Basically, I need to show that f(x)=(x+1)/(x-1) is its own inverse by computing f(f(x)). I can set up the problem, but I have run into a road block for proving it is its own inverse.
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$\begingroup$What's wrong with just computing that ? If for any $x$, $f ( f (x) ) = x$, $f$ is its own inverse function.
$$ f(f(x)) = \frac{f(x)+1}{f(x)-1} = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} = \frac{\frac{x+1+(x-1)}{x-1}}{\frac{x+1-(x-1)}{x-1}} = \frac{x+1+(x-1)}{x+1-(x-1)} =\frac{2 \cdot x}{2} = x. $$
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