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

Given:- Function $f(x)=x^{8}+6 x^{2}$

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,

$f(x)=x^{8}+6 x^{2}$

$\Rightarrow f(x)=\frac{d}{d x}\left(8 x^{7}+12 x\right)$

$\Rightarrow \mathrm{f}(\mathrm{x})=8 \mathrm{x}^{7}+12 \mathrm{x}$

$\Rightarrow \mathrm{f}(\mathrm{x})=4 \mathrm{x}\left(2 \mathrm{x}^{6}+3\right)$

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

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

$\Rightarrow 4 x\left(2 x^{6}+3\right)=0$

$\Rightarrow x\left(2 x^{6}+3\right)=0$

$\Rightarrow \mathrm{x}=0, \sqrt[\frac{1}{6}]{-\frac{3}{2}}$

Since $x=\sqrt[\frac{1}{6}]{-\frac{3}{2}}$ is a complex number, therefore only check range on 0 sides of number line.

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

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

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

and $f(x)$ is decreasing on interval $x \in(-\infty, 0)$