1

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

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

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

My Question:

What about at points $x=1,2$

Hear me out. If $f'(x) = 0$ at points (not intervals) then $f(x)$ can still be considered strictly monotone. And it 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).

Eg : Take $x= 1$. Let's say I include this point in both IOI and IOD. So IOI is now, $x \in (-∞, 1]$ and you can see that it doesn't contradict the definition of "strictly increasing function in interval" either. Take any $p,q \in (-∞, 1], p>q \Rightarrow f(p) > f(q)$ similarly my IOD, now would be $x\in [1,2]$ (notice I included $2$) and it still follows the definition.

But! There's always a but!

As I understand that, definition of monotonicity functions at a point, would now get in the way.

I can't include $x=1$ because there exists no $h>0$ such that taking $p,q \in (1-h,1+h) \nRightarrow f(p) > f(q)$ similarly, I can't add $x= 1$ in interval of decrease either.

If I'd have to pick, my intuition would lead me to pick the second one, but I can't see why I should reject the first one either since it actually follows the definition of strictly increasing function in interval.

Thank you for taking your time to read all of this. I'd very appreciate if someone could counter (with reasons) either of those arguments.

William
  • 5,151
  • "If $f'(x)=0$ at points (not intervals) then $f(x)$ can still be considered strictly monotone." Is that true? – cangrejo Apr 15 '19 at 07:12
  • @broncoAbierto Yes but there's a catch, that the point at which $f'(x)=0$ must have a deleted nbd in which $f'(x)>0$ take $x^3$ at $0$ for example, then $f$ is still strictly increasing, similar reasoning for strictly decreasing. But let us just ignore this catch (slightly) for a while and you can see we still have those conditions met. – William Apr 15 '19 at 07:28
  • @broncoAbierto Yes I'm aware it's a sufficient condition. I used it as a sufficient condition too, not a necessary condition (see my usage of "if" and "then" in the main explanation. Also "you can just put $1$ in both intervals". Did you read the second half of my explanation? Because I've written why that would be a problem. Basically I've written in the first half that I should include $1,2$ technically. In the second half I've explained, how that would create a problem. – William Apr 15 '19 at 07:46
  • @broncoAbierto Dude that is literally the definition of function strictly monotone at a point! Real Analysis : Bartle, Page no. 175.Not only that, check out, https://math.stackexchange.com/a/364619/552998 – William Apr 15 '19 at 08:15
  • Then your function is not monotonic at that point, and it is strictly increasing at the interval. What's the problem. – cangrejo Apr 15 '19 at 08:20
  • @broncoAbierto Aiyo! – William Apr 15 '19 at 08:30
  • @broncoAbierto No dude, I don't think you're understanding what I'm exactly trying to say here. So an Aiyo. Lol. Nvm let's wait for other answers. – William Apr 15 '19 at 08:55
  • I understood your problem. You think monotonicity at an interval should imply monotonicity at the points therein. With the definitions you've used, it doesn't. – cangrejo Apr 15 '19 at 09:01

1 Answers1

1

Let $J$ be any (open, halfopen, closed / bounded, unbounded) interval and $g : J \to \mathbb R$ be a function. Then $g$ is strictly increasing on $J$ if for all $x,y \in J$ such that $x < y$ we have $g(x) < g(y)$. You may also define that $g$ is strictly increasing in a point $p \in J$ if there is $\varepsilon > 0$ such that $g$ is strictly increasing on $J \cap (p - \varepsilon, p + \varepsilon)$.

You have a function $f : \mathbb R \to \mathbb R$ and ask for the (maximal) intervals $J$ on which $f$ strictly increases or decreases.

If you interpret this in the sense that $f \mid_J$ has this property, then you get the result which you call reasonable. I share this point of view.

However, you could also interpret the question in the sense that $f$ should strictly increase or decrease in all points of $J$ which gives your "but".

There is no contradiction, but only different interpretations, and you have to decide which you prefer.

Paul Frost
  • 87,968