Without using the derivative,

We have,

$f(x)=|x|=\{x, x>0$

(a)Let $\mathrm{x}_{1}, \mathrm{x}_{2} \in(0, \infty)$ and $\mathrm{x}_{1}>\mathrm{x}_{2}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)>\mathrm{f}\left(\mathrm{x}_{2}\right)$

So, $f(x)$ is increasing in $(0, \infty)$

(b) Let $x_{1}, x_{2} \in(-\infty, 0)$ and $x_{1}>x_{2}$

$\Rightarrow-\mathrm{x}_{1}<-\mathrm{x}_{2}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)>\mathrm{f}\left(\mathrm{x}_{2}\right)$

$\therefore f(x)$ is strictly decreasing on $(-\infty, 0)$.