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

$\text { Increasing interval }(-3,-1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(-1, 2)$

Given:

Here given that

$f(x)=\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7$

To find:

We have to find the increasing or decreasing interval for f(x).

Hint:

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

Solution:

We have,

$f(x)=\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7$

Differentiating w.r.t. x, we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7\right) \\ &\Rightarrow f^{\prime}(x)=\frac{4 x^{3}}{4}+\frac{2}{3}\left(3 x^{2}\right)-\frac{5}{2}(2 x)-6 \\ &\Rightarrow f^{\prime}(x)=x^{3}+2 x^{2}-5 x-6 \\ &\Rightarrow f^{\prime}(x)=(x+1)\left(x^{2}+x-6\right) \\ &\Rightarrow f^{\prime}(x)=(x+1)(x-2)(x+3) \end{aligned}

For critical points. we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x+1)(x-2)(x+3)=0 \\ &\Rightarrow x=-1, \quad x=2, \quad x=-3 \end{aligned}

$\text { The possible intervals are }(-3,-1),(-\infty,-3),(-1,2) \text { and }(2, \infty)$

$\text { Now we take }(-3,-1) . \text { i.e, }-3

$\text { In this case we have } x+1<0, x-2<0 \text { and } x+3>0 . \text { Clearly, } f^{\prime}(x)>0 \text { if }-3

\begin{aligned} &\text { Now we take the intervals }(-\infty,-3) \text { i.e, }-\infty

$\text { After that we take }(-1,2) \text { i.e, }-1

$\text { Clearly, we have } x+1>0, x-2<0 \text { and } x+3>0 . \text { So, } f^{\prime}(x)<0 \text { if }-1

$\text { Finally we take }(2, \infty) \text { i.e, } 20, x-2>0 \text { and } x+3>0 \text { . }$

$\text { Clearly }f^{\prime}(x)>0 \text { if } 2

$\text { Thus, the function is increasing on }(-3,-1) \cup(2, \infty) \text { and }$

$f(x) \text { is decreasing on interval }(-\infty,-3) \cup(-1,2) \text { . }$