Question

what is the domain of the function f(x) = \sqrt{\frac{1}{3}x + 2} ? \
\
\bigcirc \\ x \leq -6 \
\bigcirc \\ x > 6 \
\bigcirc \\ x < 6 \
\bigcirc \\ x \geq -6 \
\text{done}

Explanation:

Step1: Recall the domain of square root function

For a square root function \( f(x)=\sqrt{g(x)} \), the expression inside the square root (the radicand) must be non - negative, i.e., \( g(x)\geq0 \). In the function \( f(x)=\sqrt{\frac{1}{3}x + 2} \), we have \( g(x)=\frac{1}{3}x+2 \). So we need to solve the inequality \( \frac{1}{3}x + 2\geq0 \).

Step2: Solve the inequality

Subtract 2 from both sides of the inequality:
\( \frac{1}{3}x+2 - 2\geq0 - 2 \)
\( \frac{1}{3}x\geq - 2 \)

Multiply both sides of the inequality by 3 (since 3 is a positive number, the direction of the inequality sign remains the same):
\( 3\times\frac{1}{3}x\geq3\times(- 2) \)
\( x\geq - 6 \)

Answer:

\( x\geq - 6 \) (corresponding to the option \( x\geq - 6 \))