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    The differential of the curve f(x) = 3e^x - \frac{1}{2} \ln x - 2

    is f'(x) = 3e^x - \frac{1}{2x}

    The curve f(x) has a turning point at p, the x co-ordinate of p is \alpha

    Show that \alpha = \frac{1}{6}e^{- \alpha}

    So far what I've done

    3e^ \alpha - \frac{1}{2 \alpha} = 0

    3e^ \alpha = \frac{1}{2 \alpha}

    6 \alpha e^ \alpha = 1

    e^ \alpha = \frac{1}{6 \alpha}

    So what do I do know take the natural log of both sides ? But then how is that going to get me closer to my target which I have to show.
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    i have no ideal this is hard lol
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    (Original post by uer23)
    6 \alpha e^ \alpha = 1
    From here, you have  \alpha = \dfrac{1}{6 e^ \alpha} = \frac{1}{6} e^{-\alpha} which is what you want to show.
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    Instead of dividing by 2alpha try dividing by 3e^alpha.
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    (Original post by ttoby)
    From here, you have  \alpha = \dfrac{1}{6 e^ \alpha} = \frac{1}{6} e^{-\alpha} which is what you want to show.
    dam beat me to it
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    (Original post by ttoby)
    From here, you have  \alpha = \dfrac{1}{6 e^ \alpha} = \frac{1}{6} e^{-\alpha} which is what you want to show.
    I really don't get that at all ? Is there a particular Index law you are using to rewrite the first part.
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    (Original post by reshep)
    dam beat me to it
    But how ?
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    It's just addition, multiplication and division mate.
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    (Original post by Copacetic)
    It's just addition, multiplication and division mate.
    No It's more than that.
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    (Original post by uer23)
    No It's more than that.
    No it isn't.

    Are you wondering why 1/6e^a is the same as (1/6)e^-a ?
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    (Original post by Copacetic)
    No it isn't.

    Are you wondering why 1/6e^a is the same as (1/6)e^-a ?
    It's okay I think I've finally got it.
 
 
 
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Updated: January 16, 2010
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