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Could I get some help determining the point of inflection of a function?

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the point of inflection on the graph y = x^4 - 4x^3


    2. Relevant equations



    3. The attempt at a solution

    The second derivative of this function is 12x^2 - 24x

    I dont know how to show my work to get a point of inflection of (0,0)

    How would I go about doing that?

    Thank you for your help!
     
  2. jcsd
  3. Feb 25, 2009 #2
    Set the second derivative equal to zero and solve. You should get two values and then check a trial point on both sides of each of those in order to make sure that the concavity changes.
     
  4. Feb 25, 2009 #3
    How do I do a trial to see the concavity changes?

    I figured here my x values were x = 0 and x = 2. Do I plug these values into the second derivative to determine if the concavity changes?
     
  5. Feb 26, 2009 #4

    Mark44

    Staff: Mentor

    Yes. At each of these points y'' should change from pos to neg or neg to pos.
     
  6. Feb 26, 2009 #5
    I used the original function to check to see if it changes from positive to negative and vise versa, not y". When I used y" it told me x=2 is a point of inflection when I can see on the graph that it is not.
     
  7. Feb 26, 2009 #6
    Oh I think I got it wrong. x=2 is a point of inflection. Could someone verify for me? I get (0,0) and (2,-16) as points of infection for this function.
     
  8. Feb 26, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the second derivative is, as you said, [itex]f"(x)= 12x^2- 24x= 12x(x- 2)[/itex]. If x< 0, both x and x-2 are negative so f" is positive. If 0< x< 2, x is positive while x- 2 is still negative so f" is negative. If 2< x, both x and x-2 are positive so f" is positive. f" changes sign at both 0 and 2 so 0 and 2 are inflection points.
     
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