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 $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .