Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Practice Exercises - Page 244: 39

Answer

See graph and explanations.

Work Step by Step

Step 1. See graph for the function $y=x^{2/3}+(x-1)^{1/3}$. Step 2. We have $y'=\frac{2}{3}x^{-1/3}+\frac{1}{3}(x-1)^{-2/3}$ Step 3. It is not difficult to see that $y'=0$ does not have a solution. Thus the critical points are $x=0,1$, where $y'$ is undefined. Step 4. Checking the signs of $y'$ across the critical points, we have $..(-)..(0)..(+)..(1)..(+)..$; thus the function decreases over $(-\infty,0)$ and increases over $(0,1), (1,\infty)$. And $y(0)=-1$ is a local minimum. Step 5. As $x\to0^-, y'\to-\infty$, $x\to0^+, y'\to\infty$, the function has a cusp at $x=0$, where $y'(0)$ is undefined. Step 6. As $x\to1^-, y'\to\infty$, $x\to0^+, y'\to\infty$, the function has a vertical tangent line at $x=1$ where $y'(1)$ is undefined.
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.