Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.1 Analysis Of Functions I: Increase, Decrease, and Concavity - Exercises Set 3.1 - Page 196: 54

Answer

Decreasing: $x=-\frac{\sqrt{3}}{3}$ Increasing: $x=\frac{\sqrt{3}}{3}$

Work Step by Step

First derivative \[ \frac{-2 x}{\left(x^{2}+1\right)^{2}}=y^{\prime} \] Second derivative \[ y^{\prime \prime}=\frac{-3+6 x^{2}}{\left(x^{2}+1\right)^{3}} \] \[ \begin{array}{l} 0^{\prime}=\frac{6 x^{2}-2}{\left(1+x^{2}\right)^{3}} \\ -2+6 x^{2}=0 \\ 6 x^{2}=2 \\ x=\pm \frac{\sqrt{3}}{3} \end{array} \] The function $y$ increases/decreases most rapidly; if the function $y^{\prime}$ has a maximum/minimum, $y^{\prime \prime}$ has a zero at this value. \[ y^{\prime}=\frac{-2 \cdot \frac{\sqrt{3}}{3}}{\left(\frac{1}{9}+1\right)^{2}}<0 \] and \[ y^{\prime}=\frac{-2 \cdot-\frac{\sqrt{3}}{3}}{\left(1+\frac{1}{9}\right)^{2}}>0 \] Determine if $y^{\prime}$ is positive or negative at the roots of $y^{\prime \prime}(\text { there, } y$ is increasing most/decreasing most).
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