Posted by Juliet on Thursday, January 19, 2012 at 10:48pm.
find the relative extrema and points of inflection, if possible, of y = xln((x^4)/8)
Any help would be awesome...thanks!
(i know that the relative extrema come from critical values from y' and points of inflection come from y'')

calculus  Reiny, Friday, January 20, 2012 at 12:02am
dy/dx = y' = ln((x^4)/8) + x(4/x)
= ln ( (x^4)/8 ) + 4
y '' = 4/x
for relative extrema, y' = 0
ln ((x^4)/8) = 0
x^4 / 8 = e^0 = 1
x^4 = 8
x = ± 8^(1/4)
plug in those values into y = ...
for pts of inflection, 4/x = 0 > no solution
thus, no points of inflection

calculus  Steve, Friday, January 20, 2012 at 10:29am
And yet, y'' changes from  to + as x increases through 0, so while y is discontinuous at x=0, there appears to be an inflection point there.
To get the limit of y at x=0,
y(0) = 0*(oo)
setting t=1/x and using L'Hopital's rule a bit, we can see that the limit is 0.
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