The given function $f(x)$ is defined as:

$f(x) = -x^3 - 3x + 2$ and $f(1) = -2$. We are asked to find the value of $g'(-2)$, where $g$ is the inverse function of $f$.

To find $g'(-2)$, we use the formula for the derivative of the inverse function:

$g'(y) = \frac{1}{f'(x)}$ where $f(x) = y$. In this case, $f(1) = -2$, so we know $x = 1$ when $y = -2$. Thus, we need to compute $f'(1)$, the derivative of $f(x)$ evaluated at $x = 1$.

Step 1: Find $f'(x)$

The derivative of $f(x)$ is:

$f'(x) = \frac{d}{dx} \left( -x^3 - 3x + 2 \right) = -3x^2 - 3$

Step 2: Evaluate $f'(1)$

Now, substitute $x = 1$ into the derivative:

$f'(1) = -3(1)^2 - 3 = -3 - 3 = -6$

Step 3: Find $g'(-2)$

Finally, using the formula for the derivative of the inverse function:

$g'(-2) = \frac{1}{f'(1)} = \frac{1}{-6} = -\frac{1}{6}$

Thus, the value of $g'(-2)$ is:

$g'(-2) = -\frac{1}{6}$

Would you like further details on any part of the solution?

Here are 5 related questions for further exploration:

1. How do you find the inverse of a function algebraically?
2. What conditions must a function satisfy to have an inverse?
3. How do you interpret the meaning of the derivative of an inverse function?
4. What is the graphical relationship between a function and its inverse?
5. How would you compute the inverse of the function $f(x) = -x^3 - 3x + 2$ if possible?

Tip: The derivative of the inverse function tells us the rate of change of the original function’s output with respect to its input, but from the perspective of the inverse function.