Find a value of c such that the conclusion of the mean value theorem is satisfied for

f(x) = -2x ³ + 6x - 2

f(x) is a polynomial function and is continuous and differentiable for all real numbers. Let us evalute f(x) at x = -2 and x = 2

f(-2) = -2(-2) ³ + 6(-2) - 2 = 2

f(2) = -2(2) ³ + 6(2) - 2 = - 6

Evaluate [f(b) - f(a)] / (b - a)

[f(b) - f(a)] / (b - a) = [ -6 - 2 ] / (2 - -2) = -2

Let us now find f '(x).

f '(x) = -6x ² + 6

We now construct an equation based on f '(c) = [f(b) - f(a)] / (b - a)

-6c ² + 6 = -2

Solve for c to obtain 2 solutions

c = 2 sqrt(2/3) and c = - 2 sqrt(2/3)

Below is shown the graph of f, a secant and the two tangent corresponding to the two solutions found. The secant and the two tangents are parallel since their slopes are equal according to the mean value theorem.