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The problem you're working on involves finding the inverse of the function \( h(x) = \frac{9x - 2}{2x + 3} \) and determining the domain and range of its inverse.

### Step 1: Find the Inverse \( h^{-1}(x) \)

To find the inverse of the function \( h(x) = \frac{9x - 2}{2x + 3} \):

1. **Rewrite the equation** \( h(x) = y \), so:
   \[
   y = \frac{9x - 2}{2x + 3}
   \]

2. **Swap** \( x \) and \( y \):
   \[
   x = \frac{9y - 2}{2y + 3}
   \]

3. **Solve for** \( y \):
   - Multiply both sides by \( 2y + 3 \) to eliminate the denominator:
     \[
     x(2y + 3) = 9y - 2
     \]
   - Distribute \( x \) on the left-hand side:
     \[
     2xy + 3x = 9y - 2
     \]
   - Group the \( y \)-terms on one side:
     \[
     2xy - 9y = -3x - 2
     \]
   - Factor out \( y \) on the left-hand side:
     \[
     y(2x - 9) = -3x - 2
     \]
   - Solve for \( y \):
     \[
     y = \frac{-3x - 2}{2x - 9}
     \]

Thus, the inverse function is:
\[
h^{-1}(x) = \frac{-3x - 2}{2x - 9}
\]

### Step 2: Find the Domain and Range of \( h^{-1}(x) \)

- **Domain of** \( h^{-1}(x) \): The domain of the inverse function is the range of the original function. Since the denominator \( 2x + 3 \) in \( h(x) \) cannot be zero, solve \( 2x + 3 = 0 \) to find that \( x = -\frac{3}{2} \) is excluded from the domain of \( h(x) \). Therefore, the range of \( h(x) \) is all real numbers except \( y = -\frac{9}{2} \), which gives the domain of \( h^{-1}(x) \) as \( (-\infty, -\frac{9}{2}) \cup (-\frac{9}{2}, \infty) \).

- **Range of** \( h^{-1}(x) \): The range of \( h^{-1}(x) \) is the domain of the original function \( h(x) \). The domain of \( h(x) \) excludes \( x = -\frac{3}{2} \). Therefore, the range of \( h^{-1}(x) \) is \( (-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, \infty) \).

### Final Answer:
- The inverse function is: \( h^{-1}(x) = \frac{-3x - 2}{2x - 9} \).
- Domain of \( h^{-1}(x) \): \( (-\infty, -\frac{9}{2}) \cup (-\frac{9}{2}, \infty) \).
- Range of \( h^{-1}(x) \): \( (-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, \infty) \).

Do you want more details or have any questions about this solution?

Here are 5 related questions to expand your knowledge:
1. How can we determine if a function is one-to-one before finding its inverse?
2. What are the graphical interpretations of a function and its inverse?
3. How does the domain of a function relate to the range of its inverse?
4. What happens to the function's asymptotes when we calculate the inverse?
5. How do restrictions on the domain of a function affect its inverse?

**Tip:** To confirm if the inverse is correct, try composing \( h(h^{-1}(x)) \) and \( h^{-1}(h(x)) \) and check if both return \( x \).

La función inyectiva h(x) = (9x - 2) / (2x + 3). Hallar la inversa de h, indicar el dominio y el rango de h^-1(x).

Learn how to find the inverse of h(x) = (9x - 2) / (2x + 3) and determine the domain and range of its inverse. This problem covers key algebra concepts including inverse functions and interval notation.

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