To solve the problem, we need to find the inverse of $f(x) = (x-1)^2 + 6$ for the domain $(-\infty, 1]$. Let us proceed step by step.

1. Replace $f(x)$ with $y$: $y = (x-1)^2 + 6$

2. Switch $x$ and $y$ to find the inverse: $x = (y-1)^2 + 6$

3. Solve for $y$:

   • Subtract 6 from both sides: $x - 6 = (y-1)^2$
   • Take the square root of both sides. Since the domain is $(-\infty, 1]$, $y-1$ must be non-positive, so we use the negative square root: $y - 1 = -\sqrt{x - 6}$
   • Add 1 to both sides: $y = 1 - \sqrt{x - 6}$

Thus, the inverse function is: $f^{-1}(x) = 1 - \sqrt{x - 6}$

Looking at the answer choices, the correct option is: $\boxed{f^{-1}(x) = 1 - \sqrt{x - 6}}$

Let me know if you have any questions or need further clarification!

Here are 5 related questions for deeper understanding:

1. How do domain restrictions affect the calculation of inverse functions?
2. Why is the negative square root chosen here for the inverse function?
3. What happens to the inverse function if the domain changes?
4. How can you verify that the inverse function is correct?
5. How would this problem change if the function were $f(x) = (x-1)^2 + 6$ with a domain of $[1, \infty)$?

Tip: Always pay close attention to the domain when finding the inverse of a function, as it determines which part of the function you invert.