Answer
$y=(2x-1)^{1/3}$ is continuous on $(-\infty,\infty)$.
Work Step by Step
$$y=(2x-1)^{1/3}=\sqrt[3]{2x-1}$$
- Domain: $y$ is defined on $(-\infty,\infty)$
We know that $\lim_{x\to c}(2x-1)=2c-1$ on $(-\infty,\infty)$
So $y=2x-1$ is continuous on $(-\infty,\infty)$
Applying Theorem 8, if $f$ is continuous at $x=c$, then $\sqrt[n]f$ is also continuous at $x=c$ ($f$ must be defined on an open interval containing $c$, and $n\gt0$ and $n\in Z$)
Therefore, $y=\sqrt[3]{3x-1}=(2x-1)^{1/3}$ is continuous on $(-\infty,\infty)$.
