For \( f(x)=\frac{9}{x-5} \) and \( g(x)=\frac{2}{x} \), 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. Substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{9}{\frac{2}{x} - 5} \] 2. Simplify the expression: \[ f(g(x)) = \frac{9}{\frac{2 - 5x}{x}} = \frac{9x}{2 - 5x} \] **Domain of \( f \circ g \)**: - The function \( g(x) = \frac{2}{x} \) is defined for \( x \neq 0 \). - The expression \( \frac{2}{x} - 5 \) must not be zero: \[ \frac{2}{x} - 5 \neq 0 \implies \frac{2}{x} \neq 5 \implies 2 \neq 5x \implies x \neq \frac{2}{5} \] - Therefore, the domain of \( f \circ g \) is \( x \in \mathbb{R} \setminus \{0, \frac{2}{5}\} \). ### (b) \( g \circ f \) The composite function \( g \circ f \) is defined as \( g(f(x)) \). 1. Substitute \( f(x) \) into \( g(x) \): \[ g(f(x)) = g\left(\frac{9}{x-5}\right) = \frac{2}{\frac{9}{x-5}} = \frac{2(x-5)}{9} \] **Domain of \( g \circ f \)**: - The function \( f(x) = \frac{9}{x-5} \) is defined for \( x \neq 5 \). - The expression \( \frac{9}{x-5} \) must not be zero: \[ \frac{9}{x-5} \neq 0 \implies 9 \neq 0 \text{ (always true)} \] - Therefore, the domain of \( g \circ f \) is \( x \in \mathbb{R} \setminus \{5\} \). ### (c) \( f \circ f \) The composite function \( f \circ f \) is defined as \( f(f(x)) \). 1. Substitute \( f(x) \) into itself: \[ f(f(x)) = f\left(\frac{9}{x-5}\right) = \frac{9}{\frac{9}{x-5} - 5} \] 2. Simplify the expression: \[ f(f(x)) = \frac{9}{\frac{9 - 5(x-5)}{x-5}} = \frac{9(x-5)}{9 - 5x + 25} = \frac{9(x-5)}{34 - 5x} \] **Domain of \( f \circ f \)**: - The function \( f(x) \) is defined for \( x \neq 5 \). - The expression \( \frac{9}{\frac{9}{x-5} - 5} \) must not be zero: \[ \frac{9}{\frac{9 - 5(x-5)}{x-5}} \neq 0 \implies 9 \neq 0 \text{ (always true)} \] - The denominator \( \frac{9}{x-5} - 5 \) must not be zero: \[ \frac{9 - 5(x-5)}{x-5} \neq 0 \implies 9 - 5(x-5) \neq 0 \implies 9 \neq 5x - 25 \implies 5x \neq 34 \implies x \neq \frac{34}{5} \] - Therefore, the domain of \( f \circ f \) is \( x \in \mathbb{R} \setminus \{5, \frac{34}{5}\} \). ### (d) \( g \circ g \)