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. \( (f \circ g)(x)=\square \) (Simplify your answer.)

To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{8}{x}\right) = \frac{\frac{8}{x}}{\frac{8}{x} + 1} \] Now, we simplify the right side: \[ = \frac{\frac{8}{x}}{\frac{8}{x} + \frac{x}{x}} = \frac{\frac{8}{x}}{\frac{8 + x}{x}} = \frac{8}{8 + x} \] So, \[ (f \circ g)(x) = \frac{8}{8 + x} \] Next, we need to find the domain of \( f \circ g \). The domain of \( g(x) = \frac{8}{x} \) is \( x \neq 0 \) because we cannot divide by zero. For \( f(x) \) to be defined at \( g(x) \), \( g(x) \) should also not cause the denominator of \( f(x) = \frac{x}{x + 1} \) to be zero. Therefore, we need \( g(x) + 1 \neq 0 \): \[ \frac{8}{x} + 1 \neq 0 \implies \frac{8 + x}{x} \neq 0 \implies 8 + x \neq 0 \implies x \neq -8 \] Thus, the domain of \( f \circ g \) is: \[ \text{Domain: } x \in \mathbb{R}, x \neq 0, x \neq -8 \] So, the complete answer is: \[ (f \circ g)(x) = \frac{8}{8 + x} \] Domain: \( x \in \mathbb{R}, x \neq 0, x \neq -8 \)