Contradiction?

[/0]

Consider the graph of ... it has an infinite gradient at x = 0. But for it to have an infinite gradient, there must be two points such that their x values are the same but their y values are different, so it can't be a function? What's up with that?

[Mao]

wait what?

that reasoning doesnt make sense, because that practically goes against stationary points in general: for a turning point such as in , there is only one such point, not two.

if you are applying the first principles, you must realise that regardless how small is, it is still not 0, so there's always an infinitesimal difference between your "two points".

its really hard to explain...

[/0]

So is it the case that the limit to 0 is infinity but the gradient is at 0 is not?

[Mao]

this limit is difficult (if at all possible) to evaluate by hand

but you should keep in mind that:

gradient = derivative = tangent = =

they are all infinite.

[bigtick]

 at x=0, dy/dx = lim(h->0) [(0+h)^(1/3) - 0^(1/3)]/h = lim(h->0) 1/h^(2/3)  ->infinity

[Fitness]

so it can't be a function?

It's a 'one-to-one' graph. Of course it's a function.

Consider the graph of ... it has an infinite gradient at x = 0.

NO! It's not infinite! It's undefinable!

[Neobeo]

so it can't be a function?

It's a 'one-to-one' graph. Of course it's a function.

Consider the graph of ... it has an infinite gradient at x = 0.

NO! It's not infinite! It's undefinable!

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

[/0]

lol I think I'm just gonna take this for granted for now and see how it rooolllls...

[Glockmeister]

What's Ahmad's opinion of the topic? In otherwords, What Would Ahmad Do?

[bigtick]

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

[Fitness]

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

see

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

=> f'(0) = 1/0
Therefore it is undefined. kthx.

[Mao]

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

see

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

=> f'(0) = 1/0
Therefore it is undefined. kthx.

but infinitity is a well defined concept!

[Fitness]

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

see

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

=> f'(0) = 1/0
Therefore it is undefined. kthx.

but infinitity is a well defined concept!

Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.

Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.

[Mao]

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

see

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

=> f'(0) = 1/0
Therefore it is undefined. kthx.

but infinitity is a well defined concept!

Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.

Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.

why cant it have a tangent with an infinite gradient? [its equation is x=0]

[Fitness]

Actually the gradient is well-defined at x = 0. So it is in fact infinite.

see

Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.

=> f'(0) = 1/0
Therefore it is undefined. kthx.

but infinitity is a well defined concept!

Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.

Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.

why cant it have a tangent with an infinite gradient? [its equation is x=0]

Because 1/0 is not infinity. The end lol