## Calculus with Applications (10th Edition)

The domain is $(-\infty,-\frac{2}{5}] \cup [\frac{1}{3},\infty)$ .
$$f(x)=\sqrt {15x^{2}+x-2}$$ the values $x$ for $f(x)$ is defined when: $$15x^{2}+x-2 \geq 0$$ we can find all zeros of $f(x)$ : $$15x^{2}+x-2 = 0$$ $\Rightarrow$ $$\:x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:15\left(-2\right)}}{2\cdot \:15}$$ $\Rightarrow$ $$x_{1}=\frac{1}{3},\:x_{2}=-\frac{2}{5}$$ this numbers form the intervals $(-\infty,-\frac{2}{5}), (-\frac{2}{5} ,\frac{1}{3})$ and $(\frac{1}{3},\infty)$. we observe that only values in the intervals $(-\infty,-\frac{2}{5})$ and $(\frac{1}{3},\infty)$ satisfy the inequality. So the domain is $(-\infty,-\frac{2}{5}] \cup [\frac{1}{3},\infty)$ .