The question in the image asks the following:

1. Let $f(x) = (x - 1)^2$.
   • Find a domain where $f$ is one-to-one and non-decreasing.
   • Find the inverse of $f$ restricted to this domain.

Step 1: Finding a Domain Where $f(x)$ is One-to-One and Non-Decreasing

The function $f(x) = (x - 1)^2$ is a parabola, which opens upwards (since the coefficient of $x^2$ is positive). A parabola is not one-to-one over its entire domain because it is symmetric around its vertex. The vertex of $f(x) = (x - 1)^2$ is at $x = 1$.

To make the function one-to-one, we need to restrict the domain to either the left or right of the vertex. Since the question asks for a domain where the function is non-decreasing, we choose the interval where $f(x)$ is increasing. This happens for $x \geq 1$.

Thus, the domain on which $f(x)$ is one-to-one and non-decreasing is: $x \geq 1$

Step 2: Finding the Inverse of $f(x)$ on the Restricted Domain

To find the inverse, we start with the equation for $f(x)$: $y = (x - 1)^2$ Now, solve for $x$ in terms of $y$: $\sqrt{y} = x - 1$ $x = \sqrt{y} + 1$ Since the domain is restricted to $x \geq 1$, we take the positive square root.

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

Final Answer:

1. The domain where $f$ is one-to-one and non-decreasing: $x \geq 1$.
2. The inverse of $f$ restricted to this domain: $f^{-1}(x) = \sqrt{x} + 1$.

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

1. How do we determine the vertex of a quadratic function?
2. Why is a function not one-to-one over its entire domain if it is symmetric?
3. Can the domain restriction be $x \leq 1$ instead, and if so, how would the inverse change?
4. What is the general method for finding the inverse of a function?
5. How does restricting the domain of a function affect its graph?

Tip: When finding the inverse of a quadratic function, always be mindful of domain restrictions, as a quadratic function is not naturally one-to-one.