f(x)= ( e^ 2 x -1 e^ 2 x +1 )is

$$\begin{array}{l} \because \mathrm{f}(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right) \\ \therefore \mathrm{f}(-x)=\frac{e^{-2 x}-1}{e^{-2 x}+1}=\frac{1-e^{2 x}}{1+e^{2 x}} \\ \Rightarrow \mathrm{f}(-x)=\frac{-\left(e^{2 x}-1\right)}{e^{2 x}+1}=-\mathrm{f}(x) \end{array}$$\therefore \mathrm{f}(x)$is an odd function. Again,$\mathrmf^{\prime}(x)=\frac{4 e^{2 x}}{\left(1+e^{2 x}\right)^{2}} > 0, \forall x \in \mathbb{R}\Rightarrow \mathrm{f}(x)$ is an increasing function.