Timeline for Is $x^3$ really an increasing function for all intervals?
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2023 at 18:11 | answer | added | gnasher729 | timeline score: 0 | |
| Nov 17, 2023 at 15:46 | answer | added | Noiril | timeline score: 0 | |
| Nov 17, 2023 at 11:38 | vote | accept | Maddy | ||
| Nov 17, 2023 at 11:38 | comment | added | Maddy | @MW, got it! infinitesimal, local and global scopes... the infinitesimal information is what is determined using the local and global scopes. The local and global scopes are NOT determined using the infinitesimal, as far as the formal 'increasing' definition goes. Anyways, thank you! | |
| Nov 17, 2023 at 11:10 | comment | added | M W | @Maddy one final thought on your last comment - it is fine to say the slope (of the tangent line) for $x^3$ is $0$ at that individual point. It’s equally fine to say the rate of change is equal to $0$ at that point. Those are both “infinitesimal” concepts for lack of a better word. Increasing, on the other hand, is what I would call a local concept (and sometimes a global one as well), meaning it describes behavior over a range of inputs. I won’t digress further on the exact meaning of these words but as you move forward in mathematics it will help to keep in mind this kind of distinction. | |
| Nov 17, 2023 at 10:25 | comment | added | Maddy | Thank you, everyone, for your help and clarification. I see I've been quite stubborn with my thoughts but well, I'm still learning and am still just a kid I guess. [xD Sorry, just had a crazy thought at night so well here goes (I do realise I'm wrong): we MIGHT define increase as if imagining a car moving up a ramp with zero friction. for each height it gains, it has to accelerate. say $f(x) = x^3$ is what defines the ramp. I guess it is obvious that there comes a single instance at $x=0$ for which the car would NOT have to accelerate. This is what I mean by zero slope and no increase. huh?] | |
| Nov 17, 2023 at 10:18 | vote | accept | Maddy | ||
| Nov 17, 2023 at 10:21 | |||||
| Nov 16, 2023 at 22:30 | answer | added | Lee Mosher | timeline score: 2 | |
| Nov 16, 2023 at 20:47 | comment | added | Riemann'sPointyNose | @Maddy I have to agree with JonathanZ here and say we are getting fuzzy with words here. What exactly does it mean to say the function isn't increasing at a single point? If you want to talk about whether a function is increasing or decreasing on an interval, you need to consider it's behaviour over that entire interval. If you consider the behaviour of ${x^3}$ over the whole real line (i.e. the interval ${(-\infty,\infty)}$), then it is in fact increasing, in the precise sense that if ${a < b}$ then ${f(a) < f(b)}$ | |
| Nov 16, 2023 at 20:01 | comment | added | JonathanZ | We're getting a bit fuzzy here, but "calculus ... has the essence of capturing information about what a function does..... at intervals around a specific point" is where I think you're going wrong. Calculus tells you what's left when the interval shrinks away to nothing. (I'm speaking very loosely there, practically poetically.) And increasing/decreasing, strictly, describes behavior over an interval. There are theorems that connect the two (sometimes), but they are different things. | |
| Nov 16, 2023 at 19:21 | comment | added | Maddy | I mean come onnnnn just agree that AT x=0, the function isn't increasing, i.e the rate of change at its infinitesimal proximity IS INDEED 0. | |
| Nov 16, 2023 at 19:08 | comment | added | Maddy | @Riemann'sPointyNose, in the essence of calculus, it is! That is what my point is precisely about. For a limit of values closer and closer to 0, they will be equal. Well okay, that might be wrong, I see you do have a point there... no matter how small, there is a difference. But that means including the 0 for increasing as well as decreasing intervals for $f(x) = x^2$ would be just fine. Why don't we do it then? | |
| Nov 16, 2023 at 19:03 | comment | added | Maddy | @MetehanTuran here is one: Lim h->0 (x-h, x+h) I see everyone's point here... but well everyone says increasing or decreasing has got nothing to do with calculus. That means increase or decrease are unrelated to rates of change? huh? My point is that calculus, in itself, has the essence of capturing information about what a function does (in terms of change) at intervals around a specific point. If everyone else is in fact correct, which I'm beginning to guess they are, my only question then is, why don't we include 0 within the intervals for $f(x) = x^2$? It would satisfy both. | |
| Nov 16, 2023 at 18:25 | comment | added | JonathanZ | Just a note to add on to the many correct answers you've already gotten: People do use the phrase "increasing at point X", but it's always a loose phrase, it's not always reliable, and always needs to take a back seat to the rigorous two-point definition. | |
| Nov 16, 2023 at 17:48 | answer | added | Toby Bartels | timeline score: 2 | |
| Nov 16, 2023 at 17:48 | comment | added | Severus' Constant | if you are right, you must find an interval (a,b) where $a\neq b$ such that $ f(a) \ge f(b) $. | |
| Nov 16, 2023 at 17:42 | comment | added | Riemann'sPointyNose | Another way to think about it is as follows: if you take a tiny step in the positive direction at ${x=0}$, say ${h}$, is ${f(x+h) > f(x)}$? And the answer is yes: ${(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 > x^3}$. When it comes to asking if a function is increasing or decreasing, the definition is not based on derivatives. It's just that it's possible to use derivatives to aid you in figuring out if a function is increasing or decreasing on a particular interval (for example, if a function has ${f'(x) > 0}$ all across an interval, you can conclude it's increasing on that interval) | |
| Nov 16, 2023 at 17:30 | answer | added | Zuy | timeline score: 2 | |
| Nov 16, 2023 at 17:30 | answer | added | M W | timeline score: 10 | |
| Nov 16, 2023 at 17:29 | comment | added | whpowell96 | The key here is that you are interested in the function's behavior on intervals, but only looking at derivative information at a single point. Positive derivative at a point does not imply that the function is increasing in any interval containing that point. For the canonical example, see here math.stackexchange.com/questions/4315614/… | |
| Nov 16, 2023 at 17:25 | comment | added | DonAntonio | Your teacher and classmate are right, and the explanation of the latter (between square parentheses) is correct up to the point where he says the function's constant in that point, which makes no sense. Another way to put it in a simple way: zero is not an extreme value, but it is an inflection point. Yet the very discussion of whether a function is increasing or not at a single point seems to be rather pointless, imo. | |
| S Nov 16, 2023 at 17:19 | review | First questions | |||
| Nov 16, 2023 at 17:20 | |||||
| S Nov 16, 2023 at 17:19 | history | asked | Maddy | CC BY-SA 4.0 |