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anonymous

  • one year ago

Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When g(t) = t+ cost , -2pi less than or equal to "t" less than or equal to 2pi A) Find the intervals of increase of decrease B) Find the local max and mins C) Find the intervals of concavity and inflection points D) Use the information from parts A-C to sketch the graph.

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  1. anonymous
    • one year ago
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    So for the first derivative I got: g' (t) = 1-sint t and so the critical values would be pi/2 & -pi/2 ?

  2. DanJS
    • one year ago
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    sin(x) = 1 when x = ... in the interval of the function

  3. anonymous
    • one year ago
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    when x= -pi/2 and pi/2 ?

  4. DanJS
    • one year ago
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    just pi/2 so far

  5. DanJS
    • one year ago
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    -pi/2 is the negative y axis and the sin would be -1

  6. DanJS
    • one year ago
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    so you can subtract 2 pi from pi/2 and still be in the interval of the function

  7. anonymous
    • one year ago
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    hmm... so its jut pi/2? I get really confused when they start given me intervals where there is negatives? Does -2pi mean we go around the unit circle clockwise?

  8. DanJS
    • one year ago
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    sin(x) = 1 , when x = pi/2 + k*2pi, for any k value

  9. DanJS
    • one year ago
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    just swinging around 360 degrees on the unit circle to the same spot with the 2pi

  10. DanJS
    • one year ago
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    when k=-1 x = p/2 - 2 pi = -3pi/2

  11. DanJS
    • one year ago
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    so you have two critical points inside the interval of the function.... x = pi/2 and x = -3pi/2

  12. anonymous
    • one year ago
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    So to test out a number before -3pi/2 can I choose -pi?

  13. DanJS
    • one year ago
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    no , larger, -3/2 is -1.5

  14. DanJS
    • one year ago
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    try -2pi

  15. anonymous
    • one year ago
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    That would give me a postive value when plugged in

  16. DanJS
    • one year ago
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    right, do the same for the other 2 intervals...

  17. anonymous
    • one year ago
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    Could I plug in 0 for the numbers between -3pi/2 and pi/2 and then 2pi after pi/2?

  18. DanJS
    • one year ago
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    yes of course

  19. anonymous
    • one year ago
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    sorry, just making sure

  20. DanJS
    • one year ago
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    any number inside the interval will work ,

  21. anonymous
    • one year ago
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    thats weird, all of them give me postive values? So does that mean that there is no interval where the function is decreasing?

  22. DanJS
    • one year ago
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    |dw:1437608804982:dw|

  23. DanJS
    • one year ago
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    yep, the function is always increasing over the interval, but they do have horizontal tangents at those critical values like above

  24. DanJS
    • one year ago
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    calculate the y values, so you can add the (x,y) points for those two points on the graph

  25. DanJS
    • one year ago
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    since it is always increasing you can sketch a rough draft , you may want to calculate the x and y intercept points also

  26. DanJS
    • one year ago
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    |dw:1437609016552:dw|

  27. DanJS
    • one year ago
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    The points are like the horizontal tangent on the graph of y=x^3, increasing both sides

  28. anonymous
    • one year ago
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    I just dont get how you got the horizontal tangent?

  29. DanJS
    • one year ago
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    you took the derivative, and found that sin(x) = 1 , when the derivative is zero right?

  30. anonymous
    • one year ago
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    OH.. yea.. those values we got were the horizontal tangent ...

  31. DanJS
    • one year ago
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    yes, when you hear derivative, think "slope of a tangent line there"

  32. anonymous
    • one year ago
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    Yea, just forgot for a second there

  33. anonymous
    • one year ago
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    So is function increasing from (-2pi, 2pi)?

  34. DanJS
    • one year ago
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    no problem, so simple things to add to the graph to refine before the second derivative are, x and y intercepts, and the values of the function at the endpoints of the interval

  35. DanJS
    • one year ago
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    right, increasing everywhere

  36. DanJS
    • one year ago
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    with horizontal tangents at those two pooints you found

  37. anonymous
    • one year ago
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    so for the horizontal tangents I got points: (-pi/2, - pi/2) and the (-3pi/2, -5.712)

  38. anonymous
    • one year ago
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    and then for the second derivative, I would get: g''(t) = cos (t)

  39. anonymous
    • one year ago
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    And for its critical values I would get pi/2 and -3pi/2, the same values we got before for the first derivative?

  40. DanJS
    • one year ago
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  41. DanJS
    • one year ago
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    Those are all the points you can add before the second derivative test

  42. DanJS
    • one year ago
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    any up to there you dont understand

  43. anonymous
    • one year ago
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    Okay, and well when did the second derivative test.. I got CD: (-2pi, -3pi/2) and CD: (-3pi/2, pi/2) U (pi/2,2pi)

  44. anonymous
    • one year ago
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    For the inflection point: I got (-3pi/2, -3pi/2)

  45. anonymous
    • one year ago
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    I have a good feeling im right

  46. DanJS
    • one year ago
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    \[F(x) = x + \cos(x)\] \[F'(x) = 1 + \sin(x)\] \[F''(x) = -\cos(x) \]

  47. DanJS
    • one year ago
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    second derivative is negative cosine

  48. anonymous
    • one year ago
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    wait...but doesnt -sin = cos?

  49. anonymous
    • one year ago
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    so that would just make it cos(t)

  50. DanJS
    • one year ago
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    sorry back

  51. anonymous
    • one year ago
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    lol k you almost gave me a heartattack

  52. DanJS
    • one year ago
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    yes, so cos(x) = 0 at pi/2 and 3pi/2 plus and minus since you can move 2 pi backwards

  53. anonymous
    • one year ago
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    we dont use -3pi/2? Only 3pi/2?

  54. DanJS
    • one year ago
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  55. DanJS
    • one year ago
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    notice where cos(x) is zero in the interval

  56. anonymous
    • one year ago
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    wow, so do I use all those points?

  57. DanJS
    • one year ago
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    Cos(x) = 0 when the angle is at 90 or 270 , or any addition or subtraction of 360

  58. DanJS
    • one year ago
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    Cos(x) = 0 , x = pi/2 , x = 3pi/2, take off 2 pi from both x = -pi/2, x = -3pi/2

  59. DanJS
    • one year ago
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    It is the X coordinate on the unit circle, zero on the Y axis at 90 and 270 degrees

  60. anonymous
    • one year ago
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    yea makes sense

  61. DanJS
    • one year ago
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    So those are your 4 test points

  62. anonymous
    • one year ago
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    k thanks

  63. DanJS
    • one year ago
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    welcome, the graph of the actual function is above for reference..

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