Find the inverse of the function of f(x)= X-1/x+1 , X= 1 and verify fof^-1 is an identity function

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Solution

Dear Student,

$f (x) = \frac{x - 1}{x + 1} l e t y = \frac{x - 1}{x + 1} y x + y = x - 1 y x - x = - 1 - y x (y - 1) = - (y + 1) x = \frac{- (y + 1)}{(y - 1)} h e n c e f^{- 1} (x) = \frac{- (x + 1)}{x - 1} f o f^{- 1} (x) = f (f^{- 1} (x)) = f (\frac{- (x + 1)}{x - 1}) \Rightarrow \frac{\frac{- (x + 1)}{x - 1} - 1}{\frac{- (x + 1)}{x - 1} + 1} \Rightarrow \frac{\frac{- (x + 1) - (x - 1)}{x - 1}}{\frac{- (x + 1) + (x - 1)}{x - 1}} \Rightarrow \frac{- x - 1 - x + 1}{- x - 1 + x - 1} \Rightarrow \frac{- 2 x}{- 2} \Rightarrow x f o f^{- 1} (x) = x h e n c e i t i s i d e n t i t y f u n c t i o n .$

Regards,

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