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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?

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

1 Answer 1

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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.

Perhaps you can do the third interval case yourself and put it all together and get a solution?

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  • $\begingroup$ Thanks, could you elaborate on your second paragraph? "In the first...". Where does pi/4 come from? $\endgroup$
    – iveqy
    Jan 5, 2014 at 17:07
  • $\begingroup$ $\arctan(2x)$ is a strictly increasing function with value $\pi/4$ at $x=1/2$ $\endgroup$
    – neelp
    Jan 6, 2014 at 2:54

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