Question #63609

This question tests the First Derivative Test Trap combined with the Weierstrass substitution ($t=\tan(x/2)$).

Students often blindly apply the identities $\cos^{-1}(\frac{1-t^2}{1+t^2}) = 2\tan^{-1}t$ and $\sin^{-1}(\frac{2t}{1+t^2}) = 2\tan^{-1}t$, concluding that $f(x) = 6\tan^{-1}(\tan(x/2)) = 3x$.

This naive rationalization of the derivative yields a constant positive slope, falsely implying the function has no stationary points or extrema.

The trap lies at the boundary of the substitution domain ($x=\pi$, where $t \to \infty$).

By properly mapping the Algebraic-Trigonometric graphs, the function reveals a non-differentiable corner point at $x=\pi$ where the slope sharply transitions from positive to negative, creating a hidden local maximum that standard rote optimization completely misses.