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

Solution:

Given:- Function $f(x)=5 x^{3}-15 x^{2}-120 x+3$

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

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

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

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,

$f(x)=5 x^{3}-15 x^{2}-120 x+3$

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(5 x^{3}-15 x^{2}-120 x+3\right)$

$\Rightarrow f^{\prime}(x)=15 x^{2}-30 x-120$

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

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

$\Rightarrow 15 x^{2}-30 x-120=0$

$\Rightarrow 15\left(x^{2}-2 x-8\right)=0$

$\Rightarrow 15\left(x^{2}-4 x+2 x-8\right)=0$

$\Rightarrow x^{2}-4 x+2 x-8=0$

$\Rightarrow(x-4)(x+2)=0$

$\Rightarrow x=4,-2$

clearly, $f^{\prime}(x)>0$ if $x<-2$ and $x>4$

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

Thus, $f(x)$ increases on $(-\infty,-2) \cup(4, \infty)$

and $f(x)$ is decreasing on interval $x \in(-2,4)$