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Math Help - How do you determine the number of inflection points

  1. #1
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    How do you determine the number of inflection points

    for the function y=2x+cos(x^2) on the interval of 0\leq x \leq 5
    f\prime\prime = -2x(2xcosx^2)- 2sinx^2

    I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by swatpup32 View Post
    for the function y=2x+cos(x^2) on the interval of 0\leq x \leq 5
    f\prime\prime = -2x(2xcosx^2)- 2sinx^2

    I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
    Solve -4x^2\cos(x^2)-2\sin(x^2)=0 on [0,5]

    Let u=x^2 (this extends the domain to [0,5^2])

    -4u\cos(u)-2\sin(u)=0 \implies 2u\cos(u)+\sin(u)=0.

    Clearly, a solution is u=0 \implies x=0. There are eight others, but I only know that because I graphed the equation and counted them. I honestly don't know how to solve for the other roots analytically.

    But there are 9 zeroes and thus 9 inflection points.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    But there are 9 zeroes and thus 9 inflection points.
    Not neccessarily true. An inflection point is a point in which (1.) the slope of f' is zero or undefined vertically and (2.) there is a change of the sign of the slope of f' from - to + or from + to - . A zero can exist without the slope sign changing, therefore it'd not be an inflection point.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Ok, but this is a continuous function defined on a compact set, so it can't be unbounded, and whenever it changes from positive to negative there will be a corresponding zero by the IVT.
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  5. #5
    Behold, the power of SARDINES!
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    I have an indeterminate form ie  0\cdot \infty this is not well defined, and I checked if the limits go to zero and they do not.

    The zero factor principle does not apply becuase  0\cdot \infty \ne =

    Thanks for the graph skeeter.

    So the 2nd derivative is

    f''(x)=-4x^2\cos(x^2)-2\sin(x^2)

    This can be factored as follows

    0=-2\cos(x^2)[2x^2+\tan(x^2)]

    The 2nd factor only has x=0 as a solution so the solutions are

    \cos(x^2)=0 \iff x^2=\frac{\pi}{2}+n \pi= \frac{(2n+1) \pi}{2}

    x=\pm \sqrt{\frac{(2n+1)\pi}{2}}

    Note that in the factorization that (2x^2+\tan(x^2)) > 0 for all  x> 0

    Since \cos(x^2) changes sign at each of it zero all of the above zero's are critical points
    Last edited by mr fantastic; May 5th 2009 at 05:23 PM.
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  6. #6
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    graph of the 2nd derivative ...
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