Find \( f+g, f-g \), \( f g \), and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=\sqrt{x-9}, g(x)=\sqrt{9-x} \) \( (f+g)(x)=\sqrt{x-9}+\sqrt{9-x} \) What is the domain of \( f+g \) ? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. The domain of \( f+g \) is \( \{9\} \). (Use a comma to separate answers as needed.) B. The domain of \( f+g \) is (Type your answer in interval notation.) The domain of \( f+g \) is \( \varnothing \). \( (f-g)(x)=\square \)

To find \( (f-g)(x) \), you simply subtract the two functions: \[ (f-g)(x) = \sqrt{x-9} - \sqrt{9-x}. \] Now, let's determine its domain. For \( f(x) = \sqrt{x - 9} \) to be defined, \( x - 9 \geq 0 \), which gives \( x \geq 9 \). For \( g(x) = \sqrt{9 - x} \) to be defined, \( 9 - x \geq 0 \), which gives \( x \leq 9 \). Thus, the only value that satisfies both is \( x = 9 \), making the domain \( \{9\} \). Next, let's compute \( fg(x) \): \[ (fg)(x) = \sqrt{x-9} \cdot \sqrt{9-x} = \sqrt{(x-9)(9-x)}. \] To find the domain for \( fg(x) \), we can use the same reasoning as before. Both \( \sqrt{x-9} \) and \( \sqrt{9-x} \) must be defined, which limits us to \( x = 9 \). The output is zero at this point. Hence, the domain for \( fg \) is also \( \{9\} \). Finally, let's consider \( \frac{f}{g} \): \[ \frac{f}{g}(x) = \frac{\sqrt{x-9}}{\sqrt{9-x}}. \] Here, \( g(x) \) must not be zero, meaning \( \sqrt{9-x} \) can't equal zero, which means \( x \) cannot be \( 9 \). Therefore, the domain for \( \frac{f}{g} \) is \( \varnothing \), because there's no value \( x \) that can fulfill the requirement. In summary: \[ (f+g)(x) = \sqrt{x-9} + \sqrt{9-x}, \text{ domain: } \{9\}. \] \[ (f-g)(x) = \sqrt{x-9} - \sqrt{9-x}, \text{ domain: } \{9\}. \] \[ (fg)(x) = \sqrt{(x-9)(9-x)}, \text{ domain: } \{9\}. \] \[ \frac{f}{g}(x) = \frac{\sqrt{x-9}}{\sqrt{9-x}}, \text{ domain: } \varnothing. \]