maximum and minimum values

jeca86

Junior Member
Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= x-2cosx [-pi, pi]

So i differentiated:
f'(x)= 1+2sinx
I think this is right.
Then I ended with sinx=-1/2.
This is where I get stuck. I think the next step goes back to trig, but I can't remember. I don't know how to get the values from what I have ended up with.

ryan_kidz

Junior Member
U need to know which sin that is equal to -1/2
I think sin 210 and sin 330 are equal to -1/2

soroban

Elite Member
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: $$\displaystyle (-\frac{5\pi}{6},\,\sqrt{3} - \frac{5\pi}{6})$$