Question 948659
Naming the inverse instead as d(x), and meaning {{{c^-1(x)=d(x)}}}, your inverse would be in the equation,   {{{x=4/(d(x)+2)-2}}}.  Solve this for d(x).


{{{x+2=4/(d(x)-2)}}}


{{{1/(x+2)=(d(x)-2)/4}}}


{{{4/(x+2)=d(x)-2}}}


{{{4/(x+2)+2=d(x)}}}


{{{highlight(d(x)=4/(x+2)+2)}}}