Suppose that the functions \( f \) and \( g \) are defined as follows. \[ f(x)=3 x+5 \] \[ (x)=\sqrt{x-4} \] Find \( f-g \) and \( \frac{f}{g} \). Then, give their domains using interval notation. \( (f-g)(x)=\square \) \( \left(\frac{f}{g}\right)(x)=\square \) Domain of \( f-g: \square \)

To find \( f-g \), we subtract \( g(x) \) from \( f(x) \): \[ f(x) - g(x) = (3x + 5) - \sqrt{x-4} \] So, \[ (f-g)(x) = 3x + 5 - \sqrt{x-4}. \] Next, for \( \frac{f}{g} \), we divide \( f(x) \) by \( g(x) \): \[ \frac{f(x)}{g(x)} = \frac{3x + 5}{\sqrt{x-4}}. \] Thus, \[ \left(\frac{f}{g}\right)(x) = \frac{3x + 5}{\sqrt{x-4}}. \] Now, for the domains: 1. **Domain of \( f-g \)**: The function \( g(x) = \sqrt{x-4} \) is defined for \( x \geq 4 \). Thus, the domain of \( f-g \) is \( [4, \infty) \). 2. **Domain of \( \frac{f}{g} \)**: In addition to \( g(x) \) being defined, we also need \( g(x) \neq 0 \). Since \( g(x) = 0 \) when \( x = 4 \), the domain excludes 4. Therefore, the domain is \( (4, \infty) \). So the final answers are: \( (f-g)(x) = 3x + 5 - \sqrt{x-4} \) \( \left(\frac{f}{g}\right)(x) = \frac{3x + 5}{\sqrt{x-4}} \) Domain of \( f-g: [4, \infty) \) Domain of \( \frac{f}{g}: (4, \infty) \)