Determine whether Rolle's thecrem can be applied to fon the closed interval \( [a, b] \). (Select all that apply.) \[ f(x)=x^{2 / 3}-1, \quad[-64,64] \] Yes, Rolle's theorem can be applied. No, because \( \boldsymbol{f} \) is not continuous on the closed interval \( [a, b] \). No, because \( f \) Is not differentidble on the open Interval \( (a, b) \). No, because \( f(a) \neq f(b) \). If Rolle's theorem can be applled, find all values of \( c \) in the open interval \( (a, b) \) such that \( f^{\prime \prime}(c)=a \). (Fnter your answers as a comma-separated list, If Rolle's theorem cannot be applied, enter NA.) \[ c= \] \( \square \)

To determine whether Rolle’s theorem can be applied to the function \( f(x) = x^{2/3} - 1 \) over the closed interval \( [-64, 64] \), we need to check the three essential conditions: continuity, differentiability, and equal function values at the endpoints. First off, \( f(x) \) is continuous everywhere, including on the closed interval \( [-64, 64] \). However, the function is not differentiable at \( x = 0 \) because the derivative has a cusp there (the slope abruptly changes). Since Rolle's theorem requires differentiability on the open interval \( (-64, 64) \), we cannot apply it here. Additionally, the endpoint values \( f(-64) \) and \( f(64) \) are equal to \( -17 \), but this does not satisfy the differentiability requirement. Thus, the correct answer is: No, because \( f \) is not differentiable on the open interval \( (a, b) \). \[ c = NA \] If Rolle's theorem could be applied, it would typically involve searching for a \( c \) in \( (a, b) \) where \( f'(c) = 0 \), leading to potential points of interest. However, since it can't be applied in this scenario, we simply denote the conclusion as "NA."