To solve for $g'(1)$ where $g(x) = f^{-1}(x)$, we can use the inverse function derivative formula.

Given:

1. $f(x) = x^3 + 4x + 1$.
2. $g(x) = f^{-1}(x)$ and $g(1) = 0$.

Goal:

Find $g'(1)$.

Solution:

Since $g(x) = f^{-1}(x)$, the derivative of $g(x)$ at a point $x = a$ can be expressed as: $g'(a) = \frac{1}{f'(g(a))}$

Step 1: Identify $g(1)$

We are given that $g(1) = 0$. This implies that $f(0) = 1$, so we need $f'(0)$ to continue.

Step 2: Calculate $f'(x)$

The derivative of $f(x)$ is: $f'(x) = 3x^2 + 4$

Step 3: Evaluate $f'(g(1)) = f'(0)$

Substitute $x = 0$ into $f'(x)$: $f'(0) = 3(0)^2 + 4 = 4$

Step 4: Find $g'(1)$

Now, substitute into the formula for $g'(1)$: $g'(1) = \frac{1}{f'(0)} = \frac{1}{4}$

Answer:

$g'(1) = \frac{1}{4}$

Would you like more details on any of these steps, or do you have further questions?

Here are some related questions you might explore:

1. What happens to $g'(x)$ if $f(x)$ is a different polynomial?
2. How would the answer change if $g(1)$ were not equal to $0$?
3. What is the significance of using the inverse function derivative formula?
4. Can we verify this result by computing $g(x)$ explicitly?
5. How does this method generalize for non-polynomial functions?

Tip: For inverse functions, remember that $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ is a quick way to find derivatives without solving for the inverse explicitly.