In class we discussed Darboux’s Theorem and the oddity that it implies. Differentiability of a function does not imply the continuity of its derivative, but it does imply the Intermediate Value Theorem holds for its derivative. For this blog, I would like to explore a function for which this oddity occurs.
Consider the function 
discussed in the introduction to this chapter. As we have read, this function is differentiable in all of 
, but is not continuous at 0. What we have learned from Darboux’s Theorem, is that the Intermediate Value Theorem does apply to this function. This means that if you isolate an interval ![[a,b]](https://s0.wp.com/latex.php?latex=%5Ba%2Cb%5D&bg=ffffff&fg=000&s=0&c=20201002)
, then for any 
such that 
or 
, there exists 
such that 
. With this function, this does make sense. Without including 0 in the interval, you just have a continuous function and the IVT holds like normal. If you do include 0 in the interval, that is where some ambiguity might arise. The problem is the derivative at 0. For this function 
. This is the one that we think the IVT might not hold for. But this function simply oscillates faster and faster as it approaches 0, so any interval you pick that contains 0 will still have an oscillation of the function such that we have a point so 
.
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Thank you for the exploration of Darboux’s Theorem, this is a good example to consider as you wrangle with the theorem. There are a few minor confusions that I want to address though. First, you should be talking about the “Intermediate Value Property” (not the Theorem with a similar namesake. The theorem says that continuity on an interval implies the Intermediate Value Property holds. Second, you seem to indicate that the function is not continuous at 0, but it actually is… I think you mean to say that the derivative function is not continuous at 0…