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: 44

Answer

The tangent line at $x=\pi/2$ satisfies what the exercise asks for.
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Work Step by Step

$$y=f(x)=\cot x\hspace{1cm}\{0\lt x\lt\pi\}$$ 1) Find $f'(x)$: $$f'(x)=-\csc^2x=-\frac{1}{\sin^2x}$$ To find all the tangent lines parallel with the line $y=-x$, we rely on the fact that parallel lines have the same slope. So first, we would find all values of $x$ for which $f'(x)=-1$, which is the slope of $y=-x$. $$-\frac{1}{\sin^2x}=-1$$ $$\sin^2x=1$$ $$\sin x=\pm1$$ - For $\sin x=1$: On $(0,\pi)$, there is one value of $x$ for which $\sin x=1$, which is $x=\pi/2$. We have $f(\pi/2)=\cot(\pi/2)=0$ The equation of the tangent line is $$(y-0)=-(x-\frac{\pi}{2})=-x+\frac{\pi}{2}$$ $$y=-x+\frac{\pi}{2}$$ - For $\sin x=-1$: On $(0,\pi)$, there are no values of $x$ for which $\sin x=-1$. Therefore, overall, only the tangent line at $x=\pi/2$ satisfies the exercise.
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