How to prove $f(x)=x\cos(x)$ is a many to one function?

Question

I used Desmos to determine the graph of the function $f(x)=x\cos(x)$. Clearly, any horizontal line which is parallel to the $x$-axis cuts the graph at more than one point indicating the function is many-one. I wish to prove it is many-one by using calculus, for which I differentiated $f(x)$ to get $f'(x)= \cos(x) - x\sin(x)$. Now we need to prove that the function is not monotonic by showing the derivative changes sign. But how to show that it's not monotonic?

Answer 1

As for using a derivative of $f'(x) = \cos(x) - x\sin(x)$, note $f'\left(\left(2k + \frac{1}{2}\right)\pi\right) = -\left(2k + \frac{1}{2}\right)\pi$ and $f'\left(\left(2k - \frac{1}{2}\right)\pi\right) = \left(2k - \frac{1}{2}\right)\pi$ for all integers $k$. This shows that, for every $2\pi$ range, it goes from a continually larger negative to a larger positive value and then back to an even more negative value. Thus, it's not monotonic, as well as being unbounded.

For another, more direct way, note $f(2k\pi) = 2k\pi$ and $f((2k+1)\pi) = -(2k+1)\pi$ for all integers $k$. Next, since $f(x)$ is a continuous function, you can use this and the Intermediate value theorem to show that $f(x)$ is a many-to-one function for all real values.