text: ## Calculus with Applications (10th Edition)
  
  The domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
  $$f(x)=\frac{1}{\sqrt {3x^{2}+2x-1}}=\frac{1}{\sqrt {(3x-1)(x+1)}}$$ the values $x$ for $f(x)$ is defined when: $$(3x-1)(x+1) \gt 0$$ since the radical cannot be negative and the denominator of the function cannot be zero. So we can solve: $$(3x-1)(x+1) = 0$$ $\Rightarrow$ $$x_{1}=\frac{1}{3},\:x_{2}=-1$$ this numbers $-1$ and $\frac{1}{3}$ divide the number line into 3 intervals, $(-\infty,-1), (-1 ,\frac{1}{3})$ and $(\frac{1}{3},\infty)$. we observe that only values in the intervals $(-\infty,-1)$ and $(\frac{1}{3},\infty)$ satisfy the inequality. So the domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
explanation: The text extract provides a detailed explanation of how to determine the domain of the function \( f(x) = \frac{1}{\sqrt{3x^2 + 2x - 1}} \). It includes the steps to solve the inequality \( (3x-1)(x+1) > 0 \) and identifies the intervals where the function is defined. The extract does not contain a question or an answer to a question.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Calculus with Applications (10th Edition)
  
  The domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
  $$f(x)=\frac{1}{\sqrt {3x^{2}+2x-1}}=\frac{1}{\sqrt {(3x-1)(x+1)}}$$ the values $x$ for $f(x)$ is defined when: $$(3x-1)(x+1) \gt 0$$ since the radical cannot be negative and the denominator of the function cannot be zero. So we can solve: $$(3x-1)(x+1) = 0$$ $\Rightarrow$ $$x_{1}=\frac{1}{3},\:x_{2}=-1$$ this numbers $-1$ and $\frac{1}{3}$ divide the number line into 3 intervals, $(-\infty,-1), (-1 ,\frac{1}{3})$ and $(\frac{1}{3},\infty)$. we observe that only values in the intervals $(-\infty,-1)$ and $(\frac{1}{3},\infty)$ satisfy the inequality. So the domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract provides a detailed explanation of how to determine the domain of the function \( f(x) = \frac{1}{\sqrt{3x^2 + 2x - 1}} \). It includes the steps to solve the inequality \( (3x-1)(x+1) > 0 \) and identifies the intervals where the function is defined. The extract does not contain a question or an answer to a question.
  
  Label: no-question