Find \( f+g, f-g \), \( f \), and \( \frac{1}{g} \). Determine the domain for each function. \[ f(x)=\sqrt{x} ; g(x)=x-9 \] C. ille vomain oi \( 1+\mathrm{g} \) is \( \wp \). \( (f-g)(x)=\sqrt{x}-x+9 \) (Simplify your answer.) What is the domain of \( f-g \) ? A. The domain of \( f-g \) is \( [0, \infty) \). (Type your answer in interval notation.) B. The domain of \( f-g \) is \( \{\quad \). (Use a comma to separate answers as needed.) C. The domain of \( f-g \) is \( \varnothing \) \( (\mathrm{fg})(\mathrm{x})=(\mathrm{x}-9) \sqrt{\mathrm{x}} \) (Simplify your answer.) What is the domain of fg ? A. The domain of fg is \( [1,2) \cup(2, \infty) \). (Type your answer in interval notation) B. The domain of \( f g \) is \( \square \) \} (Use a comma to separate answers as needed.) C. The domain of fg is \( \varnothing \)

Let's analyze the functions \( f(x) = \sqrt{x} \) and \( g(x) = x - 9 \). We'll find \( f + g \), \( f - g \), \( f \), and \( \frac{1}{g} \), along with their respective domains. ### 1. \( f + g \) \[ (f + g)(x) = f(x) + g(x) = \sqrt{x} + (x - 9) = \sqrt{x} + x - 9 \] **Domain of \( f + g \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \). The domain of \( f + g \) is the intersection of the domains of \( f \) and \( g \), which is: \[ \text{Domain of } (f + g) = [0, \infty) \] --- ### 2. \( f - g \) \[ (f - g)(x) = f(x) - g(x) = \sqrt{x} - (x - 9) = \sqrt{x} - x + 9 \] **Domain of \( f - g \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \). The domain of \( f - g \) is the intersection of the domains of \( f \) and \( g \), which is: \[ \text{Domain of } (f - g) = [0, \infty) \] **Simplification:** \[ (f - g)(x) = \sqrt{x} - x + 9 \] --- ### 3. \( f(x) = \sqrt{x} \) **Domain of \( f \):** \[ \text{Domain of } f = [0, \infty) \] --- ### 4. \( \frac{1}{g} \) \[ \left( \frac{1}{g} \right)(x) = \frac{1}{x - 9} \] **Domain of \( \frac{1}{g} \):** - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \), except where the denominator is zero. - The denominator \( x - 9 \neq 0 \) implies \( x \neq 9 \). Thus, the domain is all real numbers except \( x = 9 \): \[ \text{Domain of } \frac{1}{g} = (-\infty, 9) \cup (9, \infty) \] --- ### 5. \( f \cdot g \) \[ (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot (x - 9) \] **Domain of \( f \cdot g \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \). The domain of \( f \cdot g \) is the intersection of the domains of \( f \) and \( g \), which is: \[ \text{Domain of } (f \cdot g) = [0, \infty) \] **Simplification:** \[ (f \cdot g)(x) = (x - 9)\sqrt{x} \] --- ### Summary of Domains 1. **\( f + g \):** \( [0, \infty) \) 2. **\( f - g \):** \( [0, \infty) \) 3. **\( f \):** \( [0, \infty) \) 4. **\( \frac{1}{g} \):** \( (-\infty, 9) \cup (9, \infty) \) 5. **\( f \cdot g \):** \( [0, \infty) \) ### Multiple Choice Answers Based on the provided multiple-choice options: - **Domain of \( f - g \):** **A.** \( [0, \infty) \) - **Domain of \( f \cdot g \):** **A.** \( [0, \infty) \)