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)correct?

f decrease (-infinity , e^(-1/3))

Local Minimum:

f(e^(-1/3)) = .123 so local minimum is:

.123correct?

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..