Is the Implicit Function Theorem Applicable to F=K?

[deluks917]

Can anyone check if this argument works. I just made up this problem to check if I understand what's going on.

Consider F = xy*e^x + y*e^y = K. I want to see if there is a unique solution.

F_y = x*e^x + y*e^y + e^y.

Since we are on the surface x*e^x + y*e^y = K.

So if K >= 0 then F_y ≠ 0. So given any x we can find a unique y.

Did I understand the theorem correctly?

 

[Char. Limit]

Can anyone check if this argument works. I just made up this problem to check if I understand what's going on.

Consider F = xy*e^x + y*e^y = K. I want to see if there is a unique solution.

F_y = x*e^x + y*e^y + e^y.

Since we are on the surface x*e^x + y*e^y = K.

So if K >= 0 then F_y ≠ 0. So given any x we can find a unique y.

Did I understand the theorem correctly?

You're missing a few things here. Firstly, if $F = xy e^x + y e^y = K$, then $F_y = x e^x + (y+1) e^y = 0$. That's an implicit function. And secondly, given any x, there's not necessarily a unique y. For example, take x=0.05. For this point x, there are two solutions to y satisfying the derivative F_y (namely y=-4.06616... and y=-1.16923...).

Is that what you mean?

 

[NickMusicMan]

deluks, you are correct.

The implicit function theorem implies that F=K is LOCALLY solvable for y as a function of x. That is, for each point there exists a neighbourhood of that point where y can be written as a function of x. In this case, F is only 2 variables, so "Locally solvable everywhere" is equivalent to "globally solvable" and so your conclusion is correct, but this would not be valid if F had more than 2 variables.

To the other poster: He is referring to the Curve F=K having a unique x for each y, not anything about the derivative.