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Algebra 18 Online
OpenStudy (anonymous):

Find critical values of y: yln(y+2) = 0

OpenStudy (anonymous):

for critical values you must take the first derivative and solve for dy/dx = 0

OpenStudy (anonymous):

I'm basically trying to get values of y. (for example on y-3 = 0 we know that y = +3. I just can't figure it out for this ln problem.

OpenStudy (anonymous):

disregard the part that says critical value..

OpenStudy (anonymous):

if your expression is \[y \ln(y+2)= 0\]than the answers are y = 0 and y = -1 recall that ln1 = 0

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