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Question Number 83381 by Rio Michael last updated on 01/Mar/20

Given that the function f(x) = x^3  is   differentiable in the interval (−2,2) us the mean  value theorem to find the value of x for which the   tangent to the curve is parrallel to the chord   through the points (−2,8) and (2,8).

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\mathrm{is}\: \\ $$$$\mathrm{differentiable}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left(−\mathrm{2},\mathrm{2}\right)\:\mathrm{us}\:\mathrm{the}\:\mathrm{mean} \\ $$$$\mathrm{value}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\: \\ $$$$\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{is}\:\mathrm{parrallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{chord}\: \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{2},\mathrm{8}\right)\:\mathrm{and}\:\left(\mathrm{2},\mathrm{8}\right). \\ $$

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