## 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 $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
The passage describes the process of finding the domain of a function \( f(x) = \frac{1}{\sqrt{3x^2 + 2x - 1}} \) and provides a detailed explanation of the steps involved, including solving an inequality and determining the intervals where the function is defined.
no_question

the text does not contain a mathematical question; it is a detailed explanation of a solution.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## 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 $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
user
The passage describes the process of finding the domain of a function \( f(x) = \frac{1}{\sqrt{3x^2 + 2x - 1}} \) and provides a detailed explanation of the steps involved, including solving an inequality and determining the intervals where the function is defined.

TYPE: NO_QUESTION

The text does not contain a mathematical question; it is a detailed explanation of a solution.