Question 1000687


to find inverse of {{{y=sqrt(1/(x-2))}}}, swap {{{x}}} and {{{y}}}

{{{x=sqrt(1/(y-2))}}}.....solve for {{{y}}}

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

so, inverse is {{{y^-1= (2x^2+1)/x^2}}}

domain: 

{ {{{x}}} element {{{R}}}: {{{x<>0}}} }

range:

{ {{{y}}} element {{{R}}} : {{{y>2}}} }