Question

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

Answer 1

\[f(x)=\frac{ x }{ x-2 }\]
\[x \neq 2 \]

Answer 2

you need the inverse?

Answer 3

Yes. I know I need to switch f(x) and x, but I dunno how you would simplify from there

Answer 4

you can take
\[y=\frac{x}{x-2}\] switch \(x\) and \(y\) to write
\[x=\frac{y}{y-2}\] and solve for \(y\)

Answer 5

you get
\[x=\frac{y}{y-2}\]\[x(y-2)=y\] as a first step
then multiply out

Answer 6

\[xy-2x=y\] put the \(y\)'s on one side of the equal sign, the \(2x\) on the other
\[xy-y=2x\]

Answer 7

factor out the \(y\) on the left
\[(x-1)y=2x\]

Answer 8

Okay. I've got it from there

Answer 9

I didn't think to subtract y over ^^;;

Answer 10

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 11

answer should be
\[y=\frac{2x}{x-1}\]

Answer 12

yeah you got to get the \(y\)' all on one side
good luck

Answer 13

Thanks.