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In a simple case of the IFT, consider $$f(x,y)=y-x^3:\mathbb{R}^2\to\mathbb{R}$$ Now we have that $f(0,0)=0$ and $\frac{\partial f}{\partial x}(0,0)=-3(0)^2=0$, so it would seem that we can't solve for $x(y)$ in a neighborhood of the origin.

That seems counter-intuitive, however, since we can surely write $x=\sqrt[3]{y}$, which seems to capture the essence of the function exactly, even near/at the origin.

I'm sure I am making a silly mistake, but I'd very much appreciate some help. Thanks in advance.

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In the single-variable case, $f'(a) \ne 0$ is a sufficient condition for $f$ having an inverse in some neighborhood of $a$. This is not a necessary condition, as it is possible for $f'$ to be $0$ at isolated points and still have local inverses everywhere.

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  • $\begingroup$ That makes perfect sense. I still feel somewhat unconvinced, but the wording of the theorem clearly states the conditions as sufficient, but not necessary, to ensure the outcome. Thanks! $\endgroup$
    – Nico
    Commented Apr 15, 2015 at 3:45

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