University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 94: 26


$y=\sqrt[4]{3x-1}$ 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 $[1/3,\infty)$.
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