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

$\text {Increasing interval} (-\infty,-2) \cup(4, \infty) \\ \text {Decreasing interval} (-2,4)$

Given:

Here given that

$f(x)=5x^{3}-15x^{2}-120x+3$

To find:

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

Hint:

Put f(x)=0 and solve this equation to find critical points of given function.

Solution:

We have,

$f(x)=5x^{3}-15x^{2}-120x+3$

Differentiating with respect to x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(5 x^{3}-15 x^{2}-120 x+3\right) \\ &f^{\prime}(x)=15 x^{2}-30 x-120 \end{aligned}

Now we have to find critical points for f(x).

We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 15 x^{2}-30 x-120=0\\ &\Rightarrow 15\left(x^{2}-2 x-8\right)=0\\ &\Rightarrow x^{2}-2 x-8=0\{\therefore 15>0\}\\ &\Rightarrow x^{2}-4 x+2 x-8=0\\ &\Rightarrow x(x-4)+2(x-4)=0\\ &\Rightarrow(x-4)(x+2)=0\\ &\Rightarrow x-4=0 \text { and } x+2=0\\ &\Rightarrow x=4 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>4 \text { or } x \in(-\infty,-2) \text { and } x \in(4, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}

\begin{aligned} &-2