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

$\text { Increasing interval }(-3,0) \cup(5, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(0,5)$

Given:

Here given that

$f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51$

To find:

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

Hint:

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

Solution:

We have,

$f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51$

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

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{3}{2} x^{4}-4 x^{3}-45 x^{2}+51\right) \\ &f^{\prime}(x)=\frac{3}{2}\left(4 x^{3}\right)-12 x^{2}-45(2 x) \\ &=6 x^{3}-12 x^{2}-90 x \end{aligned}

We have to find critical points, we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{3}-12 x^{2}-90 x=0 \\ &\Rightarrow 6 x\left(x^{2}-2 x-15\right)=0 \\ &\Rightarrow x\left(x^{2}-2 x-15\right)=0,\{\therefore 6=0\} \\ &\Rightarrow x\left(x^{2}-5 x+3 x-15\right)=0 \\ &\Rightarrow x(x-5)(x+3)=0 \\ &\Rightarrow x=0 ; x=5 ; x=-3 \end{aligned}

The points x=0, 5, -3 divide the real line into four disjoint intervals,

\begin{aligned} &(-\infty,-3),(-3,0),(0,5) \text { and }(5, \infty) \\ &\text { In intervals }(-3,0) \text { and }(5, \infty), f^{\prime}(x)>0 \end{aligned}

$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3

\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(0,5) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty

$\text { Thus, } f(x) \text { is increasing in the intervals }(-3,0) \cup(5, \infty) \text { and } \\ \text { decreasing in the intervals }(-\infty,-3) \cup(0,5) \text { . }$