To determine which value of [tex]\( x \)[/tex] is in the domain of the function [tex]\( f(x) = \sqrt{x - 6} \)[/tex], we need to consider when the expression inside the square root is defined and non-negative.
The square root function [tex]\( \sqrt{u} \)[/tex] is defined when [tex]\( u \geq 0 \)[/tex]. In our given function [tex]\( f(x) = \sqrt{x - 6} \)[/tex], this means that the expression [tex]\( x - 6 \)[/tex] must be greater than or equal to zero:
[tex]\[
x - 6 \geq 0
\][/tex]
Solving this inequality for [tex]\( x \)[/tex]:
[tex]\[
x \geq 6
\][/tex]
So, [tex]\( x \)[/tex] must be greater than or equal to 6 for [tex]\( f(x) \)[/tex] to be defined. Now we need to check each of the provided values:
- [tex]\( x = 0 \)[/tex]: [tex]\( 0 - 6 = -6 \)[/tex]. Since [tex]\(-6\)[/tex] is not greater than or equal to [tex]\( 0 \)[/tex], [tex]\( x = 0 \)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- [tex]\( x = -4 \)[/tex]: [tex]\( -4 - 6 = -10 \)[/tex]. Since [tex]\(-10\)[/tex] is not greater than or equal to [tex]\( 0 \)[/tex], [tex]\( x = -4 \)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- [tex]\( x = 5 \)[/tex]: [tex]\( 5 - 6 = -1 \)[/tex]. Since [tex]\(-1\)[/tex] is not greater than or equal to [tex]\( 0 \)[/tex], [tex]\( x = 5 \)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- [tex]\( x = 10 \)[/tex]: [tex]\( 10 - 6 = 4 \)[/tex]. Since [tex]\( 4 \)[/tex] is greater than or equal to [tex]\( 0 \)[/tex], [tex]\( x = 10 \)[/tex] is in the domain of [tex]\( f(x) \)[/tex].
Therefore, the value of [tex]\( x \)[/tex] that is in the domain of [tex]\( f(x)=\sqrt{x-6} \)[/tex] is:
[tex]\[
\boxed{10}
\][/tex]
