The one-to-one function $h$ is defined below. \[ h(x)=\frac{4 x-3}{7 x+9} \] Find $h^{-1}(x)$, where $h^{-1}$ is the inverse of $h$. Also state the domain and range of $h^{-1}$ in interval notation.

Step 1 ：Let's start by finding the inverse of the function \(h(x) = \frac{4x - 3}{7x + 9}\). To do this, we switch the x and y values and solve for y. This gives us the equation \(x = \frac{4y - 3}{7y + 9}\).

Step 2 ：Solving this equation for y, we get \(y = \frac{3(-3x - 1)}{7x - 4}\). Therefore, the inverse function is \(h^{-1}(x) = \frac{3(-3x - 1)}{7x - 4}\).

Step 3 ：Next, we need to find the domain and range of the inverse function. The domain of the inverse function is the range of the original function, and the range of the inverse function is the domain of the original function.

Step 4 ：The domain of the original function \(h(x)\) is all real numbers except \(-\frac{9}{7}\), and its range is all real numbers except \(\frac{3}{4}\). Therefore, the domain of \(h^{-1}(x)\) is all real numbers except \(\frac{3}{4}\), and its range is all real numbers except \(-\frac{9}{7}\).

Step 5 ：\(\boxed{\text{Final Answer: The inverse function is } h^{-1}(x) = \frac{3(-3x - 1)}{7x - 4}, \text{ the domain of } h^{-1}(x) \text{ is all real numbers except } \frac{3}{4}, \text{ and the range of } h^{-1}(x) \text{ is all real numbers except } -\frac{9}{7}.}\)