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 \( f \circ g \) of the functions \( f(x) = \frac{5}{x+2} \) and \( g(x) = \frac{1}{x-2} \), we need to substitute the expression for \( g(x) \) into \( f(x) \) and simplify. Given: \( f(x) = \frac{5}{x+2} \) \( g(x) = \frac{1}{x-2} \) Substitute \( g(x) \) into \( f(x) \): \( (f \circ g)(x) = f(g(x)) = \frac{5}{g(x)+2} \) Substitute \( g(x) = \frac{1}{x-2} \) into the expression: \( (f \circ g)(x) = \frac{5}{\frac{1}{x-2}+2} \) Simplify the expression: \( (f \circ g)(x) = \frac{5}{\frac{1}{x-2}+2} = \frac{5}{\frac{1+2(x-2)}{x-2}} = \frac{5}{\frac{1+2x-4}{x-2}} = \frac{5}{\frac{2x-3}{x-2}} = \frac{5(x-2)}{2x-3} \) Therefore, the composition \( f \circ g \) is \( \frac{5(x-2)}{2x-3} \). To find the domain of \( f \circ g \), we need to ensure that the denominator of the composition is not equal to zero. The denominator is \( 2x-3 \), so the domain of \( f \circ g \) is all real numbers except when \( 2x-3 = 0 \), which gives \( x = \frac{3}{2} \). Therefore, the domain of \( f \circ g \) is \( x \neq \frac{3}{2} \).