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Thread: 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

    $\displaystyle 4e^{-2x} - 2e^{-x} = 0$

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

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

    $\displaystyle 2X(2X - 1) = 0$

    $\displaystyle 2X = 0$ or $\displaystyle 2X - 1 = 0$

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

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


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

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

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

    $\displaystyle e^{x} = 2$

    $\displaystyle x = \ln{2}$.


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