Answer
$ y $ is continuous on
$$[\frac{1}{3},\infty)$$
Work Step by Step
Given $$ y=\sqrt[4]{3x-1}$$
Since the 4th root is always non-negative
$3x-1 \geq 0 \ \ \ \Rightarrow x\geq\frac{1}{3}$
So, $ y $ is continuous on
$$[\frac{1}{3},\infty)$$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 2: Limits and Continuity - Section 2.5 - Continuity - Exercises 2.5 - Page 84: 26
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