University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 151: 42

Answer

The graph of $y=x+2\cos x$ has two horizontal tangents on $[0,2\pi]$, which are at $x=\pi/6$ and at $x=5\pi/6$.

Work Step by Step

$$y=f(x)=x+2\cos x$$ To know whether the graph of $f(x)$ has any horizontal tangents on $[0,2\pi]$ or not, we rely on the fact that horizontal tangents are the only ones which possess the slope value $0$. So, by taking the derivative of $f(x)$, we will see whether the derivative can obtain the value $0$ or not on $[0,2\pi]$. $$f'(x)=(x+2\cos x)'=1-2\sin x$$ We have $f'(x)=0$ when $$1-2\sin x=0$$ $$\sin x=\frac{1}{2}$$ On $[0,2\pi]$, there are two values of $x$ for which $\sin x=1/2$, which are $x=\pi/6$ and $x=5\pi/6$. So the graph of $f(x)$ has two horizontal tangents on $[0,2\pi]$, which are at $x=\pi/6$ and at $x=5\pi/6$.
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