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Points of inflexion

  1. Feb 1, 2005 #1
    Generally speaking, how do I find the point of inflexion of a curve y=f(x) if there is no x such that y'=0.

    For example, say we have the curve y=arcsinh(x+1) and want to find its point of inflexion, so y'=1/[1+sqrt{1+(x+1)^2}]=0, but there are no values of x that do that. I tried to use the inverse of x and find dx/dy instead, then set it to zero, but again, I couldn't find a suitable value for y.

    How do I approach questions like this?
     
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  3. Feb 1, 2005 #2

    dextercioby

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    Do you know what an inflexion point is???My guess is that u don't...So how about read and then tackle problems...??

    Daniel.
     
  4. Feb 1, 2005 #3
    Set [tex] \frac{d^2y}{dx^2} = 0 [/tex]
     
  5. Feb 1, 2005 #4
    ...

    What do you think I'm doing?
     
  6. Feb 1, 2005 #5

    dextercioby

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    You weren't after the inflexion points...You were determining the critical points which is something totally different.

    Daniel.
     
  7. Feb 1, 2005 #6
    My book says:
    If f'(x)=0 and f''(x)=0, the turning point is a point of inflexion.

    So I did that, but I can't find values of x that make f'(x) and f''(x) equal to zero...
     
  8. Feb 1, 2005 #7
    Wait, nevermind, the problem was with my differentiation.
     
  9. Feb 1, 2005 #8

    dextercioby

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    You asked about INFLEXION POINTS...Not about turning points...
    The definition of an inflexion point is that of a point x* for which THE SECOND DERIVATIVE COMPUTED IN THAT POINT IS ZERO AND MOREOVER IT CHANGES THE SIGN...(plus-->minus or viceversa).

    Daniel.
     
    Last edited: Feb 1, 2005
  10. Feb 1, 2005 #9
    Just for the original poster's benefit, this isn't necessarily true. An inflection point is where the graph of a function changes concavity (measured by the sign of the second derivative). You should always check the behavior of the second derivative around points where f'(x)=f''(x)=0 in order to make sure that concavity changes, as it's naive to assume the second derivative (or any function) changes sign whenever the function hits 0.
    For example, the function f(x)=x^4 has f''(0)=f'(0)=0, but (0,0) is not a point of inflection. It's just very, very flat. :wink: As dexter said, I also disagree with the wording of the statement. Points where f'(x)=0 should be called critical points, not turning points.
    If your statement above was indicative of the wording of your book and this is all your book has to say about points of inflection, I would consider learning from another source, or complaining to the professor.
     
    Last edited: Feb 1, 2005
  11. Feb 2, 2005 #10

    HallsofIvy

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    That does NOT say that f'(x) and f"(x) must both be zero for x to be an inflexion point. It says that IF f'(x)= 0 and f"(x)= 0 then the point is BOTH a turning point and a point of inflexion. i.e. it is a turning point AND a point of inflexion.
     
  12. Feb 2, 2005 #11
    Hi Halls,
    The statement in the book is still false (as shown above). The book may be defining "point of inflection" this way, but it doesn't fit any other definition I've seen and has no unique graphical meaning beyond "Well, f''(x)=0 at that point.". :smile:
    To the original poster, remember that a point of inflection is a point where the second derivative (concavity) changes sign. Study the behavior of the second derivative to see whether there are any such points. As HallsofIvy said, the first derivative need not be 0 at a point of inflection. The function f(x)=cos(x) has a point of inflection at x=pi/2, for example.
     
    Last edited: Feb 2, 2005
  13. Feb 2, 2005 #12

    HallsofIvy

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    Thank you. That occured to me but I didn't say it. f"= 0 is a necessary condition but not sufficient! If y= x4 then f"(x)= 12x2.
    The second derivative is 0 at x= 0 but does not change sign there. (0,0) is NOT an inflexion point.
     
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