For \( \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+2 \) and \( \mathrm{g}(\mathrm{x})=\sqrt{4-\mathrm{x}} \), find a. \( (\mathrm{f} \circ \mathrm{g})(\mathrm{x}) \); b. the domain of \( \mathrm{f} \circ \mathrm{g} \) a. \( (\mathrm{f} \circ \mathrm{g})(\mathrm{x})=\square \) (Simplify your answer.)

To find \( (\mathrm{f} \circ \mathrm{g})(\mathrm{x}) \), we need to substitute \( g(x) \) into \( f(x) \). 1. Start with \( g(x) = \sqrt{4 - x} \). 2. Now substitute \( g(x) \) into \( f(x) \): \[ (\mathrm{f} \circ \mathrm{g})(\mathrm{x}) = f(g(x)) = f(\sqrt{4 - x}) = (\sqrt{4 - x})^2 + 2 \] 3. Simplifying that gives: \[ (\sqrt{4 - x})^2 = 4 - x \] Thus, \[ f(g(x)) = (4 - x) + 2 = 6 - x \] So, \( (\mathrm{f} \circ \mathrm{g})(\mathrm{x}) = 6 - \mathrm{x} \). Now, for the domain of \( f \circ g \): To find the domain, we need to consider the domain of \( g(x) \) and the values that \( g(x) \) can take. 1. The function \( g(x) \) is defined when \( 4 - x \geq 0 \), which means \( x \leq 4 \). 2. Moreover, \( g(x) = \sqrt{4 - x} \geq 0\), meaning \( \sqrt{4 - x} \) can take any value from \( 0 \) to \( 2 \) (as plugging in \( 0 \) and \( 4 \) results in \( g(x) = 0 \) and \( g(x) = 2 \), respectively). Thus, the output of \( g(x) \) lies in the interval \([0, 2]\), where \( f(x) \) has no restrictions. Therefore, the domain of \( f \circ g \) is all \( x \) values such that \( x \leq 4 \). Thus: Domain of \( (\mathrm{f} \circ \mathrm{g}) \) is \( (-\infty, 4] \).