By Ran Kiri


2019-01-06 22:37:15 8 Comments

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?

4 comments

@Mindlack 2019-01-06 22:51:34

I think that as long as $f^{-1}$ is well-defined on a neighborhood of $f(x_0)$, and continuous at $f(x_0)$, there is no issue.

Indeed, $f(f^{-1}(f(x_0)+h))=f(x_0)+h$ so $h=f(f^{-1}(f(x_0)+h))-f(x_0) \sim f’(x_0)(f^{-1}(f(x_0)+h)-x_0)$, and the conclusion (of differentiability and value of the derivative) follows.

@jmerry 2019-01-06 22:50:36

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.

@Ran Kiri 2019-01-06 23:01:57

Yes I can see it now. Thank you very much!

@Milo Brandt 2019-01-06 22:58:48

The continuity condition is not necessary. It's enough that $f$ be injective on some neighborhood. This said, if your function has a sequence of jump discontinuities near $x_0$, you might have that there is no open interval $U$ around $x_0$ for which $f(U)$ is also an interval. This means that $f^{-1}$ might be defined on a strange domain, though we can still technically differentiate it to get the desired result.


Formally, the statement you would need to prove is the following:

Let $A$ and $B$ be subsets of $\mathbb R$ and $f:A\rightarrow B$ and $g:B\rightarrow \mathbb R$. Suppose that $x_0\in A$ is an accumulation point of $A$ and $f(x_0)$ is an accumulation point of $B$. Then,

  • If two of the derivatives $f'(x_0)$ and $g'(f(x_0))$ and $(f\circ g)'(x_0)$ exist and are non-zero, the third exists as well.

  • If all of the derivatives exist, then $(f\circ g)'(x_0)=f'(x_0)\cdot g'(f(x_0)).$

One you have this statement, you can apply it to a pair where we take $g=f^{-1}$. Note that we can make this work even if $f$ isn't defined on an interval around $x_0$ - it's okay as long as we have enough points to define the relevant limit towards $x_0$.

Granted, it is a bit unusual to talk about derivatives on sets that aren't open, but there's no technical limitations preventing it, though the proof of the suggested lemma is a pain.

@Henno Brandsma 2019-01-06 22:51:19

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.

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