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: \(\displaystyle f'(x)\:=\:1\,+\,2\cdot\sin x\)
.
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: \(\displaystyle (-\frac{5\pi}{6},\,\sqrt{3} - \frac{5\pi}{6})\)
