Find the intervals in which the following functions are increasing or decreasing.

Given:- Function $f(x)=x^{3}-6 x^{2}-36 x+2$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

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

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$f(x)=x^{3}-6 x^{2}-36 x+2$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}-6 \mathrm{x}^{2}-36 \mathrm{x}+2\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-12 \mathrm{x}-36$

For $f(x)$ lets find critical point, we must have

$\Rightarrow f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}-12 x-36=0$

$\Rightarrow 3\left(x^{2}-4 x-12\right)=0$

$\Rightarrow 3\left(x^{2}-6 x+2 x-12\right)=0$

$\Rightarrow x^{2}-6 x+2 x-12=0$

$\Rightarrow(x-6)(x+2)=0$

$\Rightarrow x=6,-2$

clearly, $f^{\prime}(x)>0$ if $x<-2$ and $x>6$

and $f^{\prime}(x)<0$ if $-2

Thus, $f(x)$ increases on $(-\infty,-2) \cup(6, \infty)$

and $f(x)$ is decreasing on interval $x \in(-2,6)$