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)


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

$G i v e n t h a t t h e f u n c t i o n f (x) = x^{3} i s d i f f e r e n t i a b l e$ $i n t h e i n t e r v a l (- 2, 2), U s e t h e m e a n v a l u e t h e o r e m$ $t o f i n d t h e v a l u e o f x f o r w h i c h t h e t a n g e n t t o t h e c u r v e$ $i s p a r a l l e l t o t h e c h o r d t h r o u g h t h e p o i n t (- 2, 8) a n d (2, 8)$
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