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I don't know how to phrase the title better.

We say that a function is "growing" (don't know the English term) if

$f'(x)\ge0, \forall x$

However, if we want to say that the function is "strictly growing", we also need an extra statement saying that

$S=\{x | f'(x)=0\}$

contains no intervals. This means that if the derivative of a function is 0 in one isolated point, the function can still described as "strictly growing".

That got me thinking. If we had a function like the following:


It's derivative is obviously


But why wouldn't the function

$g(x)=\begin{cases} 1 &\mbox{if } x \ne 2 \\ 0 & \mbox{if } x = 2 \end{cases} $

Also be it's derivative? I've chosen the number 2 at random but the point is that only one of the points differs which I don't think is enough for the function to "change it's angle". So, what gives?

Furthermore, what does the function look like if it's derivative is something like

$g(x)=\begin{cases} 1 &\mbox{if } x \in \mathbb{Q} \\ 0 & \mbox{if } x \not\in \mathbb{Q} \end{cases} $

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See also… – oks Mar 5 '13 at 12:50
See also… – oks Mar 5 '13 at 12:58
up vote 2 down vote accepted

The derivative is the limit of $$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ When a function is differentiable the limit exists, and limits must be unique, so $g(x)$ is not the derivative.

There is no function whichs derivative is $g(x)$ as $g(x)$ is discontinuous in every point.

share|cite|improve this answer
Right. Thanks. Is there's some rule that says it can't be discontinuous in EVERY point or is it that it can't be discontinuous in ANY point? – Luka Horvat Mar 5 '13 at 12:48
For a function to be differential over an interval (a,b), it MUST be continuous in that whole interval. If the function is not continuous at any point, it is not differentiable at that point. – mardat Mar 5 '13 at 12:51

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