# derivatives – Find the interval in which \$ f (x) \$ increases and decreases.

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 $$1 so $$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.