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Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and $f(1)=1$.For every integer $n$ show that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $

Is it theorem or i can show it with basic differentiation knowledge .Please give me hint .

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    $\begingroup$ Hint: A twist of MVT - given any two numbers $a,b$ such that $0 \le a < b \le 1$ and $f(b) \ne f(a)$, there exists a $c \in (a,b)$ such that $\frac{f(b)-f(a)}{f'(c)} = b - a$. $\endgroup$ – achille hui Jan 5 '15 at 21:28
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    $\begingroup$ Hint: Use the mean value theorem repeatedly. First to get the existence of a point $\xi$ such that $f'(\xi)=1$. Then pair points $a$ and $b$ (again using the mean value theorem and the intermediate value property of derivatives) such that $1/f'(a)+1/f'(b)=2$ $\endgroup$ – mickep Jan 5 '15 at 21:28
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$f$ is continuous with $f(0) = 0$ and $f(1) = 1$. It follows from the intermediate value theorem that there are numbers $$ 0 = b_0 < b_1 < \dots < b_{n-1} < b_n = 1 $$ such that $$ f(b_j) = \frac jn \text{ for } j= 0, \ldots, n \quad . $$

Now use the fact that $f$ is differentiable and apply the mean value theorem to each interval $[b_{j-1}, b_j]$, $j = 1, \ldots n$. It follows that there are numbers $\alpha_j \in (b_{j-1}, b_j)$ such that $$ f'(\alpha_j) = \frac{f(b_j) - f(b_{j-1})}{b_j - b_{j-1}} = \frac 1{n(b_j - b_{j-1})} \quad . $$ Then $$ \sum_{j=1}^n \frac 1{f'(\alpha_j)} = n \sum_{j=1}^n (b_j - b_{j-1}) = n(b_n - b_0) = n \quad . $$ The $\alpha_j$ are necessarily distinct because they are in distinct intervals $(b_{j-1}, b_j)$.

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