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Math Help - Trig Derivative Problem

  1. #1
    Junior Member
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    Trig Derivative Problem

    Hello everyone,

    Could anyone please offer hints or tips as to how to solve this problem? Some of my work is shown below.

    Thank you.

    ---

    5. Consider the function  f(x) = x^2 + \cos(kx) , where k is the real parameter. For what values of k does the given function have no inflection points?

    ---

    I found these:

     f'(x) = 2x - k\sin(kx)

     f''(x) = 2 - k^2 \cos(kx)

    To determine the inflection points, I set  f''(x) = 0 .

    Therefore:

     k^2 \cos(kx) = 2

     cos(kx) = \frac{2}{k^2}

     x = \frac{\cos^{-1} \frac{2}{k^2}}{k}

    However, I am not sure how to proceed from here.
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  2. #2
    MHF Contributor red_dog's Avatar
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    The condition is \frac{2}{k^2}>1, (because \cos\alpha\in[-1,1], \ \forall\alpha\in\mathbf{R})

    Then k^2<2\Rightarrow k\in(-\sqrt{2},\sqrt{2})
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