#### Answer

$f^{-1}(x) = (x - 2)^{2} - 3, x\geq2$

#### Work Step by Step

$f(x) = 2 + \sqrt {3 + x}$
$y= 2 + \sqrt {3 + x}$
$y - 2 =\sqrt {3 + x}$
$(y - 2)^{2} = (\sqrt {3 + x})^{2}$
$(y - 2)^{2} = 3 + x$
$x = (y - 2)^{2} - 3$
$f^{-1}(x) = (x - 2)^{2} - 3, x\geq2$