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 find the condition that makes [tex]\( \frac{1}{3}x + 2 \geq 0 \)[/tex].
1. Set up the inequality:
[tex]\[ \frac{1}{3}x + 2 \geq 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{3}x \geq -2 \][/tex]
Multiply both sides by 3 to eliminate the fraction:
[tex]\[ x \geq -6 \][/tex]
So, the function [tex]\( f(x) = \sqrt{\frac{1}{3}x + 2} \)[/tex] is defined for all [tex]\( x \)[/tex] that satisfy [tex]\( x \geq -6 \)[/tex].
Thus, the domain of the function is [tex]\( x \geq -6 \)[/tex].
Out of the provided options, the correct one is:
[tex]\[ x \geq -6 \][/tex]
