The condition that f is always increasing or decreasing 
on (-oo,oo) is that f'(x) is either always positive, or
always negative.

So we find the derivative f'(x)

f(x)= ax³ + bx² + cx + d

f'(x) = 3ax² + 2bx + c

Then set that > 0

3ax² + 2bx + c > 0

For this to be true, 

f'(x) = 3ax² + 2bx + c 

must represent a parabola
which is always either above the x-axis
or always below the x-axis.  This means
that f'(x) can have no real zeros.

Therefore its discriminant must be
negative.  The discriminant of

Ax² + Bx + C is B²-4AC, and in our case

A = 3a, B=2b, C = c, so the discriminant is

(2b)² - 4(3a)(c) or 4b² - 12ac, so we must have

4b² - 12ac < 0 or

       4b² < 12ac or

        b² < 3ac

is the requirement.

Now for a word of caution.  There is some disagreement
among mathematicians as to whether to say that a 
function is increasing or decreasing at a horizontal 
inflection point.  If your teacher is one who 
says that the function f(x) = x³ + 3x² + 3x, graphed
below



is increasing everywhere, even at the point (-1,-1), 
where it has a horizontal inflection point, i.e. its 
derivative is 0, indicated below by the horizontal 
tangent line:

    

then you must replace all the strict inequalities
" < " by " <, and then the requirement will
be 

        b² < 3ac

and also you would have to change the initial
statement above to

The condition that f is always increasing or decreasing 
on (-oo,oo) is that f'(x) is either always nonpositive,
or always nonnegative.


So be sure to ask your teacher whether or not he or she
considers a function to be increasing (or decreasing)
at a horizontal inflection point, where the derivative
is 0, as long as it is increasing (or decreasing)
everywhere else.

Edwin