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Thread: Concavity and Inflection problem, with Trig Functions.

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

    Namaste!

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

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

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

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

    $\displaystyle -16=-32sin(4x)$

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

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

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

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

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

    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 $\displaystyle \sin^{2}4x,$ is $\displaystyle 8\sin4x \cos 4x.$
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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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