f (x) = x3 − x, [0, 2].Determine a number c ∈ (0, 2) that satisfy the conclusions of the MVT

The Mean Value Theorem (MVT) states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].

Given the function f(x) = x^3 - x on the interval [0, 2], we first need to check if the function satisfies the conditions of the MVT.

1. The function f(x) = x^3 - x is a polynomial function, which is continuous and differentiable for all real numbers. So, it is continuous on [0, 2] and differentiable on (0, 2).

2. Next, we find the average rate of change of the function over the interval [0, 2]. This is given by [f(b) - f(a)] / (b - a) = [f(2) - f(0)] / (2 - 0) = [(2^3 - 2) - (0)] / 2 = 2.

3. Now, we find the derivative of the function, f'(x) = 3x^2 - 1.

4. We set the derivative equal to the average rate of change and solve for x: 3x^2 - 1 = 2, which simplifies to 3x^2 = 3, and further simplifies to x^2 = 1. The solutions to this equation are x = 1 and x = -1.

5. However, only x = 1 lies in the interval (0, 2). So, c = 1 is the number that satisfies the conclusion of the MVT for the function f(x) = x^3 - x on the interval [0, 2].