Let's go through each part of the problem based on the function $f(x) = 2x + 1$.

a) Find $f(2)$

We substitute $x = 2$ into the given function:

$f(2) = 2(2) + 1 = 4 + 1 = 5$

Thus, $f(2) = 5$.

b) Find $f^{-1}(x)$

The inverse function $f^{-1}(x)$ undoes the original function. To find the inverse, follow these steps:

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

c) Find $f^{-1}(7)$

Now that we know $f^{-1}(x) = \frac{x - 1}{2}$, substitute $x = 7$:

$f^{-1}(7) = \frac{7 - 1}{2} = \frac{6}{2} = 3$

Thus, $f^{-1}(7) = 3$.

Would you like more details or clarifications on any part of this solution? Here are five related questions to expand on this topic:

1. What are the conditions for a function to have an inverse?
2. How can you verify that $f^{-1}(x)$ is indeed the inverse of $f(x)$?
3. What is the domain and range of $f(x)$ and $f^{-1}(x)$?
4. How do transformations (like shifts and stretches) affect the inverse function?
5. Can a quadratic function have an inverse? If so, under what conditions?

Tip: To check if two functions are inverses of each other, verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.