To find the correct expression for $g'(-8)$, where $g$ is the inverse function of $f$, we'll use the property of derivatives of inverse functions:

$g'(a) = \frac{1}{f'(g(a))}$

Here, $f(x) = x^3 - 6x^2 - 12x + 8$ and it's given that $f(4) = -8$. Therefore, $g(-8) = 4$.

We first calculate $f'(x)$, then evaluate $f'(4)$ to use in the expression for $g'(-8)$.

1. Differentiate $f(x)$: $f(x) = x^3 - 6x^2 - 12x + 8$ $f'(x) = 3x^2 - 12x - 12$

2. Evaluate $f'(4)$: $f'(4) = 3(4)^2 - 12(4) - 12 = 3 \times 16 - 48 - 12 = 48 - 48 - 12 = -12$

Using the property of the derivative of the inverse function: $g'(-8) = \frac{1}{f'(4)} = \frac{1}{-12} = -\frac{1}{12}$

Now we review the given options to match our result. The correct expression is: $g'(-8) = \frac{1}{f'(4)}$

The most closely matching option: $g'(-8) = \frac{1}{f'(4)}$ But the options provided all use a positive denominator or are misstated in terms of the specific value. Thus, the actual correct expression should be $\frac{1}{-12}$, but this is not an option provided. It seems the correct expression should be a variant of $\frac{1}{f'(4)}$, taking into account that $f'(4) = -12$.

Would you like to review any specific calculation or concept in more detail, or do you have any other questions on this topic? Here are some related questions you might consider:

1. How would you graph $f(x)$ based on its derivative?
2. Why is the derivative of the inverse function given by $\frac{1}{f'(g(a))}$?
3. What are the implications of $f'(x)$ being negative at $x = 4$?
4. How would you determine the intervals where $f(x)$ is increasing or decreasing?
5. How does the behavior of $f(x)$ around $x = 4$ affect the graph of $g(x)$?

Tip: Always remember to check the signs when computing derivatives and evaluating them at specific points to avoid errors in subsequent calculations.