For the functions \( f(x)=\frac{5}{x+2} \) and \( g(x)=\frac{1}{x-2} \), find the composition \( f \circ g \) and simp notation. \( (f \circ g)(x)=\square \) Domain of \( f \circ g: \square \)

To find the composition \( f \circ g \), we substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{1}{x-2}\right) = \frac{5}{\frac{1}{x-2} + 2} \] Calculating the denominator: \[ \frac{1}{x-2} + 2 = \frac{1 + 2(x-2)}{x-2} = \frac{1 + 2x - 4}{x-2} = \frac{2x - 3}{x-2} \] Now, substituting back into \( f(g(x)) \): \[ f(g(x)) = \frac{5}{\frac{2x - 3}{x-2}} = \frac{5(x-2)}{2x - 3} \] Thus, we have: \[ (f \circ g)(x) = \frac{5(x-2)}{2x - 3} \] For the domain: 1. The domain of \( g(x) \) requires \( x - 2 \neq 0 \) \(\Rightarrow x \neq 2\). 2. The input to \( f \) (which is \( g(x) \)) must be defined, thus: \(\frac{1}{x - 2} + 2 \neq 0 \Rightarrow \frac{1}{x - 2} \neq -2 \Rightarrow x - 2 \neq -\frac{1}{2} \Rightarrow x \neq \frac{3}{2}\). Combining these, the domain of \( f \circ g \) is: \[ \text{Domain of } f \circ g: x \in \mathbb{R} \setminus \{2, \frac{3}{2}\} \]