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tmagloire1
 one year ago
Find the xcoordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]
tmagloire1
 one year ago
Find the xcoordinates where f '(x) = 0 for f(x) = 2x + sin(2x) in the interval [0, 2π]

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0go ahead and differentiate

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2did you find \[f'(x)\]?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2\[ f(x) = 2x + \sin(2x) \] right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2what is the derivative of \(2x\)?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2now how about the derivative of \(\sin(2x)\)?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2so the derivative is not \(4\cos^2(x)\) is it?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2it is \[f'(x)=2+2\cos(2x)\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2final job is to set \[2+2\cos(2x)=0\] and solve for \(x\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2you good from there?

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Yes thank you! I was just getting the wrong derivative then my bad!

misty1212
 one year ago
Best ResponseYou've already chosen the best response.2\[\color\magenta\heartsuit\]

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0actually can someone help me find 2+2cos(2x)=0

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1`subtract` 2cos(2x)=2 `divide 2` cos(2x)=1

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1cosine of angle is 1 when the angle isssss...... pi, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1careful though! the angle is (2x). that leads to 2x=pi still need to go further to solve for x.

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0im still confused though. if it is 2x=pi than how do you get the x coordinates for the original problem @zepdrix

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0oh.my.god.im retarded lolol. so x=pi/2

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Wow how could i not see that. gr8 moments

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh oh we should be careful though...\[\large\rm \cos(2x)=1\qquad\to\qquad 2x=\pi+2k \pi,\qquad k\in \mathbb Z\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1It'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
 one year ago
Best ResponseYou've already chosen the best response.1This will be important because we might get several solutions within this interval.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Dividing by 2,\[\large\rm x=\frac{\pi}{2}+k \pi\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1we'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
 one year ago
Best ResponseYou've already chosen the best response.0k=1 will give us 3pi/2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good! :) Here is a graph to help you see what is going on. https://www.desmos.com/calculator/xm7sfacgsm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1That'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
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay so critical points is a another way of looking at it! Thank you so much!
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