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Is there any function $f$ which is differentiable on an open interval $(a,b)$ but is not continuous on (and also cannot be extended continuously to) the closed interval $[a,b]$?

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    $\begingroup$ $f(x)=1$ if $x\in(0,1)$, $f(0)=f(1)=0$. $\endgroup$ Nov 26, 2014 at 12:38
  • $\begingroup$ I have written the very same example... $\endgroup$
    – ajotatxe
    Nov 26, 2014 at 12:39
  • $\begingroup$ @Anne You should tell us what attempts you have made to find a function like this $\endgroup$
    – terrace
    Nov 26, 2014 at 14:35
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    $\begingroup$ "not continuous on $[a,b]$" and "can't be extended continuously to $[a,b]$" are not the same things - which one did you mean? (the answers here seem to suppose the first meaning) an example of the latter kind: $sin(1/x)$ on $(0,1)$ $\endgroup$
    – Dániel G.
    Nov 26, 2014 at 15:26

5 Answers 5

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$$\Large\dot{}\!\!\underline{\qquad\qquad\qquad}\!\!\Large\dot{}$$

$$$$

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  • $\begingroup$ What is it? Some sort of joke? $\endgroup$
    – ajotatxe
    Nov 26, 2014 at 12:45
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    $\begingroup$ It is supposed to be a graph of such a function :) $\endgroup$ Nov 26, 2014 at 12:46
  • $\begingroup$ ._.$\phantom{2}$ $\endgroup$
    – Upc
    Mar 30, 2018 at 19:37
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$$f(x)=\frac1{(x-a)(x-b)},\qquad f(a)=f(b)=0.$$

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The easiest function I can think of is

$$f(x)=\begin{cases}\sqrt x&,\;\;x\in (0,1]\\{}\\18&,\;\;x=0\end{cases}$$

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  • $\begingroup$ @Timbuc How did you think of it (I mean, rationale)? $\endgroup$
    – hola
    Nov 26, 2014 at 14:53
  • $\begingroup$ @pushpen.paul, the square root is my box example of function defined and continuous on a certain interval but not differentiable in one point there. $\endgroup$
    – Timbuc
    Nov 26, 2014 at 15:10
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Consider $f(x)=x-\lfloor x\rfloor$ at any interval $(n,n+1)$ with integer endpoints.

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Differentiability implies continuity, but the intervals $(a,b)$ and $[a,b]$ were not the same; the first was open second was closed. This means at the points $a$ and $b$ it can be not continuous and it will still be differentiable on open $a,b$.

Thusly you can have a function that does what you said.


For example, you could have $f(a) = 5$ and $f(x) =2$ otherwise (when $x \neq a$). This function will have a discontinuity at $x=a$.

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