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+--- Thread: Local maxima/minima and removable discontinuities (/showthread.php?tid=39668)
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Local maxima/minima and removable discontinuities - Five Second Pose - 2011-03-23
Say I've got the function (x^3)/(x). This is a standard parabola with a removable discontinuity at (0, 0), which is the vertex. Is there a minimum there? For every point close to the vertex, there's one closer, so I'm not sure and my calculus teacher didn't know for sure either.
Local maxima/minima and removable discontinuities - Corn - 2011-03-23
Well, taking the derivative...it equals 2x. Set 2x equal to 0, there should be a minimum/maximum at x = 0. So I'm assuming yes.
Local maxima/minima and removable discontinuities - Kalovale - 2011-03-23
f(x) = x^3 * x^-1 (x != 0)
f'(x) = (3x^2 * x^-1) + (x^3 * -1x^-2) = (3x^2)/(x) - (x^3/x^2) <-- product rule
f'(x) = 0 <=> (3x^2)/(x) - (x^3/x^2) = 0 <--- bullshit distribution
<=> x^2/x * (3 - x/x) = 0
<=> x^2 = 0 or 3 - x/x = 0
<=> x = 0, or x(3-1)/x = 0
<=> x = 0, or x = 0, or 3-1 = 0
So no, it doesn't have a minimum.
Local maxima/minima and removable discontinuities - Kabanaw - 2011-03-23
Pretty simply:
x^2, x!=0
2x=0
x=0
The one local extreme is x=0, and since that is undefined there are no relative extrema. That's the only point it could be, and since it's undefined it can't be.
Local maxima/minima and removable discontinuities - 2147483647 - 2011-03-24
, you'll find that the minimum is very close to 0, but if you search on the interval [0,∞, you'll hit the problematic hole.Never mind. There's no minimum. See Russt's post.
Local maxima/minima and removable discontinuities - Noah - 2011-03-24
There are no minimas at the interval (-∞, ∞
for this function, yes. However, for any interval [ε, ∞] or [-∞, -ε], 0 < ε, there would be a local minima. Removing the infinities from the intervals will also give a local maxima.Noah
Local maxima/minima and removable discontinuities - Cyadd - 2011-03-24
Noah Wrote:There are no minimas at the interval (-∞, ∞
Noah
I would disagree with you on this point. x^2 clearly has a global minimum at 0.
Local maxima/minima and removable discontinuities - Kalovale - 2011-03-24
Noah Wrote:There are no minimas at the interval (-∞, ∞
Noah
Wouldn't this be like saying: "Give me a situation where everybody else in a group is shorter than me (and chase everyone taller than me out of the group), then I will be the tallest in that group." ?
Though it definitely makes sense, I find it rather pointless.
Local maxima/minima and removable discontinuities - hadriel - 2011-03-24
f'(x) = x^2 = 0 implies a global minimum at x = 0, but you cannot forget that the meaning of derivatives are true only for delta x --> 0. Just because a global minimum exists by analysis of derivatives does not mean that it actually exists! What about x^2/x? That's even more problematic.
214 got it right, or nearly - you have a "local minima" if you pick a point very very close to (but not equal to) 0, but it is not a true minima because I can argue by a next-step-closer-towards-0 PoV, which just goes back to the meaning of "infinity". By l'Hopital's rule the limit clearly exists at x = 0, but if the function fails to exist at x = 0, then there clearly isn't a local minima.
My 2 pennies. Generally I shy away from math where possible, especially analysis.
Hadriel
Local maxima/minima and removable discontinuities - shouri - 2011-03-24
Noah Wrote:There are no minimas at the interval (-∞, ∞
Noah
This is correct ^. Except shame on noah for the bolded part
(x^3)/x is only defined from (-inf, 0) U (0,inf)
y = x^3/x
dy/dx = 2x which can only equal zero if x = 0. Since x=0 isn't in the domain of this function, it cannot be considered for maxima/minima. Since critical points have failed to find max/mins then the only other candidates for max/min are endpoints of the domain. But none of those endpoints are included, thus there are no points which are candidates for max/mins.
TL;DR: To be a local maximum or a local minimum of a function, you have to be in the domain of the function. So no, it doesn't have a min at x=0.
