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: 41

Answer

The graph of $y=x-\cot x$ does not have any horizontal tangents on $[0,2\pi]$.
1538240726

Work Step by Step

$$y=f(x)=x-\cot x$$ 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 $[0,2\pi]$. $$f'(x)=(x-\cot x)'=1-(-\csc^2 x)=1+\csc^2x$$ We have $f'(x)=0$ when $$1+\csc^2 x=0$$ $$\csc^2 x=-1$$ We know that $\csc^2x\ge0$ for all $x$. That means there is no value of $x$ for which $\csc^2 x=-1$. So the graph of $f(x)$ does not have any horizontal tangents on $[0,2\pi]$.
Small 1538240726
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.