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

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

Given:

Here given that

$f(x)=6+12x+3x^{2}-2x^{3}$

To find:

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

Hint:

Put f ‘(x) = 0 to find the critical points and then use increasing and decreasing property.

Solution:

We have,

$f(x)=6+12x+3x^{2}-2x^{3}$

Differentiating w.r.t. x we get,

$f'(x)=12+6x-6x^{2}$

For f(x), we have to find critical points.

We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12+6 x-6 x^{2}=0\\ &\Rightarrow 6\left(2+x-x^{2}\right)=0\\ &\Rightarrow 2+x-x^{2}=0\\ &\Rightarrow x^{2}-x-2=0\\ &\Rightarrow x^{2}-2 x+x-2=0\\ &\Rightarrow(x-2)(x+1)=0\\ &\Rightarrow x-2=0 \text { and } x+1=0\\ &\Rightarrow x=2 \text { and } x=-1\\ &\text { Clearly, } f^{\prime}(x)>0, \text { if }-12 \text { or } x \in(-\infty,-1) \text { and } x \in(2, \infty) \end{aligned}

$\text { Thus, } f(x) \text { is increasing on the interval (-1,2) and } \\ \text { decreasing on the interval } (-\infty,-1) \cup(2, \infty)$