Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]

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Find the x-coordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]

Mathematics
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go ahead and differentiate
HI!!
did you find \[f'(x)\]?

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Other answers:

f'(x) =4cos^2(x)
hmm no
\[ f(x) = 2x + \sin(2x) \] right?
Yep
what is the derivative of \(2x\)?
2
ok good
now how about the derivative of \(\sin(2x)\)?
2cos2x
ok
so the derivative is not \(4\cos^2(x)\) is it?
it is \[f'(x)=2+2\cos(2x)\]
final job is to set \[2+2\cos(2x)=0\] and solve for \(x\)
you good from there?
Yes thank you! I was just getting the wrong derivative then my bad!
\[\color\magenta\heartsuit\]
actually can someone help me find 2+2cos(2x)=0
`subtract` 2cos(2x)=-2 `divide 2` cos(2x)=-1
cosine of angle is -1 when the angle isssss...... pi, ya?
Ohh ok thank you!
careful though! the angle is (2x). that leads to 2x=pi still need to go further to solve for x.
im still confused though. if it is 2x=pi than how do you get the x coordinates for the original problem @zepdrix
divide... by 2 0_o
oh.my.god.im retarded lolol. so x=pi/2
Wow how could i not see that. gr8 moments
Oh oh we should be careful though...\[\large\rm \cos(2x)=-1\qquad\to\qquad 2x=\pi+2k \pi,\qquad k\in \mathbb Z\]
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.
This will be important because we might get several solutions within this interval.
Dividing by 2,\[\large\rm x=\frac{\pi}{2}+k \pi\]
So pi/2 and 5pi/2
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 :)
k=1 will give us 3pi/2
Good! :) Here is a graph to help you see what is going on. https://www.desmos.com/calculator/xm7sfacgsm
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!
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.
Ohh okay so critical points is a another way of looking at it! Thank you so much!
yay team \c:/

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