Local maxima/minima and removable discontinuities - OB3LISK - 2011-03-24
Dammit don't you nerds sleep. I came in about to dazzle everyone and all you pineappleing people are in here already. *MAD*
Gosh. Are you guys using the quotient rule when you're doing your derivative of (x^3)/x?? Or are you just doing the derivative of x^2...
If you do the quotient rule it turns out to be 3x - x...Which is what Kalovale got by bringing the x up to -1 power and using product rule.
Dammit 21whatshisface is right. The minimum is close to 0 but not 0.
Local maxima/minima and removable discontinuities - shouri - 2011-03-24
OB3LISK Wrote:Dammit don't you nerds sleep. I came in about to dazzle everyone and all you pineappleing people are in here already. *MAD*
Gosh. Are you guys using the quotient rule when you're doing your derivative of (x^3)/x?? Or are you just doing the derivative of x^2...
If you do the quotient rule it turns out to be 3x - x...Which is what Kalovale got by bringing the x up to -1 power and using product rule.
Dammit 21whatshisface is right. The minimum is close to 0 but not 0.
That's quite interesting cause i don't spot that using the quotient rule. Though it's technically right.
dy/dx = [x*(3x^2)- (x^3)*1] / x^2 = [3x^3 - x^3] / x^2 = 2 x^3/x^2 = 2x. Or you can keep it as the fraction... either way.
Edit: Oh wait i see it... why would you split up the fraction though? xd
Local maxima/minima and removable discontinuities - OB3LISK - 2011-03-24
Real gangsters split the fractions up.
Local maxima/minima and removable discontinuities - Corn - 2011-03-24
OB3LISK Wrote:Dammit don't you nerds sleep. I came in about to dazzle everyone and all you pineappleing people are in here already. *MAD*
Gosh. Are you guys using the quotient rule when you're doing your derivative of (x^3)/x?? Or are you just doing the derivative of x^2...
If you do the quotient rule it turns out to be 3x - x...Which is what Kalovale got by bringing the x up to -1 power and using product rule.
Dammit 21whatshisface is right. The minimum is close to 0 but not 0.
I used Wolfram, honestly. I was too lazy to calculate the quotient rule by hand.
Local maxima/minima and removable discontinuities - shouri - 2011-03-24
Wolfram tells you it has a minimum at x=0 though, since it first reduces the fraction and then pretends you told it to work with the reduced fraction... which is retarded
Local maxima/minima and removable discontinuities - Kabanaw - 2011-03-24
OB3LISK Wrote:Dammit don't you nerds sleep. I came in about to dazzle everyone and all you pineappleing people are in here already. *MAD*
Gosh. Are you guys using the quotient rule when you're doing your derivative of (x^3)/x?? Or are you just doing the derivative of x^2...
If you do the quotient rule it turns out to be 3x - x...Which is what Kalovale got by bringing the x up to -1 power and using product rule.
Dammit 21whatshisface is right. The minimum is close to 0 but not 0.
So if you use the quotient rule you end up with 3x-x... which equals 2x. Oh, so it's the same solution as using x^2, just more complicated?
Anyways, unless you can actually define a number for a local minimum, you can't use it. Can you define the number that is close to 0 but not 0 using the statement "the lowest value of f(x) is f© at x=c?"
Local maxima/minima and removable discontinuities - OB3LISK - 2011-03-24
Psh. Raquaza#### bringing up this question is more complicated. You had to try it using his wayyy.
Local maxima/minima and removable discontinuities - Shidoshi - 2011-03-24
The function has no local minimum. You can't find a value for x different than zero for which the function has a minimum value (i.e. no other x in the vicinity of the first one exists for which the funcion has a lower value). There is always a value for x that you will give you a lower value of the function.
Local maxima/minima and removable discontinuities - Russt - 2011-03-24
2147483647 Wrote:I think there is a minimum. It's just not at 0. Just because one point is removed from the function doesn't mean that the next best point(s) aren't minima. If you search on the interval (0,∞
Not by a strict definition of minimum: f has a (global) minimum at x* if f(x*) ≤ f(x) for all x (in the domain of f).
For any x* > 0, f(x*) = x*^2. But then f(x*/2) = x*^2/4 = f(x*)/4 < f(x*), so x* is not a minimum.
Local maxima/minima and removable discontinuities - Stereo - 2011-03-24
0 is a lower bound, yes. But x* that satisfies the conditions of the def. of minimum is not in the domain of the function, so it's not a minimum.