maximum and minimum values

jeca86

Junior Member
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Sep 9, 2005
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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.
 
U need to know which sin that is equal to -1/2
I think sin 210 and sin 330 are equal to -1/2
 
Hello, jeca86!

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

f(x)=x2cosx,    [π,π]\displaystyle f(x)\:=\:x\,-\,2\cdot\cos x,\;\;[-\pi,\,\pi]

So i differentiated: f(x)=1+2sinx\displaystyle f'(x)\:=\:1\,+\,2\cdot\sin x
.
Then I ended with: sinx=12\displaystyle \sin x\,=\,-\frac{1}{2}

This is where I get stuck.
Where is the sine equal to -1/2 ?

. . On that interval: .x=π6,5π6\displaystyle x\,=\,-\frac{\pi}{6},\,-\frac{5\pi}{6}


Second derivative test: .f(x)=2cosx\displaystyle f''(x)\,=\,2\cdot\cos x

At x=π6:    f(π6)=2cos(π6)=2(32)=+3\displaystyle x = -\frac{\pi}{6}:\;\;f''(-\frac{\pi}{6}) \:= \:2\cdot\cos(-\frac{\pi}{6}) \:= \:2\left(\frac{\sqrt{3}}{2}\right) \:= \:+\sqrt{3}
. . f(x)\displaystyle f''(x) is positive, concave up: \displaystyle \cup . . . minimum

At x=5π6:    f(5π6)=2cos(5π6)=2(32)=3\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}
. . f(x)\displaystyle f''(x) is negative, concave down: \displaystyle \cap . . . maximum


f(π6)=π62cos(π6)=π62(32)=π63\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: (π6,π63)\displaystyle (-\frac{\pi}{6},\,-\frac{\pi}{6} - \sqrt{3})

f(5π6)=5π62cos(5π6)=5π62(32)=5π6+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: (5π6,35π6)\displaystyle (-\frac{5\pi}{6},\,\sqrt{3} - \frac{5\pi}{6})
 
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