For \( f(x)=\frac{6}{x-5} \) and \( g(x)=\frac{7}{x} \), find the following composite functions and state the domain of each. \( \begin{array}{llll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f & \text { (d) } g \circ g\end{array} \) \( \begin{array}{l}\text { (a) }(f \circ g)(x)=\square \\ \text { (Simplify your answer. Use integers or fractions for any numbers in the expression.) }\end{array} \)

To find the composite functions \( f \circ g \), \( g \circ f \), \( f \circ f \), and \( g \circ g \), we will follow these steps: 1. **Calculate each composite function.** 2. **Determine the domain of each composite function.** ### (a) \( (f \circ g)(x) \) The composite function \( (f \circ g)(x) \) is defined as \( f(g(x)) \). 1. Substitute \( g(x) \) into \( f(x) \): \[ g(x) = \frac{7}{x} \] Therefore, \[ f(g(x)) = f\left(\frac{7}{x}\right) = \frac{6}{\frac{7}{x} - 5} \] 2. Simplify the expression: \[ f\left(\frac{7}{x}\right) = \frac{6}{\frac{7 - 5x}{x}} = \frac{6x}{7 - 5x} \] Thus, \[ (f \circ g)(x) = \frac{6x}{7 - 5x} \] ### Domain of \( f \circ g \) The domain of \( f \circ g \) is determined by the restrictions from both \( g(x) \) and \( f(x) \): - \( g(x) \) is undefined when \( x = 0 \). - \( f(g(x)) \) is undefined when \( g(x) = \frac{7}{x} = 5 \), which gives \( x = \frac{7}{5} \). Thus, the domain of \( f \circ g \) is: \[ \text{Domain: } x \in \mathbb{R} \setminus \{0, \frac{7}{5}\} \] ### (b) \( (g \circ f)(x) \) The composite function \( (g \circ f)(x) \) is defined as \( g(f(x)) \). 1. Substitute \( f(x) \) into \( g(x) \): \[ f(x) = \frac{6}{x - 5} \] Therefore, \[ g(f(x)) = g\left(\frac{6}{x - 5}\right) = \frac{7}{\frac{6}{x - 5}} = \frac{7(x - 5)}{6} \] Thus, \[ (g \circ f)(x) = \frac{7(x - 5)}{6} = \frac{7x - 35}{6} \] ### Domain of \( g \circ f \) The domain of \( g \circ f \) is determined by the restrictions from both \( f(x) \) and \( g(x) \): - \( f(x) \) is undefined when \( x = 5 \). - \( g(f(x)) \) is defined for all \( f(x) \) except when \( f(x) = 0 \), which does not occur since \( f(x) \) is never zero. Thus, the domain of \( g \circ f \) is: \[ \text{Domain: } x \in \mathbb{R} \setminus \{5\} \] ### (c) \( (f \circ f)(x) \) The composite function \( (f \circ f)(x) \) is defined as \( f(f(x)) \). 1. Substitute \( f(x) \) into itself: \[ f(f(x)) = f\left(\frac{6}{x - 5}\right) = \frac{6}{\frac{6}{x - 5} - 5} \] 2. Simplify the expression: \[ f\left(\frac{6}{x - 5}\right) = \frac{6}{\frac{6 - 5(x - 5)}{x - 5}} = \frac{6(x - 5)}{6 - 5x + 25} = \frac{6(x - 5)}{31 - 5x} \] Thus, \[ (f \circ f)(x) = \frac{6(x - 5)}{31 - 5x} \] ### Domain of \( f \circ f \) The domain of \( f \circ f \) is determined by the restrictions from \( f(x) \): - \( f(x) \) is undefined when \( x = 5 \). - \( f(f(x)) \) is undefined when \( f(x) = 5 \), which gives \( x = 6 \). Thus, the domain of \( f \circ f \) is: \[ \text{Domain: } x \in \mathbb{R} \setminus \{5, 6\} \] ### (d) \( (g \circ g)(x) \) The composite function \( (g \circ g)(x) \) is defined as \( g(g(x)) \). 1. Substitute \( g(x) \) into itself: \[ g(g(x)) = g\left(\frac{7}{x}\right) = \frac{7}{\frac{7}{x}} = x \] Thus, \[ (g \circ g)(x) = x \] ### Domain of \( g \circ g \) The domain of \( g \circ g \) is determined by the restrictions from \( g(x) \): - \( g(x) \) is undefined when \( x = 0 \). Thus, the domain of \( g \circ g \) is: \[ \text{Domain: } x \in \mathbb{R} \setminus \{0\} \] ### Summary of Results - \( (f \circ g)(x) = \frac{6x}{7 - 5x} \) with domain \( x \in \mathbb{R} \setminus \{0, \frac{7}{5}\} \) - \( (g \circ f)(x) = \frac{7x - 35}{6} \) with domain \( x \in \mathbb{R} \setminus \{5\} \) - \( (f \circ f)(x) = \frac{6(x - 5)}{31 - 5x} \) with domain \( x \in \mathbb{R} \setminus \{5, 6\} \) - \( (g \circ g)(x) = x \) with domain \( x \in \mathbb{R} \setminus \{0\} \)