Question

what is the domain of the function

\\(f(x) = \sqrt{\frac{1}{3}x + 2}\\) ?

\\(x \le -6\\)
\\(x > 6\\)
\\(x < 6\\)
\\(x \ge -6\\)

Explanation:

Set the radicand non-negative

For a square root function to be defined in real numbers, the expression inside the radical must be greater than or equal to zero.

$$ \frac{1}{3}x + 2 \ge 0 $$

Solve the inequality for x

Subtract \(2\) from both sides of the inequality.

$$ \frac{1}{3}x \ge -2 $$

Multiply both sides by \(3\) to isolate \(x\).

$$ x \ge -6 $$

Answer:

• (A) \(x \le -6\)
• (B) \(x > 6\)
• (C) \(x < 6\)
• (D) \(x \ge -6\) (Correct answer)