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Math Help - Concavity and Inflection problem, with Trig Functions.

  1. #1
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    Concavity and Inflection problem, with Trig Functions.

    Namaste!

    Let f(x)=8x^2+sin^2(4x) where x \in \left[0,\frac{\pi}{4} \right]

    Find the interval where f is concave up, and where f is concave down. Finally, list any points of inflection (if there are any).

    I began by taking the first derivative of f(x):
    f'(x)=16x+8cos(4x) (not 100% sure this is correct)
    Then I found the second derivative of f(x) and got:
    f''(x)=16-32sin(4x)

    To find the critical points I (attempt to) set the second derivative to zero to find where the graph of f(x) is concave up or concave down.
    Broken into steps:
    0=16-32sin(4x)

    -16=-32sin(4x)

    \frac{-16}{-32}=sin(4x)

    \frac{1}{2}=sin(4x)

    I try to think of where the graph of sin(x)=\frac{1}{2}

    \frac{\pi}{6}=4x

    Finally: x=\frac{\pi}{24}

    So this is the critical point, and possibly a point of inflection.

    I test points on a number line on either side of this to find where the second derivative is positive or negative to find where the original function is concave up, or concave down.
    A hint was given to use a double angle formula, however I am not very familiar with how to implement these.

    However, all of my answers so far have been incorrect. I don't want the solution, I just want to know where I went wrong through this process. Any help would be greatly appreciated. Thanks in advance.
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  2. #2
    Newbie Esteban's Avatar
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    Re: Concavity and Inflection problem, with Trig Functions.

    The first derivate is not correct
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    Re: Concavity and Inflection problem, with Trig Functions.

    As you suspect as a possibility, your first derivative is incorrect.

    The derivative of \sin^{2}4x, is  8\sin4x \cos 4x.
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  4. #4
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    Re: Concavity and Inflection problem, with Trig Functions.

    A 20 minute post, answered in 20 seconds...(sigh).
    But in all seriousness, thanks.
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