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: 25


$y=\sqrt{2x+3}$ is continuous on $[-3/2,\infty)$.

Work Step by Step

$$y=\sqrt{2x+3}$$ - Domain: $y$ is defined where $2x+3\ge0$, which means $x\ge-3/2$ So our domain here is $[-3/2,\infty)$ We know that $\lim_{x\to c}(2x+3)=2c+3$ on $[-3/2,\infty)$ So $y=2x+3$ is continuous on $[-3/2,\infty)$ 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, $y=\sqrt{2x+3}$ is continuous on $[-3/2,\infty)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.