Prove the following

(i)

$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1}$

Substituting $\mathrm{x}$ by $1 / \mathrm{x}$, we get

$f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$

$=\frac{\frac{1-x}{x}}{\frac{1+x}{x}}$

$=\frac{1-x}{1+x}$

$=\frac{-(x-1)}{1+x}$

$=-\frac{x-1}{x+1}$

Therefore,

We get,

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

Hence proved

(ii)

$f(x)=\frac{x-1}{x+1}$

Substituting x by – 1/x, we get

$f\left(-\frac{1}{x}\right)=\frac{\left(-\frac{1}{x}\right)-1}{\left(-\frac{1}{x}\right)+1}$

$=\frac{\frac{-1-x}{x}}{\frac{-1+x}{x}}$

$=\frac{-1-x}{-1+x}$

$=\frac{-(x+1)}{x-1}$

$=\frac{-1}{\frac{x-1}{x+1}}$

Therefore,

$\mathrm{f}\left(-\frac{1}{\mathrm{x}}\right)=\frac{-1}{\mathrm{f}(\mathrm{x})}$

Hence proved