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: 32

Answer

See graph and explanations.

Work Step by Step

Step 1. Given $y=x\sqrt {4-x^2}=x(4-x^2)^{1/2}, -2\leq x\leq 2$, we have $y'=(4-x^2)^{1/2}-x^2(4-x^2)^{-1/2}=2(2-x^2)(4-x^2)^{-1/2}$ and $y''=-4x(4-x^2)^{-1/2}+2x(2-x^2)(4-x^2)^{-3/2}=2x(x^2-6)(4-x^2)^{-3/2}$ Step 2. Letting $y'=0$, we have $2(2-x^2)(4-x^2)^{-1/2}=0$, which gives $x=\pm\sqrt 2$; also $x=\pm2$ are endpoints. Thus $x=\pm\sqrt 2, \pm2$ are possible critical points. We have $y(-2)=0,y(-\sqrt 2)=-2,y(\sqrt 2)=2 y(2)=0$. Step 3. Check signs of $y'$ to get $(-2)..(-)..(-\sqrt 2)..(+)..(\sqrt 2)..(-)..(2)$; thus the function decreases on $(-2,-\sqrt 2)$, $(\sqrt 2,2)$ and increases on $(-\sqrt 2,\sqrt 2)$ . This also means that $y(-\sqrt 2)=-2, y(2)=0$ are local minima and $y(-2)=0,,y(\sqrt 2)=2$ are local maxima. Step 4. Letting $y''=0$, we have $x=0, \pm\sqrt 6$, but only $x=0$ is in the domain. Check the signs of $y''$ to get $.(-2)..(+)..(0)..(-)..(2)$, indicating $x=0$ is an inflection point with the function concave up in $(-2,0)$ and concave down in $(0,2)$. Step 5. Graph the function based on the above information as shown in the figure.
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