#### Answer

$f^{-1}(x) = \sqrt {4 - y^{2}}, 0\leq x\leq 2$

#### Work Step by Step

$f(x) = \sqrt {4 - x^{2}}, 0\leq x\leq 2$
$y = \sqrt {4 - x^{2}}$
$y^{2} = 4 - x^{2}$
$x^{2} = 4 - y^{2}$
$x = \sqrt {4 - y^{2}}$
$f^{-1}(x) = \sqrt {4 - y^{2}}, 0\leq x\leq 2$