Show that


Show that $f(x)=\frac{1}{1+x^{2}}$ decreases in the interval $[0, \infty)$ and increases in the interval $(-\infty, 0]$.


We have,


Case 1

When $x \in[0, \infty)$

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

$\Rightarrow \mathrm{x}_{1}^{2}>\mathrm{x}_{2}^{2}$

$\Rightarrow 1+\mathrm{x}_{1}^{2}>1+\mathrm{x}_{2}^{2}$

$\Rightarrow \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}}$

$\Rightarrow f\left(x_{1}\right)

$\therefore f(x)$ 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 $R$.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now