Find \( f+g, f-g \), fg and \( \frac{1}{g} \) Determine the domain for each function. \( f(x)=5 x+8, g(x)=x+6 \) What is the domain of \( f+g \) ? A. The domain of \( f+g \) is \( \{\quad\} \) (Use a comma to separate answers as needed.) C. The domain of \( f+g \) is \( \varnothing \). ( \( f-g)(x)=4 x+2 \) (Simplify your answer.) What is the domain of \( f-g \) ? B. The domain of \( f-g \) is \( (-\infty, \infty) \). (Type your answer in interval notation.) C. Type your answer in interval notation.) C. The domain of \( f-g \) is \( \varnothing \) ( \( f g)(x)=\square \) (Simplify your answer.)

Let's analyze the given functions and perform the required operations step by step. **Given Functions:** - \( f(x) = 5x + 8 \) - \( g(x) = x + 6 \) ### 1. \( f + g \) **Calculation:** \[ (f + g)(x) = f(x) + g(x) = (5x + 8) + (x + 6) = 6x + 14 \] **Domain:** Both \( f(x) \) and \( g(x) \) are linear functions defined for all real numbers. \[ \text{Domain of } f + g: \quad (-\infty, \infty) \] **Answer:** **A.** The domain of \( f + g \) is \( (-\infty, \infty) \). --- ### 2. \( f - g \) **Calculation:** \[ (f - g)(x) = f(x) - g(x) = (5x + 8) - (x + 6) = 4x + 2 \] **Domain:** Both \( f(x) \) and \( g(x) \) are defined for all real numbers. \[ \text{Domain of } f - g: \quad (-\infty, \infty) \] **Answer:** **B.** The domain of \( f - g \) is \( (-\infty, \infty) \). --- ### 3. \( f \cdot g \) **Calculation:** \[ (fg)(x) = f(x) \cdot g(x) = (5x + 8)(x + 6) \] Expanding the expression: \[ = 5x \cdot x + 5x \cdot 6 + 8 \cdot x + 8 \cdot 6 = 5x^2 + 30x + 8x + 48 = 5x^2 + 38x + 48 \] **Domain:** Both \( f(x) \) and \( g(x) \) are defined for all real numbers. \[ \text{Domain of } f \cdot g: \quad (-\infty, \infty) \] **Answer:** \[ (fg)(x) = 5x^2 + 38x + 48 \] --- ### 4. \( \frac{1}{g} \) **Calculation:** \[ \frac{1}{g}(x) = \frac{1}{x + 6} \] **Domain:** The function \( \frac{1}{g}(x) \) is undefined when \( g(x) = 0 \): \[ x + 6 \ne 0 \implies x \ne -6 \] Thus, the domain excludes \( x = -6 \). \[ \text{Domain of } \frac{1}{g}: \quad (-\infty, -6) \cup (-6, \infty) \] **Answer:** \[ \frac{1}{g}(x) = \frac{1}{x + 6}, \quad \text{Domain: } (-\infty, -6) \cup (-6, \infty) \] --- ### Summary of Domains: - **\( f + g \):** \( (-\infty, \infty) \) - **\( f - g \):** \( (-\infty, \infty) \) - **\( f \cdot g \):** \( (-\infty, \infty) \) - **\( \frac{1}{g} \):** \( (-\infty, -6) \cup (-6, \infty) \) If you have specific multiple-choice options you'd like to verify, feel free to provide them, and I can help determine the correct answers based on the above explanations.