Monotonicity of f(x) = 2x - tan(x) using the derivative

First differentiate: $f'(x) = 2 - \sec^2(x)$. Setting $f'(x) = 0$ gives $\sec^2(x) = 2$, so $\cos^2(x) = \tfrac{1}{2}$, which means $\cos(x) = \pm\tfrac{1}{\sqrt{2}}$. Within $\left(-\pi/2, \pi/2\right)$, this yields critical numbers $x = -\pi/4$ and $x = \pi/4$.

These two critical numbers split the interval into three pieces: $(-\pi/2, -\pi/4)$, $(-\pi/4, \pi/4)$, and $(\pi/4, \pi/2)$. Testing a point in each: at $x = -\pi/3$, $\sec^2(-\pi/3) = 4$ so $f'(-\pi/3) = 2-4 = -2 < 0$; at $x=0$, $\sec^2(0)=1$ so $f'(0) = 2-1 = 1 > 0$; at $x = \pi/3$, $\sec^2(\pi/3) = 4$ so $f'(\pi/3) = 2-4 = -2 < 0$. So $f$ decreases on $(-\pi/2, -\pi/4)$, increases on $(-\pi/4, \pi/4)$, and decreases on $(\pi/4, \pi/2)$.

By the First Derivative Test, since $f'$ changes from negative to positive at $x=-\pi/4$, that point is a local minimum; since $f'$ changes from positive to negative at $x=\pi/4$, that point is a local maximum.

It is a local minimum. As shown in part a, f prime changes from negative to positive at x equals negative pi over four, since f decreases just before this point and increases just after it. By the First Derivative Test, a negative-to-positive sign change in the derivative indicates a local minimum.