Answer :
To determine which value of \( x \) is in the domain of \( f(x) = \sqrt{x - 6} \), we need to ensure that the expression inside the square root is non-negative. This means:
[tex]\[ x - 6 \geq 0 \][/tex]
Solving this inequality:
1. Add 6 to both sides:
[tex]\[ x \geq 6 \][/tex]
So, for \( f(x) = \sqrt{x - 6} \) to be defined, \( x \) must be greater than or equal to 6.
Now, let's check each option:
A. \( x = 0 \):
[tex]\[ 0 - 6 = -6 \][/tex]
\(\sqrt{-6}\) is not defined for real numbers, so \( x = 0 \) is not in the domain.
B. \( x = 10 \):
[tex]\[ 10 - 6 = 4 \][/tex]
\(\sqrt{4}\) is defined and equals 2, so \( x = 10 \) is in the domain.
C. \( x = -4 \):
[tex]\[ -4 - 6 = -10 \][/tex]
\(\sqrt{-10}\) is not defined for real numbers, so \( x = -4 \) is not in the domain.
D. \( x = 5 \):
[tex]\[ 5 - 6 = -1 \][/tex]
\(\sqrt{-1}\) is not defined for real numbers, so \( x = 5 \) is not in the domain.
Therefore, the only value of \( x \) that is in the domain of \( f(x) = \sqrt{x - 6} \) is:
[tex]\[ \boxed{10} \][/tex]
[tex]\[ x - 6 \geq 0 \][/tex]
Solving this inequality:
1. Add 6 to both sides:
[tex]\[ x \geq 6 \][/tex]
So, for \( f(x) = \sqrt{x - 6} \) to be defined, \( x \) must be greater than or equal to 6.
Now, let's check each option:
A. \( x = 0 \):
[tex]\[ 0 - 6 = -6 \][/tex]
\(\sqrt{-6}\) is not defined for real numbers, so \( x = 0 \) is not in the domain.
B. \( x = 10 \):
[tex]\[ 10 - 6 = 4 \][/tex]
\(\sqrt{4}\) is defined and equals 2, so \( x = 10 \) is in the domain.
C. \( x = -4 \):
[tex]\[ -4 - 6 = -10 \][/tex]
\(\sqrt{-10}\) is not defined for real numbers, so \( x = -4 \) is not in the domain.
D. \( x = 5 \):
[tex]\[ 5 - 6 = -1 \][/tex]
\(\sqrt{-1}\) is not defined for real numbers, so \( x = 5 \) is not in the domain.
Therefore, the only value of \( x \) that is in the domain of \( f(x) = \sqrt{x - 6} \) is:
[tex]\[ \boxed{10} \][/tex]