Let's walk through the problem:

Given the function: $f(x) = (x - 8)^2$

1. Finding a domain where $f$ is one-to-one and non-decreasing:

   • $f(x) = (x - 8)^2$ is a quadratic function with its vertex at $x = 8$.
   • The function is symmetric about the vertex $x = 8$.
   • Quadratic functions like this are increasing on the interval $[8, \infty)$ and decreasing on $(-\infty, 8]$.
   • To make the function one-to-one (a requirement for inverse functions), we need to restrict the domain to an interval where the function is either strictly increasing or decreasing.

   Answer for the first part:

   • A valid domain is $[8, \infty)$ because the function is non-decreasing (increasing) on this interval.

2. Finding the inverse of $f$ restricted to this domain:

   • Start with: $y = (x - 8)^2$
   • To solve for $x$, take the square root of both sides: $\sqrt{y} = |x - 8|$
   • Since we are only considering the interval $[8, \infty)$, $x - 8 \geq 0$, so: $\sqrt{y} = x - 8$
   • Solving for $x$: $x = \sqrt{y} + 8$
   • Now, switch $x$ and $y$ to get the inverse function: $f^{-1}(x) = \sqrt{x} + 8$

   Answer for the second part:
   $f^{-1}(x) = \sqrt{x} + 8$

Let me know if you need more details or further clarifications!

Here are five related questions to explore further:

1. What would the domain of the inverse function $f^{-1}(x) = \sqrt{x} + 8$ be?
2. How does restricting the domain affect the behavior of quadratic functions?
3. How can you verify whether a function is one-to-one?
4. What happens if the domain is restricted to $(-\infty, 8]$ instead?
5. How do you find the inverse of functions that are not quadratic?

Tip: Always check the derivative of a function to determine if it is increasing or decreasing on a given interval.