Answer
The graph is shown in below.
Work Step by Step
$f\left( x \right)=\sqrt{x}$
Evaluate the inverse of the function $f\left( x \right)=\sqrt{x}$.
Replace the function $f\left( x \right)$ with y.
$y=\sqrt{x}$
Interchange the variables x and y.
$x=\sqrt{y}$
Solve for y value.
$y={{x}^{2}}$
Replace y with ${{f}^{-1}}\left( x \right)$ as follows.
${{f}^{-1}}\left( x \right)={{x}^{2}}$
Thus, the inverse function is ${{f}^{-1}}\left( x \right)={{x}^{2}}$.
Since the function is $f\left( x \right)=\sqrt{x}$, the inverse is ${{f}^{-1}}\left( x \right)={{x}^{2}}\left( x>0 \right)$.
$f\left( x \right)=\sqrt{x}$
Substitute $x=9,4,1$ in the function $f\left( x \right)=\sqrt{x}$.
For $x=9$, the value of $f\left( x \right)$ is,
$\begin{align}
& f\left( x \right)=\sqrt{9} \\
& =3
\end{align}$
For $x=4$, the value of $f\left( x \right)$ is,
$\begin{align}
& f\left( x \right)=\sqrt{4} \\
& =2
\end{align}$
For $x=1$, the value of $f\left( x \right)$ is,
$\begin{align}
& f\left( x \right)=\sqrt{1} \\
& =1
\end{align}$
Tabulate for obtained values as shown below.
$\begin{matrix}
x & f\left( x \right)=\sqrt{x} \\
9 & 3 \\
4 & 2 \\
1 & 1 \\
\end{matrix}$
${{f}^{-1}}\left( x \right)={{x}^{2}}$
Substitute $x=1,2,3$ in the function ${{f}^{-1}}\left( x \right)={{x}^{2}}$.
For $x=1$, the value of ${{f}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{f}^{-1}}\left( 1 \right)={{1}^{2}} \\
& =1
\end{align}$
For $x=2$, the value of ${{f}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{f}^{-1}}\left( 2 \right)={{2}^{2}} \\
& =4
\end{align}$
For $x=3$, the value of ${{f}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{f}^{-1}}\left( 3 \right)={{3}^{2}} \\
& =9
\end{align}$
Tabulate for obtained values as shown below.
$\begin{matrix}
x & {{f}^{-1}}\left( x \right)={{x}^{2}} \\
1 & 1 \\
2 & 4 \\
3 & 9 \\
\end{matrix}$
Plot these points and sketch the graphs of the functions $f\left( x \right)=\sqrt{x}$ and ${{f}^{-1}}\left( x \right)={{x}^{2}}$ as shown in the figure below.
