To determine which value of [tex]\( x \)[/tex] is in the domain of the function [tex]\( f(x) = \sqrt{x} - 2 \)[/tex], we need to consider the domain of the square root function, [tex]\( \sqrt{x} \)[/tex].
The square root function, [tex]\( \sqrt{x} \)[/tex], is defined only for [tex]\( x \geq 0 \)[/tex]. Therefore, [tex]\( f(x) = \sqrt{x} - 2 \)[/tex] is defined for [tex]\( x \geq 0 \)[/tex].
Let's analyze each given value:
- Option A: [tex]\( x = -2 \)[/tex]
Since the square root of a negative number is not defined in the set of real numbers, [tex]\( \sqrt{-2} \)[/tex] does not exist in the reals. Hence, [tex]\( x = -2 \)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- Option B: [tex]\( x = 0 \)[/tex]
For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = \sqrt{0} - 2 = 0 - 2 = -2
\][/tex]
Thus, [tex]\( f(0) \)[/tex] is defined and [tex]\( x = 0 \)[/tex] is in the domain of [tex]\( f(x) \)[/tex].
- Option C: [tex]\( x = 1 \)[/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = \sqrt{1} - 2 = 1 - 2 = -1
\][/tex]
Thus, [tex]\( f(1) \)[/tex] is defined and [tex]\( x = 1 \)[/tex] is in the domain of [tex]\( f(x) \)[/tex].
- Option D: [tex]\( x = 2 \)[/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = \sqrt{2} - 2
\][/tex]
The exact value of [tex]\( \sqrt{2} \)[/tex] is approximately [tex]\( 1.414 \)[/tex], so:
[tex]\[
f(2) = 1.414 - 2 = -0.586 \text{ (approximately)}
\][/tex]
Thus, [tex]\( f(2) \)[/tex] is defined and [tex]\( x = 2 \)[/tex] is in the domain of [tex]\( f(x) \)[/tex].
Now, summarizing our findings:
- [tex]\( x = -2 \)[/tex] is not in the domain.
- [tex]\( x = 0 \)[/tex] is in the domain.
- [tex]\( x = 1 \)[/tex] is in the domain.
- [tex]\( x = 2 \)[/tex] is in the domain.
Therefore, the values of [tex]\( x \)[/tex] that are in the domain of [tex]\( f(x) = \sqrt{x} - 2 \)[/tex] are:
[tex]\[ \boxed{0, 1, 2} \][/tex]
