Question:
If $f(x)=\frac{x-1}{x+1}$, then show that
(i) $f\left(\frac{1}{x}\right)=-f(x)$
(ii) $f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}$
Solution:
Given: $f(x)=\frac{x-1}{x+1}$ ...(1)
(i) Replacing $x$ by $\frac{1}{x}$ in $(1)$, we get
$f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$
$=\frac{1-x}{1+x}$
$=-\frac{x-1}{x+1}$
$=-f(x)$
(ii) Replacing $x$ by $-\frac{1}{x}$ in (1), we get
$f\left(-\frac{1}{x}\right)=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}$
$=\frac{-1-x}{-1+x}$
$=-\frac{x+1}{x-1}$
$=-\frac{1}{\left(\frac{x-1}{x+1}\right)}$
$=-\frac{1}{f(x)}$
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