Question 1048296
{{{f(x)=  (2x+a)/(bx-2)}}}

===> {{{f(f^(-1) (x))= x =  (2f^(-1)(x)+a)/(bf^(-1)(x)-2)}}}

===> {{{x(bf^(-1)(x)-2) =  (2f^(-1)(x)+a)}}}

===> {{{f^(-1)(x) = (2x+a)/(bx-2)}}}, after some more algebraic manipulations.

===> {{{f(x) = f^(-1)(x)}}} , for all x common to their domains.

===> There are no restrictions on the values a and b.