To solve for $(f \circ g)(x)$, we substitute $g(x)$ into $f(x)$. Then we calculate the domain based on the restrictions of both $f(x)$ and $g(x)$.

Step 1: Definitions

• $f(x) = x^2 + 3$
• $g(x) = \sqrt{5 - x}$
• $(f \circ g)(x)$ means $f(g(x))$.

Step 2: Substitute $g(x)$ into $f(x)$

$(f \circ g)(x) = f(g(x)) = f\left(\sqrt{5 - x}\right)$ $f\left(\sqrt{5 - x}\right) = \left(\sqrt{5 - x}\right)^2 + 3$ $(f \circ g)(x) = 5 - x + 3 = 8 - x$

Step 3: Domain of $(f \circ g)(x)$

The domain of $f(g(x))$ is restricted by $g(x)$, which includes a square root. For $\sqrt{5 - x}$, the expression inside the square root must be non-negative: $5 - x \geq 0$ $x \leq 5$ Thus, the domain of $(f \circ g)(x)$ is: $x \in (-\infty, 5]$

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Final Answer for $(f \circ g)(x)$ and Domain:

$(f \circ g)(x) = 8 - x, \quad \text{Domain: } x \in (-\infty, 5]$

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Would you like to proceed with Part 2 for $(g \circ f)(x)$? Or do you have questions about this part?

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Related Questions:

1. What is $(g \circ f)(x)$ for the given functions?
2. How is the domain of a composite function determined?
3. What are the general rules for combining square roots with other functions?
4. How does the domain of $g(x)$ affect $(f \circ g)(x)$?
5. Could the domain change if $f(x)$ had additional restrictions?

Tip: Always check the restrictions on the inner function when evaluating composite functions!