Find the intervals in which the following functions are increasing or decreasing.
$f(x)=2 x^{3}-24 x+107$
Given:- Function $f(x)=2 x^{3}-24 x+107$
Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.
(i) If $f^{\prime}(x)>0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
(ii) If $f^{\prime}(x)<0$ for $a l l x \in(a, b)$, 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,
$f(x)=2 x^{3}-24 x+107$
$\Rightarrow f(x)=\frac{d}{d x}\left(2 x^{3}-24 x+107\right)$
$\Rightarrow f^{\prime}(x)=6 x^{2}-24$
For $f(x)$ lets find critical point, we must have
$\Rightarrow f^{\prime}(x)=0$
$\Rightarrow 6 x^{2}-24=0$
$\Rightarrow 6\left(x^{2}-4\right)=0$
$\Rightarrow(x-2)(x+2)=0$
$\Rightarrow x=-2,2$
clearly, $f^{\prime}(x)>0$ if $x<-2$ and $x>2$
and $f^{\prime}(x)<0$ if $-2 Thus, $f(x)$ increases on $(-\infty,-2) \cup(2, \infty)$ and $f(x)$ is decreasing on interval $x \in(-2,2)$