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I studied derivative of function ${f^{-1}}'(y)=\frac1{{f}'(x)}$

When I tried above proof , it needs continuity of inverse function

At this point , I have a question

  1. $f$ is continuous on D , then what condition implies continuity of inverse? (on D or at $x$)

  2. If $f$ is differentiable at $x$ then inverse function is continuous at $f(x)$?

user128766
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2 Answers2

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(1) You know nothing about the continuity of the inverse function. For example the function $\displaystyle f(x)= \left\{\begin{matrix} x &, &x \in [0, 1] \\ & & \\ x-1& , & x \in (2, 3) \end{matrix}\right.$ is continuous whereas its inverse defined as

$$f^{-1}(x)= \left\{\begin{matrix} x &, &x \in (0, 1] \\ x+1&, & x \in (1, 2] \end{matrix}\right.$$

is not continuous. Of course if $f$ is continuous in a compact interval , that is an interval $[a, b]$ then , yes, the inverse function is also continuous. (It is a theorem)

Update: In order to answer the initial question , conditions that may impliy contiunity are:

1. A function $f$ is strictly monotonic and continuous on an interval $[a, b]$ then the inverse is continuous. This also holds if $[a, b]$ is replaced with a random $I \subseteq \mathbb{R}$.

2. A function $f$ is defined on an open interval $I$ and there it is continuous and bijective , then the inverse is also continuous.

(2) Not necessarily. It holds that:

$$\left ( f^{-1}(y) \right )' = \frac{1}{f'(x)}$$

Ok, $f$ is differentiable but no-one can guarantee you that $f'$ is continuous. Therefore you cannot extract such a result. See my counter example above. (of course the theorem you proved holds under the assumption that $f'(x) \neq 0$.)

Tolaso
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  • Sorry but i said 1. what condition , 2. continuity of inverse – user128766 Jul 09 '15 at 07:15
  • Well, one condidition would be $f$ to be defined in a compact interval , another condition would be to defined in an open interval and be injective in that interval.

    In order to have continuity of inverse you need continuity of the derivative.

    I apologize. Although I read the question I wrote a wrong answer for the first part. Anyway, I leave the answer as it. If needed I can always update it.

     – Tolaso Jul 09 '15 at 07:18
  • What theorem allows your inverse to be continuous if your are defined on a compact interval. Continuity seems quite weak. You do need injectivity moreover a least. – Lærne Jul 09 '15 at 07:29
  • @Lærne Thank you.Also if $f:[a, b] \rightarrow \mathbb{R}$ be a strictly monotone and continuous function then the inverse is also continuous. Same applies if $f$ is defined in $I \subseteq \mathbb{R}$. – Tolaso Jul 09 '15 at 07:34
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If you suppose $f$ injective and if $f(U)$ is an open set for $U$ an open set, then the inverse function is continuous.

Euler88 ...
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