To determine the domain of the function [tex]\( f(x) = \sqrt{\frac{1}{3}x + 2} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] for which the expression under the square root is non-negative (since the square root of a negative number is not defined in the set of real numbers).
Here are the steps to find the domain:
1. Set the expression inside the square root to be non-negative:
[tex]\[
\frac{1}{3}x + 2 \geq 0
\][/tex]
2. Solve the inequality for [tex]\( x \)[/tex]:
- First, isolate [tex]\( x \)[/tex] by subtracting 2 from both sides:
[tex]\[
\frac{1}{3}x \geq -2
\][/tex]
- Next, multiply both sides of the inequality by 3 to solve for [tex]\( x \)[/tex]:
[tex]\[
x \geq -6
\][/tex]
3. Interpret the result:
- The inequality [tex]\( x \geq -6 \)[/tex] means that [tex]\( x \)[/tex] must be greater than or equal to [tex]\(-6\)[/tex] for the expression inside the square root to be non-negative.
Therefore, the domain of the function [tex]\( f(x) = \sqrt{\frac{1}{3}x + 2} \)[/tex] is:
[tex]\[ x \geq -6 \][/tex]
So the correct answer is:
[tex]\[ x \geq -6 \][/tex]
