# How many solutions has this equation?

I've the equation $x - \arctan(2x) = a$ and the question is, how many solution has the equation for different values of $a$, where $a$ is a real number.

I've plotted the graph and found the extreme values at $x = \pm \frac{1}{2}$.

so naturally I would say:

1 solution for $a < -\frac{1}{2}$ and $a > \frac{1}{2}$

2 solutions for $a = \pm \frac{1}{2}$

and

3 solutions for $-\frac{1}{2} < a < \frac{1}{2}$

However that's wrong. The solution contains $\pi/4$ too, why? How to solve this?

• $-\inf < x < \inf$ and x is reell. Jan 5, 2014 at 14:52
• $\pm.5$ are only the preimage of the extremes. The local maximum and minimum are $\pm(.5-\arctan1)$ Jan 5, 2014 at 14:54
• It seems that you have interpreted the extreme values of $x$ as boundary values of $a$, while it should be for $f(x_{extreme})$. Jan 5, 2014 at 15:03

Differentiating $f(x) = x - \arctan(2x)-a$, we get ${f}'(x) = 1-\frac{2}{1+4x^{2}} = 0$ when $x = \frac{1}{2}$ or $x= -\frac{1}{2}$. If $|x| > -\frac{1}{2}$, then function is increasing. Otherwise, it is decreasing. Thus, $f(x)$ has at most 3 roots and we just have to check the intervals $x > \frac{1}{2}$, $x < - \frac{1}{2}$, and $-\frac{1}{2} \leq x \leq \frac{1}{2}$.
In the first interval, $\arctan(2x) > \pi/4$ and $x > \frac{1}{2}$ and $f(x)$ is increasing. Thus, if $a$ is greater than $\frac{1}{2} -\pi/4$ then you get a solution on that interval. Otherwise, you do not.
In second interval, $\arctan(2x) < -\pi/4$ and $x < -\frac{1}{2}$ and $f(x)$ is increasing. So we want $f(\frac{1}{2}) > 0$ for us to have a solution on this interval. Hence, $a < \pi/4 -1/2$ gives you a root on this interval. Otherwise, you do not have a root.
• $\arctan(2x)$ is a strictly increasing function with value $\pi/4$ at $x=1/2$ Jan 6, 2014 at 2:54