For the functions \( f(x)=\frac{5}{x+2} \) and \( g(x)=\frac{1}{x-2} \), find the composition \( f \circ g \) and simplify your notation. \( \begin{array}{l}(f \circ g)(x)=\square \\ \text { Domain of } f \circ g: \square\end{array} \)

To find the composition \( f \circ g \), we need to substitute \( g(x) \) into \( f(x) \). First, we have: \[ g(x) = \frac{1}{x-2} \] Now, substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left( \frac{1}{x-2} \right) = \frac{5}{\frac{1}{x-2} + 2} \] To simplify the denominator: \[ \frac{1}{x-2} + 2 = \frac{1}{x-2} + \frac{2(x-2)}{x-2} = \frac{1 + 2(x-2)}{x-2} = \frac{1 + 2x - 4}{x-2} = \frac{2x - 3}{x-2} \] So now, we rewrite \( f(g(x)) \): \[ f(g(x)) = \frac{5}{\frac{2x - 3}{x-2}} = 5 \cdot \frac{x-2}{2x - 3} = \frac{5(x-2)}{2x - 3} \] Thus, we have: \[ (f \circ g)(x) = \frac{5(x-2)}{2x - 3} \] Next, we need to find the domain of \( f \circ g \). The domain of \( g(x) \) is \( x \neq 2 \) because of the denominator \( x-2 \). Additionally, for \( f(x) \), its domain excludes \( x = -2 \), but when we apply \( g \), we must check if \( g(x) = -2 \): \[ \frac{1}{x-2} = -2 \Rightarrow 1 = -2(x - 2) \Rightarrow 1 = -2x + 4 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2} \] Thus, \( x = \frac{3}{2} \) is another restriction for the domain. So, the complete domain of \( f \circ g \) is: \[ \text{Domain of } f \circ g: x \neq 2 \text{ and } x \neq \frac{3}{2} \] Final Result: \[ \begin{array}{l}(f \circ g)(x)=\frac{5(x-2)}{2x - 3} \\ \text { Domain of } f \circ g: x \neq 2 \text{ and } x \neq \frac{3}{2} \end{array} \]