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: 58

Answer

At $x=-2$, $y=-1/2$ minimum. At $x=0$, $y=1/2$ maximum.

Work Step by Step

Step 1. Given the function $y=\frac{x+1}{x^2+2x+2}$, we can find its derivative as $y′=\frac{x^2+2x+2-(x+1)(2x+2)}{(x^2+2x+2)^2}=\frac{-x^2-2x}{(x^2+1)^2}$ Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x^2+2x=0$ and we have $x=-2,0$ Step 3. At $x=-2$, $y=-1/2$ and we can identify this as a local and absolute minimum (use local test points) as shown in the figure. Step 4. At $x=0$, $y=1/2$ and we can identify this as a local and absolute maximum (use local test points) as shown in the figure.
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