For \( f(x)=x^{2}+6 \) and \( g(x)=\sqrt{5-x} \), find a. \( (f \circ g)(x) ; \quad \) b. the domain of \( f \circ g \)

To solve the problem, we will follow these steps: 1. **Find \( (f \circ g)(x) \)**, which means we will substitute \( g(x) \) into \( f(x) \). 2. **Determine the domain of \( f \circ g \)** by considering the domain of \( g(x) \) and ensuring that the output of \( g(x) \) is within the domain of \( f(x) \). ### Step 1: Find \( (f \circ g)(x) \) The composition \( (f \circ g)(x) \) is defined as \( f(g(x)) \). Given: - \( f(x) = x^2 + 6 \) - \( g(x) = \sqrt{5 - x} \) Now, we substitute \( g(x) \) into \( f(x) \): \[ (f \circ g)(x) = f(g(x)) = f(\sqrt{5 - x}) = (\sqrt{5 - x})^2 + 6 \] Calculating this gives: \[ (f \circ g)(x) = 5 - x + 6 = 11 - x \] ### Step 2: Determine the domain of \( f \circ g \) To find the domain of \( f \circ g \), we need to consider the domain of \( g(x) \) first. The function \( g(x) = \sqrt{5 - x} \) is defined when the expression under the square root is non-negative: \[ 5 - x \geq 0 \implies x \leq 5 \] Thus, the domain of \( g(x) \) is \( (-\infty, 5] \). Next, we need to ensure that the output of \( g(x) \) is within the domain of \( f(x) \). The function \( f(x) = x^2 + 6 \) is defined for all real numbers, so there are no additional restrictions from \( f(x) \). Therefore, the domain of \( f \circ g \) is the same as the domain of \( g(x) \): \[ \text{Domain of } (f \circ g) = (-\infty, 5] \] ### Final Answers a. \( (f \circ g)(x) = 11 - x \) b. The domain of \( f \circ g \) is \( (-\infty, 5] \)