To find the inverse function $g^{-1}(x)$ for the given function $g(x) = \left( \frac{6x - 7}{2x + 3} \right)^2$, we need to follow these steps:

Step-by-step solution:

1. Start with the equation for $g(x)$: $y = \left( \frac{6x - 7}{2x + 3} \right)^2$ We want to solve for $x$ in terms of $y$, and then swap $x$ and $y$ to get $g^{-1}(x)$.

2. Take the square root of both sides: $\sqrt{y} = \frac{6x - 7}{2x + 3}$ (Note that technically, we would need $\pm \sqrt{y}$, but we'll handle this by considering the valid domain at the end).

3. Cross-multiply to eliminate the fraction: $\sqrt{y}(2x + 3) = 6x - 7$

4. Expand the left-hand side: $2x\sqrt{y} + 3\sqrt{y} = 6x - 7$

5. Group the terms involving $x$ on one side: $2x\sqrt{y} - 6x = -7 - 3\sqrt{y}$

6. Factor out $x$ on the left-hand side: $x(2\sqrt{y} - 6) = -7 - 3\sqrt{y}$

7. Solve for $x$: $x = \frac{-7 - 3\sqrt{y}}{2\sqrt{y} - 6}$

8. Replace $y$ with $x$ to express the inverse: $g^{-1}(x) = \frac{-7 - 3\sqrt{x}}{2\sqrt{x} - 6}$

Thus, the inverse function is:

$g^{-1}(x) = \frac{-7 - 3\sqrt{x}}{2\sqrt{x} - 6}$

Would you like more details or further explanation? Here are some related questions to deepen your understanding:

1. How do we determine the domain and range of $g(x)$ and $g^{-1}(x)$?
2. What happens if we include both $+\sqrt{x}$ and $-\sqrt{x}$ when solving for the inverse?
3. Can the inverse function be simplified further?
4. How do we verify that the inverse we found is correct?
5. What role does the denominator play in limiting the domain of $g^{-1}(x)$?

Tip: When solving for the inverse of a function involving a square, always pay attention to domain restrictions, as not all values may be valid for the inverse.