Let $f: R \to R$ be a function defined by $f\left(x\right)=\frac{e^{\left|x\right|}-e^{-x}}{e^x+e^{-x}}$. Then 

$f$ is not one-one as f (0) = 0 and f (-1) = 0.

 $f$ is also not onto as for$y = 1$ there is no$x \in R$ such that f (x) = 1. 

If there is such a$x \in R$ then $e^{\left|x\right|} – e^{–x }= e^{ x }+ e^{–x} ,$ clearly $x\ne 0.$

For x > 0, this equation gives $-e^{-x}=e^{-x}$ which is not possible. For x < 0, 

the above equation gives $e^{ x }= - e^{–x}$ which is also not possible .