Chapter 2, Functions - Section 2.8 - One-to-One Functions and their Inverses - 2.8 Exercises - Page 263: 70

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

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