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Thread: mix/max problem

  1. #1
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    mix/max problem

    I am working on a mix/max problem and am stuck.

    $\displaystyle y=xe^(-\frac{x^2}{8})$ within the interval of [-1,4]. The exponent of e is $\displaystyle -\frac{x^2}{8}$, can't get Latex to cooperate with me.

    Using the product rule to differentiate and I get:

    y'= $\displaystyle e^(-\frac{x^2}{8})+x(e^(-\frac{x^2}{8})(\frac{x}{4})$

    Factor out the $\displaystyle e^(-\frac{x^2}{8}$ and I get

    $\displaystyle e^(-\frac{x^2}{8})(1+\frac{x^2}{4})$

    For all real numbers, the 2nd term will never equal 0, so the only one that can equal 0 is $\displaystyle e^(-\frac{x^2}{8})$.

    I don't know how to solve for x when $\displaystyle e^(-\frac{x^2}{8})=0$. Any pointers?
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    Re: mix/max problem

    $\displaystyle \displaystyle \begin{align*} e^{-\frac{x^2}{8}} > 0 \end{align*}$ for all x, so there are not any critical points to this function.
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    Re: mix/max problem

    Quote Originally Posted by baldysm View Post
    I am working on a mix/max problem and am stuck.

    $\displaystyle y=xe^{-\frac{x^2}{8}}$ within the interval of [-1,4]. The exponent of e is $\displaystyle -\frac{x^2}{8}$, can't get Latex to cooperate with me.

    Using the product rule to differentiate and I get:

    y'= $\displaystyle e^(-\frac{x^2}{8})+x(e^(-\frac{x^2}{8})(\frac{x}{4})$
    . Any pointers?
    You missed a minus sign y'= $\displaystyle e^{-\frac{x^2}{8}}+x(e^{-\frac{x^2}{8}})({\frac{-x}{4})$

    Look at this graph

    About your LaTeX, If you have more than one character in an exponent the you must use {} around the entire exponent.
    [TEX] e^{-\frac{x^2}{8}} [/TEX] gives $\displaystyle e^{-\frac{x^2}{8}} $.
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    Re: mix/max problem

    Quote Originally Posted by Plato View Post
    You missed a minus sign y'= $\displaystyle e^{-\frac{x^2}{8}}+x(e^{-\frac{x^2}{8}})({\frac{-x}{4})$

    Look at this graph

    About your LaTeX, If you have more than one character in an exponent the you must use {} around the entire exponent.
    [TEX] e^{-\frac{x^2}{8}} [/TEX] gives $\displaystyle e^{-\frac{x^2}{8}} $.
    Please disregard my previous post as Plato has pointed out the error I and the OP have both made.
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  5. #5
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    Re: mix/max problem

    Yup, I see my algebra error, it should be $\displaystyle -\frac{x}{4}$.

    I also screwed up the interval. It should be [-4,1].

    The derivative is

    $\displaystyle y'=e^{\frac{-x^2}{8}}+x(e^{\frac{-x^2}{8}})(\frac{-x}{4})$

    Some factoring and I get

    $\displaystyle y'=e^{\frac{-x^2}{8}}(1-\frac{x^2}{4})$

    The e term can't be 0 for all of x (although my TI-84 gives an error when x=-100, not sure why), and the only number that makes the 2nd term zero is either -2 or 2. Only -2 is in the interval, so I need to find out f(x).for x= -4, -2, and 1. Biggest number is max, smallest is min. I might just understand this yet.

    I appreciate the help.
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