# Verify Mean value theorem of f(x) = x3 – 3x + 2 for [-1, 2].

Solution

Let    f(x) = x3 – 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) = 3x3 – 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.