Question
फलन $f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$
Now,
$ \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$
Thus $\mathrm{f}(\mathrm{x})$ is decreasing in
$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
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