\( 1 \leftarrow \quad \) Find \( f+g, f-g, f g \), and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=\sqrt{x-7} ; g(x)=\sqrt{x-9} \) \( (f+g)(x)=\square \)

To find \( (f+g)(x) \), we combine the two functions: \[ (f+g)(x) = f(x) + g(x) = \sqrt{x-7} + \sqrt{x-9} \] Now, let's determine the domain. Function \( f(x) \) is defined when \( x - 7 \geq 0 \) (or \( x \geq 7 \)), and function \( g(x) \) is defined when \( x - 9 \geq 0 \) (or \( x \geq 9 \)). Thus, the combined function \( (f+g)(x) \) is only defined for \( x \geq 9 \). For \( (f-g)(x) \): \[ (f-g)(x) = f(x) - g(x) = \sqrt{x-7} - \sqrt{x-9} \] The domain remains \( x \geq 9 \) for the same reason as before. For \( (fg)(x) \): \[ (fg)(x) = f(x) \cdot g(x) = \sqrt{x-7} \cdot \sqrt{x-9} = \sqrt{(x-7)(x-9)} \] The domain is again \( x \geq 9 \). For \( \left(\frac{f}{g}\right)(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x-7}}{\sqrt{x-9}} \] This function is defined when \( g(x) \neq 0 \), which occurs when \( x > 9 \), so the domain is \( x > 9 \). In summary: \((f + g)(x) = \sqrt{x-7} + \sqrt{x-9}\), defined for \(x \geq 9\). \((f - g)(x) = \sqrt{x-7} - \sqrt{x-9}\), defined for \(x \geq 9\). \((fg)(x) = \sqrt{(x-7)(x-9)}\), defined for \(x \geq 9\). \(\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x-7}}{\sqrt{x-9}}\), defined for \(x > 9\).