Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 34


$x=\frac{\pi}{4}+\pi k$

Work Step by Step

First, note that to have a horizontal tangent, we must have a slope of zero. Since the derivative of a function gives us the slope at that point we will determine when $f'(x)=0$. If $f(x)=\mathrm{e}^{x}\cos{x}$ then $f'(x)=\mathrm{e}^{x}\cos{x}-\mathrm{e}^{x}\sin{x}=\mathrm{e}^{x}(\cos{x}-\sin{x})$ using the product rule. $f'(x)=\mathrm{e}^{x}(\cos{x}-\sin{x})=0 \Rightarrow \mathrm{e}^{x}=0$ or $\cos{x}-\sin{x}=0$. Since $\mathrm{e}^{x}\ne0$ we will solve $\cos{x}-\sin{x}=0$. If $\cos{x}-\sin{x}=0 \Rightarrow \sin{x}=\cos{x} \Rightarrow \tan{x}=1$ (dividing both sides by $\cos{x}$). Thus, $x=\frac{\pi}{4}+2\pi k, \frac{5\pi}{4}+2\pi k$ or simply $\frac{\pi}{4}+\pi k$.
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