## University Calculus: Early Transcendentals (3rd Edition)

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