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 4.4 - Concavity and Curve Sketching - Exercises 4.4 - Page 213: 38

Answer

$y(0)=2\sqrt 2$ is a local maximum and $y(-\sqrt 2)=0, y(\sqrt 2)=0$ is a local minima; absolute minimum is at $y(-\sqrt 2)=0, y(\sqrt 2)=0$ and the absolute maximum is at $y(0)=2\sqrt 2$; concave up on $(-\sqrt 2,-1), (1,\sqrt 2)$ and concave down on $(-1,1)$; $x=\pm1$ are inflection points. See graph.

Work Step by Step

Step 1. Given the function $y=(2-x^2)^{3/2}, -\sqrt2\leq x\leq\sqrt2$, we have $y'=\frac{3}{2}(2-x^2)^{1/2}(-2x)=3x(2-x^2)^{1/2}$ and $y''=3(2-x^2)^{1/2}+\frac{3x}{2}(2-x^2)^{-1/2}(-2x)=\frac{6-3x^2}{\sqrt {2-x^2}}-\frac{3x^2}{\sqrt {2-x^2}}=\frac{6-6x^2}{\sqrt {2-x^2}}$ Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=-\sqrt 2, 0, \sqrt 2$ as critical points with $y(-\sqrt 2)=0, y(0)=2\sqrt 2, y(\sqrt 2)=0$ Step 3. Examine the sign change of $y'$ across the critical points: $(-\sqrt 2)..(+)..(0)..(-)..(\sqrt 2)$; we can identify $ y(0)=2\sqrt 2$ as a local maximum and $y(-\sqrt 2)=0, y(\sqrt 2)=0$ as a local minima. The absolute minimum is at $y(-\sqrt 2)=0, y(\sqrt 2)=0$ and the absolute maximum is at $y(0)=2\sqrt 2$ . Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=\pm1$ as possible inflection points within the domain. Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $(-\sqrt 2)..(+)..(-1). (-)..(1)..(+)..(\sqrt 2)$, Step 6. The function is concave up on $(-\sqrt 2,-1), (1,\sqrt 2)$ and concave down on $(-1,1)$, this means that $x=\pm1$ are inflection points. Step 7. See graph.
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