If $f(x)=\frac{x-1}{x+1}$ then show that
(i) $f\left(\frac{1}{x}\right)=-f(x)$
(ii) $f\left(\frac{-1}{x}\right)=\frac{-1}{f(x)} f$
Given: $f(x)=\frac{x-1}{x+1}$
(i) Need to prove: $f\left(\frac{1}{x}\right)=-f(x)$
Now replacing $x$ by $\frac{1}{x}$ we get,
$f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$
$\Rightarrow f\left(\frac{1}{x}\right)=\frac{1-x}{1+x}$
$\Rightarrow f\left(\frac{1}{x}\right)=\frac{-(x-1)}{(x+1)}=-f(x)$ [Proved]
(ii) Need to prove: $f\left(\frac{-1}{x}\right)=\frac{-1}{f(x)}$
Now replacing $\mathrm{x}$ by $-\frac{1}{x}$ we get,
$f\left(\frac{-1}{x}\right)=\frac{\frac{-1}{x}-1}{\frac{-1}{x}+1}$
$\Rightarrow f\left(\frac{-1}{x}\right)=\frac{-1-x}{-1+x}$
$\Rightarrow f\left(\frac{-1}{x}\right)=\frac{-(x+1)}{x-1}$
$\Rightarrow f\left(\frac{-1}{x}\right)=\frac{-1}{\frac{x-1}{x+1}}=\frac{-1}{f(x)}$ [Proved]
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