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: 27


$y=(2x-1)^{1/3}$ 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 $(-\infty,\infty)$.
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