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Differentiation Problem

  1. Sep 27, 2007 #1
    The problem

    The Diagram shows the graph of y=x^3-12x+17
    A is the maximum point on the curve
    C is the minimum point on the curve
    The curve crosses the y axis at B

    For the equation find dy/dx, y=x^3-12x+17 (DONE)

    Heres the problem

    find the gradient of the curve at B

    now what am I supposed to do here?

    this is what i tried but im not sure if its right


    make x the subject therefore giving


    so is that right or..?
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2


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    For b) you need to find the value of the derivative at the point B. What are the coordinates of the point B ?
  4. Sep 27, 2007 #3
    im guessing the coordinates at point B are (0,17), but how does that help me in finding the gradient at that point
  5. Sep 27, 2007 #4


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    I presume you found that the derivative is 3x^2- 12. By why are you asserting that the derivative is 17? You are supposed to be finding the value of the derivative!

    No, it's not right- you weren't even asked for a value of x!

    You know that f'(x)= 3x^2- 12. You are asked to find its value at point B. Of course, to do that you need to know what x is there. Saying the graph crosses the x-axis tells you that y= 0. Okay y= x^3- 12x+ 17= 0. Can you determine what x is from that?

  6. Sep 27, 2007 #5
    Sorry, I don't seem to be understand what you are saying. Do you want me to find the solutions of the equation x^3-12x+17?
  7. Oct 13, 2007 #6
    Remember. The curve crosses y-axis at point B. So you know the y = 0 but you have to find the x-value for the point B:

    In other words. B = (x,0)

    Also, you know that f'(x) = 3x^2-12

    Let f'(x) = 0 and then find a value for x. Then you have got point B! :)
  8. Oct 13, 2007 #7


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    Yes, that's exactly what I said! Then find the derivative at that value of x.

    ?? No, the curve is y= f(x), not y= f'(x).
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