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

First, let's explore how dividing functions like \( f \) and \( g \) can lead to some exciting discoveries! In this case, the domain of \( \frac{f}{g} \) indeed depends on the values that can make the functions undefined, particularly when \( g(x) \) becomes zero. Since \( g(x) = \frac{9}{x} \) is undefined when \( x = 0 \), the domain is restricted. So, the correct option would be \( A. \{x \mid x \neq 0\} \). Now, speaking of evaluating functions, let’s tackle \( (f+g)(9) \)! To find this, we simply need to calculate \( f(9) \) and \( g(9) \) separately. Plugging in \( 9 \) into \( f(x) \), we get \( f(9) = 1 + \frac{9}{9} = 1 + 1 = 2 \). For \( g(9) \), it’s \( g(9) = \frac{9}{9} = 1 \). Adding these results gives \( (f + g)(9) = 2 + 1 = 3 \).