Find \( f+g, f-g, f g \) and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=\frac{3 x+8}{x^{2}-81}, g(x)=\frac{2 x-1}{x^{2}-81} \) A. The domain of \( f+g \) is \( (-\infty,-9) \cup(-9,9) \cup(9, \infty) \) (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 \( \varnothing \). What is the domain of \( f-g \) ? ( \( f-g)(x)=\frac{1}{x-9} \) (Simplify your answer.) The domain of \( f-g \) is \( (-\infty,-9) \cup(-9,9) \cup(9, \infty) \). (Type your answer in interval notation.) B. The domain of \( f-g \) is \( \{ \). (Use a comma to separate answers as needed.) C. The domain of \( f-g \) is \( \varnothing \). (fg)(x) \( =\square \) (Simplify your answer.)

To solve the problem, we will first find \( f+g \), \( f-g \), \( fg \), and \( \frac{f}{g} \) for the given functions \( f(x) \) and \( g(x) \). We will also determine the domain for each function. Given: \[ f(x) = \frac{3x + 8}{x^2 - 81}, \quad g(x) = \frac{2x - 1}{x^2 - 81} \] ### Step 1: Find \( f + g \) \[ f + g = \frac{3x + 8}{x^2 - 81} + \frac{2x - 1}{x^2 - 81} = \frac{(3x + 8) + (2x - 1)}{x^2 - 81} = \frac{5x + 7}{x^2 - 81} \] ### Step 2: Find the domain of \( f + g \) The domain of \( f + g \) is determined by the denominator \( x^2 - 81 \neq 0 \): \[ x^2 - 81 = 0 \implies x^2 = 81 \implies x = \pm 9 \] Thus, the domain is: \[ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \] ### Step 3: Find \( f - g \) \[ f - g = \frac{3x + 8}{x^2 - 81} - \frac{2x - 1}{x^2 - 81} = \frac{(3x + 8) - (2x - 1)}{x^2 - 81} = \frac{x + 9}{x^2 - 81} \] ### Step 4: Find the domain of \( f - g \) The domain of \( f - g \) is the same as that of \( f + g \): \[ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \] ### Step 5: Find \( fg \) \[ fg = f \cdot g = \left(\frac{3x + 8}{x^2 - 81}\right) \cdot \left(\frac{2x - 1}{x^2 - 81}\right) = \frac{(3x + 8)(2x - 1)}{(x^2 - 81)^2} \] ### Step 6: Find the domain of \( fg \) The domain of \( fg \) is also determined by the denominator \( (x^2 - 81)^2 \neq 0 \): \[ x^2 - 81 = 0 \implies x = \pm 9 \] Thus, the domain is: \[ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \] ### Step 7: Find \( \frac{f}{g} \) \[ \frac{f}{g} = \frac{\frac{3x + 8}{x^2 - 81}}{\frac{2x - 1}{x^2 - 81}} = \frac{3x + 8}{2x - 1} \] ### Step 8: Find the domain of \( \frac{f}{g} \) The domain of \( \frac{f}{g} \) is determined by both the denominator \( 2x - 1 \neq 0 \) and \( x^2 - 81 \neq 0 \): 1. \( 2x - 1 = 0 \implies x = \frac{1}{2} \) 2. \( x^2 - 81 = 0 \implies x = \pm 9 \) Thus, the domain is: \[ (-\infty, -9) \cup (-9, \frac{1}{2}) \cup (\frac{1}{2}, 9) \cup (9, \infty) \] ### Summary of Results 1. \( f + g = \frac{5x + 7}{x^2 - 81} \) - Domain: \( (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \) 2. \( f - g = \frac{x + 9}{x^2 - 81} \) - Domain: \( (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \) 3. \( fg = \frac{(3x + 8)(2x - 1)}{(x^2 - 81)^2} \) - Domain: \( (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \) 4. \( \frac{f}{g} = \frac{3x + 8}{2x - 1} \) - Domain: \( (-\infty, -9) \cup (-9, \frac{1}{2}) \cup (\frac{1}{2}, 9) \cup (9, \infty) \)