Question

Show that the function $f:R\to \{x\in R:-1<x<1\}$ defined by $f\left(x\right)=\frac{x}{1+|x|},x\in R$ is one one and onto function.

Answer 1

It is given that
$f\left(x\right)\in (-1,1)$
$f\left(x\right)=\frac{x}{1+|x|}$
we know that : $|x|=\{\begin{array}{cc}x& \text{if}x\ge 0\\ -x& \text{if}x<0\end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+x}& \text{if}x\ge 0\\ \frac{x}{1-x}& \text{if}x<0\end{array}$

Checking - one-one
case 1: for $x\ge 0$
$f\left(x_1\right)=\frac{x_1}{1+x_1}$
$f\left(x_2\right)=\frac{x_2}{1+x_2}$

on putting $f\left(x_1\right)=f\left(x_2\right)$, we have:
$\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$
$\Rightarrow x_1(1+x_2)=x_2(1+x_1)\Rightarrow x_1+x_1{x}_2=x_2+x_2{x}_1\Rightarrow x_1=x_2$

Similarly,
In case 2: for $x<0$, we get:
$x_1=x_2$

Hence, if $f\left(x_1\right)=f\left(x_2\right)$, then $x_1=x_2$
$\therefore f$ is one-one.

Checking onto
case 1: for $x\ge 0$
$f\left(x\right)=\frac{x}{1+x}\text{Let}f\left(x\right)=y\Rightarrow y=\frac{x}{1+x}\Rightarrow y(1+x)=x\Rightarrow y=x-xy\Rightarrow x(1-y)=y\Rightarrow x=\frac{y}{1-y}$

Similarly, in case 2: for $x<0$ , we have:
$x=\frac{y}{1+y}$
$\because y\in (-1,1)$
$\therefore \forall y\in (-1,1)$, $x$ is defined.
$\therefore f$ is onto.

Hence, $f\left(x\right)$ is one-one and onto (proved).