Question 1061793
With the function {{{f(x)}}} as written,
there is an inverse relation (not an inverse function),
{{{y =" " +- sqrt((2x-1)/2)}}} ,
that can be found from {{{y=x^2+1/2}}}
by interchanging {{{x}}} and {{{y}}} ,
and then "solving for y."
If we define the function with a restricted domain, as
{{{f(x)=x^2+1 / 2}}}{{{" ,"}}}{{{"for"}}}{{{x>=0)}}} ,
then it has an inverse function, which is
{{{f^(-1)}}}{{{(x)=sqrt((2x-1)/2)}}} or {{{f^(-1)}}}{{{(x)=sqrt(x-1/2)}}} .
When we interchange {{{x}}} and {{{y}}} from {{{y=x^2+1/2}}}{{{" ,"}}}{{{"for"}}}{{{x>=0)}}} ,
we get
{{{x=y^2+1/2}}}{{{" ,"}}}{{{"with"}}}{{{y>=0)}}} ,
{{{x=(2y^2+1)/2}}}{{{" ,"}}}{{{"with"}}}{{{y>=0)}}} ,
{{{2x=2y^2+1}}}{{{" ,"}}}{{{"with"}}}{{{y>=0)}}} ,
{{{2x-1=2y^2}}}{{{" ,"}}}{{{"with"}}}{{{y>=0)}}} ,
{{{(2x-1)/2=y^2}}}{{{" ,"}}}{{{"with"}}}{{{y>=0)}}} , and
{{{y=sqrt((2x-1)/2)}}} <--> {{{y=sqrt(x-1/2)}}} .