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)

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)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Domain of Combined Functions", "Solving Linear Inequalities" ], "new_concepts": [ "Domain of Square Root Functions" ], "current_concepts": [ "Domain of Combined Functions", "Solving Linear Inequalities", "Domain of Square Root Functions" ] } ## Set up the domain inequality $$ \frac{1}{3}x + 2 \ge 0 $$ ## Solve the inequality for x $$ \begin{aligned} \frac{1}{3}x &\ge -2 \ x &\ge -6 \end{aligned} $$

$x \le -6$

$x > 6$

$x < 6$

$x \ge -6$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Domain of Square Root Functions" ] }

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QUESTION IMAGE

Question

Explanation:

Set up the domain inequality

Solve the inequality for x

Answer: