how to prove it is a cusp!?

[/0]

SAY YOU HAVE THE GRAPHS

FOR

AND FOR

THE GRADIENT AT X = 0 IS THE SAME FOR BOTH GRAPHS, YET THAT POINT IS A CUSP................................

HOW DOES ONE GO ABOUT PROOFING THAT IT IS A CUSP???

[Mao]

if

then f(x) a cusp at x=a

[/0]

But the limits are the same aren't they???

[Mao]

for a smooth join they are the same (hence differentiable)

but for a sharp turn they are two different values approaching from left and right.

[/0]

But for the equation I put up, the equations are differentiable at x = O, and they approach the same values from left and right. HOWEVER, there is a cusp at x = O , so how does that work???

[Mao]

neither of the two graphs have cusps. where is the question from?

[/0]

Soz what I mean is, if you graph the two graphs as I have described in the first post, then visually, it will seem like there is a cusp at .
i.e. you could think of the relation as

[evaporade]

you look at lim (invf)'(y) as y approaches 0+ and 0-

[kamil9876]

By using the basics of Mao's argument, you see that:

must NOT occur.

However: doesn't even exist. Since doesn't exist, where

In fact f isn't even a function.

If you define a cusp as a non-differentiable point the above is true.

However I wasn't sure about a formal definition of a cusp so i did some research and found: http://mathworld.wolfram.com/Cusp.html
Which is contrary to the popular statement y=|x| has a cusp at the origin... Meh lately these quasi-geometrical terms seem like crap to me... although if this is your hw I can see that you should be worried

Anyways, according to source above:

let
let


, but is not satisfied(because both limits do not exist) hence the two curves meet with same tangents however they are 'branches'(link uses this term) as they do not extend beyond this intersection.

You decide which definition of cusp to believe. I agree with the math but I'm not the best linguist