Let's solve the problem step-by-step.

Given Functions:

1. $f(x) = \sqrt{x - 3}$
2. $g(x) = 2 - 4x$

We need to find $(f \circ g)(x)$, which represents the composite function $f(g(x))$, and then find the domain of this composite function.

Step 1: Finding $f(g(x))$

The composite function $f(g(x))$ is defined as: $f(g(x)) = \sqrt{g(x) - 3}$

Substitute $g(x) = 2 - 4x$ into $f(x)$: $f(g(x)) = \sqrt{(2 - 4x) - 3}$ $f(g(x)) = \sqrt{-4x - 1}$

Step 2: Finding the Domain of $f(g(x))$

For the square root to be defined, the expression inside the square root must be non-negative: $-4x - 1 \geq 0$ Solve for $x$: $-4x \geq 1$ $x \leq -\frac{1}{4}$

Thus, the domain of $f(g(x))$ is all $x$ values for which $x \leq -\frac{1}{4}$.

Final Answers:

1. The composite function $(f \circ g)(x)$:
   $f(g(x)) = \sqrt{-4x - 1}$

2. The domain of $f \circ g$:
   $\left( -\infty, -\frac{1}{4} \right]$

Would you like any further clarification on this problem?

Related Questions:

1. How do you determine the domain of a composite function in general?
2. Why is the domain of $f(g(x))$ restricted to $x \leq -\frac{1}{4}$?
3. What would happen to the domain if the functions $f(x)$ or $g(x)$ were different?
4. Can $f \circ g$ be defined if $g(x)$ outputs a value not in the domain of $f$?
5. How does the composition of functions work if one function is not one-to-one?

Tip:

When finding the domain of a composite function $f(g(x))$, always ensure that the output of $g(x)$ lies within the domain of $f(x)$.