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