The problem states:

Let $f$ and $g$ be functions that are differentiable everywhere. If $g$ is the inverse function of $f$ and $g(-2) = 5$, and $f'(5) = -\frac{1}{2}$, then find $g'(-2)$.

Solution:

Since $g$ is the inverse of $f$, we use the formula for the derivative of the inverse function: $g'(x) = \frac{1}{f'(g(x))}$

Given:

• $g(-2) = 5$
• $f'(5) = -\frac{1}{2}$

Substitute $x = -2$ into the inverse derivative formula:

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

Since $g(-2) = 5$, this simplifies to:

$g'(-2) = \frac{1}{f'(5)}$

Substituting $f'(5) = -\frac{1}{2}$:

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

Answer:

$\boxed{-2}$

Would you like more details or have any other questions?

Here are some related questions to explore:

1. How do you find the derivative of an inverse function in general?
2. What is the geometric interpretation of the inverse function's derivative?
3. How do we interpret the signs of $f'(x)$ and $g'(x)$ in terms of function behavior?
4. What are the conditions for a function to have an inverse function?
5. Can the formula $g'(x) = \frac{1}{f'(g(x))}$ be used if $f'(g(x)) = 0$? Why or why not?

Tip: Always check that the derivative $f'(g(x))$ is non-zero when using the inverse function derivative formula to avoid undefined expressions.