\( 1 \leftarrow \) For \( f(x)=\frac{x}{x-4} \) and \( g(x)=-\frac{9}{x} \), find the following composite functions and state the domain of each. \( \begin{array}{llll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f & \text { (d) } g \circ g\end{array} \) (a) \( (f \circ g)(x)=\square \) (Simplify your answer. Use integers or fractions for any numbers in the expression.)

To find the composite functions \( f \circ g \) and the others, we first start with their definitions and the functions provided: 1. **Calculating \( f \circ g \)**: \[ (f \circ g)(x) = f(g(x)) = f\left(-\frac{9}{x}\right) \] Now substituting \( g(x) \) into \( f(x) \): \[ f\left(-\frac{9}{x}\right) = \frac{-\frac{9}{x}}{-\frac{9}{x} - 4} = \frac{-\frac{9}{x}}{-\frac{9 + 4x}{x}} = \frac{-9}{-9 - 4x} = \frac{9}{9 + 4x} \] Therefore, \[ (f \circ g)(x) = \frac{9}{9 + 4x} \] 2. **Finding the domain of \( f \circ g \)**: - The function \( g(x) \) is defined for all \( x \neq 0 \). - The function \( f(x) \) is defined as long as \( x \neq 4 \). - Here \( g(x) = -\frac{9}{x} \) must not equal 4, i.e., \( g(x) \neq 4 \). - Thus, solving \( -\frac{9}{x} = 4 \) gives \( x = -\frac{9}{4} \). So the domain for \( f \circ g \) is \( x \in \mathbb{R}, x \neq 0, x \neq -\frac{9}{4} \). --- 3. **Calculating \( g \circ f \)**: \[ (g \circ f)(x) = g(f(x)) = g\left(\frac{x}{x-4}\right) \] Now substituting \( f(x) \) into \( g(x) \): \[ g\left(\frac{x}{x-4}\right) = -\frac{9}{\frac{x}{x-4}} = -\frac{9(x-4)}{x} = -\frac{9x - 36}{x} = -9 + \frac{36}{x} \] Therefore, \[ (g \circ f)(x) = -9 + \frac{36}{x} \] 4. **Finding the domain of \( g \circ f \)**: - The function \( f(x) \) is defined for \( x \neq 4 \). - The function \( g(x) \) is defined as long as \( g(x) \neq 0 \). In this case, \( g(f(x)) \) is defined when \( \frac{x}{x-4} \neq 0 \), which holds for all \( x \neq 0 \). - The restrictions are that \( x \neq 4 \) and \( x \neq 0 \). So the domain for \( g \circ f \) is \( x \in \mathbb{R}, x \neq 0, x \neq 4 \). --- 5. **Calculating \( f \circ f \)**: \[ (f \circ f)(x) = f(f(x)) = f\left(\frac{x}{x-4}\right) \] Now substituting into \( f(x) \): \[ f\left(\frac{x}{x-4}\right) = \frac{\frac{x}{x-4}}{\frac{x}{x-4} - 4} = \frac{\frac{x}{x-4}}{\frac{x - 4(x-4)}{x-4}} = \frac{\frac{x}{x-4}}{\frac{-3x + 16}{x-4}} = \frac{x}{-3x + 16} \] Therefore, \[ (f \circ f)(x) = \frac{x}{-3x + 16} \] 6. **Finding the domain of \( f \circ f \)**: - The initial restriction is \( x \neq 4 \). - We also check for when \( f(f(x)) \) is undefined (denominator equals zero): \( -3x + 16 \neq 0 \) implies \( x \neq \frac{16}{3} \). Thus, the domain for \( f \circ f \) is \(