To determine the domain of the function [tex]\( f(x) = \sqrt{\frac{1}{3}x + 2} \)[/tex], we need to ensure that the expression inside the square root is non-negative because the square root of a negative number is not defined in the set of real numbers.
Let's set up the inequality to find when the expression inside the square root is greater than or equal to zero:
[tex]\[
\frac{1}{3}x + 2 \geq 0
\][/tex]
Next, we solve this inequality for [tex]\(x\)[/tex].
1. Subtract 2 from both sides:
[tex]\[
\frac{1}{3}x + 2 - 2 \geq -2
\][/tex]
[tex]\[
\frac{1}{3}x \geq -2
\][/tex]
2. Multiply both sides by 3 to eliminate the fraction:
[tex]\[
x \geq -6
\][/tex]
Thus, the inequality simplifies to:
[tex]\[
x \geq -6
\][/tex]
This tells us that [tex]\(x\)[/tex] must be greater than or equal to -6 for the expression under the square root to be non-negative.
Therefore, the domain of the function [tex]\( f(x) = \sqrt{\frac{1}{3}x + 2} \)[/tex] is all real numbers [tex]\( x \)[/tex] such that [tex]\( x \geq -6 \)[/tex].
Among the given options, the correct choice that matches our solution is:
[tex]\[ \boxed{x \geq -6} \][/tex]
