Solution

Let f(x) = x^{3} – 3x + 2 for [-1, 2]

Clearly f(x) is a polynomial function and we know the polynomial function is continuous everywhere. Therefore, f(x) is continuous on [-1, 2].

And

f^{‘}(x) = 3x^{3} – 3

Clearly f^{‘}(x) is again a polynomial function, so f^{‘}(x) is continuous on (-1, 2).

This means f(x) is differentiable on (-1, 2)

Thus, f(x) satisfies both conditions of Mean Value Theorem (MVT), so by this theorem there is c in (-1, 2) such that

\(f^{‘}(c) = \frac{f(2) – f(-1)}{(2) – (-1)}\)

\(\Rightarrow 3c^2 – 3 = \frac{(8 – 6 + 3) – (-1 + 3 + 3)}{2 + 1}\)

\(\Rightarrow 3(c^2 – 1) = \frac{5 – 5}{3}\)

\(\Rightarrow 3(c^2 – 1) = 0\)

\(\Rightarrow c^2 = 1\)

\(\Rightarrow c = \pm 1\)

Clearly \(c = 1 \in (-1, 1)\). This means f(x) verifies the Mean Value Theorem.

## Discussion