Answer
$f^{-1}(x)=(x-2)^{2}-3$
Work Step by Step
$f(x)=2+\sqrt{3+x}$
Rewrite this expression as $y=2+\sqrt{3+x}$ and solve for $x$:
$y=2+\sqrt{3+x}$
Take $2$ to the left side:
$y-2=\sqrt{3+x}$
Square both sides:
$(y-2)^{2}=(\sqrt{3+x})^{2}$
$(y-2)^{2}=3+x$
$3+x=(y-2)^{2}$
Take $3$ to the right side:
$x=(y-2)^{2}-3$
Interchange $x$ and $y$:
$y=(x-2)^{2}-3$
The inverse of the original function is $f^{-1}(x)=(x-2)^{2}-3$