Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 40

The mean value theorem may be applied to the given function since all polynomial functions are continuous and differentiable on R, hence, the given function is continuous on [0,2] and differentiable on (0,2).
If the function is continuous and differentiable over the given interval, then, there exists a point c in (0,2), such that:
f(2) - f(0) = f'(c)(2 - 0)
You need to evaluate f(2) and f(0):
f(2) = 2^4 - 8*2 => f(2) = 16 - 16 = 0
f(0) = 0^4 - 8*0 = 0
You need to evaluate f'(c) = (c^4 - 8c)' => f'(c) = 4c^3 - 8
Replace the found values in equation f(2) - f(0) = f'(c)(2 - 0):
0 - 0 = 2(4c^3 - 8) => 4c^3 - 8 = 0 => 4c^3= 8 => c^3 = 8/4 => c^3 = 2 => c = root(3)(2)
Hence, evaluating if the mean value theorem is applicable, yields that it is. The value of c in (0,2) is c = root(3)(2).