closed interval and differentiable on the open interval, and the function takes the same values at the endpoints of the interval, then there exists at least one point within the interval where the derivative of the function is zero.
In this case, the function in question is f(x) = x^(1/3) defined on the interval [-1, 1]. To determine whether Rolle's theorem can be applied, we need to check the conditions of the theorem.
First, we need to check if the function is continuous on the closed interval [-1, 1]. The function f(x) = x^(1/3) is defined and continuous for all x, including the endpoints -1 and 1 in the interval [-1, 1].
Next, we need to check if the function is differentiable on the open interval (-1, 1). To do this, we find the derivative of f(x): f'(x) = (1/3)x^(-2/3). The derivative is defined and exists for all x in the open interval (-1, 1), so the function is differentiable on this interval.
Finally, we need to check if the function takes the same values at the endpoints of the interval. Evaluating the function at the endpoints: f(-1) = (-1)^(1/3) = -1 and f(1) = 1^(1/3) = 1. So, f(-1) does not equal f(1).
Based on these findings, we can conclude that Rolle's theorem cannot be applied to the function f(x) = x^(1/3) on the interval [-1, 1] because f is not differentiable on the interval [–1, 1], and f(-1) does not equal f(1). Therefore, the correct answer is "f is not differentiable on the interval [–1, 1] and f(-1) does not equal f(1)."