Mean value theorem related question

Question

Let $f:[a,b]\to\mathbb{R}$ be a differentiable function such that $f(a)=a$ and $f(b)=b$. Prove that there exists a $c\in(a,b)$ such that $f'(c)=1$ and that for every $n\in\mathbb{N}$ there exist different $c_1,c_2,\ldots,c_n$ such that $f'(c_1)+f'(c_2)+\ldots+f'(c_n)=n$.

First part is direct from Mean value theorem.

Could you give me any hints for second part?

Answer 1

This is only a hint. Divide the interval $(a,b)$ into $n$ equal intervals. The average of average slope of these $n$ intervals must equal the average slope of the interval $(a,b)$.

Answer 2

Let $a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b$ where $x_i := a + \frac{i}{n}(b-a)$.

For each $i \ge 1$ there exist $c_i$ such that $f'(c_i) = \frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}$. Then $$\sum_{i=1}^n f'(c_i) = \sum_{i=1}^n \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}} = \frac{1}{(b-a)/n} \sum_{i=1}^n (f(x_i) - f(x_{i-1})) = n.$$