Question 462903
If the function is {{{f(x)=3/(x-7)}}}, then replace f(x) with y, then swap x and y to get {{{x=3/(y-7)}}} and solve for y 



{{{x=3/(y-7)}}}



{{{x(y-7)=3}}}



{{{y-7=3/x}}}



{{{y=3/x+7}}}



So the inverse function is *[Tex \LARGE f^{-1}(x)=\frac{3}{x}+7]



Domain of f: *[Tex \LARGE \left(-\infty,7\right)\cup\left(7,\infty\right)] (ie you can plug in any number but 7 since this value results in a division by zero)




Range of f: *[Tex \LARGE \left(-\infty,0\right)\cup\left(0,\infty\right)] (this is equal to the horizontal asymptote)



Now simply flip the domain and range of f(x) to get the domain and range for the inverse *[Tex \LARGE f^{-1}(x)]




Domain of *[Tex \LARGE f^{-1}(x)]: *[Tex \LARGE \left(-\infty,0\right)\cup\left(0,\infty\right)] 




Range of *[Tex \LARGE f^{-1}(x)]: *[Tex \LARGE \left(-\infty,7\right)\cup\left(7,\infty\right)]