[tex]f(x) =\text{arctg}\,x+\text{arctg}\,\dfrac{1}{x} \\ \\ f'(x) = \dfrac{1}{1+x^2}+\dfrac{\Big(\dfrac{1}{x}\Big)'}{1+\Big(\dfrac{1}{x}\Big)^2} = \dfrac{1}{1+x^2}+\dfrac{-\dfrac{1}{x^2}}{\dfrac{x^2+1}{x^2}} = \dfrac{1}{1+x^2}-\dfrac{1}{1+x^2} = 0\\ \\ f(x) = \text{arctg}\,x+\text{arctg}\,\dfrac{1}{x} = \left\{\begin{array}{III}-\dfrac{\pi}{2},\quad x<0\\\\\dfrac{\pi}{2},\quad x\geq 0 \end{array}\right[/tex]
