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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.
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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