Given the function [tex]\( f(x) = \sqrt{\frac{1}{3} x + 2} \)[/tex], we need to determine the domain. The domain of a function is the set of all possible values of [tex]\( x \)[/tex] for which the function is defined.
For the function [tex]\( f(x) \)[/tex] to be defined, the expression inside the square root must be non-negative (since the square root of a negative number is not a real number).
Thus, the condition we need to satisfy is:
[tex]\[ \frac{1}{3} x + 2 \geq 0 \][/tex]
Let's solve this inequality step by step:
1. Isolate [tex]\( x \)[/tex] by first subtracting 2 from both sides:
[tex]\[ \frac{1}{3} x + 2 - 2 \geq 0 - 2 \][/tex]
[tex]\[ \frac{1}{3} x \geq -2 \][/tex]
2. Eliminate the fraction by multiplying both sides of the inequality by 3:
[tex]\[ 3 \cdot \frac{1}{3} x \geq -2 \cdot 3 \][/tex]
[tex]\[ x \geq -6 \][/tex]
Thus, the inequality [tex]\( \frac{1}{3} x + 2 \geq 0 \)[/tex] simplifies to [tex]\( x \geq -6 \)[/tex].
The domain of the function [tex]\( f(x) \)[/tex] is therefore all [tex]\( x \)[/tex] such that:
[tex]\[ x \geq -6 \][/tex]
So, the correct answer is:
[tex]\[ x \geq -6 \][/tex]
