A function with a non-zero derivative, with an inverse function that has no derivative.

While studying calculus, I encountered the following statement: "Given a function $f(x)$ with $f'(x_0)\neq 0$, such that $f$ has an inverse in some neighborhood of $x_0$, and such that $f$ is continuous on said neighborhood, then $f^{-1}$ has a derivative at $f(x_0)$ given by: $${f^{-1}}'(x_0)=\frac{1}{f'(x_0)}$$

My questions is - why does $f$ have to be continuous on a whole neighborhood of $x_0$ and not just at $x_0$? Is there some known counter-example for that?

Solution 1:

The suggestion in the title isn't how it'll work. Instead of having an inverse that doesn't have a derivative, we'll fail to have a continuous inverse. Also, the required condition for the theorem isn't just that $f$ is continuous on an interval - it's that $f'$ is continuous on an interval around the key point.

Example: $f(x)=\begin{cases}x+2x^2\sin\frac 1x&x\neq 0\\0&x = 0\end{cases}$. This $f$ is differentiable everywhere, with derivative $1$ at zero, but it doesn't have an inverse in any neighborhood of zero. Why? Because it isn't monotone on any neighborhood of zero. We have $f'(x)=1+4x\sin\frac1x-2\cos\frac1x$ for $x\neq 0$, which is negative whenever $\frac1x\equiv 0\mod 2\pi$. We can find a one-sided inverse $g$ with $f(g(x))=x$, but this $g$ will necessarily have infinitely many jump discontinuities near zero.

The calculation of the derivative of $f^{-1}$ is just an application of the chain rule. The real meat of the inverse function theorem is the existence of a differentiable inverse.

Solution 2:

First off, any function has an inverse "at $x_0$" because we just assign $f(x_0)$ the value $x_0$; it's really meaningless to talk at an inverse existing at a point. We need a whole neighbourhood because then we can use the derivative, which is defined by a limit, so we must be able to "approach" $x_0$ arbitarily closely.