Answer
$ f^{-1}(x)=\sqrt {x-4}$
See graph.
Work Step by Step
Step 1. $f(x)=x^2+4, x\ge0\Longrightarrow y=x^2+4 \Longrightarrow x=y^2+4, y\ge0 \Longrightarrow y=\sqrt {x-4}\Longrightarrow f^{-1}(x)=\sqrt {x-4}$
Step 2. Check $f( f^{-1}(x))=(\sqrt {x-4})^2+4=x$ and $f^{-1}(f(x))=\sqrt {(x^2+4)-4} =x$
Step 3. See graph for both $f$ and $f^{-1}$.