Question 921397
Domain of given y is {{{x<=11}}} and range is {{{y>=0}}}.


The inverse h(x) of this y has domain {{{x>=0}}} and range {{{h(x)<=11}}}.


FINDING THE INVERSE, h(x):
{{{x=(h(x)-11)^2}}}, vertical and horizontal components change places.
{{{h(x)-11=0+- sqrt(x)}}}
{{{highlight(h(x)=11+- sqrt(x))}}}
---And again, be reminded of the domain and the range.  You need to use the MINUS square root form and reject the other branch.