Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 41

The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.
The mean value theorem states:
f(b) - f(a) = f'(c)(b-a)
Replacing 1 for b and 0 for a, yields:
f(1) - f(0) = f'(c)(1-0)
Evaluating f(1) and f(0) yields:
f(1) = 1^(2/3) => f(1) =1
f(0) = 0
You need to evaluate f'(c):
f'(c) = (c^(2/3))' => f'(c) = (2/3)c^(2/3 - 1) => f'(c) = (2/3)*(c^(-1/3)) => f'(c) = 2/(3 root(3) c)
Replacing the found values in equation f(1) - f(0) = f'(c)(1-0):
1 - 0 = (2/(3 root(3) c))(1-0) => 1 = 2/(3 root(3) c) =>3 root(3) c = 2 => root(3) c = 2/3 => c = (2/3)^3 => c = 8/27 in [0,1]
Hence, in this case, the mean value theorem can be applied and the value of c is c = 8/27 .