Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 37

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 -2 for a, yields:
f(1) - f(-2) = f'(c)(1+ 2)
Evaluating f(1) and f(-2) yields:
f(1) = 1^2 => f(1) =1
f(-2) = (-2)^2 => f(-2) = 4
You need to evaluate f'(c):
f'(c) = (c^2)' => f'(c) = 2c
Replacing the found values in equation f(1) - f(-2) = f'(c)(1 + 2):
1 - 4 = 2c(1+2) => 6c = -3 => c = -3/6 => c = -1/2 in [-2,1]
Hence, in this case, the mean value theorem can be applied and the value of c is c = -1/2 .