This is my lecturer's Darboux Theorem, however I have a problem with one of the steps there, I signed it wih ***. He said xmin is not equal to a and b. Buf if it is equal then what? I mean intuitively when I think about it if it is equal then I cant get c such that f ' (c)=0 but still can I just say xmin is not equal to a and b? Isn't that specific case? Thank you.
So he had to say that if f'(a)<0 then there exists h>0 such that a-h<x<a+h where the function is decreasing. so there exsist t such that a<t<a+h with f(t)<f(a). And since it's an interval you can find a point that is smaller than a and therefore it can't be xmin I get it ! Thanks
The following example is taken from the book "Counterexamples in Analysis" (chapter 3, example 5). Let
Thus $f'(1)=1>0$, but the term $2\cos(1/x)$ causes oscillations around $x=0$. The result is that $f'(x)$ takes both positive and negative values, and so $f(x)$ increases and decreases, in each neighborhood of 0. As expected, $f'(x)$ is not continuous at 0.
What I was saying is that if $f'(a)<0$, then using the definition of derivative it is easy to show that there exists (at least) one point $t>a$ such that $f(t)<f(a)$.
For every epsilon there exists delta such that, |x-a|<delta then |f(x)-f(a))/((x-a))-L|<epsilion, and this means that there is a neighbourhood, otherwise we couldn't talk about the limit. Since f'(a) exists so does the limit and therefore according to the definition of limits there has to be a neighbourhood. Is my logic wrong?
If $f'(a)<0$, then there does indeed exist a neighborhood of $a$ where $f(x)>f(a)$ for $x<a$ and $f(a)>f(x)$ for $x>a$. However, there may not be a neighborhood of $a$ where $f(x)$ is decreasing. Do you see the difference between the two claims?
My previous post was wrong. Continuity of f’(x) is sufficient but not necessary to prove Darboux’s theorem by intermediate value theorem.
So I took OP proof seriously. Having done so, the OP attachment seems quite straight-forward, and I don’t see all the fuss about ***:
The OP begins:
f(x) continuous on [a,b], therefore f(x) achieves a min on [a,b], and
a<xmin<b by definition of right and left hand derivative.*
*f(x)<f(a) for points near a to the right of a (f’(a)<0) and f(x)<f(b) for points near b to the left of b (f’(b)>0).
Continuity of derivative is irrelevant to this proof.