How do I show that $x - \tan x$ has exactly 3 solutions on $(-2\pi, 2\pi)$

Question

Is there an easy way to show this without graphing the function? I am really looking for how 12th grader you would solve this.

Answer 1

I will leave the original solution here, for anyone who wants to read it over quickly.

Define $I_n:=(-\pi/2,\pi/2)+\pi n$. The function $f(x)=\tan x-x$ is strictly increasing and continuous on $I_n$ with $f((-\pi/2+\pi n)^+)=-\infty,f((\pi/2+\pi n)^-)=\infty$. Hence $f$ has exactly one root in $I_n$.

Now, we have exactly three such intervals included in $(-2\pi,2\pi)$. It remains only to prove that there are no roots on $(-2\pi,-3\pi/2)$ and $(3\pi/2,2\pi)$. For example, over $(3\pi/2,2\pi)$, since $f(x)$ is increasing we have:

$$f(x)<f(2\pi)=-2\pi<0$$

so no real roots over $(3\pi/2,2\pi)$. The case $(-2\pi,-3\pi/2)$ is similar. In conclusion, there are exactly three real roots over $(-2\pi,2\pi)$.

Now I'll try to explain it more clearly. We are asked to prove that the equation $\tan x = x$ has exactly three roots over $[-2\pi,2\pi]$. This is the same as saying the function

$$f:(-2\pi, 2\pi)\to \mathbb{R},\ f(x)=\tan x -x$$

has exactly three zeroes. However, the domain is not correct because $\tan$ is not defined at $k\frac{\pi}{2}, k\in \mathbb{Z}^*$. So, we should actually define $f$ as:

$$f:\left(-2\pi,-\frac{3\pi}{2}\right) \cup \left(-\frac{3\pi}{2},-\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\frac{3\pi}{2}\right) \cup \left(\frac{3\pi}{2},2\pi\right)\to \mathbb{R}$$

The intervals $\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right), \left(-\frac{\pi}{2},\frac{\pi}{2}\right),\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ are the intervals I denoted by $I_n$ in the original solution. The function $\tan x$ is periodic with period $\pi$ so its behavior is the same over each of these intervals. So let's pick one of them, say $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$. I'll add a picture of the functions over $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, so that it's easier to grasp the arguments.

With red, it's $\tan x$ and with green it's $f(x)=\tan x - x$.

Now, as you can see in the picture, the function $f$ is continuous over $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, of course this is because both $\tan x$ and $x$ are continuous over $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ and since $f$ is the difference of two continuous function, it is continuous. Also, from the picture, you can see that both functions are increasing. To support this, we can compute the first derivative: $(\tan x)'=\tan^2 x+1>0$ and $f'(x)=\tan^2x\geq 0$, so clearly they are both increasing.

Now, let's see from and toward what do they increase. For this, we must compute the limit to the right of the lower bound and the limit to the left of the upper bound. We have:

$$\lim_{x\to \frac{\pi}{2}^+} f(x)=\lim_{x\to \frac{\pi}{2}^+} \tan x-\lim_{x\to \frac{\pi}{2}^+}x=-\infty-\frac{\pi}{2}=-\infty$$

$$\lim_{x\to \frac{3\pi}{2}^-} f(x)=\lim_{x\to \frac{3\pi}{2}^-} \tan x-\lim_{x\to \frac{3\pi}{2}^-}x=\infty-\frac{3\pi}{2}=\infty$$

Now, since the function is continuous and increases from $-\infty$ to $\infty$, this means it covers all the possible values from $\mathbb{R}$ exactly once. That means, it intersects the $x$-axis exactly once over the interval $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, i.e. we have exactly one root over this interval (the $B$ point in the picture).

By exactly the same arguments, there is exactly one solution over $\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)$ and exactly one solution over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

So we have our three roots. It remains to prove that there are none other over the rest of $f$'s domain.

Let's take for example $\left(\frac{3\pi}{2},2\pi\right)$.

By the same arguments as previously, $\tan x- x$ is increasing and continuous over $\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)$, which means it is increasing over $\left(\frac{3\pi}{2},2\pi\right]$. Thus, for any $x\in \left(\frac{3\pi}{2},2\pi\right)$, we have

$$f(x) < f(2\pi)=\tan 2\pi-2\pi=-2\pi < 0$$

This means $f(x)$ can't be zero for any $x\in \left(\frac{3\pi}{2},2\pi\right)$ (since it is smaller than zero), which means no roots over this interval. You can also see in the picture that $f(x)$ is way below the $x-axis$. To be more specific, $f$'s root over the interval $\left(\frac{3\pi}{2},\frac{5\pi}{2}\right)$ lies in the right half $\left(2\pi, \frac{5\pi}{2}\right)$.

So, we're good with the interval $\left(\frac{3\pi}{2},2\pi\right)$, because we don't have any other roots here. For the interval $\left(-2\pi,-\frac{3\pi}{2}\right)$ the argument is almost identical (you have $f(x) > f(-2\pi)=2\pi >0$), so no solutions there either.

In conclusion, there are exactly three roots over the interval $(-2\pi, 2\pi)$.