This is a challenge question:
Let with differentiable
Show there exists distinct points such that .
Moderator edit: Approved Challenge question.
From Lagrange theorem it follows that there exist such that
so we can assume or for all
(otherwise the proof it trivial). Suppose we have and for some .
If then we're done.
If then and thus the existence of such
that is guaranteed by the intermediate value theorem for derivatives and we're done.
Otherwise and so either or . Either
way, using the intermediate value theorem for derivatives we're done again.
Now suppose the previous case is not true. That is, for all or for all
Therefore the function is either monotone increasing or monotone decreasing (by looking at the derivative).
Either way, it means that . But from the conditions on it follows that , a contradiction.
My solution was a little shorter:
We may assume . (otherwise there's nothing to prove).
By the fundamental theorem of calculus we have . Hence we can't have or for all x. (since we also assumed )
But then there exist points such that f'(a)<1 and f'(b)>1 and the intermediate value theorem now gives that f'(x) reaches all points in (f'(a),f'(b)) and trivially we can find p,q in this interval such that pq= 1.
Check Fundamental theorem of calculus - Wikipedia, the free encyclopediaThe fundumental theorem of calculus has the assumption that is continuous,
which is not given, and there are example of derivatives which are not continuous, so I think your argument is
not perfectly justified.
Second Fundamental Theorem of Calculus
We don't have to assume f'(x) is continuous.
The examples given in your link are of oscillating functions that are unbounded. (at least from what I read) From say, ?
I hope you didn't forget our function is and is by definition bounded.
And I remember that a function can still be integrable, even though it may have a countable set of problem-points.
The second last post gives a function, from where they agree it is integrable....(oscillating but still bounded)
But, maybe we need help of an expert here.