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Calculate the inverse of $s(x) = \frac{1+f(x)}{1-f(x)}$ in terms of $f^{-1}$,f is a $1-1$ function with inverse $f^{-1}$

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What have you tried? –  Matthew Conroy Oct 29 '12 at 5:33
Can you say why you're interested in this question? –  Jonah Sinick Oct 29 '12 at 6:57

2 Answers 2

Assumption: $s^{-1}$ function exists, then,

Let $$y=s(x)=\frac{1+f(x)}{1-f(x)}\implies y-yf(x)=1+f(x)\implies f(x)=\frac{y-1}{y+1}\implies x=f^{-1}(\frac{y-1}{y+1})\implies s^{-1}(y)=f^{-1}(\frac{y-1}{y+1})$$ or $$s^{-1}(x)=f^{-1}(\frac{x-1}{x+1})$$

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Thank you soo much! You're a life saver. –  Jack Oct 29 '12 at 5:49
i am glad, i could help you :) –  Aang Oct 29 '12 at 5:52
question, how did you get x=f−1(y−1/y+1) You lost me for a moment there. –  Jack Oct 29 '12 at 5:55
@Jack: Whenevwer you have $y=f(x)$ then you can write $x=f^{-1}(y)$. –  Babak S. Oct 29 '12 at 6:52
@Jack: If $f^{-1}$ exists then if $y=f(x)\implies $ $x=f^{-1}(y)$ –  Aang Oct 29 '12 at 10:42

A thought: IF we consider the function $$g(x) = \frac{1 + x}{1 - x} $$ Then obviusly, $s = g \circ f$. Is $g$ injective??? What do you know about composition of injective functions? Assume $x \neq 1$.

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At this point in my course, absolutely nothing. We were just introduced to inverse functions so this on is really getting me. –  Jack Oct 29 '12 at 5:41

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