Question

if $f(x)=\sqrt{x - 3}$, which inequality can be used to find the domain of $f(x)$?
\bigcirc\\ $\sqrt{x - 3}\geq0$
\bigcirc\\ $x - 3\geq0$
\bigcirc\\ $\sqrt{x - 3}\leq0$
\bigcirc\\ $x - 3\leq0$

Explanation:

Step1: Recall domain of square root function

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

Step2: Identify the radicand in \( f(x)=\sqrt{x - 3} \)

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

Step3: Determine the inequality for the domain

To find the domain of \( f(x)=\sqrt{x - 3} \), we need to ensure that the radicand \( x-3\) is non - negative. So the inequality that we use to find the domain is \( x - 3\geq0 \).

Answer:

B. \( x - 3\geq0 \) (where the option is labeled as the one with \( x - 3\geq0 \))