f(x) = x^3 lnx
f'(x) = x^2 (3lnx +1)
x= e^(-1/3)
left side of e^(-1/3) is negative, right side is positive so
f increase at (e^(-1/3), infinity)
f decrease (-infinity , e^(-1/3)) correct?
Local Minimum:
f(e^(-1/3)) = .123 so local minimum is:
.123 correct?
Inflection point:
f'(x) = x^2 (3lnx +1)
f "(X) = 6xlnx + 3x + 2x
= x (6lnx + 5) = 0
x= e^(-5/6)
----- e^(-5/6) ++++
I don't know what to do from here..