## Calculus: Early Transcendentals 8th Edition

$Q(x)$ is continuous on its domain $(-\infty,\sqrt[3]2)\cup(\sqrt[3]2,\infty)$
$$Q(x)=\frac{\sqrt[3]{x-2}}{x^3-2}$$ 1) Consider the numerator: The domain of $\sqrt[3]{x-2}$ is $(-\infty,\infty)$ According to Theorem 7, $\sqrt[3]{x-2}$ is continuous on $(-\infty,\infty)$ 2) Consider the denominator: The domain of $x^3-2$ is $(-\infty,\infty)$ According to Theorem 7, $x^3-2$ is continuous on $(-\infty,\infty)$ Overall, both $\sqrt[3]{x-2}$ and $x^3-2$ are continuous on $(-\infty,\infty)$ 3) According to part 5 of Theorem 4, $Q(x)$ is continuous on $(-\infty,\infty)$ except where $(x^3-2)=0$ We see that $(x^3-2)=0$ $\Leftrightarrow x^3=2$ $\Leftrightarrow x=\sqrt[3]2$ Therefore, $Q(x)$ is continuous on its domain $(-\infty,\sqrt[3]2)\cup(\sqrt[3]2,\infty)$