Let's solve the inverse of each function one by one.

7. $f(x) = x + 3$

To find $f^{-1}(x)$, follow these steps:

1. Replace $f(x)$ with $y$: $y = x + 3$
2. Swap $x$ and $y$: $x = y + 3$
3. Solve for $y$: $y = x - 3$ Thus, the inverse function is: $f^{-1}(x) = x - 3$

8. $f(x) = x + 5$

1. Start with $y = x + 5$.
2. Swap $x$ and $y$: $x = y + 5$
3. Solve for $y$: $y = x - 5$ Thus, the inverse function is: $f^{-1}(x) = x - 5$

9. $f(x) = 2 - x$

1. Start with $y = 2 - x$.
2. Swap $x$ and $y$: $x = 2 - y$
3. Solve for $y$: $y = 2 - x$ Thus, the inverse function is: $f^{-1}(x) = 2 - x$

10. $f(x) = 3 - x$

1. Start with $y = 3 - x$.
2. Swap $x$ and $y$: $x = 3 - y$
3. Solve for $y$: $y = 3 - x$ Thus, the inverse function is: $f^{-1}(x) = 3 - x$

11. $f(x) = \frac{x}{x+2}$

1. Let $y = \frac{x}{x+2}$.
2. Multiply both sides by $x + 2$: $y(x + 2) = x$
3. Distribute $y$: $yx + 2y = x$
4. Bring all terms involving $x$ on one side: $yx - x = -2y$ Factor out $x$: $x(y - 1) = -2y$
5. Solve for $x$: $x = \frac{-2y}{y - 1}$ Thus, the inverse function is: $f^{-1}(x) = \frac{-2x}{x - 1}$

12. $f(x) = \frac{2x+3}{5x+4}$

1. Let $y = \frac{2x+3}{5x+4}$.
2. Multiply both sides by $5x + 4$: $y(5x + 4) = 2x + 3$
3. Distribute $y$: $5yx + 4y = 2x + 3$
4. Bring all terms involving $x$ on one side: $5yx - 2x = 3 - 4y$ Factor out $x$: $x(5y - 2) = 3 - 4y$
5. Solve for $x$: $x = \frac{3 - 4y}{5y - 2}$ Thus, the inverse function is: $f^{-1}(x) = \frac{3 - 4x}{5x - 2}$

If you have any questions or want further details, feel free to ask!

Here are some follow-up questions:

1. How can you verify if two functions are inverses of each other?
2. What is the graphical relationship between a function and its inverse?
3. Can every function have an inverse? Why or why not?
4. How does the horizontal line test relate to the invertibility of functions?
5. How can you apply the concept of inverse functions in real-life scenarios?

Tip: A function has an inverse if it is one-to-one (bijective), meaning each input corresponds to exactly one output and vice versa.