Local maxima/minima and removable discontinuities

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.

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.

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.

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.

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,∞, 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.

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

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

I would disagree with you on this point. x^2 clearly has a global minimum at 0.

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

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.

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

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

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.

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.

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

Real gangsters split the fractions up.

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.

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

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?"

Psh. Raquaza#### bringing up this question is more complicated. You had to try it using his wayyy.

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.

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,∞, 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.

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.

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.