Locating and classifying critical points using the Second Derivative Test

Compute $f'(x) = \cos(x) - \sin(x)$. Setting this equal to zero gives $\cos(x) = \sin(x)$, which happens at $x = \pi/4$ and $x = 5\pi/4$ in $(0, 2\pi)$.

Compute $f''(x) = -\sin(x) - \cos(x)$. At $x = \pi/4$: $f''(\pi/4) = -\sin(\pi/4) - \cos(\pi/4) = -\sqrt{2} < 0$, so this is a local maximum. At $x = 5\pi/4$: $f''(5\pi/4) = -\sin(5\pi/4) - \cos(5\pi/4) = \sqrt{2} > 0$, so this is a local minimum.

Set $f''(x) = -\sin(x) - \cos(x) = 0$, which simplifies to $\sin(x) = -\cos(x)$, or $\tan(x) = -1$. In $(0, 2\pi)$, this gives $x = 3\pi/4$ and $x = 7\pi/4$.

Testing sign of $f''(x)$ around $x = 3\pi/4$: it goes from negative (concave down) to positive (concave up), confirming a true inflection point. Around $x = 7\pi/4$: $f''$ goes from positive (concave up) to negative (concave down), confirming another true inflection point.