To find the domain of the function $f(x) = \sqrt{x^2 - 3x}$, we need to determine the set of $x$-values for which the expression under the square root is non-negative (since the square root is defined only for non-negative values).

Step 1: Set the inequality

$x^2 - 3x \geq 0$

Step 2: Factorize the inequality

Factorize $x^2 - 3x$: $x(x - 3) \geq 0$

Step 3: Solve the inequality

The critical points are $x = 0$ and $x = 3$. These points divide the number line into three intervals:

1. $(-\infty, 0)$
2. $(0, 3)$
3. $(3, \infty)$

Test each interval to determine where $x(x - 3) \geq 0$:

• For $x \in (-\infty, 0)$: Both $x < 0$ and $x - 3 < 0$, so their product $x(x - 3) > 0$ does not hold.
• For $x \in (0, 3)$: $x > 0$ and $x - 3 < 0$, so their product $x(x - 3) > 0$ does not hold.
• For $x \in (3, \infty)$: Both $x > 0$ and $x - 3 > 0$, so their product $x(x - 3) > 0$ holds.
• At $x = 0$ and $x = 3$: $x(x - 3) = 0$, so it satisfies $\geq 0$.

Step 4: Write the domain

The domain is: $[0, 3] \cup [3, \infty) = [0, \infty)$

Final Answer:

The domain of $f(x) = \sqrt{x^2 - 3x}$ is: $[0, \infty)$

Let me know if you'd like a detailed explanation of any step! Here are some related questions you might explore:

1. How would the domain change if the function were $f(x) = \frac{1}{\sqrt{x^2 - 3x}}$?
2. What is the range of $f(x) = \sqrt{x^2 - 3x}$?
3. How do you sketch the graph of $f(x) = \sqrt{x^2 - 3x}$?
4. What is the derivative of $f(x)$, and where is it defined?
5. Can we rewrite $f(x)$ as a piecewise function?

Tip: When analyzing square root functions, always check for non-negativity of the radicand (expression under the square root).