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I'm trying to verify whether or not $f$ has the IVP on the interval [0,1].

$f(x)= \begin{cases} x&\text{if}\, x\neq 1\\ 0&\text{if}\, x = 1\\ \end{cases}$

The function is not continuous on the closed interval $[0,1]$ so I can't use an argument involving continuity. Is there any way to make the argument using the definition of the IVT?

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    $\begingroup$ On the interval $[0.5, 1]$, does $f$ take every value between $f(0.5)$ and $f(1)$? $\endgroup$ – angryavian Oct 22 '18 at 20:00
  • $\begingroup$ @angryavian No, it doesn't. So this function would not have the IVT. Is there anyway to modify the function so that it would have the IVT yet not be continuous at x = 1? $\endgroup$ – Is12Prime Oct 22 '18 at 20:07
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    $\begingroup$ I think the standard counterexample is $f(x) = \sin(1/x)$ for $x > 0$ and $f(0)=0$. $\endgroup$ – angryavian Oct 22 '18 at 20:15
  • $\begingroup$ Or even more fun (but a bit messy), the Conway base 13 function takes every single real value on every single open interval yet is nowhere continuous $\endgroup$ – bitesizebo Oct 22 '18 at 20:24

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