0
$\begingroup$

Is there example of let's say a two variable function that can be rewritten as an explicit function of one variable but that does not satisfy the assumptions of the implicit function theorem?

In fact, if the theorem only provides sufficient conditions, we should say that there exists some implicit function on which, however, the theorem may fail to be applied.

| cite | improve this question | | | | |
$\endgroup$
  • 2
    $\begingroup$ Do you mean something like $f(x,y)=y-x^3$ around $(x,y)=(0,0)$? $\endgroup$ – popoolmica Dec 21 '19 at 12:06
  • $\begingroup$ yes, thank you very much! $\endgroup$ – Shootforthemoon Dec 21 '19 at 16:08
  • 1
    $\begingroup$ Note that $x=\sqrt[3]{y}$ isn't differentiable at $y=0$, though. $\endgroup$ – Hans Lundmark Dec 21 '19 at 16:11
  • $\begingroup$ @HansLundmark that's true $\endgroup$ – Shootforthemoon Dec 21 '19 at 17:04
1
$\begingroup$

The equation $F(x,y)=(x-y)^2=0$ can be solved for $y=f(x)=x$ (a smooth function) despite $\partial F/\partial y$ being zero at every point on the curve $F=0$.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks for the answer! $\endgroup$ – Shootforthemoon Dec 21 '19 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.