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

Given:- Function $f(x)=2 x^{3}-24 x+7$

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,

$f(x)=2 x^{3}-24 x+7$

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-24 x+7\right)$

$\Rightarrow f^{\prime}(x)=6 x^{2}-24$

For $f(x)$ to be increasing, we must have

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

$\Rightarrow 6 x^{2}-24>0$

$\Rightarrow x^{2}<\frac{24}{6}$

$\Rightarrow x^{2}<4$

$\Rightarrow x<-2,+2$

$\Rightarrow x \in(-\infty,-2)$ and $x \in(2, \infty)$

Thus $f(x)$ is increasing on interval $(-\infty,-2) \cup(2, \infty)$

Again, For $f(x)$ to be increasing, we must have

$f^{\prime}(x)<0$

$\Rightarrow 6 x^{2}-24<0$

$\Rightarrow x^{2}>\frac{24}{6}$

$\Rightarrow x^{2}<4$

$\Rightarrow x>-1$

$\Rightarrow x \in(-1, \infty)$

Thus $f(x)$ is decreasing on interval $x \in(-1, \infty)$