Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises - Page 268: 65

Answer

The tangent line is horizontal at the points $~~(\frac{\pi}{4},\sqrt{2})~~$ and $~~(\frac{5\pi}{4},-\sqrt{2})$

Work Step by Step

$y = sin~x+cos~x$ $y' = cos~x-sin~x$ If the tangent line is horizontal, then $y'=0$ We can find the values of $x$ where $y' = 0$: $y' = cos~x-sin~x = 0$ $cos~x = sin~x$ $tan~x = 1$ $x = \frac{\pi}{4}, \frac{5\pi}{4}$ When $x = \frac{\pi}{4}$: $y = sin~\frac{\pi}{4}+cos~\frac{\pi}{4}$ $y = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}$ $y = \sqrt{2}$ When $x = \frac{5\pi}{4}$: $y = sin~\frac{5\pi}{4}+cos~\frac{5\pi}{4}$ $y = -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}$ $y = -\sqrt{2}$ The tangent line is horizontal at the points $~~(\frac{\pi}{4},\sqrt{2})~~$ and $~~(\frac{5\pi}{4},-\sqrt{2})$
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