Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 180: 6


0 and $\frac{1}{12}$

Work Step by Step

$ f'(t)$ exists for all t. Therefore, the only critical points are the solutions of $ f'(t)=0$. $ f'(t)=24t^{2}-2t=0$ $\implies t=0$ or $24t^{2}=2t\implies 12t=1\implies t=\frac{1}{12}$ Thus, the critical points are the roots $ c= 0$ and $ c=\frac{1}{12}$
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