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What do you mean? If we have any continuous function, then we do not have necessarily a derivative. So how could we make a statement about derivatives?~~theorem~~without requiring the derivative function to be continuous. Why is this property not true for any continuous function in its intermediate value theorem?

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This is to compare intermediate value thm and Darboux thm. Clear?

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Not at all. You asked:This is to compare intermediate value thm and Darboux thm. Clear?

[Darboux: ##f'## takes every value between ##f'(a)## and ##f'(b)##]Why is this property ...

Continuous functions in general do not have derivatives. Hence the property cannot hold since the statement is void.... not true for anycontinuousfunction in its intermediate value theorem?

Did you mean why is it possible that ##f'## isn't continuous?

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The reason is, that ##f'## doesn't jump! ##f## is differentiable, so it may have kinks, but it does not have gaps. It goes roughly along the following lines:

differentiability of ##f## > continuity of ##f## > (Weierstraß) maximal point ##x_0## > (Fermat) ##f'(x_0)=0## > application on ##g:=f-c## where ##c## is the desired intermediate value for ##g##

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why can f' not have any gaps while f' can have kinks?##f## is differentiable, so it may have kinks, but it does not have gaps.

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This is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.why can f' not have any gaps while f' can have kinks?

You take a value in the gap and construct a contradiction to Darboux. It is a bit messy with all the theorems that basically look like one another.

The reference I found is not really well written, but if you like then have a look:

https://pnp.mathematik.uni-stuttgart.de/iadm/Weidl/ana1-13-14/skript/skriptse41.html

It's in the wrong language but if you open it with Chrome you can use its translation function. Not perfect, but it should do.

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is there any counter example?

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yes, is there any intuition to understand this consequenceThis is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.

You take a value in the gap and construct a contradiction to Darboux. It is a bit messy with all the theorems that basically look like one another.

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is there any counter example?

##x^2\sin(1/x)##

Basically the idea is that it is bounded by a parabola, so the derivative at ##x=0## is 0, but it wiggles around really fast at you get near 0, so you can get not close to 0 derivatives no matter how close to 0 you get.

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this example is to show that its derivative jumps or not?##x^2\sin(1/x)##

Basically the idea is that it is bounded by a parabola, so the derivative at ##x=0## is 0, but it wiggles around really fast at you get near 0, so you can get not close to 0 derivatives no matter how close to 0 you get.

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this example is to show that its derivative jumps or not?

No, the derivative does not jump, but it is discontinuous at x=0. You should try graphing this function and its derivative and thinking for a while about what is going on.

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Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?No, the derivative does not jump, but it is discontinuous at x=0. You should try graphing this function and its derivative and thinking for a while about what is going on.

For a discontinuous function, the function won't in general be able to take any intermediate values demanded unless specially designed.

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Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?

For a discontinuous function, the function won't in general be able to take any intermediate values demanded unless specially designed.

Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0 yourself.

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yes, the derivative is discontinuous at x=0Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0 yourself.

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yes, the derivative is discontinuous at x=0

So this is what the darboux theorem is saying. The only way a derivative can be discontinuous is it can wiggle around and take a bunch of different values near x=a, but still be defined at x=a

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why is this feature not in continuous functions?So this is what the darboux theorem is saying. The only way a derivative can be discontinuous is it can wiggle around and take a bunch of different values near x=a, but still be defined at x=a

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If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we havewhy is this feature not in continuous functions?

$$

f(h)=f(0) + \varphi (h) + r(h) = f(0)+c\cdot h +r(h)

$$

with a small remainder term ##r(h)## and a linear function ##\varphi \, : \,x\longmapsto c\cdot x .## Now ##c\cdot h## is still small for small ##h## and our function value ##f(h)## is still close to ##f(0)##. I.e. there is no way to get a gap of ##+2## for small and smaller ##h.## We can have as derivative ##\varphi (x)=cx## from the left and ##\varphi' (x)=c'x## from the right, which is discontinuous if ##c\neq c'.## This means we have a discontinuity of the derivative. However, ##ch## and ##c'h## are close for small ##h##.

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why is this feature not in continuous functions?

This is the definition of continuity, that ##f(x)\approx f(a)## if ##x\approx a##.

If you have a sequence of points ##x_n## that converge to ##a##, then if ##f## is continuous we must have ##f(x_n)\to f(a)##

But for the function I described above, you can find a sequence ##x_n\to 0## such that ##f'(x_n)=1/2##, yet ##f'(0)=0##.

In some sense, it would be nicer if ##f'(0)## didn't exist, because then the derivative would be continuous on its domain.

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but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we have

$$

f(h)=f(0) + \varphi (h) + r(h) = f(0)+c\cdot h +r(h)

$$

with a small remainder term ##r(h)## and a linear function ##\varphi \, : \,x\longmapsto c\cdot x .## Now ##c\cdot h## is still small for small ##h## and our function value ##f(h)## is still close to ##f(0)##. I.e. there is no way to get a gap of ##+2## for small and smaller ##h.## We can have as derivative ##\varphi (x)=cx## from the left and ##\varphi' (x)=c'x## from the right, which is discontinuous if ##c\neq c'.## This means we have a discontinuity of the derivative. However, ##ch## and ##c'h## are close for small ##h##.

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Yes. My example has (excuse the sloppy notation): ##f'(0^+)=c## and ##f'(0^-)=c'##. So we have a gap of ##|c'-c|.## But the function values ##f(h)## and ##f(-h)## are close to ##f(0)## because the derivative is multiplied by ##\pm h.## This means that there is still some control on ##f(x)## by the simple fact that it is differentiable. A property we do not have for only continuous functions.but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?

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