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.)

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### 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 $$ (-\infty,\ 4] $$.
a. $$ (f \circ g)(x) = 6 - x $$ b. The domain of $$ f \circ g $$ is $$ (-\infty, 4] $$.

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