Let's go through the problem step by step.

Given Functions:

• $f(x) = \sqrt{x}$
• $g(x) = x + 9$

(a) Find $(f \circ g)(x)$:

This is the composition $f(g(x))$.

• $g(x) = x + 9$, so $f(g(x)) = f(x + 9) = \sqrt{x + 9}$.
• The simplified result is $\sqrt{x + 9}$.

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

For $f(g(x)) = \sqrt{x + 9}$, the expression under the square root must be non-negative:

• $x + 9 \geq 0$
• $x \geq -9$

Thus, the domain of $f(g(x))$ is $[-9, \infty)$.

(b) Find $(g \circ f)(x)$:

This is the composition $g(f(x))$.

• $f(x) = \sqrt{x}$, so $g(f(x)) = g(\sqrt{x}) = \sqrt{x} + 9$.
• The simplified result is $\sqrt{x} + 9$.

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

For $g(f(x)) = \sqrt{x} + 9$, the square root $\sqrt{x}$ requires $x \geq 0$.

Thus, the domain of $g(f(x))$ is $[0, \infty)$.

Corrected Results:

• $(f \circ g)(x) = \sqrt{x + 9}$, with domain $[-9, \infty)$.
• $(g \circ f)(x) = \sqrt{x} + 9$, with domain $[0, \infty)$.

Would you like further details on the steps, or do you have any questions?

Here are 5 related questions for deeper understanding:

1. How do you determine the domain of a square root function?
2. What is the difference between $f \circ g$ and $g \circ f$?
3. How would the domain change if the square root were replaced by a cubic root?
4. Can you compose functions when one is not defined at certain points?
5. How does the range of $f(x) = \sqrt{x}$ affect the domain of $g(f(x))$?

Tip: When composing functions, always ensure that the output of the inner function fits the domain of the outer function.