The function given is $f(x) = \sqrt{3 - 2x}$. To find the domain of this function, we need to determine the values of $x$ for which the expression under the square root is non-negative. This ensures that the function produces real values.

Step 1: Set the expression under the square root greater than or equal to zero

$3 - 2x \geq 0$

Step 2: Solve the inequality

$-2x \geq -3$ $x \leq \frac{3}{2}$

Step 3: Write the domain

Since the function is defined for values of $x$ less than or equal to $\frac{3}{2}$, the domain is: $(-\infty, \frac{3}{2}]$

Thus, the domain of the function $f(x) = \sqrt{3 - 2x}$ is $(-\infty, \frac{3}{2}]$.

Would you like further details or have any other questions? Here are some additional related questions to expand your understanding:

1. How would the domain change if the function was $f(x) = \sqrt{3 + 2x}$?
2. What is the range of the function $f(x) = \sqrt{3 - 2x}$?
3. How would the domain be affected if the square root was squared?
4. What happens to the domain if we replace $\sqrt{3 - 2x}$ with $\frac{1}{\sqrt{3 - 2x}}$?
5. Can you explain how the domain would change if the expression inside the square root were a quadratic?

Tip: Always check for non-negative values inside a square root when determining the domain of a function involving radicals!