Application Of Derivatives Question 4
Question 4 - 2024 (29 Jan Shift 2)
The function $f(x)=\frac{x}{x^{2}-6 x-16}, x \in \mathbb{R}-{-2,8}$
(1) decreases in $(-2,8)$ and increases in
$(-\infty,-2) \cup(8, \infty)$
(2) decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
(3) decreases in $(-\infty,-2)$ and increases in $(8, \infty)$
(4) increases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
Show Answer
Answer (2)
Solution
$f(x)=\frac{x}{x^{2}-6 x-16}$
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 $f(x)$ is decreasing in
$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
