#### Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion ii

$f(x) \text { is increasing on the interval }(-1, \infty) \text { and decreasing on the interval }(-\infty,-)$

Given:

Here given that

$f(x)=x^{2}+2x-5$

To find:

We have to find out the intervals in which function is increasing and decreasing.

Hint:

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

Solution:

Here given that

$f(x)=x^{2}+2x-5$

On differentiating we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{2}+2 x-5\right) \\ &\Rightarrow f^{\prime}(x)=2 x+2 \end{aligned}

For f(x) to be increasing, we must have

\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow 2 x+2>0 \\ &\Rightarrow 2 x>-2 \\ &\Rightarrow x>-\frac{2}{2} \\ &\Rightarrow x>-1 \\ &x \in(-1, \infty) \end{aligned}

$\text { So, } f(x) \text { is increasing on the interval }(-1, \infty) \text { . }$

\text { Now, } \\ \begin{aligned} &\text { For } \mathrm{f}(\mathrm{x}) \text { to be decreasing, we must have }\\ &f^{\prime}(x)<0 \end{aligned}

\begin{aligned} &\Rightarrow 2 x+2<0 \\ &\Rightarrow 2 x<-2 \\ &\Rightarrow x<-\frac{2}{2} \\ &\Rightarrow x<-1 \\ &\Rightarrow x \in(-\infty,-1) \end{aligned}

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