To solve the given problem, we need to determine the domains of $g(x)$ and $(f \circ g)(x)$.

Given functions: $f(x) = x^2 - 3$ $g(x) = \sqrt{3 - x}$

i) Find the domain of $g(x)$:

For $g(x) = \sqrt{3 - x}$ to be defined, the expression inside the square root must be non-negative: $3 - x \geq 0$ $x \leq 3$

So, the domain of $g(x)$ is: $\text{Domain of } g(x) = (-\infty, 3]$

ii) Find the domain of $(f \circ g)(x)$:

The composition $(f \circ g)(x) = f(g(x))$ requires that $g(x)$ be defined and that the result $g(x)$ be within the domain of $f(x)$.

1. From part (i), $g(x)$ is defined for $x \in (-\infty, 3]$.
2. We then need to ensure that the output of $g(x)$ falls within the domain of $f(x)$, which is all real numbers because $f(x) = x^2 - 3$ is defined for all $x$.

Since $g(x)$ produces values within $[0, \sqrt{3}]$ and the domain of $f(x)$ is all real numbers, there are no additional restrictions beyond those already found for $g(x)$.

Thus, the domain of $(f \circ g)(x)$ is the same as the domain of $g(x)$: $\text{Domain of } (f \circ g)(x) = (-\infty, 3]$

Summary:

1. The domain of $g(x)$ is $(-\infty, 3]$.
2. The domain of $(f \circ g)(x)$ is $(-\infty, 3]$.

Would you like further details or have any questions about this solution? Here are some related questions you might find interesting:

1. What is the range of $g(x) = \sqrt{3 - x}$?
2. How do you determine the range of $(f \circ g)(x)$?
3. Can you find the inverse function of $g(x)$?
4. What are the critical points of $f(x) = x^2 - 3$?
5. How does the graph of $f(x)$ change when $g(x)$ is applied?
6. What is the composite function $(g \circ f)(x)$?
7. How would the domain of $(g \circ f)(x)$ differ from $(f \circ g)(x)$?
8. Can you find the derivative of $f(g(x))$ using the chain rule?

Tip: When finding the domain of a composite function, ensure the output of the inner function lies within the domain of the outer function.