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Math Help - More max/min (sort of)

  1. #1
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    More max/min (sort of)

    We need to find:

    -Find the interval where the function is decreasing.
    -Find the inflection point.
    -Find the interval where the function is concave up/down

    F(x) = ln(4-ln(x))
    F'(x) = \frac{1}{4-ln(x)} * \frac{-1}{x} = \frac{-1}{4x-xln(x)}

    I thought this was right, but that gave me a decreasing interval of (e^3,\infty)

    F'(x) = \frac{-1}{4x-xln(x)}
    F''(x) = \frac{0-(5-ln(x))(-1)}{(4x-xln(x))^2} = \frac{5-ln(x)}{(4x-xln(x))^2}

    Which is pretty much where I've gotten stuck.

    Is it one of my derivatives that's gone askew? Or did I do those right and I'm just getting the wrong intervals?
    Last edited by Open that Hampster!; October 22nd 2009 at 06:44 PM.
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  2. #2
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    in F'(x) didn't you forget the negative . I believe you timed it by 1/x instead of -1/x.
    right?
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  3. #3
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    Sorry. I was doing that by memory.
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  4. #4
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    Still don't have an answer on this on.

    Does that mean I did it correct?
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  5. #5
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    check your work on the second derivative.

    remember that (4x-xln(x))' = 3-ln(x)
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