tmagloire1
  • tmagloire1
Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
go ahead and differentiate
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
did you find \[f'(x)\]?

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More answers

tmagloire1
  • tmagloire1
f'(x) =4cos^2(x)
misty1212
  • misty1212
hmm no
misty1212
  • misty1212
\[ f(x) = 2x + \sin(2x) \] right?
tmagloire1
  • tmagloire1
Yep
misty1212
  • misty1212
what is the derivative of \(2x\)?
tmagloire1
  • tmagloire1
2
misty1212
  • misty1212
ok good
misty1212
  • misty1212
now how about the derivative of \(\sin(2x)\)?
tmagloire1
  • tmagloire1
2cos2x
misty1212
  • misty1212
ok
misty1212
  • misty1212
so the derivative is not \(4\cos^2(x)\) is it?
misty1212
  • misty1212
it is \[f'(x)=2+2\cos(2x)\]
misty1212
  • misty1212
final job is to set \[2+2\cos(2x)=0\] and solve for \(x\)
misty1212
  • misty1212
you good from there?
tmagloire1
  • tmagloire1
Yes thank you! I was just getting the wrong derivative then my bad!
misty1212
  • misty1212
\[\color\magenta\heartsuit\]
tmagloire1
  • tmagloire1
actually can someone help me find 2+2cos(2x)=0
zepdrix
  • zepdrix
`subtract` 2cos(2x)=-2 `divide 2` cos(2x)=-1
zepdrix
  • zepdrix
cosine of angle is -1 when the angle isssss...... pi, ya?
tmagloire1
  • tmagloire1
Ohh ok thank you!
zepdrix
  • zepdrix
careful though! the angle is (2x). that leads to 2x=pi still need to go further to solve for x.
tmagloire1
  • tmagloire1
im still confused though. if it is 2x=pi than how do you get the x coordinates for the original problem @zepdrix
zepdrix
  • zepdrix
divide... by 2 0_o
tmagloire1
  • tmagloire1
oh.my.god.im retarded lolol. so x=pi/2
tmagloire1
  • tmagloire1
Wow how could i not see that. gr8 moments
zepdrix
  • zepdrix
Oh oh we should be careful though...\[\large\rm \cos(2x)=-1\qquad\to\qquad 2x=\pi+2k \pi,\qquad k\in \mathbb Z\]
zepdrix
  • zepdrix
It's not just pi, we can also add any multiple of 2pi to that angle, ya? It's like spinning an extra time around the circle and landing in the same location.
zepdrix
  • zepdrix
This will be important because we might get several solutions within this interval.
zepdrix
  • zepdrix
Dividing by 2,\[\large\rm x=\frac{\pi}{2}+k \pi\]
tmagloire1
  • tmagloire1
So pi/2 and 5pi/2
zepdrix
  • zepdrix
we're using integers for k. So k=0 gives us pi/2, looks good! k=1 gives us ... ? k=2 gives us 5pi/2 <- this is larger than 2pi, out of our interval, so we can throw it away :)
tmagloire1
  • tmagloire1
k=1 will give us 3pi/2
tmagloire1
  • tmagloire1
@zepdrix
zepdrix
  • zepdrix
Good! :) Here is a graph to help you see what is going on. https://www.desmos.com/calculator/xm7sfacgsm
zepdrix
  • zepdrix
I graphed the function 2x+sin(2x) from 0 to 2pi. You can clearly see that at pi/2 and 3pi/2 we have critical points!
zepdrix
  • zepdrix
That's what f'(x)=0 gives us, critical points. locations where the slope of the function is zero. graphically it looks like the top of a hill, or the bottom of a valley.
tmagloire1
  • tmagloire1
Ohh okay so critical points is a another way of looking at it! Thank you so much!
zepdrix
  • zepdrix
yay team \c:/

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