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

To find \( f + g \), \( f - g \), \( fg \), and \( \frac{f}{g} \): 1. **Sum**: \[ f + g = \sqrt{x} + (x - 9) \] 2. **Difference**: \[ f - g = \sqrt{x} - (x - 9) \] 3. **Product**: \[ fg = \sqrt{x} \cdot (x - 9) = x \sqrt{x} - 9\sqrt{x} \] 4. **Quotient**: \[ \frac{f}{g} = \frac{\sqrt{x}}{x - 9} \] Next, we determine the domains for each function: For \( f(x) = \sqrt{x} \), the domain is \( [0, \infty) \) since the square root function is defined for non-negative values. For \( g(x) = x - 9 \), the domain is \( (-\infty, \infty) \) since there are no restrictions. For \( f + g \) and \( f - g \), they inherit the domain of \( f(x) \), which is \( [0, \infty) \). For \( fg\), the domain is also \( [0, \infty) \) as \( \sqrt{x} \) is defined there. Now, for \( \frac{f}{g} \), we must consider where the denominator does not equal zero. Thus: - \( g(x) = x - 9 \neq 0 \) implies \( x \neq 9 \). - The domain is constrained by \( f(x) \) and the restriction \( x \neq 9\). Combining these, the domain of \( \frac{f}{g} \) is \( [0, 9) \cup (9, \infty) \). To summarize: A. The domain of \( \frac{f}{g} \) is \( [0, 9) \cup (9, \infty) \). B. The domain of \( \frac{f}{g} \) is \( [0, 9), (9, \infty) \).