Although this problem may look easy, but I am very much confused over this. Consider the function

$$f(x) = \begin{cases}(1-x)/|x|, & x\neq 0\\1, & x=0\end{cases}$$

Does $f$ satisfy the Intermediate Value Property on $[-2,2]$?


  • $\begingroup$ Would you mind and tell us what you've tried? You know that the intermediate value property holds for continuous functions and that your $f$ is continuous except for one point? $\endgroup$ – Dirk Oct 31 '13 at 13:02
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    $\begingroup$ Graphing the function would be a good first step; the picture leads fairly easily to counterexamples like the one that tetori gives. $\endgroup$ – Brian M. Scott Oct 31 '13 at 13:09
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    $\begingroup$ @Dirk: It's worth noting that, while continuous functions satisfy the IVP, not all functions satisfying the IVP are continuous. In fact there are functions (such as Conway's Base 13 function) satisfying IVP that are everywhere discontinuous. $\endgroup$ – Cameron Buie Oct 31 '13 at 13:50

$f$ does not satisfy intermediate value property. By definition $f(1/4)=3$. If $f$ satisfy IVP, then there is $0<a<1/4$ s.t. $f(a)=2$. But $f(x)=1/x-1$ for $x>0$ and $1/x-1> 3$ if $0<x<1/4$.

  • $\begingroup$ thanks a lot...i missed that small point $\endgroup$ – wanderer Oct 31 '13 at 15:14

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