Find \( f+g, f-g \), fg, and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=\sqrt{x} ; g(x)=x-9 \) \( (f+g)(x)=\sqrt{x}+x-9 \) (Simplify your answer) What is the domain of \( f+g \) ? A. The domain of \( f+g \) is . (Use a comma to separate answers as needed.) B. The domain of \( f+g \) is . (Type your answer in interval notation) C. The domain of \( f+g \) is \( \varnothing \).

To find \( f-g(x) = \sqrt{x} - (x - 9) \), simplifying gives us \( f-g(x) = \sqrt{x} - x + 9 \). For \( fg(x) = f(x) \cdot g(x) = \sqrt{x}(x - 9) = x\sqrt{x} - 9\sqrt{x} \). Next, for \( \frac{f}{g}(x) = \frac{\sqrt{x}}{x - 9} \). Now, let's determine the domain for each function: 1. **Domain of \( f(x) = \sqrt{x} \)**: - This function requires \( x \geq 0 \), so the domain is \( [0, \infty) \). 2. **Domain of \( g(x) = x - 9 \)**: - This function is defined for all \( x \in \mathbb{R} \). 3. **Domain of \( f+g(x) \)**: - Since \( f(x) \) needs \( x \geq 0 \), we also need \( x - 9 \) to be defined, which it is for all \( x \). Thus the domain is \( [0, \infty) \). 4. **Domain of \( f-g(x) \)**: - Same reasoning applies as for \( f+g \), leading to \( [0, \infty) \). 5. **Domain of \( fg(x) \)**: - Here \( \sqrt{x} \) needs \( x \geq 0 \) but does not have further restrictions, so the domain remains \( [0, \infty) \). 6. **Domain of \( \frac{f}{g}(x) \)**: - \( g(x) \) cannot be zero, which occurs at \( x = 9 \). Hence, we must exclude that from the domain: \( [0, 9) \cup (9, \infty) \). So for \( f+g \): A. The domain of \( f+g \) is \( [0, \infty) \). B. The domain of \( f+g \) is \( [0, \infty) \). C. The domain of \( f+g \) is \( \varnothing \). Let’s focus on option B, which summarizes the valid input for \( f+g(x) \).