Posted by Chelsea on Sunday, March 23, 2008 at 5:55pm.
y = ln (x e^-x)
y' = (1/(x e^-x) ) * [ x (-e^-x) + e^-x ]
= -1 + 1/x
that derivative is zero when x = 1
then xe^-x = .36788
what is the second derivative?
y'' = 0 - 1/x^2
humm, when x = 1
curvature is - so that is a maximum and the only one I see
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