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

$\text { Increasing interval }(-\infty, 2) \cup(6, \infty) \\ \text { Decreasing interval(2.6) }$

Given:

Here given that

$f(x)=x^{3}-12x^{2}+36x+17$

To find:

We have to find the intervals in which f(x) is increasing or decreasing.

Hint:

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

Solution:

We have,

$f(x)=x^{3}-12x^{2}+36x+17$

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

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-12 x^{2}+36 x+17\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-24 x+36 \end{aligned}

For f(x) we have to find critical points, for this we must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-24 x+36=0\\ &\Rightarrow 3\left(x^{2}-8 x+12\right)=0\\ &\Rightarrow x^{2}-8 x+12=0\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x-2 x+12=0\\ &\Rightarrow(x-6)(x-2)=0\\ &\Rightarrow x-6=0 \text { and } x-2=0\\ &\Rightarrow x=6 \text { and } x=2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<2 \text { and } x>6 \text { or } x \in(-\infty, 2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &2

$\text { So, } f(x) \text { is increasing on the interval }(-\infty, 2) \cup \: (6, \infty) \text { and } f(x) \text { is decreasing on interval }(2,6) \text {.}$