Mean Value Theorems

Rolle's theorem applies to a function continuous on a closed interval, differentiable on the open interval, and taking equal values at the endpoints, guaranteeing at least one interior point where the derivative vanishes. The polynomial f is continuous and differentiable everywhere, and f(-\sqrt{3}) = (-\sqrt{3})^3 - 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0 = f(\sqrt{3}), so all hypotheses hold. Differentiate: f'(x) = 3x^2 - 3, set to zero to get x^2 = 1, so x = \pm 1. Both x = 1 and x = -1 lie strictly inside (-\sqrt{3}, \sqrt{3}) since 1 < \sqrt{3} \approx 1.732, and both satisfy f'(c) = 0, so there are exactly two values of c that fulfil Rolle's conclusion. Option 0 is impossible because the hypotheses hold and f' does vanish inside. Option 1 under-counts by reporting only Rolle's minimum guarantee of one point rather than the actual number of interior stationary points. Option 3 over-counts beyond the two roots of the quadratic f'. The governing pattern is Rolle's theorem on a symmetric cubic, whose derivative is a quadratic with two real roots. Plausibility check: f' = 3x^2 - 3 is an upward parabola crossing zero at x = -1 and x = 1, both well within the symmetric window (-\sqrt{3}, \sqrt{3}), giving exactly two Rolle points.