For the given functions \( f \) and \( g \), complete parts (a)-(h). For parts (a)-(d) also find the domain. \[ f(x)=1+\frac{9}{x} ; g(x)=\frac{9}{x} \] B. The domain is \( \{x \mid x \) is any real number \( \} \). (d) Find \( \left(\frac{f}{g}\right)(x) \). \[ \left(\frac{f}{g}\right)(x)=\frac{x+9}{9} \] What is the domain of \( \frac{f}{g} \) ? Select the correct choice below and, if recessary, fill in the answer box to complete your choice. A. The domain is \( \{x \mid x \neq 0\} \) (Use integers or fractions for any numbers in the expression.) B. The domain is \( \{x \mid x \) is any real number \( \} \). (e) Find \( (f+g)(9) \). \[ (f+g)(9)=\square \] (Type an integer or a simplified fraction.) Clear all

Let's solve the problem step by step, starting with the functions provided: 1. **Functions:** \[ f(x) = 1 + \frac{9}{x}, \quad g(x) = \frac{9}{x} \] 2. **Domain of \( f(x) \) and \( g(x) \):** - The function \( f(x) \) is defined for all \( x \) except \( x = 0 \) (since division by zero is undefined). - The function \( g(x) \) is also defined for all \( x \) except \( x = 0 \). Therefore, the domain for both functions is: \[ \{x \mid x \neq 0\} \] 3. **Finding \( \left(\frac{f}{g}\right)(x) \):** \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{1 + \frac{9}{x}}{\frac{9}{x}} = \frac{x + 9}{9} \] 4. **Domain of \( \frac{f}{g} \):** - The function \( \frac{f}{g} \) is also undefined at \( x = 0 \) (due to \( g(x) \) in the denominator). - Thus, the domain of \( \frac{f}{g} \) is: \[ \{x \mid x \neq 0\} \] The correct choice is **A**. 5. **Finding \( (f+g)(9) \):** \[ (f+g)(x) = f(x) + g(x) = \left(1 + \frac{9}{x}\right) + \frac{9}{x} = 1 + \frac{9}{x} + \frac{9}{x} = 1 + \frac{18}{x} \] Now, substituting \( x = 9 \): \[ (f+g)(9) = 1 + \frac{18}{9} = 1 + 2 = 3 \] Thus, the final answers are: - The domain of \( \frac{f}{g} \) is \( \{x \mid x \neq 0\} \) (Choice A). - \( (f+g)(9) = 3 \).