Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(x)=0.

f(x) = x^(2/3) - 1 [-8,8]

I plugged in both values and found out that they both equal 4 so I thought Rolle's theorem applied.

However, the answer is that this function is not differentiable at 0.

Can someone explain to me why it is not differentiable even though f(a)=f(b).

1 answer

correct. f'(x) = 2/3∛x
which is not defined at x=0.

so, the requirements of the theorem are not met. There is, in fact no value of x where f'(x) = 0.