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Question Number 73542 by Rio Michael last updated on 13/Nov/19

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

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

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