Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 38

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 6 for b and 0 for a, yields:
f(6) - f(0) = f'(c)(6 - 0)
Evaluating f(6) and f(0) yields:
f(6) = 2*6^3 => f(6) = 2*216 = 432
f(0) = 0
You need to evaluate f'(c):
f'(c) = (2c^3)' => f'(c) = 6c^2
Replacing the found values in equation f(6) - f(0) = f'(c)(6 - 0):
432 - 0 = 6c^2(6 - 0) => 36c^2 = 432 => c^2 = 12 => c = sqrt12 => c = 2sqrt3
Hence, in this case, the mean value theorem can be applied and the value of c is c = 2sqrt3.