Let's solve the problem step by step.

Step 1: Find the inverse function $g^{-1}(x)$.

Given $g(x) = \frac{5x}{9x - 5}$, we need to find the inverse function. To do this, we follow these steps:

1. Start by replacing $g(x)$ with $y$:

$y = \frac{5x}{9x - 5}$

2. Swap $x$ and $y$ (this is a key step in finding the inverse):

$x = \frac{5y}{9y - 5}$

3. Solve for $y$. Multiply both sides by $9y - 5$ to get rid of the denominator:

$x(9y - 5) = 5y$

4. Expand the left-hand side:

$9xy - 5x = 5y$

5. Rearrange the equation to group terms involving $y$ on one side:

$9xy - 5y = 5x$

6. Factor out $y$ on the left-hand side:

$y(9x - 5) = 5x$

7. Solve for $y$:

$y = \frac{5x}{9x - 5}$

So, the inverse function is:

$g^{-1}(x) = \frac{5x}{9x - 5}$

Step 2: Find the domain and range of the inverse function.

The domain of the inverse function $g^{-1}(x)$ is the same as the range of the original function $g(x)$, and the range of the inverse function is the same as the domain of $g(x)$.

1. Domain of $g(x)$:

   To find the domain of $g(x)$, we must ensure the denominator does not equal zero:

$9x - 5 \neq 0 \implies x \neq \frac{5}{9}$

Thus, the domain of $g(x)$ is:

$(-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty)$

2. Range of $g(x)$:

   The range of $g(x)$ is all real numbers since $g(x)$ is a rational function with no horizontal asymptotes restricting the range.

Thus, the range of $g(x)$ is $(-\infty, \infty)$.

Step 3: State the domain and range of $g^{-1}(x)$.

• The domain of $g^{-1}(x)$ is the range of $g(x)$, which is $(-\infty, \infty)$.
• The range of $g^{-1}(x)$ is the domain of $g(x)$, which is $(-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty)$.

Final Answer:

$g^{-1}(x) = \frac{5x}{9x - 5}$

• Domain of $g^{-1}(x)$: $(-\infty, \infty)$
• Range of $g^{-1}(x)$: $(-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty)$

Let me know if you'd like any clarifications, or further details on any step.

Here are 5 follow-up questions to expand your understanding:

1. How do you verify that two functions are inverses of each other?
2. What role does the domain restriction play in one-to-one functions?
3. Can you find the asymptotes of both $g(x)$ and $g^{-1}(x)$?
4. What happens to the range if we restrict the domain of $g(x)$?
5. How does the graph of $g(x)$ compare to the graph of $g^{-1}(x)$?

Tip: Always check the function's domain and range before calculating its inverse, as certain values may make the function undefined.