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tmagloire1

  • one year ago

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

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  1. IrishBoy123
    • one year ago
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    go ahead and differentiate

  2. misty1212
    • one year ago
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    HI!!

  3. misty1212
    • one year ago
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    did you find \[f'(x)\]?

  4. tmagloire1
    • one year ago
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    f'(x) =4cos^2(x)

  5. misty1212
    • one year ago
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    hmm no

  6. misty1212
    • one year ago
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    \[ f(x) = 2x + \sin(2x) \] right?

  7. tmagloire1
    • one year ago
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    Yep

  8. misty1212
    • one year ago
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    what is the derivative of \(2x\)?

  9. tmagloire1
    • one year ago
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    2

  10. misty1212
    • one year ago
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    ok good

  11. misty1212
    • one year ago
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    now how about the derivative of \(\sin(2x)\)?

  12. tmagloire1
    • one year ago
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    2cos2x

  13. misty1212
    • one year ago
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    ok

  14. misty1212
    • one year ago
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    so the derivative is not \(4\cos^2(x)\) is it?

  15. misty1212
    • one year ago
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    it is \[f'(x)=2+2\cos(2x)\]

  16. misty1212
    • one year ago
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    final job is to set \[2+2\cos(2x)=0\] and solve for \(x\)

  17. misty1212
    • one year ago
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    you good from there?

  18. tmagloire1
    • one year ago
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    Yes thank you! I was just getting the wrong derivative then my bad!

  19. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  20. tmagloire1
    • one year ago
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    actually can someone help me find 2+2cos(2x)=0

  21. zepdrix
    • one year ago
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    `subtract` 2cos(2x)=-2 `divide 2` cos(2x)=-1

  22. zepdrix
    • one year ago
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    cosine of angle is -1 when the angle isssss...... pi, ya?

  23. tmagloire1
    • one year ago
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    Ohh ok thank you!

  24. zepdrix
    • one year ago
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    careful though! the angle is (2x). that leads to 2x=pi still need to go further to solve for x.

  25. tmagloire1
    • one year ago
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    im still confused though. if it is 2x=pi than how do you get the x coordinates for the original problem @zepdrix

  26. zepdrix
    • one year ago
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    divide... by 2 0_o

  27. tmagloire1
    • one year ago
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    oh.my.god.im retarded lolol. so x=pi/2

  28. tmagloire1
    • one year ago
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    Wow how could i not see that. gr8 moments

  29. zepdrix
    • one year ago
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    Oh oh we should be careful though...\[\large\rm \cos(2x)=-1\qquad\to\qquad 2x=\pi+2k \pi,\qquad k\in \mathbb Z\]

  30. zepdrix
    • one year ago
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    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.

  31. zepdrix
    • one year ago
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    This will be important because we might get several solutions within this interval.

  32. zepdrix
    • one year ago
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    Dividing by 2,\[\large\rm x=\frac{\pi}{2}+k \pi\]

  33. tmagloire1
    • one year ago
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    So pi/2 and 5pi/2

  34. zepdrix
    • one year ago
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    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 :)

  35. tmagloire1
    • one year ago
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    k=1 will give us 3pi/2

  36. tmagloire1
    • one year ago
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    @zepdrix

  37. zepdrix
    • one year ago
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    Good! :) Here is a graph to help you see what is going on. https://www.desmos.com/calculator/xm7sfacgsm

  38. zepdrix
    • one year ago
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    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!

  39. zepdrix
    • one year ago
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    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.

  40. tmagloire1
    • one year ago
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    Ohh okay so critical points is a another way of looking at it! Thank you so much!

  41. zepdrix
    • one year ago
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    yay team \c:/

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