College Algebra, Chapter 3, 3.7, Section 3.7, Problem 56

Suppose that $f(x) = 16 - x^2, x \geq 0$. (a) Sketch the graph of $f$. (b) Use the graph of $f$ to sketch the graph of $f^{-1}$. (c) Find $f^{-1}$.

a.) The graph of $f(x) = 16 - x^2$ is obtained by reflecting the graph of $y = x^2$ about $x$-axis and the result is shifted 16 units upward.







b.) To sketch the graph of $f^{-1}$, we interchange the values of $x$ and $y$.







c.) To find $f^{-1}$, we set $y = f(x)$


$
\begin{equation}
\begin{aligned}

y =& 16 - x^2
&& \text{Solve for $x$, add $x^2$ and subtract $y$}
\\
\\
x^2 =& 16 - y
&& \text{Take the square root}
\\
\\
x =& \pm \sqrt{16 - y}
&& \text{Interchange $x$ and $y$}
\\
\\
y =& \pm \sqrt{16 - x}
&& \text{Apply restrictions, } x \geq 0
\\
\\
y =& \sqrt{16 - x}
&&


\end{aligned}
\end{equation}
$


Thus, the inverse of $f(x) = 16 - x^2$ is $f^{-1} (x) = \sqrt{16 - x}$.