College Algebra, Chapter 3, 3.7, Section 3.7, Problem 52

Find the inverse of $\displaystyle f(x) = \sqrt{9 - x^2}$ from $0 \leq x \leq 3$

To find the inverse of $f(x)$, we write $y = f(x)$


$
\begin{equation}
\begin{aligned}

y =& \sqrt{9 - x^2}
&& \text{Solve for $x$, square both sides}
\\
\\
y^2 =& 9 - x^2
&& \text{Add $x^2$ and subtract } y^2
\\
\\
x^2 =& 9 - y^2
&& \text{Take the square root}
\\
\\
x =& \pm \sqrt{9 - y^2}
&& \text{Interchange $x$ and $y$}
\\
\\
y =& \pm \sqrt{9 - x^2}
&& \text{Apply restrictions, from } 0 \leq x < 3
\\
\\
y =& \sqrt{9 - x^2}
&&

\end{aligned}
\end{equation}
$


Thus, the inverse of $f(x) = \sqrt{9 - x^2}$ is $\displaystyle f^{-1} (x) = \sqrt{9 - x^2}$.

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