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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.
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