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

Solution:

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

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.

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

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

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

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

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

$\Rightarrow(x-2)(x+1)=0$

$\Rightarrow x=-1,2$

clearly, $f^{\prime}(x)>0$ if $-1

and $f^{\prime}(x)<0$ if $x<-1$ and $x>2$

Thus, $f(x)$ increases on $x \in(-1,2)$

and $f(x)$ is decreasing on interval $(-\infty,-1) \cup(2, \infty)$