Why cannot we find a constant for function $f(x) = \operatorname{arccot} x - \arctan\frac{1}{x}$ even though its derivative is $0$?

Question

I came across this exercise from Apostol book Volume 1 Ex 6.22 Q11.

The function $f(x) = \operatorname{arccot} x - \arctan\frac{1}{x}$ has derivative $0$ when $x\neq 0$. But in the meantime, we cannot find a constant number $C$ such that $f(x) = C$.

I am interested to know what is happening here.

Answer 1

There are two commonly-used definitions of $\mathrm{arccot}:$

1. $$\mathrm{arccot}(x)=\begin{cases} \arctan\left(\frac1x\right) &\text{ when }x\neq0;\\ \frac {\pi}2 &\text{ when }x=0. \end{cases}$$

   With Definition 1, $f(x)$ is indeed identically zero for nonzero $x.$

   

2. $$\mathrm{arccot}(x)=\begin{cases} \arctan\left(\frac1x\right)+\pi &\text{ when }x<0;\\ \frac {\pi}2 &\text{ when }x=0;\\ \arctan\left(\frac1x\right) &\text{ when }x>0. \end{cases}$$

   If Apostle is using Definition 2, then what's going on is this:

   

   Even though the Zero Derivative Theorem says that $$f'(x)=0\implies f(x)=C,$$ this is with reference to $f$ being continuous on the interval in question. Since $(-\infty,0)\cup(0,\infty)$ isn't an interval, the theorem is inapplicable.