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.
${{x}^{2}}=y$
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 $F\left( x \right)=-\sqrt{x}$, the inverse function is ${{F}^{-1}}\left( x \right)={{x}^{2}}\left( x<0 \right)$.
Consider the function.
$F\left( x \right)=-\sqrt{x}$
Substitute $x=0,4,9$ in the function $F\left( x \right)=-\sqrt{x}$.
For $x=0$, the value of $F\left( x \right)$ is,
$\begin{align}
& F\left( x \right)=-\sqrt{0} \\
& =0
\end{align}$
For $x=4$, the value of $F\left( x \right)$ is,
$\begin{align}
& F\left( x \right)=-\sqrt{4} \\
& =-2
\end{align}$
For $x=9$, the value of $F\left( x \right)$ is,
$\begin{align}
& F\left( x \right)=-\sqrt{9} \\
& =-3
\end{align}$
Tabulate for obtained values as shown below.
$\begin{matrix}
x & F\left( x \right)=-\sqrt{x} \\
0 & 0 \\
4 & -2 \\
9 & -3 \\
\end{matrix}$
Consider the function.
${{F}^{-1}}\left( x \right)={{x}^{2}}\left( x<0 \right)$
Substitute $x=-3,-2,0$ in the function ${{F}^{-1}}\left( x \right)={{x}^{2}}\left( x<0 \right)$.
For $x=-3$, the value of ${{F}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{F}^{-1}}\left( -3 \right)={{\left( -3 \right)}^{2}} \\
& =9
\end{align}$
For $x=-2$, the value of ${{F}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{F}^{-1}}\left( -2 \right)={{\left( -2 \right)}^{2}} \\
& =4
\end{align}$
For $x=0$, the value of ${{F}^{-1}}\left( x \right)$ is,
$\begin{align}
& {{F}^{-1}}\left( 0 \right)={{\left( 0 \right)}^{2}} \\
& =0
\end{align}$
Tabulate for obtained values as shown below.
$\begin{matrix}
x & {{F}^{-1}}\left( x \right)={{x}^{2}} \\
-3 & 9 \\
-2 & 4 \\
0 & 0 \\
\end{matrix}$
Plot the 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.
