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 local maximum value of the function $f$ in ($-4,4)$, then M =
Solution
$$
\begin{aligned}
& f(x)=8 x^3-36 x+8 \\\\
& =4\left(2 x^3-9 x+2\right) \\\\
& =4(x-2)\left(2 x^2+4 x-1\right) \\\\
& =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right)
\end{aligned}
$$<br/><br/>
Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$<br/><br/>
$f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2}$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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