#### Answer

$f^{-1}(x) = \frac{1}{\sqrt x}, x\gt0$

#### Work Step by Step

$f(x) = \frac{1}{x^{2}}, x\gt0$
$y = \frac{1}{x^{2}}$
$x^{2} = \frac{1}{y}$
$x = \sqrt {\frac{1}{y}}$
$x = \frac{1}{\sqrt y}$
$f^{-1}(x) = \frac{1}{\sqrt x}, x\gt0$

Published by
Brooks Cole

ISBN 10:
1305115546

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

$f^{-1}(x) = \frac{1}{\sqrt x}, x\gt0$

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