For \( f(x)=\frac{x}{x+1} \) and \( g(x)=\frac{8}{x} \), find a. \( (f \circ g)(x) ; \quad \) b. the domain of \( f \circ g \) a. \( (f \circ g)(x)=\frac{8}{8+x} \) (Simplify your answer.) b. What is the domain of \( f \circ g \) ? The domain is \( \square \). (Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the expressic

To find the composition \( (f \circ g)(x) \), we start by substituting \( g(x) \) into \( f(x) \). So: \[ f(g(x)) = f\left(\frac{8}{x}\right) = \frac{\frac{8}{x}}{\frac{8}{x}+1} \] To simplify, we need a common denominator in the denominator's fraction: \[ = \frac{\frac{8}{x}}{\frac{8 + x}{x}} = \frac{8}{8 + x} \] Thus, \( (f \circ g)(x) = \frac{8}{8+x} \). For the domain of \( f \circ g \), we first consider the domain of \( g(x) = \frac{8}{x} \), which is \( x \neq 0 \). Next, since \( f(x) \) requires its input to be valid, we also need to check \( g(x) \). We set up the conditions: 1. \( x \neq 0 \) (to satisfy \( g(x) \)). 2. \( \frac{8}{x} + 1 \neq 0 \Rightarrow \frac{8 + x}{x} \neq 0 \Rightarrow x \neq -8 \). Therefore, the domain for \( f \circ g \) combines both conditions as: \[ \text{Domain: } (-\infty, -8) \cup (-8, 0) \cup (0, \infty). \]