College Algebra, Chapter 3, 3.7, Section 3.7, Problem 34

Use the inverse function property to show that $\displaystyle f(x) = \sqrt{4 - x^2}, 0 \leq x \leq 2$ and $\displaystyle g(x) = \sqrt{4 - x^2}, 0 \leq x \leq 2$ are inverses of each other.

By using the Property of Inverse Function, we let $\displaystyle f^{-1} (x) = \sqrt{4 - x^2}$, so..


$
\begin{equation}
\begin{aligned}

f^{-1} (f(x)) =& f^{-1} \left(\sqrt{4 - x^2}\right)
&&
\\
\\
=& \sqrt{4 - \left(\sqrt{4 - x^2}\right)^2}
&& \text{Substitute } \sqrt{4 - x^2}
\\
\\
=& \sqrt{4 - (4 - x^2)}
&& \text{Simplify}
\\
\\
=& \sqrt{4 - 4 + x^2}
&&
\\
\\
=& \sqrt{x^2}
&&
\\
\\
=& x^{\frac{2}{2}}
&&
\\
\\
=& x
&&


\end{aligned}
\end{equation}
$


Thus, $f$ and $g$ are inverses of each other.