Hello, jeca86!

Find the absolute maximum and absolute minimum values of \(\displaystyle f\) on the given interval.

\(\displaystyle f(x)\:=\:x\,-\,2\cdot\cos x,\;\;[-\pi,\,\pi]\)

So i differentiated: \(\displaystyle f'(x)\:=\:1\,+\,2\cdot\sin x\)

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Then I ended with: \(\displaystyle \sin x\,=\,-\frac{1}{2}\)

This is where I get stuck.

Where is the sine equal to -1/2 ?

. . On that interval:

.\(\displaystyle x\,=\,-\frac{\pi}{6},\,-\frac{5\pi}{6}\)

Second derivative test:

.\(\displaystyle f''(x)\,=\,2\cdot\cos x\)

At \(\displaystyle x = -\frac{\pi}{6}:\;\;f''(-\frac{\pi}{6}) \:= \:2\cdot\cos(-\frac{\pi}{6}) \:= \:2\left(\frac{\sqrt{3}}{2}\right) \:= \:+\sqrt{3}\)

. . \(\displaystyle f''(x)\) is positive, concave up: \(\displaystyle \cup\) . . . minimum

At \(\displaystyle x=-\frac{5\pi}{6}:\;\;f''(-\frac{5\pi}{6}) \:= \:2\cdot\cos(-\frac{5\pi}{6}) \:= \:2\left(-\frac{\sqrt{3}}{2}\right) \:= \:-\sqrt{3}\)

. . \(\displaystyle f''(x)\) is negative, concave down: \(\displaystyle \cap\) . . . maximum

\(\displaystyle f(-\frac{\pi}{6})\:=\:-\frac{\pi}{6} - 2\cdot\cos(-\frac{\pi}{6}) \:= \:-\frac{\pi}{6} - 2(-\frac{\sqrt{3}}{2}) \:= \:-\frac{\pi}{6} - \sqrt{3}\)

. . Absolute minimum at: \(\displaystyle (-\frac{\pi}{6},\,-\frac{\pi}{6} - \sqrt{3})\)

\(\displaystyle f(-\frac{5\pi}{6})\:=\:-\frac{5\pi}{6} - 2\cdot\cos(-\frac{5\pi}{6}) \:=\:-\frac{-5\pi}{6} - 2(-\frac{\sqrt{3}}{2}) \:= \:-\frac{5\pi}{6} + \sqrt{3}\)

. . Absolute maximum at: \(\displaystyle (-\frac{5\pi}{6},\,\sqrt{3} - \frac{5\pi}{6})\)