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: 41

Answer

$ x=3$

Work Step by Step

Given $$ y=x^{2}-6 x+7$$ Then to find the critical points \begin{align*} y'&=0\\ 2x-6&=0 \end{align*} Hence the critical point is $ x=3$
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