F(x)= 9-x^2, x ≥ 0. f^-1(x)=??

1 answer:

7 0

$y=9-x^{2},\ x\ge0$

Reverse $x$ and $y$ :
$x=9-y^{2},\ y\ge0$
Solve for $y$ :
$x=9-y^{2}$
$\rightarrow y^{2}=9-x$
$\rightarrow y=\sqrt{9-x}$ since $y\ge0$

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