To find the inverse of the given function $f(x) = (x - 1)^2$ with the domain $(-\infty, 1]$:

Steps:

1. Write the function: $y = (x - 1)^2$

2. Switch $x$ and $y$: $x = (y - 1)^2$

3. Solve for $y$: Take the square root on both sides: \sqrt{x} = y - 1 \quad \text{(since domain is restricted to (-\infty, 1]$, use the negative square root for$y)}

   $y = 1 - \sqrt{x}$

4. Conclusion: The inverse function is: $f^{-1}(x) = 1 - \sqrt{x}$

Thus, the correct answer is $f^{-1}(x) = 1 - \sqrt{x}$.

Let me know if you need further clarification!

Related Questions:

1. What happens to the range of a function when we invert it?
2. Why is the square root negative in this case?
3. How does the domain of $f(x)$ affect the inverse function?
4. Can we graph both $f(x)$ and its inverse to confirm correctness?
5. What are other scenarios where the domain affects the inverse process?

Tip: Always check the domain and range before inverting a function!