/sci/ - Science & Math

>>11744584
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>>11744584
Thing is, x^3 is not strictly increasing at 0. Draw a tangent line to x^3 at x=0, and you’ll see that it is horizontal.

>>11744628
f(x)=x^3 is strictly increasing for all x. Just cause the derivative is instantaneously zero doesn't mean it's not strictly increasing

what does it mean the "strictly"

>>11744584
You disproved your assumption.
Good job, that's more than what most people try to do.
Now figure out on your own why you were wrong and modify your guess to make it work.

>>11745209
For x1<x2
Increasing: f(x1)<=f(x2)
Strictly Increasing f(x1)<f(x2)

I assume you mean first derivative?

I hate you nerds, like, go watch basketball

>>11744584
roughly speaking a function behaves like its first non-zero term in the taylor expansion. if the first derivative of f is zero, it doesn't mean it that f looks like constant, it means that the first order approximation is not good enough to conclude anything.

>>11744584
OP is right that this function is strictly increasing because the definition is that for all x values where an x value a is less than an x value b, f(a) < f(b).
I didnt know this, nicw job OP. The first derivative being positive does mean a function is increasing, but it is not the only situation where this is true.
Basically, x^3 is strictly increasing brcause all points are higher than their previous points, which is quite clearly what the definition for always increasing should be.

>>11747183
>The first derivative being positive does mean a function is increasing
Not necessarily. Take the function f(x)=2(x^2)sin(1/x)+x, with f(0)=0. Its first derivative at 0 is 1, but it's not increasing in any interval containing 0.

>>11744735
>f(x)=x^3 is strictly increasing for all x
Except 0.

>>11750332
>Except 0.
what ? you saying that x^3 is not increasing at 0 ?

>>11748184
Cool anon, I never considered that. If anyone is interested I found this paper that proves the existence of functions which are everywhere differentiable but nowhere monotonic, i.e. they are not increasing or decreasing in any interval

>>11751033
https://www.ams.org/journals/proc/1976-056-01/S0002-9939-1976-0396870-2/

>>11748184
I dont believe this. If you take the limit of (f(x) - f(x+h))/h as h approaches zero there has to be points coming from both sides, and the positive derivative shows the tangent line created by those converging points is positive, i.e. that the closest possible earlier point would be just below and that the closest possible later point would be just above. I can understand why the OP function above leads to what it does because there isnt anything other than a literal single point that is a flat line so, logically all other points would be above or below it.

>>11752659
it's true, the function is not increasing at 0

>>11752659
The points left of the origin are negative and right of the origin are positive, but any interval around the origin is not increasing because the functions oscillates infinitely quickly as you approach zero. Functions are strictly increasing on intervals, yet any interval (a,b) with a<0<b is not strictly increasing.

>>11750448
Yes, because at 0 its rate of change is 0.

>>11753329
Can you list an a and b in the domain with a<b such that f(a) is not less than f(b)?

>>11753329
0 is positive so the increase is positive :)

>>11753329
I don't think you know the definition of increasing function

increasing depends on the radius of the neighborhood of the partition of the function you define. read apostol. x^3 is increasing from neg jnf to 0, 0-0 is 0, and Inc from 0 to pos inf. these are the 3 largest neighborhoods within which u partition x3 into n equal closed intervals within which u can say it is increasing.

>>11753533

>>11744584
You thought something that was false, then came across a counterexample. What needs to be explained?

>>11752659
The lesson this function teaches is that there is no "closest" point. It's intuitive to think of functions as stepping from one point to the next, but that's not how real numbers work. Between any two points are an infinity of points.
If you graph it with something like Desmos you can see that no matter how far you zoom in, the function is always going up and down.

>>11753329
In any interval containing zero, choose two numbers x and y. If x <y then x^3 < y^3. It is increasing.

>>11744584
0 is positive *and* negative (in a certain sense at least). There's your explanation.

>>11753659
You cant be increasing on an interval thats oscillating everywhere. I dont believe this to be possible.

>>11753821
I'm not sure what you mean. Yeah, you're right, it's not increasing because it's oscillating. What don't you believe?

>>11753821
Yes, that's correct. It's not increasing at 0 but its derivative is positive at 0.

>>11744584
Limits don't preserve strict inequalities. For example, if [math]x_n>0[/math] then [math]\lim\limits_{n \rightarrow \infty} x_n\geq0[/math]. If a function [math]f(x)[/math] is strictly increasing then [math]\frac{\Delta f}{\Delta x}>0[/math], but [math]\lim\limits_{\Delta x \rightarrow 0}\frac{\Delta f}{\Delta x}\geq 0[/math].

>>11744735
>Just cause the derivative is instantaneously zero doesn't mean it's not strictly increasing
but thats exactly what that means...

>>11754977
No it doesn't. Consider x^3 as a counterexample.

>>11754977
try proving "if derivative of f at x is zero, then f is not increasing at x"
you will not succeed

What is up with all the mathlets on my board? Gtfo.

>>11754996
if fx was strictly increasing on [a, b] then for All x in reals there would be no g of z, in S ele of P, where S is the domain of f' and P is a neighborhood of f on [a, Px/n] , where f' always maps a single f'x st f'x > 0 for all x in P. there would be no z, st, for a < z < b, in P, f'z = 0 for all z in s
f' of z = 0 when x = z = 0
thus there exists a z in s where the function f' maps a number equal to zero breaking the strict inequality of for all x in ab x is strictly increasing iff f'z dne 0, which is not an element in S for all neighborhoods P in S, on ab, of which z is a subset

if there exists such a z, then f' must have plural domains and instead is f' on [a, b] st b < a < 0 and 0 < c < d and is strictly monotonic constant on [0, 0]
three. separate. functions.

>>11756560
well, can't you didn't try

Isn't stricly increasing defined in an interval, not a point

Any interval that contains 0 in that function is strictly increasing

>>11753533
>>11753535
None of this implies x^3 isn't increasing on the whole real line.

>>11758785
*strictly increasing, I mean.

>>11753340
No, just 0 and 0. I'm being very precise about 'for all x, which includes 0. Yes, the definition of increasing function has a<b, but I'm saying JUST for ONE point.

>>11759142
>Yes, the definition of increasing function has a<b, but I'm saying JUST for ONE point.
>but
the definition speaks clearly, there's no "but".

>>11759204
increasING, something that's continuous.
I'm saying that AT zero, not continuous, it isn't increasing. Not covered in the definition.

>>11760382
Absolute nonsense that's unrelated to what it means for a function to be increasing.

x^3 is strictly increasing; its derivative is 3x^2+h^2. At 0, the slope is h^2, which is positive.

x^3 isn't increasing at zero, but it's only at x=0. The slope is zero from 0 to 0. But the slope from 0 to any other point is positive.

>>11760382
f is increasing at a point x if there exists a neighborhood of x such that for all t in this neighborhood t < x implies f(t) < x and t > x implies f(t) > f(x).
this is the definition, x^3 at 0 doesn't satisfiy it, ergo x^3 is not increasing at 0. period.

>>11760619
>this is the definition, x^3 at 0 doesn't satisfiy it, ergo x^3 is not increasing at 0. period.
obviously meant to say the opposite