Answer
$y=\sqrt[4]{3x-1}$ is continuous on $[1/3,\infty)$
Work Step by Step
$$y=\sqrt[4]{3x-1}$$
- Domain: $y$ is defined where $3x-1\ge0$, which means $x\ge1/3$
So our domain here is $[1/3,\infty)$
We know that $\lim_{x\to c}(3x-1)=3c-1$ on $[1/3,\infty)$
So $y=3x-1$ is continuous on $[1/3,\infty)$
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, $y=\sqrt[4]{3x-1}$ is continuous on $[1/3,\infty)$.
