Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 38


$ y=\ \frac{1}{36}$ is the minimum, when $ x=\frac{1}{18}$

Work Step by Step

Given $$ y=-9x^{2}+ x $$ So, we have \begin{aligned} y&=-9x^{2}+x\\ &=-9(x^2-\frac{1}{9}x)\\ &=-9(x^2-\frac{1}{9}x+\frac{1}{18^2}-\frac{1}{18^2})\\ &=-9((x-\frac{1}{18})^2-\frac{1}{324})\\ \end{aligned} The lowest value for the squared term is zero. So, we see that $ y=\frac{9}{324}=\frac{1}{36}$ is the minimum, when $ x=\frac{1}{18}$.
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