# More max/min (sort of)

• Oct 22nd 2009, 06:49 PM
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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?
• Oct 22nd 2009, 07:09 PM
Arturo_026
in F'(x) didn't you forget the negative . I believe you timed it by 1/x instead of -1/x.
right?
• Oct 22nd 2009, 07:44 PM
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Sorry. I was doing that by memory.
• Oct 23rd 2009, 01:55 PM
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Still don't have an answer on this on.

Does that mean I did it correct? (Wondering)
• Oct 23rd 2009, 03:39 PM
Arturo_026
check your work on the second derivative.

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