#### Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxviii maths

$\text { Increasing interval }(2, \infty) \\ \text { Decreasing interval }(-\infty , 2)$

Given:

Here given that

$f(x)=log(2+x)-\frac{2x}{2+x}$

To find:

We have to find the increasing and decreasing intervals of f(x).

Hint:

$\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { and If } f^{\prime}(x)<0 \text { for all } x \in \text { (a., } b) \text { , then } f(x) \text { is decreasing on }(a, b) \text { . }$

Solution:

We have,

$f(x)=log(2+x)-\frac{2x}{2+x}$

Differentiating w.r.t. x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left\{\log (2+x)-\frac{2 x}{2+x}\right\} \\ &=\frac{1}{2+x}-\frac{(2+x) 2-2 x \times 1}{(2+x)^{2}} \\ &=\frac{1}{2+x}-\frac{4+2 x-2 x}{(2+x)^{2}} \\ &=\frac{1}{2+x}-\frac{4}{(2+x)^{2}} \\ &=\frac{2+x-4}{(2+x)^{2}} \\ &=\frac{x-2}{(2+x)^{2}} \end{aligned}

For critical points. We must have,

\begin{aligned} &f'(x)> 0\\ &\Rightarrow \frac{x-2}{(2+x)^{2}}>0 \\ &\Rightarrow x-2>0 \\ &\Rightarrow x>2 \\ &x \in(2, \infty) \end{aligned}

Thus, f(x) is increasing on the interval (2,$\infty$)

\begin{aligned} &f^{\prime}(x)<0 \\ &\Rightarrow \frac{x-2}{(2+x)^{2}}<0 \\ &\Rightarrow x-2<0 \\ &\Rightarrow x<2 \\ &x \in(-\infty, 2) \end{aligned}

So,f(x) is decreasing on the interval ($-\infty$,2).