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Math Help - Critical numbers

  1. #1
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    Critical numbers

    I'm asked to find the critical numbers of f(x)=2((e^-x)-(e^-2x))

    I know you have to find the derivative first which I got:

    f'(x)=2((-e^-x)+(2e^-2x))
    f'(x)= (-2e^-x)+(4e^-2x)

    But now, how do I go about finding one of the x? I believe you must rewrite it as a quadratic function but since the exponent is -2 indicating it is a reciprocal.

    So:

    (4(e^x)^-2)-(2(e^x)^-1)=0

    Factoring: Divided by 4
    (1/e^x)((1/e^x)-(1/2))=0

    I can set ((1/e^x)-(1/2))=0 which will be x = ln 2

    But when it comes to (1/e^x)=0 I'm at a COMPLETE LOSS. How do I do this?

    Thank you in advance
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  2. #2
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    You have

    4e^{-2x} - 2e^{-x} = 0

    4(e^{-x})^2 - 2e^{-x} = 0

    4X^2 - 2X = 0 where X = e^{-x}

    2X(2X - 1) = 0

    2X = 0 or 2X - 1 = 0

    X = 0 or X = \frac{1}{2}.

    Therefore e^{-x} = 0 or e^{-x} = \frac{1}{2}.


    But since e^{-x} > 0 for all x, that means

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

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

    e^{x} = 2

    x = \ln{2}.


    So there is a critical value at x = \ln{2}. Use the second derivative test to determine its nature, and then substitute it back into the original function to find the corresponding y value.
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