The function $f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$
Solution
<p>$f(x)=\frac{x}{x^2-6 x-16}$</p>
<p>Now,</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\
& \mathrm{f}^{\prime}(\mathrm{x})<0
\end{aligned}$$</p>
<p>Thus $f(x)$ is decreasing in</p>
<p>$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Increasing and Decreasing Functions
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