Prove there are $\xi$, $\eta$ with $f'(\xi)f'(\eta)=1$ Let $f:[0,1]\to\mathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $\xi,\eta\in(0,1)$ so that $f'(\xi)f'(\eta)=1$.
I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.
 A: Here is one way to do this:
Let $g(x) = f(x)^2 - x^2$.  Applying Rolle's theorem to $g(x)$, we obtain a value $\eta \in (0,1)$ such that $2f(\eta)f^\prime(\eta) - 2\eta = 0$, or $f(\eta) = \frac{\eta}{f^\prime(\eta)}$.  Now apply the mean value theorem to $f$ on the domain $[0, \eta]$ to obtain a value $\xi \in (0,\eta)$ such that $f^\prime(\xi) = \frac{f(\eta)}{\eta}$.  Then $\displaystyle f^\prime(\xi)f^\prime(\eta) = \frac{f(\eta)}{\eta}f^\prime(\eta) = \frac{\eta}{\eta f^\prime(\eta)}f^\prime(\eta) = 1$.
A: Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $\xi \in (0, c), \eta \in (c, 1)$ such that $f'(\xi) = 1 = f'(\eta)$ and we are done.
Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x \in (0, 1)$ or $f(x) < x$ for all $x \in (0, 1)$.
We take the case when $f(x) > x$ for all $x \in (0, 1)$. Consider the point $d \in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p \in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q \in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q \in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.
Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.
More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $\prod f'(x_{i}) = 1 $.
