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Math Help - Finding the min value

  1. #1
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    Finding the min value

    Hello everyone, I'm stuck on this question:

    The minimum value of e^x+2e^x occurs where x = ?

    Any help appreciated, thx
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Finding the min value

    Are you sure you have the expression typed correctly? As given it is 3e^x, and as such has no extrema.
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  3. #3
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    Re: Finding the min value

    Exactly, that's what I thought. My textbook says the answer is -0.5log(e)^2
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Finding the min value

    In that case, if I am interpreting the given critical value correctly, the function should be:

    y=e^{-x}+2e^x

    Hence, differentiating and equating to zero, we find:

    \frac{dy}{dx}=-e^{-x}+2e^x=\frac{2e^{2x}-1}{e^x}=0

    This implies:

    2e^{2x}-1=0

    e^{2x}=\frac{1}{2}

    Converting from exponential to logarithmic form, we obtain:

    2x=\ln\left(\frac{1}{2} \right)=-\ln(2)

    x=-\frac{\ln(2)}{2}
    Thanks from HallsofIvy
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