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Cubic differenation problem

  1. Jan 2, 2008 #1
    Hi

    The curve C has equation y = f(x) and the point P(3,5) lies on C

    Given that

    f'(x) = 3x² - 8x + 6

    The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P

    Find the x-coordinate of Q


    So

    if there parallel the gradients are equal

    but i cant really seem to get anywhere after that :(
     
  2. jcsd
  3. Jan 2, 2008 #2

    CompuChip

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    Indeed, the gradients of y = f(x) at P and Q are equal. What is the value of that slope at P? Then, what other x gives the same slope?
     
  4. Jan 3, 2008 #3
    You factorise the derative of f(x) which is f'(x)
    giving you
    (3x+1)(x-3)

    x = 3 or x = -1/3

    we already know x = 3 so the other is -1/3

    i believe that is correct?
     
  5. Jan 3, 2008 #4

    HallsofIvy

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    Yes, that is correct but the way you phrased it confused me a bit! You did not factorize f '(x). You can easily calculate that the derivative at x= 3 is 9 so at the other point, you must have [itex]3x^2- 8x+ 6= 9[/itex] which then gives [itex]3x^3- 8x- 3= 0[/itex]. That is what you factored, not the derivative of f.
     
    Last edited: Jan 4, 2008
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