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Thread: inflection points

  1. #1
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    inflection points

    f(x)=x+sinx [0,2pi]
    f'(x)=1 +cosx
    f''(x)= -sinx
    0=-sinx
    x= 0, pi and 2pi
    i dont know if i can call 0 and 2pi inflection points since they are end points of the closed interval....
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  2. #2
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    Quote Originally Posted by zintore View Post
    f(x)=x+sinx [0,2pi]
    f'(x)=1 +cosx
    f''(x)= -sinx
    0=-sinx
    x= 0, pi and 2pi
    i dont know if i can call 0 and 2pi inflection points since they are end points of the closed interval....
    Critical points are found by solving f'(x) = 0, NOT f''(x) = 0.

    Note also, that a point has coordinates, that is, an x-coordinate AND a y-coordinate.
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  3. #3
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    sorrry still not answered

    but i was asking for the inflection points, where the graph changes concavity. can i say it changes concavity at an end point??
    Last edited by zintore; Jul 27th 2008 at 08:49 PM. Reason: still not answered!
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  4. #4
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    1) The derivative does not exist at an endpoint.

    2) Nothing can "change" at an endpoint, since there is only one side.

    3) Extrema can occur at endpoints, but the calculus will find them only by design or coincidence. You must check the endpoints yourself.
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  5. #5
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    i thought by definition inflection points are where f''(x)=0. is that not true?
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