Let x=2 be a local minima of the function f(x | vedclass

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(A) Given $f(x) = 2x^4 - 18x^2 + 8x + 12$.First,find the derivative $f'(x) = 8x^3 - 36x + 8$.Setting $f'(x) = 0$,we get $8x^3 - 36x + 8 = 0$,which sim

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$12\sqrt{6}-\frac{33}{2}$

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(A) Given $f(x) = 2x^4 - 18x^2 + 8x + 12$.First,find the derivative $f'(x) = 8x^3 - 36x + 8$.Setting $f'(x) = 0$,we get $8x^3 - 36x + 8 = 0$,which simplifies to $2x^3 - 9x + 2 = 0$.We are given that $x=2$ is a local minimum,so $(x-2)$ is a factor of $2x^3 - 9x + 2$.Dividing $2x^3 - 9x + 2$ by $(x-2)$,we get $(x-2)(2x^2 + 4x - 1) = 0$.The roots are $x=2$ and $x = \frac{-4 \pm \sqrt{16 - 4(2)(-1)}}{2(2)} = \frac{-4 \pm \sqrt{24}}{4} = -1 \pm \frac{\sqrt{6}}{2}$.The critical points are $x=2$,$x = -1 + \frac{\sqrt{6}}{2}$,and $x = -1 - \frac{\sqrt{6}}{2}$.Using the second derivative test $f''(x) = 24x^2 - 36$:For $x=2$,$f''(2) = 24(4) - 36 = 60 > 0$ (Local Minima).For $x = -1 - \frac{\sqrt{6}}{2}$,$f''(x) > 0$ (Local Minima).For $x = -1 + \frac{\sqrt{6}}{2}$,$f''(x) Thus,the local maximum occurs at $x = \frac{\sqrt{6}-2}{2}$.Substituting $x = \frac{\sqrt{6}-2}{2}$ into $f(x)$,we calculate $M = f\left(\frac{\sqrt{6}-2}{2}\right) = 12\sqrt{6} - \frac{33}{2}$.

Let $x=2$ be a local minima of the function $f(x)=2x^4-18x^2+8x+12$,$x \in (-4,4)$. If $M$ is the local maximum value of the function $f$ in $(-4,4)$,then $M =$

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