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 - Section 4.1 - Extreme Values of Functions - Exercises 4.1 - Page 192: 59

Answer

Critical points: $x=-4/5, 0$, no endpoint. At $x=-4/5$, $y=\frac{12\sqrt[3] 10}{25}$ local maximum. At $x=0$, $y=0$ local minimum.

Work Step by Step

Step 1. Given the function $y=x^{2/3}(x+2)$, we can find its derivative as $y'=\frac{2}{3}x^{-1/3}(x+2)+x^{2/3}=\frac{2x+4}{3\sqrt[3] x}+x^{2/3}=\frac{2x+4+3x}{3\sqrt[3] x}=\frac{5x+4}{3\sqrt[3] x}$ Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=-4/5, 0$. The domain of the function is $(-\infty,\infty)$ and there is no endpoint. Step 3. At $x=-4/5$, $y=(-4/5)^{2/3}(-4/5+2)=(\frac{6}{5})\frac{2\sqrt[3] 2}{\sqrt[3] {25}}=\frac{12\sqrt[3] 2}{5\sqrt[3] {25}}=\frac{12\sqrt[3] 10}{25}$ and we can identify this is a local maximum (use local test points as necessary) as shown in the figure. Step 4. At $x=0$, $y=0$ and we can identify this is a local minimum (use local test points as necessary) as shown in the figure.
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