For \( f(x)=\frac{x}{x-5} \) and \( g(x)=\frac{-9}{x} \), find the following composite functions and state the domain of each. \( \begin{array}{llll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f & \text { (d) } g \circ g\end{array} \)

To find the composite functions \( f \circ g \), \( g \circ f \), \( f \circ f \), and \( g \circ g \), we will first compute each composite function and then determine the domain for each. ### (a) \( f \circ g \) The composite function \( f \circ g \) is defined as \( f(g(x)) \). 1. Start with \( g(x) = \frac{-9}{x} \). 2. Substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{-9}{x}\right) = \frac{\frac{-9}{x}}{\frac{-9}{x} - 5} \] 3. Simplify the denominator: \[ \frac{-9}{x} - 5 = \frac{-9 - 5x}{x} \] 4. Thus, \[ f(g(x)) = \frac{\frac{-9}{x}}{\frac{-9 - 5x}{x}} = \frac{-9}{-9 - 5x} = \frac{9}{9 + 5x} \] **Domain of \( f \circ g \)**: - \( g(x) \) is undefined for \( x = 0 \). - \( f(g(x)) \) is undefined when \( 9 + 5x = 0 \) (i.e., \( x = -\frac{9}{5} \)). - Therefore, the domain of \( f \circ g \) is \( x \in \mathbb{R} \setminus \{0, -\frac{9}{5}\} \). ### (b) \( g \circ f \) The composite function \( g \circ f \) is defined as \( g(f(x)) \). 1. Start with \( f(x) = \frac{x}{x-5} \). 2. Substitute \( f(x) \) into \( g(x) \): \[ g(f(x)) = g\left(\frac{x}{x-5}\right) = \frac{-9}{\frac{x}{x-5}} = \frac{-9(x-5)}{x} = \frac{-9x + 45}{x} \] **Domain of \( g \circ f \)**: - \( f(x) \) is undefined for \( x = 5 \). - \( g(f(x)) \) is undefined for \( x = 0 \) (since \( g(x) \) is undefined for \( x = 0 \)). - Therefore, the domain of \( g \circ f \) is \( x \in \mathbb{R} \setminus \{0, 5\} \). ### (c) \( f \circ f \) The composite function \( f \circ f \) is defined as \( f(f(x)) \). 1. Substitute \( f(x) \) into itself: \[ f(f(x)) = f\left(\frac{x}{x-5}\right) = \frac{\frac{x}{x-5}}{\frac{x}{x-5} - 5} \] 2. Simplify the denominator: \[ \frac{x}{x-5} - 5 = \frac{x - 5(x-5)}{x-5} = \frac{x - 5x + 25}{x-5} = \frac{-4x + 25}{x-5} \] 3. Thus, \[ f(f(x)) = \frac{\frac{x}{x-5}}{\frac{-4x + 25}{x-5}} = \frac{x}{-4x + 25} \] **Domain of \( f \circ f \)**: - \( f(x) \) is undefined for \( x = 5 \). - \( f(f(x)) \) is undefined when \( -4x + 25 = 0 \) (i.e., \( x = \frac{25}{4} \)). - Therefore, the domain of \( f \circ f \) is \( x \in \mathbb{R} \setminus \{5, \frac{25}{4}\} \). ### (d) \( g \circ g \) The composite function \( g \circ g \) is defined as \( g(g(x)) \). 1. Substitute \( g(x) \) into itself: \[ g(g(x)) = g\left(\frac{-9}{x}\right) = \frac{-9}{\frac{-9}{x}} = \frac{-9 \