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: 38

Answer

The graphs of the curve and its tangent lines are below.
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Work Step by Step

$$y=f(x)=1+\cos x\hspace{1cm}\{-3\pi/2\le x\le 2\pi\}$$ The derivative of $f(x)$: $$f'(x)=0-\sin x=-\sin x$$ The tangent lines to $f(x)$ are: 1) At $x=-\pi/3$: $f(x)=1+\cos(-\pi/3)=1+\frac{1}{2}=\frac{3}{2}$ $f'(x)=-\sin(-\pi/3)=-(-\frac{\sqrt3}{2})=\frac{\sqrt3}{2}$ $$y-\frac{3}{2}=\frac{\sqrt3}{2}(x+\frac{\pi}{3})=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi}{6}$$ $$y=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi}{6}+\frac{3}{2}=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi+9}{6}$$ 2) At $x=3\pi/2$: $f(x)=1+\cos(3\pi/2)=1+0=1$ $f'(x)=-\sin(3\pi/2)=-(-1)=1$ $$y-1=1(x-\frac{3\pi}{2})=x-\frac{3\pi}{2}$$ $$y=x-\frac{3\pi}{2}+1=x+\frac{2-3\pi}{2}$$ The graphs of the curve and its tangent lines are below.
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