Question

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

Answer 1

$f:Rrightarrow(x\in R:-1<x<1)$

$f\left(x\right)=\frac{x}{1+\left|x\right|},x\in R$

Let,s check it for $\pm dfrac12\in (-1,1)$

$\Rightarrow f\left(\frac{-1}{2}\right)=\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}$

and $f\left(\frac{-1}{2}\right)=\frac{\frac{-1}{2}}{1-\frac{1}{2}}=-1$

$f\left(x\right)$ is different for each $x\in (-1,1)\Rightarrow f$ is one -one and $f\left(x\right)$ exist and gives real value for $\forall x\in (-1,1)\Rightarrow f$ is onto .

$\Rightarrow f$ is one one onto.