Show that the equation $\tan x = −x$ has a single root on the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$.

Show that the equation $\tan x = −x$ has a single root on the interval $(\frac{-\pi}{2}, \frac{\pi}{2})$.

We have $0 \in (\frac{-\pi}{2}, \frac{\pi}{2})$ then $\tan(0)=-0$

But its unique because $f(x)=\tan(x)+x$ is differentiable in $(\frac{-\pi}{2}, \frac{\pi}{2})$ then $f'(x)= 1+\frac{1}{\cos^2(x)}=0$ we have $-1=\cos^{2}(x)$ its a contradiction. Am I right?

Solution 1:

As you said there is at least one root to the function $f(x) = \tan x + x$. Let $a, b$ two roots of $f$ (with $-\pi/2 \le a < b \le \pi/2$). By the mean value theorem, there is $c \in (a, b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a} = 0.$$ But as you said, $f'(c) > 0$ for all $c$, which is a contradiction.

Solution 2:

Alternative approach:

Assuming that $-\pi/2 < x < \pi/2$ and that $x \neq 0$:

• $x$ is positive if and only if $\tan(x)$ is positive.
• $x$ is negative if and only if $\tan(x)$ is negative.

The above two bullet points follow, since the $\cos(x)$ is positive throughout the interval, the $\sin(x)$ is positive for $x$ positive and the $\sin(x)$ is negative for $x$ negative.

Therefore, $x = 0$ is the only possible solution to the original problem.

Solution 3:

To be a little more precise, since $f'(x)=1+\sec^2 x>0$ the function $f(x)$ is monotonically increasing. This means that if for $a,b\in (-\pi/2,\pi/2)$ we have $a<b$ then $f(a)<f(b).$ Then for $x<0$ we must have $f(x)<0$ and for $x>0$ we have $f(x)>0.$