No calculus allowed.

Suppose we have the vectors $\left \langle \arctan x, \text{arccot} x \right \rangle$ and $\left \langle 1, 1 \right \rangle$ . Then, by the Cauchy-Schwarz inequality, we have $\sqrt{\arctan^2 x + \text{arccot}^2 x } \sqrt{2} \geqslant \arctan x + \text{arccot} x$ . Dividing on both sides, we obtain that $f(x) \leqslant \boxed{\sqrt{2}}$ .