Let's go through the problem step by step.

The given function:

$y = f(x) = -2x^3 - 9$

a) Find $\frac{df}{dx}$ at $x = -2$:

To solve this, we first need to compute the derivative of $f(x)$.

$f(x) = -2x^3 - 9$ The derivative of $f(x)$ is: $f'(x) = \frac{d}{dx}[-2x^3 - 9] = -6x^2$ Now, evaluate this at $x = -2$: $f'(-2) = -6(-2)^2 = -6(4) = -24$

Thus, $\frac{df}{dx}$ at $x = -2$ is $-24$.

b) Find a formula for $x = f^{-1}(y)$:

To find the inverse function, $f^{-1}(y)$, we first express $x$ in terms of $y$.

Starting with: $y = -2x^3 - 9$ Solve for $x$: $y + 9 = -2x^3$ $x^3 = \frac{-(y + 9)}{2}$ $x = \left( \frac{-(y + 9)}{2} \right)^{\frac{1}{3}}$

Thus, the inverse function is: $f^{-1}(y) = \left( \frac{-(y + 9)}{2} \right)^{\frac{1}{3}}$

c) Find $\frac{df^{-1}}{dy}$ at $y = f(-2)$:

From part a, we know that $f(-2) = -2(-2)^3 - 9 = -2(-8) - 9 = 16 - 9 = 7$. So we need to evaluate $\frac{df^{-1}}{dy}$ at $y = 7$.

To find $\frac{df^{-1}}{dy}$, we use the relationship between the derivative of the inverse function and the original function: $\frac{df^{-1}}{dy} = \frac{1}{f'(x)}$ At $y = f(-2) = 7$, $x = -2$, and from part a, we found that $f'(-2) = -24$.

Thus: $\frac{df^{-1}}{dy}\bigg|_{y=7} = \frac{1}{-24}$

So, $\frac{df^{-1}}{dy} = \frac{-1}{24}$ at $y = 7$.

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

Here are some related questions you might find useful:

1. What is the geometric interpretation of $f^{-1}(y)$?
2. How do we compute derivatives of higher-order inverse functions?
3. How does the derivative of the inverse relate to the original function's slope?
4. What are some properties of cubic functions and their inverses?
5. How can we verify if a function is invertible?

Tip: When finding the derivative of an inverse function, remember that $\frac{df^{-1}}{dy} = \frac{1}{f'(x)}$, where $x = f^{-1}(y)$.