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


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

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.
Small 1538243283
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.