Let x = 2 be a local minima of the function f(x) = 2 x ^ 4 - 18 x ^...

Given,

$x = 2$ be a local minima of the function $f\left(x\right) = 2{x}^{4} - 18{x}^{2} + 8x + 12, x \in \left(- 4, 4\right)$

Now differentiating the above function we get,

${f}^{'}\left(x\right) = 8{x}^{3} - 36x + 8$

$⇒ {f}^{'}\left(x\right) = 4\left(2{x}^{3} - 9x + 2\right)$

Now equating it to zero we get, ${f}^{'}\left(x\right) = 0$

$⇒ 2{x}^{3} - 9x + 2 = 0$

$⇒ \left(x - 2\right)\left(2{x}^{2} + 4x - 1\right) = 0$

$⇒ x = \dfrac{- 4 \pm \sqrt{24}}{4}$

$⇒ x = \dfrac{- 2 \pm \sqrt{6}}{2}$

$∴ x = \dfrac{\sqrt{6} - 2}{2}$ {by checking sign change of ${f}^{'}\left(x\right)$ for maxima}

Now rewriting $f\left(x\right) = 2{x}^{4} - 18{x}^{2} + 8x + 12$ we get,

$f\left(x\right) = \left({x}^{2} - 2x - \dfrac{9}{2}\right)\left(2{x}^{2} + 4x - 1\right) + 24x + 7 . 5$

Hence, $f\left(\dfrac{\sqrt{6} - 2}{2}\right) = M = 12\sqrt{6} - \dfrac{33}{2}$