$f(x)=\frac{x-1}{x+1}$, then show that:
(i) $f\left(\frac{1}{x}\right)=-f(x)$
(ii) $f\left(-\frac{1}{x}\right)=\frac{-1}{f(x)}$
(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
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