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Find f^-1(x) and state any restrictions on the domain of f^-1(x)

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\[f(x)=\frac{ x }{ x-2 }\] \[x \neq 2 \]
you need the inverse?
Yes. I know I need to switch f(x) and x, but I dunno how you would simplify from there

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Other answers:

you can take \[y=\frac{x}{x-2}\] switch \(x\) and \(y\) to write \[x=\frac{y}{y-2}\] and solve for \(y\)
you get \[x=\frac{y}{y-2}\]\[x(y-2)=y\] as a first step then multiply out
\[xy-2x=y\] put the \(y\)'s on one side of the equal sign, the \(2x\) on the other \[xy-y=2x\]
factor out the \(y\) on the left \[(x-1)y=2x\]
Okay. I've got it from there
I didn't think to subtract y over ^^;;
and finally divide the domain will be all real numbers except \(x=1\) which is not surprising because that was the range of \(f\)
answer should be \[y=\frac{2x}{x-1}\]
yeah you got to get the \(y\)' all on one side good luck

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