1. The problem asks us to find the domain of the function $$g(x) = \sqrt{3 - x}$$. 2. The domain of a function involving a square root requires the expression inside the root to be non-negative because the square root of a negative number is not a real number. 3. Set the inside of the square root greater than or equal to zero: $$3 - x \geq 0$$ 4. Solve the inequality for $$x$$: $$3 - x \geq 0$$ $$\Rightarrow -x \geq -3$$ 5. When dividing or multiplying an inequality by a negative number, reverse the inequality sign: $$\Rightarrow x \leq 3$$ 6. Therefore, the domain of $$g(x)$$ is all real numbers $$x$$ such that $$x \leq 3$$. 7. In interval notation, this is: $$(-\infty, 3]$$