# Why are the conditions of the IFT not necessary

The Inverse Function Theorem states sufficient conditions for a function to have a continuous inverse.

When, if it all, are these conditions necessary conditions? Is there a nice counterexample?

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## 1 Answer

Consider $f(x)=x^3$, which has zero derivative at $x=0$.

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That's embarrassingly easy. I suppose this didn't occur to me because in one-dimension I have seen the IFT stated as saying that if a function is continuous and strictly monotonic (on some interval), then it is invertible. This seems to me to be a necessary condition as well. – user21725 Jan 25 '13 at 2:11
@EricGregor Yes, that is a necessary condition. Somewhere on this site it is proven that if $f$ is not monotonic we either have a "vee" or "wedge", that is some $x<y<z$ with $f(x)\le f(y)\ge f(z)$ or $f(x)\ge f(y)\le f(z)$. If we take $c$ between $\min\{f(x),f(z)\}$ (or $\max\{f(x),f(z)\}$ in the second case) and $f(y)$ then by the IVT we see that $f$ achieves $c$ on $(x,y)$ and $(y,z)$, so $f$ is not invertible. – Alex Becker Jan 25 '13 at 2:20