Question

if $f(x) = \sqrt{x - 3}$, which inequality can be used to find the domain of $f(x)$?
\bigcirc $\sqrt{x - 3} \geq 0$
\bigcirc $x - 3 \geq 0$
\bigcirc $\sqrt{x - 3} \leq 0$
\bigcirc $x - 3 \leq 0$

Explanation:

Step1: Recall domain of square root

The square root function \( \sqrt{u} \) is defined when the radicand \( u \) is non - negative (i.e., \( u\geq0 \)).

Step2: Identify the radicand

For the function \( f(x)=\sqrt{x - 3} \), the radicand is \( x-3 \).

Step3: Determine the inequality

To find the domain of \( f(x) \), we need to ensure that the radicand \( x - 3 \) is non - negative. So the inequality that can be used to find the domain is \( x-3\geq0 \).

Answer:

B. \( x - 3\geq0 \) (assuming the second option is labeled B, if the original options were labeled with A, B, C, D as per the order: A. \( \sqrt{x - 3}\geq0 \), B. \( x - 3\geq0 \), C. \( \sqrt{x - 3}\leq0 \), D. \( x - 3\leq0 \))