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

Let's analyze the functions \( f(x) = \sqrt{x} \) and \( g(x) = x - 9 \) by performing various operations on them: addition, subtraction, multiplication, and division. We'll also determine the domain for each resulting function. ### 1. Addition: \( f + g \) \[ (f + g)(x) = f(x) + g(x) = \sqrt{x} + (x - 9) = \sqrt{x} + x - 9 \] **Domain of \( f + g \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \). Therefore, the domain of \( f + g \) is the intersection of the domains of \( f \) and \( g \): \[ \text{Domain of } f + g = [0, \infty) \] **Answer:** **B.** The domain of \( f + g \) is \( [0, \infty) \). --- ### 2. Subtraction: \( f - g \) \[ (f - g)(x) = f(x) - g(x) = \sqrt{x} - (x - 9) = \sqrt{x} - x + 9 \] **Domain of \( f - g \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** is defined for all real numbers \( x \). Therefore, the domain of \( f - g \) is also the intersection of the domains of \( f \) and \( g \): \[ \text{Domain of } f - g = [0, \infty) \] **Answer:** **A.** The domain of \( f - g \) is \( [0, \infty) \). --- ### 3. Multiplication: \( f \cdot g \) \[ (fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot (x - 9) = \sqrt{x}(x - 9) \] **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 \( x \). Therefore, the domain of \( fg \) is: \[ \text{Domain of } fg = [0, \infty) \] --- ### 4. Division: \( \frac{f}{g} \) \[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x - 9} \] **Domain of \( \frac{f}{g} \):** - **\( f(x) = \sqrt{x} \)** is defined for \( x \geq 0 \). - **\( g(x) = x - 9 \)** cannot be zero to avoid division by zero. So, \( x - 9 \neq 0 \Rightarrow x \neq 9 \). Combining these conditions: \[ \text{Domain of } \frac{f}{g} = [0, 9) \cup (9, \infty) \] --- ### Summary of Results 1. **\( f + g \):** - **Expression:** \( \sqrt{x} + x - 9 \) - **Domain:** \( [0, \infty) \) 2. **\( f - g \):** - **Expression:** \( \sqrt{x} - x + 9 \) - **Domain:** \( [0, \infty) \) 3. **\( fg \):** - **Expression:** \( \sqrt{x}(x - 9) \) - **Domain:** \( [0, \infty) \) 4. **\( \frac{f}{g} \):** - **Expression:** \( \frac{\sqrt{x}}{x - 9} \) - **Domain:** \( [0, 9) \cup (9, \infty) \)