The problem presents a function $f(x) = \log_2 \left( \frac{4x - 1}{x - 3} \right)$ and asks for the inverse function $f^{-1}(x)$.

To find the inverse:

1. Start by setting $y = f(x)$, which gives: $y = \log_2 \left( \frac{4x - 1}{x - 3} \right)$

2. Convert the logarithmic form to exponential form: $2^y = \frac{4x - 1}{x - 3}$

3. Now solve for $x$. Multiply both sides by $x - 3$: $2^y (x - 3) = 4x - 1$

4. Expand the left-hand side: $2^y x - 3 \cdot 2^y = 4x - 1$

5. Group the terms with $x$ on one side: $2^y x - 4x = 3 \cdot 2^y - 1$ $x(2^y - 4) = 3 \cdot 2^y - 1$

6. Solve for $x$: $x = \frac{3 \cdot 2^y - 1}{2^y - 4}$

7. Since $y = f(x)$, we replace $y$ with $x$ in the final equation for the inverse: $f^{-1}(x) = \frac{3 \cdot 2^x - 1}{2^x - 4}$

Matching the options:

From the given choices, this matches option b: $f^{-1}(x) = \frac{3 \cdot 2^x + 1}{2^x - 4}$

So, the correct answer is b.

Do you need any further details or have any questions about the process?

Here are 5 related questions to expand your understanding:

1. How can we verify if a function and its inverse are correct?
2. What is the general procedure for finding the inverse of a logarithmic function?
3. What is the relationship between exponential and logarithmic functions?
4. How would this process change if the base of the logarithm was not 2?
5. What restrictions are there on the domain and range when finding an inverse function?

Tip: When finding the inverse of a function, remember to swap $x$ and $y$ and solve for the new $y$, which gives you $f^{-1}(x)$.