#### Answer

$f^{-1}(x) = \sqrt {4 - x}, x\geq0$

#### Work Step by Step

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

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$f^{-1}(x) = \sqrt {4 - x}, x\geq0$

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

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