The one-to-one function $h$ is defined below.
\[
h(x)=\frac{4 x-5}{x+9}
\]

Find $h^{-1}(x)$, where $h^{-1}$ is the inverse of $h$.
Also state the domain and range of $h^{-1}$ in interval notation.
\begin{tabular}{|}
$h^{-1}(x)=\square$ \\
Domain of $h^{-1}=\square$ \\
Range of $h^{-1}=\square$
\end{tabular}
$\left[\begin{tabular}{ccc}
\frac{\square}{\square}\right.$ & $(\square, \square)$ & {$[\square, \square]$} \\
$\square \cup \square$ & $(\square, \square]$ & {$[\square, \square)$} \\
$\varnothing$ & $\infty$ & $-\infty$ \\
$x$ & 5
\end{tabular}

Step 1 ：Replace \(h(x)\) with \(y\) to get \(y = \frac{4x - 5}{x + 9}\)

Step 2 ：Swap \(x\) and \(y\) to find the inverse: \(x = \frac{4y - 5}{y + 9}\)

Step 3 ：Solve for \(y\) to get \(xy + 9x = 4y - 5\)

Step 4 ：Rearrange the equation to get \(xy - 4y = -9x - 5\)

Step 5 ：Factor out \(y\) to get \(y(x - 4) = -9x - 5\)

Step 6 ：Divide by \(x - 4\) to get \(y = \frac{-9x - 5}{x - 4}\)

Step 7 ：So, the inverse function is \(h^{-1}(x) = \frac{-9x - 5}{x - 4}\)

Step 8 ：The domain of \(h^{-1}(x)\) is all real numbers except 4, so the domain is \((-\infty, 4) \cup (4, \infty)\)

Step 9 ：The range of \(h^{-1}(x)\) is all real numbers except -9, so the range is \((-\infty, -9) \cup (-9, \infty)\)

Step 10 ：\(\boxed{h^{-1}(x) = \frac{-9x - 5}{x - 4}}\)

Step 11 ：Domain of \(h^{-1}\) is \(\boxed{(-\infty, 4) \cup (4, \infty)}\)

Step 12 ：Range of \(h^{-1}\) is \(\boxed{(-\infty, -9) \cup (-9, \infty)}\)