Answer:
[tex]x\geq -6[/tex]
Step-by-step explanation:
The domain of a function is all of the values that [tex]x[/tex] can be under that specific function. In this case, we're asking what values of [tex]x[/tex] allow [tex]\sqrt{\frac{1}{3} x+2}[/tex] to exist.
In order for square roots to exist, the quantity under the square root must be greater than or equal to [tex]0[/tex], because you can't take the square root of a negative number. Therefore, we can write the following inequality to solve for [tex]x[/tex]:
[tex]\frac{1}{3}x+2\geq 0[/tex]
Solving this inequality, we get:
[tex]\frac{1}{3}x+2\geq 0[/tex]
[tex]\frac{1}{3}x\geq -2[/tex] (Subtract [tex]2[/tex] from both sides of the inequality to isolate [tex]x[/tex])
[tex]x\geq -6[/tex] (Multiply both sides of the inequality by [tex]3[/tex] to get rid of [tex]x[/tex]'s coefficient)
Hope this helps!
