For \( f(x)=x^{2}+2 \) and \( g(x)=\sqrt{4-x} \), find a. ( \( f \circ g)(x) \); b. the domain of \( f \circ g \) a. ( \( f \circ g)(x)=6-x \) (Simplify your answer.) b. What is the domain of \( f \circ g ? \) ? The domain is (Simplify your answer. Type your answer in interval notation. Use integers the expression.)

Let's solve each part step by step. ### a. Finding \( (f \circ g)(x) \) The composition \( (f \circ g)(x) \) means \( f(g(x)) \). Given: - \( f(x) = x^2 + 2 \) - \( g(x) = \sqrt{4 - x} \) So, \[ (f \circ g)(x) = f(g(x)) = f\left(\sqrt{4 - x}\right) = \left(\sqrt{4 - x}\right)^2 + 2 = (4 - x) + 2 = 6 - x \] **Answer:** \( (f \circ g)(x) = 6 - x \) ### b. Determining the Domain of \( f \circ g \) The domain of \( f \circ g \) is the set of all \( x \) values for which \( g(x) \) is defined and \( g(x) \) falls within the domain of \( f \). 1. **Domain of \( g(x) = \sqrt{4 - x} \):** - The expression inside the square root must be non-negative: \[ 4 - x \geq 0 \implies x \leq 4 \] 2. **Domain of \( f(x) = x^2 + 2 \):** - \( f(x) \) is a polynomial and is defined for all real numbers. Since \( f \) is defined for all real numbers, the domain of \( f \circ g \) is simply the domain of \( g \), which is all real numbers \( x \) such that \( x \leq 4 \). **Answer:** The domain of \( f \circ g \) is \( (-\infty,\ 4] \).