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 to one and onto function.

Answer 1

$f\left(x\right)=\frac{x}{1-x}-1<x<0andf\left(x\right)=\frac{x}{1+x},0\le x<1$

$now\left(i\right)f\left(x\right)=\frac{x}{1-x}-1<x<0$

Let $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}$

$\Rightarrow x_1-x_1{x}_2=x_2-x_1{x}_2$

$\Rightarrow x_1=x_2$

$\Rightarrow$f is one-one

Let $y=\frac{x}{1-x}$

$\Rightarrow y-xy=x$

$\Rightarrow y=x+xy$

$\Rightarrow y=x(1+y)$

$\Rightarrow x=\frac{y}{1+y}$

$\Rightarrow \exists x=\frac{y}{1+y}$ for all valves of y$\ne -1$

s.t $f\left(x\right)=y$

$\Rightarrow f\left(x\right)$ is onto

(ii)$f\left(x\right)=\frac{x}{1+x}0\le x\le 1$

Let $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$

$\Rightarrow x_1+x_1{x}_2=x_2+x_1{x}_2$

$\Rightarrow x_1=x_2$

$\Rightarrow$ f is one -one

Let $y=\frac{x}{1+x}$

$\Rightarrow y+xy=x$

$\Rightarrow y=x-xy$

$\Rightarrow y=x(1-y)$

$\Rightarrow x=\frac{y}{1-y}$

$\Rightarrow$ for all values of $y\ne 1\exists x=\frac{y}{1-y}$ s.t $f\left(x\right)=y$

$\Rightarrow$ f is onto

hence f(x) is one-one and onto proved