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Math Help - Stationary points

  1. #1
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    Stationary points

    Hi i am having issues with a question from a textbook.
    The normal distribution with a mean 0 and standard deviation Q(i dont know how to do the correct sign) is given by y= e^ -(x/Q)^2 . Find and classify all stationary points and points of inflection for this function Q>0.

    The main issue i have occurred is finding the first and second derivatives of this function.

    My first derivative is coming out to be dy/dx = - 2xy/Q using logarithmic and implicit differentiation.
    However this does not appear to be right. I dont see how i could find when f'(x)=0 to find my stationary points.

    Thank you.
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  2. #2
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    Quote Originally Posted by agent josh View Post
    Hi i am having issues with a question from a textbook.
    The normal distribution with a mean 0 and standard deviation Q(i dont know how to do the correct sign) is given by y= e^ -(x/Q)^2 . Find and classify all stationary points and points of inflection for this function Q>0.

    The main issue i have occurred is finding the first and second derivatives of this function.

    My first derivative is coming out to be dy/dx = - 2xy/Q using logarithmic and implicit differentiation.
    However this does not appear to be right. I dont see how i could find when f'(x)=0 to find my stationary points.

    Thank you.
     <br />
y = e^{-\left(\frac{x}{\sigma}\right)^2}<br />

    \ln{y} = -\left(\frac{x}{\sigma}\right)^2

     <br />
\frac{1}{y} \cdot \frac{dy}{dx} = -\frac{2x}{\sigma^2}<br />

     <br />
\frac{dy}{dx} = -\frac{2x}{\sigma^2} \cdot e^{-\left(\frac{x}{\sigma}\right)^2}<br />

    \frac{dy}{dx} = 0 at x = 0
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  3. #3
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    To find the second derivative, start from \frac{1}{y}\frac{dy}{dx}= y'/y= \frac{-2x}{\sigma^2}

    Differentiating both sides, \frac{yy''- y'^2}{y^2}= -\frac{2}{\sigma^2}. Since the problem is to find the points of inflection, just put y"= 0 and solve \frac{y'^2}{y^2}= \left(\frac{-2x}{\sigma^2}\right)^2= \frac{2}{\sigma^2}
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