If [tex]$f(x)=\sqrt{x-3}[tex]$[/tex], which inequality can be used to find the domain of [tex]$[/tex]f(x)$[/tex]?

A. [tex]\sqrt{x-3} \geq 0[/tex]
B. [tex]x-3 \geq 0[/tex]
C. [tex]\sqrt{x-3} \leq 0[/tex]
D. [tex]x-3 \leq 0[/tex]



Answer :

To determine the domain of the function \( f(x) = \sqrt{x-3} \), we need to consider the conditions under which the expression inside the square root is defined.

The square root function \( \sqrt{y} \) is only defined for \( y \geq 0 \). Therefore, for \( f(x) \) to be defined, the expression inside the square root \( x - 3 \) must be non-negative.

This leads to the inequality:
[tex]\[ x - 3 \geq 0 \][/tex]

This inequality can be solved by isolating \( x \). Adding 3 to both sides, we get:
[tex]\[ x \geq 3 \][/tex]

Thus, the inequality that ensures the function \( f(x) = \sqrt{x-3} \) is defined is:
[tex]\[ x - 3 \geq 0 \][/tex]

So, the correct inequality is:
[tex]\[ x - 3 \geq 0 \][/tex]