Let $ f (x) = 2x ^ 3 -9x ^ 2 + 12 x + 6 $ so $ f (x) = 6x ^ 2-18x + 12 = 6 (x-1) (x-2) $

I need intervals in which $ f (x) $ **strictly increases**,

$ f (x)> $ 0 when $ x <1 $ and $ x> $ 2 And so $ f (x) $ increases strictly in these intervals and $ f (x) <0 $ when $ 1 <x <2 $ so $ f (x) $ decreases strictly in this interval.

My question :

What about points $ x = $ 1.2

Listen to me.

Yes $ f (x) = 0 $ at points (not intervals) then $ f (x) $ can always be considered strictly monotonous. And he also seems **reasonable** (I'll add the reason below) to include the points $ 1.2 in the intervals of increase (IOI, for short) **and** decreases (IOD).

Ex: take $ x = $ 1. Let's say that I included this point in IOI and IOD. So, IOI is now, $ x in (-∞, 1]$ and you can see that this does not contradict either the definition of "strictly increasing function in the interval". Take everything $ p, q in (-∞, 1], p> q Rightarrow f (p)> f (q) $ likewise my IOD, would now $ x in [1,2]$ (note I included $ 2 $) and it always follows the definition.

But! There is always a but!

If I understand correctly, the definition of monotonicity works at a given moment, which would bother now.

I can not include $ x = $ 1 because there is no $ h> $ 0 in such a way that $ p, q in (1h, 1 + h) nRightarrow f (p)> f (q) $ Similarly, I can not add $ x = $ 1 in decrease interval either.

If I had to choose, my intuition would lead me to choose the second, but I do not see why I should reject the first either, because it corresponds to the definition of function increasing strictly in the interval.

Thank you for taking the time to read all this. I would appreciate very much that someone can counter (with reasons) one or the other of these arguments.