Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 35


The critical points are $y=0, 2$.

Work Step by Step

Given, $g(y) = \frac{y-1}{y^2-y+1}$. Differentiating $g$ with respect to $y$ using quotient's rule, $$\frac{dg}{dy}=\frac{(1)(y^2-y+1)-(y-1)(2y-1)}{(y^2-y+1)^2}$$ The term $\frac{dg}{dy}$ is equal to $0$ at critical numbers. Thus, $$\frac{(1)(y^2-y+1)-(y-1)(2y-1)}{(y^2-y+1)^2}=0$$ $$y^2-y+1-2y^2+y+2y-1=0$$ $$-y^2+2y=0$$ $$y=0, 2.$$
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