Given functions \( f \) and \( g \), find \( (a)(f \circ g)(x) \) and its domain, and \( (b)(g \circ f)(x) \) and its domain. \( f f(x)=\sqrt{x}, g(x)=x+10 \) (a) \( (f \circ g)(x)=\square \) (Simplify your answer. Type an exact answer, using radicals as needed.) The domain of \( f \circ g \) is \( \square \). (Type your answer in interval notation.) (b) (g of)(x) \( =\square \) (Simplify your answer. Type an exact answers, using radicals as needed.) The domain of \( g \) of is \( \square \). (Type your answer in interval notation.)

To find \( (f \circ g)(x) \) and its domain, we first compute \( f(g(x)) \). Given that \( g(x) = x + 10 \), we have: \[ f(g(x)) = f(x + 10) = \sqrt{x + 10} \] Now, the domain of \( f \) is restricted because \( f(x) = \sqrt{x} \), which is defined only for \( x \geq 0 \). Thus, we need: \[ x + 10 \geq 0 \implies x \geq -10 \] So, we write this in interval notation as \( [-10, \infty) \). For part (b), we need to find \( (g \circ f)(x) \), which involves computing \( g(f(x)) \): \[ g(f(x)) = g(\sqrt{x}) = \sqrt{x} + 10 \] The domain of \( g \) is unrestricted since it is simply \( g(x) = x + 10 \). However, for \( f(x) = \sqrt{x} \), the function is only defined for \( x \geq 0 \). Therefore, the domain of \( g \circ f \) is \( [0, \infty) \). To summarize: (a) \( (f \circ g)(x) = \sqrt{x + 10} \) The domain of \( f \circ g \) is \( [-10, \infty) \). (b) \( (g \circ f)(x) = \sqrt{x} + 10 \) The domain of \( g \circ f \) is \( [0, \infty) \).