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Math Help - Monotonically increasing

  1. #1
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    Monotonically increasing

    Heres 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? Ive 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?

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