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?

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 $f(x)>0.$
