Concept:Rolle's Theorem states that if f (x) is a function that satisfies:(i) f (x) is continuous on the closed interval [a, b](ii) f (x) is differentiable on the open interval (a, b)(iii) f (a) = f (b)then there exists a point c in the open interval (a, b) such that f' (c) = 0.Calculation:(i) f (x) is continuous on [-6, 3](ii) f'(x) = 2x + 3, f' (x) exists in the interval (-6, 3) so, f (x) is differentiable(iii) f (-6) = f (3) All the three conditions of Rolle's Theorem Satisified.Therefore, there exists at least one c \(\rm \epsilon\) [-6, 3] such that f' (c) = 0which implies 2c + 3 = 0 = c = - 3/2 = -1.5