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

$f(x) \text { is increasing on the interval }\left(-\infty, \frac{-3}{2}\right) \text { and decreasing on the interval }\left(\frac{-3}{2}, \infty\right)$

Given:

Here given that

$f(x)=10-6x-2x^{2}$

To find:

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

Hint:

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

Solution:

Here given that

$f(x)=10-6x-2x^{2}$

On differentiating we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(10-6 x-2 x^{2}\right)\\ &\Rightarrow f^{\prime}(x)=-6-4 x\\ &\text { For } f(x) \text { to be increasing, we must have }\\ &f^{\prime}(x)>0\\ &\Rightarrow-6-4 x>0\\ &\Rightarrow-4 x>6\\ &\Rightarrow x<\frac{-6}{4} \end{aligned}

By applying (-) ve sign change comparison sign.

\begin{aligned} &\Rightarrow x<\frac{-3}{2}\\ &x \in\left(-\infty, \frac{-3}{2}\right)\\ &\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{-3}{2}\right) \end{aligned}

\begin{aligned} &\text { For } \mathrm{f}(\mathrm{x}) \text { to be decreasing, we must have }\\ &f^{\prime}(x)<0\\ &\Rightarrow-6-4 x<0\\ &\Rightarrow-4 x<6\\ &\Rightarrow x>\frac{-6}{4} \end{aligned}

\begin{aligned} &\text { By applying }(-) \text { ve sign change comparison sign. }\\ &\Rightarrow x>\frac{-3}{2} \end{aligned}

\begin{aligned} &\Rightarrow x \in\left(\frac{-3}{2}, \infty\right)\\ &\text { So, } f(x) \text { is decreasing on the interval }\left(\frac{-3}{2},+\infty\right) \end{aligned}