The function $f(x)=\log x-\frac{2 x}{x+2}$
The function $f(x)=\log x-\frac{2 x}{x+2}$
- $x \in(-\infty, 1)$
- $x \in(-1, \infty)$
- $x \in(-\infty, 0)$
- $x\in(0, \infty)$
Solution
$\begin{aligned}
\text { Given } f(x) &=\log x-\frac{2 x}{x+2} \\
\therefore f^{\prime}(x) &=\frac{1}{x}-\left[\frac{(x+2)(2)-(2 x)(1)}{(x+2)^{2}}\right] \\
&=\frac{1}{x}-\left[\frac{4}{(x+2)^{2}}\right]=\frac{x^{2}+4 x}{x(x+2)^{2}} \\
&=\frac{x(x+4)}{x(x+2)^{2}}
\end{aligned}$
For $f^{\prime}(x)>0, x \neq 0$ and $x>-4$
Hence $x \in(0, \infty)$ is permissible.
Asked in: MHT CET 2020 (14 Oct Shift 2)
Practice more Applications of Derivatives questions on Aicharya