#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion v maths textbook solution

$\text {Increasing interval}\: (-2,3) \\ \text {Decreasing interval}\: (-\infty,-2) \cup(3, \infty)$

Given:

Here given that

$f(x)=5+36x+3x^{2}-2x^{3}$

To find:

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

Hint:

First, we will find critical points and then use increasing and decreasing property.

Solution:

Given that

$f(x)=5+36x+3x^{2}-2x^{3}$

On differentiating we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(5+36 x+3 x^{2}-2 x^{3}\right) \\ &\Rightarrow f^{\prime}(x)=36+6 x-6 x^{2} \end{aligned}

For f(x), Firstly we will find critical points.

For this we have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 36+6 x-6 x^{2}=0 \\ &\Rightarrow 6\left(6+x-x^{2}\right)=0 \\ &\Rightarrow 6+x-x^{2}=0 \\ &\Rightarrow-x^{2}+x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-x-6=0 \\ &\Rightarrow x^{2}-3 x+2 x-6=0 \\ &\Rightarrow x(x-3)+2(x-3)=0 \\ &\Rightarrow(x-3)(x+2)=0 \end{aligned}

\begin{aligned} &\Rightarrow x-3=0 \text { and } x+2=0 \\ &\Rightarrow x=3 \text { and } x=-2 \\ &\text { Clearly, } f^{\prime}(x)>0, f-2

$\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)$

$\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)$