Question 1071330
Fix your first one.



Only dealing with f(x)=(x-3)^2+2,  and upon inverting, taking the  upper branch ---


(The 'quick' way, but not the formally correct way)
{{{(y-3)^2+2=x}}}
{{{y-3=0+- sqrt(x-2)}}}
{{{y=3+- sqrt(x-2)}}}
{{{highlight(y=3+sqrt(x-2))}}}------the upper branch, range {{{y>=3}}}.


This should be shown, 
{{{highlight(f^-1(x)=3+sqrt(x-2))}}}