Asked 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'')
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'')
Answers
Answered by
Reiny
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
= 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
Answered by
Steve
And yet, y'' changes from - to + as x increases through 0, so while y is discontinuous at x=0, there <b>appears</b> 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.
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.