The function $\mathrm{f}:(-\mathrm{\infty },\mathrm{\infty })\to (-\mathrm{\infty },1)$, defined by $\mathrm{f}\left(\mathrm{x}\right)=\frac{2^{\mathrm{x}}-2^{-\mathrm{x}}}{2^{\mathrm{x}}+2^{-\mathrm{x}}}$ is :

$\begin{array}{l}\begin{array}{rl} & \mathrm{f}\left(\mathrm{x}\right)=\frac{2^{2\mathrm{x}}-1}{2^{2\mathrm{x}}+1}\\ & =1-\frac{2}{2^{2\mathrm{x}}+1}\end{array}\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{2}{{\left(2^{2\mathrm{x}}+1\right)}^2}\cdot 2\cdot 2^{2\mathrm{x}}\cdot \ln 2\text{ i.e always }+\mathrm{ve}\end{array}$

so f(x) is ↑ function
$\begin{array}{l}\therefore \mathrm{f}(-\mathrm{\infty })=-1\\ \mathrm{f}\left(\mathrm{\infty }\right)=1\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\in (-1,1)\ne \text{ co-domain }\end{array}$
so function is one-one but not onto