Express the domain of f(x) = sqrt(4x - 12) in interval notation.

Step-by-step solution

1. 1

   State the condition for a real square root. An even root is real only when its radicand is nonnegative:

   $4x - 12 \ge 0$

   There is no denominator here, so this is the only restriction.

2. 2

   Solve the inequality. Adding $12$ and dividing by the positive $4$ leaves the direction unchanged:

   $4x \ge 12 \quad\Longrightarrow\quad x \ge 3$

3. 3

   Decide whether the endpoint belongs. At $x = 3$ the radicand is $0$ and $\sqrt0 = 0$ is perfectly defined, so $3$ is in the domain. That is why the interval is closed on the left.

4. 4

   Write it in interval notation.

   $[3,\ \infty)$

   The right end is always a parenthesis: $\infty$ is not a number that can be attained.

5. 5

   Spot-check either side. At $x = 3$: $f(3) = \sqrt0 = 0$. At $x = 7$: $f(7) = \sqrt{16} = 4$. At $x = 2$: the radicand is $-4$, so $f(2)$ is not a real number — correctly outside the domain.