calculus
posted by Juliet .
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'')

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 
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.