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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