$f [0, \infty) \to [4, \infty) is defined by f (x) = x^{2} + 4 then f^{- 1} (13) =$

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Detailed Solution

$f (x) = y \Rightarrow x = f^{- 1} (y) \to (1) x^{2} + 4 = y x^{2} = y - 4$
$x = \sqrt{y - 4} \to (1)$ $from (1) & (2) f^{- 1} (y) = \sqrt{y - 4} f^{- 1} (x) = \sqrt{x - 4}$
$f^{- 1} (13) = \sqrt{13 - 4} = 3$

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