Answer
The graph is shown below.
Work Step by Step
$f\left( x \right)={{4}^{x}}$and ${{f}^{-1}}\left( x \right)={{\log }_{4}}x$
$f\left( x \right)={{4}^{x}}$
Substitute $x=0$ in $f\left( x \right)={{4}^{x}}$:
$\begin{align}
& f\left( x \right)={{4}^{0}} \\
& =1
\end{align}$
Substitute $x=1$ in $f\left( x \right)={{4}^{x}}$:
$\begin{align}
& f\left( x \right)={{4}^{1}} \\
& =4
\end{align}$
Substitute $x=2$ in $f\left( x \right)={{4}^{x}}$:
$\begin{align}
& f\left( x \right)={{4}^{2}} \\
& =16
\end{align}$
Substitute $x=-1$ in $f\left( x \right)={{4}^{x}}$:
$\begin{align}
& f\left( x \right)={{4}^{-1}} \\
& =\frac{1}{4}
\end{align}$
Substitute $x=-2$ in $f\left( x \right)={{4}^{x}}$:
$\begin{align}
& f\left( x \right)={{4}^{-2}} \\
& =\frac{1}{16}
\end{align}$
$\begin{matrix}
x & f\left( x \right) \\
0 & 1 \\
1 & 4 \\
2 & 16 \\
-1 & \frac{1}{4} \\
-2 & \frac{1}{16} \\
\end{matrix}$
Consider the second function,
${{f}^{-1}}\left( x \right)={{\log }_{4}}x$
Assume, ${{f}^{-1}}\left( x \right)=y$
The function $y={{\log }_{4}}x$ can be written as ${{4}^{y}}=x$.
Substitute $y=0$ in $x={{4}^{y}}$:
$\begin{align}
& x={{4}^{0}} \\
& =1
\end{align}$
Substitute $y=1$ in $x={{4}^{y}}$:
$\begin{align}
& x={{4}^{1}} \\
& =2
\end{align}$
Substitute $y=2$ in $x={{4}^{y}}$:
$\begin{align}
& x={{4}^{2}} \\
& =16
\end{align}$
Substitute $y=-1$ in $x={{4}^{y}}$:
$\begin{align}
& x={{4}^{-1}} \\
& =\frac{1}{4}
\end{align}$
Substitute $y=-2$ in $x={{4}^{y}}$:
$\begin{align}
& x={{4}^{-2}} \\
& =\frac{1}{16}
\end{align}$
$\begin{matrix}
x & y \\
1 & 0 \\
4 & 1 \\
16 & 2 \\
\frac{1}{4} & -1 \\
\frac{1}{16} & -2 \\
\end{matrix}$
Now, draw the graph of both functions.
