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

Solution:

Given:- Function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{4}-4 \mathrm{x}^{3}+4 \mathrm{x}^{2}+15$

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,

$f(x)=x^{4}-4 x^{3}+4 x^{2}+15$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{4}-4 x^{3}+4 x^{2}+15\right)$

$\Rightarrow f^{\prime}(x)=4 x^{3}-12 x^{2}+8 x$

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

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

$\Rightarrow 4 x^{3}-12 x^{2}+8 x=0$

$\Rightarrow 4\left(x^{3}-3 x^{2}+2 x\right)=0$

$\Rightarrow x\left(x^{2}-3 x+2\right)=0$

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

$\Rightarrow x(x-2)(x-1)$

$\Rightarrow x=0,1,2$

clearly, $f^{\prime}(x)>0$ if $02$

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

Thus, $f(x)$ increases on $(0,1) \cup(2, \infty)$

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