Let $f$ be a differentiable between $(0,1)$ and take $f(1)=1, f(0)=0$. Prove that then there exist $q_1,q_2,...,q_n$ distinct points such that $f'(q_1)+f'(q_2)+...+f'(q_n)=n$ for every natural n. By the mean value theorem I can find a point $a$ such that f'(a)=1 and fullfill the $1$ but I am stuck thinking about the rest.
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1$\begingroup$ Does this answer your question? Rolle's Theorem (apply it to $g(x):=f(x)-x$, which satisfies $g(1)=g(0)$ and $f'=1+g'$). $\endgroup$– Anne BauvalCommented May 13 at 16:26
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$\begingroup$ More generally, if $b_1,\dots,b_n$ are positive and $\sum_{k=1}^{n} b_k=1,$ there is $q_1,\dots,q_n$ such that $\sum_{k=1}^n b_kf'(q_k)=1.$ In your case, all $b_k=\frac 1n.$ $\endgroup$– Thomas AndrewsCommented May 13 at 16:30
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$\begingroup$ I looked at the other question, it's not a duplicate; it only considered the special case $n=2012$, I answered the problem in full generality. $\endgroup$– Jonathan HoleCommented May 13 at 16:54
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For $k=1, \dots ,n$ applying the Mean-Value-Theorem to $f$ on the interval $[\frac{k-1}{n}, \frac{k}{n}]$ gives a $q_k\in (\frac{k-1}{n}, \frac{k}{n})$ such that $f'(q_k)=n(f\big(\frac{k}{n}\big)-f\big(\frac{k-1}{n}\big))$. Then $$\sum\limits_{k=1}^{n} f'(q_k)=n\sum\limits_{k=1}^{n} f\Big(\frac{k}{n}\Big)-f\Big(\frac{k-1}{n}\Big)=n(f(1)-f(0))=n.$$
answered May 13 at 16:37