Find \( f+g, f-g, f g \), and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=\sqrt{x-9}, g(x)=\sqrt{9-x} \) \( (f+g)(x)=\square \)

To find \( f+g, f-g, fg, \) and \( \frac{f}{g} \) for the functions \( f(x) = \sqrt{x-9} \) and \( g(x) = \sqrt{9-x} \), we will proceed step by step. ### Step 1: Find \( f+g \) \[ (f+g)(x) = f(x) + g(x) = \sqrt{x-9} + \sqrt{9-x} \] ### Step 2: Find \( f-g \) \[ (f-g)(x) = f(x) - g(x) = \sqrt{x-9} - \sqrt{9-x} \] ### Step 3: Find \( fg \) \[ (fg)(x) = f(x) \cdot g(x) = \sqrt{x-9} \cdot \sqrt{9-x} = \sqrt{(x-9)(9-x)} \] ### Step 4: Find \( \frac{f}{g} \) \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x-9}}{\sqrt{9-x}} \] ### Step 5: Determine the domain for each function 1. **Domain of \( f(x) = \sqrt{x-9} \)**: - The expression under the square root must be non-negative: \[ x - 9 \geq 0 \implies x \geq 9 \] - Therefore, the domain of \( f \) is \( [9, \infty) \). 2. **Domain of \( g(x) = \sqrt{9-x} \)**: - The expression under the square root must be non-negative: \[ 9 - x \geq 0 \implies x \leq 9 \] - Therefore, the domain of \( g \) is \( (-\infty, 9] \). 3. **Domain of \( f+g \)**: - The domain of \( f+g \) is the intersection of the domains of \( f \) and \( g \): \[ [9, \infty) \cap (-\infty, 9] = \{9\} \] - Therefore, the domain of \( f+g \) is \( \{9\} \). 4. **Domain of \( f-g \)**: - The domain of \( f-g \) is also the intersection of the domains of \( f \) and \( g \): \[ [9, \infty) \cap (-\infty, 9] = \{9\} \] - Therefore, the domain of \( f-g \) is \( \{9\} \). 5. **Domain of \( fg \)**: - The expression \( (x-9)(9-x) \) must be non-negative: \[ (x-9)(9-x) \geq 0 \] - This is true when \( x = 9 \) or \( x = 9 \) (the product is zero). Thus, the domain of \( fg \) is \( \{9\} \). 6. **Domain of \( \frac{f}{g} \)**: - The denominator \( g(x) \) must be non-zero: \[ \sqrt{9-x} \neq 0 \implies 9-x \neq 0 \implies x \neq 9 \] - The domain of \( \frac{f}{g} \) is \( (-\infty, 9) \cup (9, \infty) \). ### Summary of Results - \( (f+g)(x) = \sqrt{x-9} + \sqrt{9-x} \) with domain \( \{9\} \) - \( (f-g)(x) = \sqrt{x-9} - \sqrt{9-x} \) with domain \( \{9\} \) - \( (fg)(x) = \sqrt{(x-9)(9-x)} \) with domain \( \{9\} \) - \( \left(\frac{f}{g}\right)(x) = \frac{\sqrt{x-9}}{\sqrt{9-x}} \) with domain \( (-\infty, 9) \cup (9, \infty) \)