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: 50

Answer

At $x=-\frac{\sqrt 6}{3}$, $y=4+\frac{4\sqrt 6}{9}$ local maximum, At $x=\frac{\sqrt 6}{3}$, $y=4-\frac{4\sqrt 6}{9}$ local minimum.

Work Step by Step

Step 1. Given the function $y=x^3−2x+4$, take its derivative to get $y′=3x^2−2$ Step 2. The critical points can be found when $y'=0$ or undefined; we have $y′=3x^2−2=0$ which gives $x=\pm\frac{\sqrt 6}{3}$ Step 3. At $x=-\frac{\sqrt 6}{3}$, $y=(-\frac{\sqrt 6}{3})^3−2(-\frac{\sqrt 6}{3})+4=-\frac{2\sqrt 6}{9}+\frac{2\sqrt 6}{3}+4=4+\frac{4\sqrt 6}{9}$ and we can identify this as a local maximum (use local test points) as shown in the figure. Step 4. At $x=\frac{\sqrt 6}{3}$, $y=(\frac{\sqrt 6}{3})^3−2(\frac{\sqrt 6}{3})+4=\frac{2\sqrt 6}{9}-\frac{2\sqrt 6}{3}+4=4-\frac{4\sqrt 6}{9}$ and we can identify this as a local minimum (use local test points) as shown in the figure.
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