Question 890063
Rolle's theorem CANNOT be applied on the interval [-1,1], because the function is not differentiable at x = 0.  The derivative function is {{{df(x)/dx = 2/(3x^(1/3))}}} which is undefined at x = 0.  Thus differentiability on (-1,1) is violated. 

Note that the function {{{f(x) = x^(2/3)}}} is continuous on the interval [-1,1]. (since the squared cube-root function is continuous), and f(-1) = f(1) = 1.