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### Given Function:
\[ h(x) = \frac{6x - 7}{7x + 8} \]

### To Find:
1. The inverse function \( h^{-1}(x) \).
2. The domain of \( h^{-1}(x) \).
3. The range of \( h^{-1}(x) \).

### Steps to Solve:

1. **Find the Inverse Function \( h^{-1}(x) \):**

   Start by setting \( y = h(x) \):

   \[
   y = \frac{6x - 7}{7x + 8}
   \]

   To find the inverse, solve for \( x \) in terms of \( y \):

   \[
   y(7x + 8) = 6x - 7
   \]
   \[
   7xy + 8y = 6x - 7
   \]
   \[
   7xy - 6x = -8y - 7
   \]
   \[
   x(7y - 6) = -8y - 7
   \]
   \[
   x = \frac{-8y - 7}{7y - 6}
   \]

   So the inverse function is:
   \[
   h^{-1}(x) = \frac{-8x - 7}{7x - 6}
   \]

2. **Determine the Domain of \( h^{-1}(x) \):**

   The domain of \( h^{-1}(x) \) is the range of the original function \( h(x) \). 
   
   Since \( h(x) \) is a rational function, it has vertical asymptotes where the denominator is zero. So let's find where the denominator of \( h(x) \) is zero:

   \[
   7x + 8 = 0 \quad \Rightarrow \quad x = -\frac{8}{7}
   \]

   The function \( h(x) \) is defined for all \( x \) except \( x = -\frac{8}{7} \). This means that the domain of \( h(x) \) is \( (-\infty, -\frac{8}{7}) \cup (-\frac{8}{7}, \infty) \). 

   The range of \( h(x) \) will be the set of all \( y \)-values except the horizontal asymptote. To find the horizontal asymptote of \( h(x) \), consider the behavior as \( x \) approaches infinity:

   \[
   \lim_{x \to \pm\infty} h(x) = \frac{6x - 7}{7x + 8} \approx \frac{6x}{7x} = \frac{6}{7}
   \]

   Thus, \( y = \frac{6}{7} \) is a horizontal asymptote and is not included in the range. Hence, the range of \( h(x) \) is \( (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \).

   Therefore, the **domain of \( h^{-1}(x) \)** is \( (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \).

3. **Determine the Range of \( h^{-1}(x) \):**

   The range of \( h^{-1}(x) \) is the domain of \( h(x) \), which we found earlier:

   The **range of \( h^{-1}(x) \)** is \( (-\infty, -\frac{8}{7}) \cup (-\frac{8}{7}, \infty) \).

### Final Answers:

- **Inverse Function:** \( h^{-1}(x) = \frac{-8x - 7}{7x - 6} \)
- **Domain of \( h^{-1}(x) \):** \( (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \)
- **Range of \( h^{-1}(x) \):** \( (-\infty, -\frac{8}{7}) \cup (-\frac{8}{7}, \infty) \)

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### Related Questions:

1. How can we graphically interpret the inverse of this rational function?
2. What are the steps to find the horizontal and vertical asymptotes of a rational function?
3. How does the process of finding an inverse differ for quadratic functions?
4. What is the significance of the horizontal asymptote in determining the range of the function?
5. How do you verify if the inverse function is correct?
6. What are the implications if the denominator of \( h(x) \) has more roots?
7. How would the domain change if \( h(x) \) had a different numerator?
8. What are some common mistakes to avoid when finding the inverse of a function?

**Tip:** When finding the inverse of a rational function, always consider the behavior of the function at infinity and the points where the original function is undefined, as these will impact the domain and range of the inverse.

Learn how to find the inverse function, domain, and range of the rational function h(x) = (6x - 7) / (7x + 8). This problem covers inverse functions, rational functions, and their domain and range analysis. Suitable for advanced high school students and beyond.

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