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What is the domain of the function [tex][tex]$f(x)=\sqrt{\frac{1}{3}x+2}$[/tex][/tex]?

A. [tex][tex]$x \leq -6$[/tex][/tex]
B. [tex][tex]$x \ \textgreater \ 6$[/tex][/tex]
C. [tex][tex]$x \ \textless \ 6$[/tex][/tex]
D. [tex][tex]$x \geq -6$[/tex][/tex]

Sagot :

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]