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Thread: Monotonically increasing

  1. #1
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    Monotonically increasing

    Here´s the question.
    Monotonically increasing-picture-1.png


    Ok so the derivative of f

    {\frac {d}{dx}}f \left( x \right) =\cos \left( x \right) -1+3\,\alpha<br />
\,{x}^{2}<br />

    The function is monotonically increasing if the derivative of f is bigger than zero in the interval. Or if \alpha<1/3\,{\frac {1-\cos \left( x \right) }{{x}^{2}}}

    So my task should be to determine the lowest value of the right hand side of the inequality and say that alpha should be less than that? Is it so? I´ve tried that but ended up with zeros in the denominator and stuff. Any ideas?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    min(cosx-1) in [0,pi/2] = -1

    max(cosx-1) in [0,pi/2] = 0
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    min(cosx-1) in [0,pi/2] = -1

    max(cosx-1) in [0,pi/2] = 0
    Yes. min(cosx-1) in [0,pi/2] =-1 and this happens as x --> pi/2 and therefore f´(0) larger than zero if

    4/3\,{\pi }^{-2}<\alpha

    But what happens when x-->0 and how about all the other values between 0 and pi/2?

    I´m still not sure how to choose alpha so that f´(x) is larger than zero in the interval.
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