Let's do it graphically first, and then we'll do it
algebraically.

First we'll draw the graph of 



That is a parabola with vertex (0,6).
However because of the 
restriction, we will have to eliminate
the left half of the graph to have the
graph of f(x), which is this:



The domain of  is    
The range of  is 

Now let's draw what is called "the identity line",
which is the 45°-line through the origin whose 
equation is .  I'll draw the identity line
in green dotted:



Now let's draw the graph of ,
the inverse of  by reflecting it across the 
green dotted identity line like this:



The domain of  is 
The range of  is 


Notice that the range of  is the domain of ,
and the domain of  is the range of .
That's doing it graphically. Now let's do it
algebraically:



Replace  by 



Interchange  and :



Solve for y:



We take the square roots of both sides, and 
since , we do not need "±".



Now we replace  by 



What is under the square root radical must not be negative,
so  or  or 

So, because of ,

The domain of  is 

and because of the ,

The range of  is 

Edwin