To solve this problem:

1. Understand the question: We need to calculate $(g \circ f)^{-1}(-2)$, where $f(x) = 3x + 1$ and $g(x) = \frac{2x-1}{x+3}$.

2. Step 1 - Compute $g \circ f(x)$: Substituting $f(x) = 3x + 1$ into $g(x)$, we get: $g(f(x)) = g(3x + 1) = \frac{2(3x + 1) - 1}{(3x + 1) + 3}$ Simplify: $g(f(x)) = \frac{6x + 2 - 1}{3x + 4} = \frac{6x + 1}{3x + 4}.$

3. Step 2 - Find $(g \circ f)^{-1}(x)$: To find the inverse of $g \circ f(x)$, let: $y = \frac{6x + 1}{3x + 4}.$ Swap $x$ and $y$, then solve for $y$: $x = \frac{6y + 1}{3y + 4}.$ Multiply through by $3y + 4$: $x(3y + 4) = 6y + 1.$ Expand and rearrange: $3xy + 4x = 6y + 1 \implies 3xy - 6y = 1 - 4x.$ Factor out $y$: $y(3x - 6) = 1 - 4x \implies y = \frac{1 - 4x}{3x - 6}.$ So, $(g \circ f)^{-1}(x) = \frac{1 - 4x}{3x - 6}$.

4. Step 3 - Calculate $(g \circ f)^{-1}(-2)$: Substitute $x = -2$ into the inverse function: $(g \circ f)^{-1}(-2) = \frac{1 - 4(-2)}{3(-2) - 6}.$ Simplify: $(g \circ f)^{-1}(-2) = \frac{1 + 8}{-6 - 6} = \frac{9}{-12} = -\frac{3}{4}.$

The correct answer is a. -3/4.

Let me know if you would like further clarification or additional explanations. Here are five related questions:

1. What is the domain of $g \circ f(x)$ based on the given functions?
2. How do we verify the inverse of a composite function?
3. What happens if $g(x)$ is not defined for certain values?
4. How does the composition of functions affect their inverses?
5. How can we use graphing to verify our solution?

Tip: Always check for restrictions in the domain when working with