Question 1131458
{{{y=x^2-3}}}....swap {{{x}}} and {{{y}}}


{{{x=y^2-3}}}....solve for {{{y}}}

{{{x+3=y^2}}}

{{{y^-1=sqrt(x+3)}}}-> inverse





{{{y=(x-2)^3+1}}}....swap {{{x}}} and {{{y}}}

{{{x=(y-2)^3+1}}}

{{{x-1=(y-2)^3}}}

{{{y-2=root(3,x-1)}}}

{{{y^-1=root(3,x-1)+2}}}