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Thread: Finding Critical Numbers for a derivative with euler's number

  1. #1
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    Finding Critical Numbers for a derivative with euler's number

    This is a problem I came across in my Calculus textbook: g(x)=e^(-x)+e^(3x)

    I am asked to find the open intervals over which this function increases and decreases. The original function isn't undefined for any value of x so I proceeded to find the derivative: g'(x)=3e^(3x)-e^(-x)

    For the critical numbers there are no values of x for which this function is undefined (where the original is defined) but there seems to be one value of x for which the derivative will be equal to zero.

    This is an odd-numbered problem in the book, so I checked the website to see what the critical number would be. The website stated that g'(x)=3e^(3x)-e^(-x)=0
    when x=-(1/4)ln(3).

    However the website did not show the steps to arrive at such a result. Would anyone be so kind as to explain the steps it takes to arrive at this possible critical number?

    This website: http://www.calcchat.com/book/Calculus-ETF-5e/ and the input section 4.3 # 17 is the problem I am dealing with.
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    Re: Finding Critical Numbers for a derivative with euler's number

    $\displaystyle 3 e^{3 x}=e^{-x}$

    $\displaystyle 3 e^{4 x}=1$
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    Re: Finding Critical Numbers for a derivative with euler's number

    I understand how you set the equation up at first, but not how you arrived at 3e^(4x)=1 or how I would get from that to -(1/4)ln3 as an x-value that makes the derivative equal to zero.
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  4. #4
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    Re: Finding Critical Numbers for a derivative with euler's number

    $\displaystyle 3 e^{3 x}=e^{-x}$

    multiply both sides by $\displaystyle e^x$ ...

    $\displaystyle 3e^{3x} \cdot e^x = e^{-x} \cdot e^x$

    $\displaystyle 3e^{3x+x} = e^{-x+x}$

    $\displaystyle 3e^{4x} = e^0$

    $\displaystyle 3 e^{4 x}=1$

    solve for x ...

    $\displaystyle e^{4x} = \frac{1}{3}$

    $\displaystyle \ln(e^{4x}) = \ln\left(\frac{1}{3}\right)$

    $\displaystyle 4x = \ln{1} - \ln{3}$

    $\displaystyle 4x = -\ln{3}$

    $\displaystyle x = -\frac{1}{4}\ln{3}$
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    Re: Finding Critical Numbers for a derivative with euler's number

    Thank you skeeter. That was cool
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