Show that $f(x)=\frac{1}{1+x^{2}}$ is neither increasing nor decreasing on $R$.
We have,
$\mathrm{f}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{2}}$
Case 1
When $x \in[0, \infty)$
Let $\mathrm{x}_{1}>\mathrm{x}_{2}$
$\Rightarrow \mathrm{x}_{1}^{2}>\mathrm{x}_{2}^{2}$
$\Rightarrow 1+\mathrm{x}_{1}^{2}>1+\mathrm{x}_{2}^{2}$
$\Rightarrow \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}}$
$\Rightarrow f\left(x_{1}\right)<f\left(x_{2}\right)$
$\Rightarrow \therefore f(x)$ is decreasing on $[0, \infty)$
Case 2
When $x \in(-\infty, 0]$
Let $\mathrm{x}_{1}>\mathrm{x}_{2}$
$\Rightarrow \mathrm{x}_{1}^{2}<\mathrm{x}_{2}^{2}$
$\Rightarrow 1+\mathrm{x}_{1}^{2}<1+\mathrm{x}_{2}^{2}$
$\Rightarrow \frac{1}{1+\mathrm{x}_{1}^{2}}>\frac{1}{1+\mathrm{x}_{2}^{2}}$
$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)>\mathrm{f}\left(\mathrm{x}_{2}\right)$
$\therefore f(x)$ is increasing on $(-\infty, 0] .$
Thus, $f(x)$ is neither increasing nor decreasing on $R$.