Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 15

The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if f(-1) = f(3).
f(-1) = ((-1)^2-2*(-1)-3)/(-1+2) = (1+2-3)/1 = 0
f(3) = ((3)^2-2*(3)-3)/(3+2) = (9-6-3)/5 = 0
Since all the three conditions are valid, you may apply Rolle's theorem:
f'(c)(b-a) = 0
Replacing 3 for b and -1 for a, yields:
f'(c)(3+1) = 0
You need to evaluate f'(c), using quotient rule:
f'(c) = ((c^2-2*c-3)'*(c+2) - (c^2-2*c-3)*(c+2)')/((c+2)^2) => f'(c) = ((2c-2)(c+2) -c^2 + 2c + 3)/((c+2)^2)
f'(c) = (2c^2 + 4c - 2c - 4 - c^2 + 2c + 3)/((c+2)^2)
f'(c) = (c^2 + 4c - 1)/((c+2)^2)
Replacing the found values in equation 4f'(c) = 0
4(c^2 + 4c - 1)/((c+2)^2) = 0 => c^2 + 4c - 1 = 0
c_(1,2) = (-4+-sqrt(16+4))/2 => c_(1,2) = (-4+-2sqrt5)/2 => c_(1,2) = (-2+-sqrt5)
Since c = (-2-sqrt5) does not belong to (-1,3), only c =(-2+sqrt5) is a valid value.
Hence, in this case, the Rolle's theorem may be applied for c = (-2+sqrt5).