Find the inverse of $f$. Then sketch the graphs of $f$ and $f ^ { - 1 }$ on the same set of axes.
$f ( x ) = \sqrt { 16 - x ^ { 2 } } , x \geq 0$
A) $f^{-1}(x)=\sqrt{4+x^{2}}, x \geq 0$
B) $f ^ { - 1 } ( x ) = - \sqrt { 16 - x ^ { 2 } } , x \geq 0$
C) $f ^ { - 1 } ( x ) = \sqrt { 16 - x ^ { 2 } } , x \geq 0$
D) $f ^ { - 1 } ( x ) = - \sqrt { 4 + x ^ { 2 } } , x \geq 0$

Question

Find the inverse of $f$. Then sketch the graphs of $f$ and $f ^ { - 1 }$ on the same set of axes.
$f ( x ) = \sqrt { 16 - x ^ { 2 } } , x \geq 0$
A) $f^{-1}(x)=\sqrt{4+x^{2}}, x \geq 0$ 
B) $f ^ { - 1 } ( x ) = - \sqrt { 16 - x ^ { 2 } } , x \geq 0$ 
C) $f ^ { - 1 } ( x ) = \sqrt { 16 - x ^ { 2 } } , x \geq 0$ 
D) $f ^ { - 1 } ( x ) = - \sqrt { 4 + x ^ { 2 } } , x \geq 0$

Quizplus · Accepted Answer

To find the inverse of $f(x)$, we replace $f(x)$ with $y$ and then solve for $x$.
$$y = \sqrt{16-x^2}$$
$$y^2 = 16-x^2$$
$$x^2 = 16-y^2$$
$$x = \sqrt{16-y^2}$$
Since the domain of $f(x)$ is $x\geq 0$, the range of $f^{-1}(x)$ must be $y\geq 0$. Therefore, we take the positive square root and obtain:
$$f^{-1}(x) = \sqrt{16-x^2}, x \geq 0$$
Now, to sketch the graphs of $f(x)$ and $f^{-1}(x)$, we plot some points:

\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|}
\hline
$x$ & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 15 & 15.5 & 15.9 \ \hline
$f(x)=\sqrt{16-x^2}$ & 4 & $\sqrt{12}$ & 0 & $\sqrt{20}$ & $\sqrt{48}$ & 0 & $-\sqrt{80}$ & 0 & 1 & 0.5 & 0.1 \ \hline
$f^{-1}(x)=\sqrt{16-x^2}$ & 0 & $\sqrt{12}$ & 4 & $\sqrt{8}$ & 0 & $\sqrt{6}$ & 8 & $\sqrt{12}$ & 1 & 0.5 & 0.1 \ \hline
\end{tabular}

Note that the graph of $f$ is half of a circle centered at the origin with radius 4, only in the first quadrant.

Therefore, the correct choice is (C).

Find the Inverse Of fff Then Sketch the Graphs Of Then

Find the Inverse Of $f$ Then Sketch the Graphs Of Then