To solve the given problem, let’s analyze it step by step.

1. Find a domain where $f(x) = (x - 3)^2$ is one-to-one and non-decreasing:

• The function $f(x) = (x - 3)^2$ is a parabola opening upwards. It is symmetric about $x = 3$. To make it one-to-one, we must restrict the domain to a part where it is either increasing or decreasing.

• Since the question asks for $f(x)$ to be non-decreasing, we restrict the domain to the right of the vertex $x = 3$, where $f(x)$ is increasing. Thus: $\text{Domain: } x \geq 3$

2. Find the inverse $f^{-1}(x)$ restricted to this domain:

• To find the inverse, start with $y = (x - 3)^2$. Swap $x$ and $y$ to get: $x = (y - 3)^2$

• Solve for $y$: $y - 3 = \sqrt{x} \quad \text{(since we are considering the non-decreasing branch, take the positive square root)}$ $y = \sqrt{x} + 3$

Thus, the inverse function is: $f^{-1}(x) = \sqrt{x} + 3, \quad \text{for } x \geq 0.$

Final Answers:

1. Domain where $f(x)$ is one-to-one and non-decreasing: $x \geq 3$

2. Inverse function $f^{-1}(x)$: $f^{-1}(x) = \sqrt{x} + 3, \quad \text{for } x \geq 0.$

Would you like a detailed explanation or a graph of the function to clarify further?

Here are five related questions to expand your understanding:

1. What would the domain and range of $f(x) = (x-3)^2$ be without restrictions?
2. How does restricting the domain of a function make it invertible?
3. What happens if you restrict $f(x)$ to $x \leq 3$ instead?
4. Why do we only take the positive square root when finding the inverse?
5. Can you verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ for the restricted domain and range?

Tip: Always check for symmetry and monotonicity when determining invertibility of functions.