To find the domain (or field of definition) of the function $f(x) = \sqrt{x - 1} - \sqrt[3]{2 - x}$, we need to consider the constraints for which the function is defined.

1. First term: $\sqrt{x - 1}$

• For the square root to be defined, the expression inside the square root must be non-negative.
• $x - 1 \geq 0$
• Solving this inequality: $x \geq 1$ So, $x \geq 1$ is a condition for the first term.

2. Second term: $\sqrt[3]{2 - x}$

• The cube root is defined for all real numbers, meaning there are no restrictions on $x$ for this term.

Combining the conditions:

• The only restriction is $x \geq 1$ from the first term.

Thus, the domain of $f(x)$ is: $\boxed{[1, \infty)}$

Would you like more details or have any questions?

Here are five related questions:

1. How would the domain change if it were a fourth root instead of a square root?
2. What is the range of this function given its domain?
3. How do cube roots differ from square roots in terms of their domains and ranges?
4. Can you graph this function and interpret the behavior for large values of $x$?
5. What happens to the function when $x = 1$ or $x = 2$?

Tip: When dealing with square roots or any even roots, always check that the argument is non-negative.