The given function is $f(x) = (x - 8)^2$.

Step 1: Identify when $f(x)$ is one-to-one and non-decreasing.

To ensure that a function is one-to-one and non-decreasing, its derivative must be non-negative ($f'(x) \geq 0$).

Derivative of $f(x)$:

$f'(x) = 2(x - 8)$ The derivative $f'(x) = 2(x - 8)$ is non-negative when $x \geq 8$.

Thus, $f(x)$ is non-decreasing for $x \geq 8$. We will restrict the domain to $x \geq 8$ to make $f(x)$ one-to-one and non-decreasing.

Step 2: Restrict the domain of $f(x)$.

The domain of $f(x)$ where it is one-to-one and non-decreasing is $[8, \infty)$.

Step 3: Find the inverse of $f(x)$ on this restricted domain.

To find the inverse of $f(x)$, follow these steps:

1. Start with $y = (x - 8)^2$.

2. Solve for $x$ in terms of $y$: $\sqrt{y} = x - 8$ $x = \sqrt{y} + 8$

   Since we are considering the domain $x \geq 8$, we only take the positive square root.

Final Answer:

• Domain: $[8, \infty)$
• Inverse function: $f^{-1}(x) = \sqrt{x} + 8$, for $x \geq 0$.

Would you like further details or have any questions?

Here are some related questions to explore:

1. What does it mean for a function to be one-to-one?
2. How does restricting a domain affect the invertibility of a function?
3. Why did we choose the positive square root when finding the inverse?
4. What are the conditions for a function to have an inverse?
5. How would the inverse change if the function had a different form?

Tip: When solving for an inverse, always check if the original function needs domain restrictions to ensure it is one-to-one.