Apply Rolle's theorem if possible and find c if y=x^(2/3)-16

Rolle's Theorem says that if a<b and f(a) = f(b) then there is a c such that a < c < b and f'(c) = 0.

Strictly, the domain is x>=0, so the theorem does not apply. No such interval exists.

However, if we fudge a bit and say that
x^2/3 = (x^1/3)^2, then
y(-64) = y(64) = 0 so we might be ok.

However, y' = 2/3 x^(-1/3) so f'(0) is not defined. Hence y is not differentiable on (-64,64).

So, for two reasons the theorem does not apply.