To solve the problem of finding the inverse functions, let's go through each function step by step.

Problem 1: $f(x) = \frac{5}{4}x - 3$

To find the inverse, follow these steps:

1. Replace $f(x)$ with $y$: $y = \frac{5}{4}x - 3$

2. Swap $x$ and $y$ to find the inverse: $x = \frac{5}{4}y - 3$

3. Solve for $y$: $x + 3 = \frac{5}{4}y$ $y = \frac{4}{5}(x + 3)$

So, the inverse function is: $f^{-1}(x) = \frac{4}{5}(x + 3)$

Problem 2: $f(x) = \frac{3}{x - 2}$

For this function:

1. Replace $f(x)$ with $y$: $y = \frac{3}{x - 2}$

2. Swap $x$ and $y$: $x = \frac{3}{y - 2}$

3. Solve for $y$: $x(y - 2) = 3$ $xy - 2x = 3$ $xy = 3 + 2x$ $y = \frac{3 + 2x}{x}$

So, the inverse function is: $f^{-1}(x) = \frac{3 + 2x}{x}$

Problem 3: $f(x) = \frac{2x + 3}{x - 7}$

Finally, for this function:

1. Replace $f(x)$ with $y$: $y = \frac{2x + 3}{x - 7}$

2. Swap $x$ and $y$: $x = \frac{2y + 3}{y - 7}$

3. Solve for $y$: $x(y - 7) = 2y + 3$ $xy - 7x = 2y + 3$ $xy - 2y = 7x + 3$ $y(x - 2) = 7x + 3$ $y = \frac{7x + 3}{x - 2}$

So, the inverse function is: $f^{-1}(x) = \frac{7x + 3}{x - 2}$

Summary of Inverse Functions:

1. $f^{-1}(x) = \frac{4}{5}(x + 3)$
2. $f^{-1}(x) = \frac{3 + 2x}{x}$
3. $f^{-1}(x) = \frac{7x + 3}{x - 2}$

Would you like more details on any step or have any questions?

---

Here are some related questions to deepen your understanding:

1. How can you verify that the function found is indeed the inverse?
2. What happens if the function is not one-to-one? Can it have an inverse?
3. How would you approach finding the inverse of a quadratic function?
4. Can you derive the inverse function for a trigonometric function, such as $\sin(x)$?
5. How does the domain and range of a function change when finding its inverse?

Tip: Always remember that not all functions have an inverse. For a function to have an inverse, it must be one-to-one (bijective), meaning each output is produced by exactly one input.