x^3 < 3x^2
would become x(x^2-3x)
then how would I graph this?
x(x^2-3x) < 0
x^2(x-3) < 0
critical values of x=0 , x=3
on a number line, mark x=0 and x=3, splitting the line into 3 parts
1. x<0
pick any number < 0, say x=-5
(+)(-) < 0 , that's what we want
2. 0 < x < 3 , pick a value, let's pick x = 1
(+)(-) < 0 , good !
3. x> 3, let x=5
(+)(+) > 0 , no good
so x < 3 will work.
another way is know what a general cubic looks like
you would have
y = x^2(x-3)
there would be double root at x=0, so it touches at x=0 without crossing, and it crosses at x = 3
easy to sketch, one can see that the curve is below the x-axis for x< 0 as well as between 0 and 3.
so that is our domain, namely x < 3