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

Solution:

Given:- Function $f(x)=3 x^{4}-4 x^{3}-12 x^{2}+5$

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,

$f(x)=3 x^{4}-4 x^{3}-12 x^{2}+5$

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(3 x^{4}-4 x^{3}-12 x^{2}+5\right)$

$\Rightarrow f^{\prime}(x)=12 x^{3}-12 x^{2}-24 x$

$\Rightarrow f^{\prime}(x)=12 x\left(x^{2}-x-2\right)$

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

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

$\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}-\mathrm{x}-2\right)>0$

$\Rightarrow \mathrm{x}\left(\mathrm{x}^{2}-2 \mathrm{x}+\mathrm{x}-2\right)>0$

$\Rightarrow \mathrm{x}(\mathrm{x}-2)(\mathrm{x}+1)>0$

$\Rightarrow-1<\mathrm{x}<0$ and $\mathrm{x}>2$

$\Rightarrow \mathrm{x} \in(-1,0) \cup(2, \infty)$

Thus $f(x)$ is increasing on interval $(-1,0) \cup(2, \infty)$

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

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

$\Rightarrow 12 x\left(x^{2}-x-2\right)<0$

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

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

$\Rightarrow-\infty

$\Rightarrow x \in(-\infty,-1) \cup(0,2)$

Thus $f(x)$ is decreasing on interval $(-\infty,-1) \cup(0,2)$