Let's break down this problem step by step:

We are given two functions:

• $f(x) = \sqrt{x - 1}$
• $g(x) = \sqrt{x - 2}$

1. Domain of $f(x)$ and $g(x)$:

For the square root functions, the expressions under the square root must be non-negative.

• For $f(x) = \sqrt{x - 1}$, we need $x - 1 \geq 0$, so $x \geq 1$. Therefore, the domain of $f(x)$ is $[1, \infty)$.

• For $g(x) = \sqrt{x - 2}$, we need $x - 2 \geq 0$, so $x \geq 2$. Therefore, the domain of $g(x)$ is $[2, \infty)$.

2. Operations on $f$ and $g$:

We now find the domain for the following operations: $f + g$, $f - g$, $f \cdot g$, and $\frac{f}{g}$.

• For $f + g$ and $f - g$:

  The domain will be the intersection of the domains of $f(x)$ and $g(x)$, which is $[2, \infty)$.

• For $f \cdot g$:

  The domain is also the intersection of $[1, \infty)$ and $[2, \infty)$, which is $[2, \infty)$.

• For $\frac{f}{g}$:

  The domain is $[2, \infty)$, but we must exclude any points where $g(x) = 0$ because division by zero is undefined. However, $g(x) = \sqrt{x - 2}$ is only zero when $x = 2$. So, the domain of $\frac{f}{g}$ is $(2, \infty)$.

Final answer:

• The domain of $f + g$, $f - g$, and $f \cdot g$ is $[2, \infty)$.
• The domain of $\frac{f}{g}$ is $(2, \infty)$.

It looks like the answer marked in the image (for $\frac{f}{g}$) is correct: $(2, \infty)$.

Would you like more details or have any further questions?

5 Related Questions:

1. How do you find the domain of a square root function?
2. Why does $g(x) = \sqrt{x - 2}$ impact the domain of $\frac{f}{g}$?
3. What happens if $g(x) = 0$ when dividing functions?
4. Can domains of different types of functions overlap in more complex cases?
5. How would the domain change if $f(x)$ or $g(x)$ included additional terms (like a polynomial)?

Tip: Always check for division by zero when determining the domain of rational functions!