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iop360

  • 3 years ago

solve the following for x...

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  1. iop360
    • 3 years ago
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    \[\frac{ f(x) - (-33) }{ x - 0 }\]

  2. iop360
    • 3 years ago
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    f(x) = \[f(x) = 36xsin(17x) + 306\sqrt{3}x^2 -33\]

  3. josiahh
    • 3 years ago
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    x=0

  4. iop360
    • 3 years ago
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    \[\frac{ f(x) - (-33)}{ x-0 } = f'(x)\]

  5. josiahh
    • 3 years ago
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    that is the derivative formula at x=0; X=0 is your answer

  6. iop360
    • 3 years ago
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    is it? the answers says that x = 5Pi/102 and 7Pi/102

  7. josiahh
    • 3 years ago
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    alright Plug in either 5pi/102 or 7pi/102 in f(x) and see if you get -33.

  8. iop360
    • 3 years ago
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    the original question was find the x coordinates of all points on the curve \[y = 36xsin(17x) + 306\sqrt{3}x^2 -33\] with a domain of \[\frac{ \Pi }{ 34 } < x < \frac{ 3\Pi }{ 34 }\] where to tangent line passes through the point P(0, -33)

  9. iop360
    • 3 years ago
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    so im not sure if the rise/run formula is the right thing to do

  10. iop360
    • 3 years ago
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    also the point (0, -33) is actually not on the curve, but this line is touching points of the curve somewhere

  11. iop360
    • 3 years ago
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    which is where answer x= 5Pi/102 and 7Pi/102 comes from

  12. josiahh
    • 3 years ago
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    alright. let's see if we can write the equation of the tangent line

  13. iop360
    • 3 years ago
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    maybe?

  14. iop360
    • 3 years ago
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    with one m being rise/ivrun and the other m being the derivativer equate the two toeeach other

  15. iop360
    • 3 years ago
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    dont think it works though

  16. josiahh
    • 3 years ago
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    yea the derivative is not defined at 0

  17. iop360
    • 3 years ago
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    m = m \[\frac{ f(x) - (-33) }{ x - 0} = f'(x)\] \[\frac{ (36xsin(17x) + 306\sqrt{3}x^2 -33) - (-33) }{x } = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]

  18. iop360
    • 3 years ago
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    i tried doing this but im still not getting the correct answer

  19. iop360
    • 3 years ago
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    did you try getting f'(x), maybe i did it wrong

  20. josiahh
    • 3 years ago
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    is the x^2 in the square root?

  21. iop360
    • 3 years ago
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    no

  22. josiahh
    • 3 years ago
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    oh, then I was doing it all wrong... let me start over

  23. josiahh
    • 3 years ago
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    the slope is 0; you just have to use the x in the denominator to cancel the xs in the numerator and take the limit of that as x approaches 0

  24. iop360
    • 3 years ago
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    so take the limit after the rise/run formula after you factor it?

  25. iop360
    • 3 years ago
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    of the rise/run*

  26. josiahh
    • 3 years ago
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    yea

  27. iop360
    • 3 years ago
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    this gives you the slope, and then what do you do to get the answer

  28. iop360
    • 3 years ago
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    \[\lim_{x \rightarrow0}36\sin(17x) + 306\sqrt{3}x^2\]

  29. iop360
    • 3 years ago
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    slope equals to 0, and then...?

  30. iop360
    • 3 years ago
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    0 = f'(x) 0 = \[0 = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]

  31. iop360
    • 3 years ago
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    maybe solving for x here would be the answer?

  32. josiahh
    • 3 years ago
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    the answer is 5pi/102

  33. iop360
    • 3 years ago
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    oh did you get it?

  34. josiahh
    • 3 years ago
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    and 7pi/102... you took the derivative wrong

  35. iop360
    • 3 years ago
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    oh..

  36. josiahh
    • 3 years ago
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    you shouldn't have 712, its 612

  37. iop360
    • 3 years ago
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    what is f'(x)?

  38. iop360
    • 3 years ago
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    ohh

  39. josiahh
    • 3 years ago
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    and set them equal as you did before, 36sin(17x)+306(3)^(1/2)x=36sin(17x)+612xcos(17x)+612sqrt(3)x

  40. iop360
    • 3 years ago
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    solve for x, and thats what you get?

  41. josiahh
    • 3 years ago
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    \[36\sin(17x)+306sqrt(3)x=36\sin(17x)+612xcos(17x)+612\sqrt(3)x\]

  42. josiahh
    • 3 years ago
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    yea, x= (1/102)(12*k*pi+5pi) or x = (1/102)(12*k*pi-5pi)

  43. josiahh
    • 3 years ago
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    n is an integer, pick n so that x is between pi/34 and 3pi/34

  44. iop360
    • 3 years ago
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    oh ok, thanks!

  45. josiahh
    • 3 years ago
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    np

  46. iop360
    • 3 years ago
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    whats n?

  47. josiahh
    • 3 years ago
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    hold on a second I will attach a file soon

  48. josiahh
    • 3 years ago
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    there you go

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  49. josiahh
    • 3 years ago
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    on the pdf file, it should not be 5pi + or - 12kpi; it should be 5pi+12kpi or x=-5pi+12kpi since cos(-a)=cos(a); it is 2kpi + or - a with a being 5pi/6 in our case... any questions??

  50. iop360
    • 3 years ago
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    what is k?

  51. josiahh
    • 3 years ago
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    I used k instead of n; its the same thing

  52. iop360
    • 3 years ago
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    oh ok

  53. iop360
    • 3 years ago
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    when you isolated for k , what are you finding

  54. iop360
    • 3 years ago
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    is that like the period

  55. josiahh
    • 3 years ago
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    no; im trying to find k. x should be in a certain range. so if x is between pi/34 and 3pi/34 then (pi/34)<x<(3pi/34). this will help you find an interval for k so that your x is in the right range or domain if you want to call it that... now k should be an integer so you look at the interval of k and decide which integers are inside that interval... if you do this for 12kpi-5pi; you should get k=1... pi/34<=(1/102)(12kpi-5pi)<=3pi/4... try it

  56. iop360
    • 3 years ago
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    ok i finally got it

  57. iop360
    • 3 years ago
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    thanks for your help!

  58. josiahh
    • 3 years ago
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    np...

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