Answered

If [tex]\( f(x) = \sqrt{x-3} \)[/tex], which inequality can be used to find the domain of [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 [tex]\( f(x) = \sqrt{x-3} \)[/tex], we must ensure that the expression under the square root is non-negative. This is because the square root function is only defined for non-negative numbers (i.e., zero or positive values).

Let's examine the expression inside the square root: [tex]\( x - 3 \)[/tex].

For [tex]\( \sqrt{x-3} \)[/tex] to be defined, the quantity [tex]\( x - 3 \)[/tex] must be greater than or equal to zero:

[tex]\[ x - 3 \geq 0 \][/tex]

Thus, this inequality tells us the condition that [tex]\( x \)[/tex] must satisfy for the function [tex]\( f(x) \)[/tex] to be defined.

Now, let's check which of the given inequalities corresponds to this condition:

1. [tex]\( \sqrt{x-3} \geq 0 \)[/tex]: This is not the correct inequality because it describes a condition on the value of the function itself, not the domain of [tex]\( x \)[/tex].
2. [tex]\( x-3 \geq 0 \)[/tex]: This is the correct inequality. It directly tells us the condition that [tex]\( x \)[/tex] must satisfy for [tex]\( f(x) \)[/tex] to be defined.
3. [tex]\( \sqrt{x-3} \leq 0 \)[/tex]: This is not correct because the square root of a non-negative number cannot be less than or equal to zero (except in the trivial case when it's zero, but that doesn't cover the domain).
4. [tex]\( x-3 \leq 0 \)[/tex]: This is not correct because this would imply [tex]\( x \leq 3 \)[/tex], which would make [tex]\( x - 3 \)[/tex] negative and therefore [tex]\( \sqrt{x-3} \)[/tex] undefined.

Therefore, the correct inequality to determine the domain of [tex]\( f(x) = \sqrt{x-3} \)[/tex] is:

[tex]\[ x - 3 \geq 0 \][/tex]