If f(x)=
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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