The given function is:

$f(x) = 3 \sqrt{2x + 1}$

Step 1: Identify the domain restriction

Since we have a square root function, the expression inside the square root must be non-negative because the square root of a negative number is not defined in real numbers.

$2x + 1 \geq 0$

Step 2: Solve for $x$

$2x \geq -1$

$x \geq -\frac{1}{2}$

Step 3: Write the domain in interval notation

$\left[ -\frac{1}{2}, \infty \right)$

Thus, the domain of $f(x)$ is $x \geq -\frac{1}{2}$, or $\left[ -\frac{1}{2}, \infty \right)$ in interval notation.

Would you like a graphical representation or further explanation?

Related Questions:

1. What is the range of $f(x) = 3 \sqrt{2x + 1}$?
2. How would the domain change if the function were $f(x) = \frac{3}{\sqrt{2x + 1}}$?
3. What happens to the domain if we replace the square root with a cube root?
4. How does multiplying by 3 affect the function?
5. What is the inverse function of $f(x)$?

Tip: Always check for restrictions in the denominator and even roots when finding the domain of a function.