Assume $f:[a,b]\to[a,b]$ be continuous and differentiable on $(a,b)$ and $f(a)=a$, $f(b)=b$. How to prove that exists distinct $x_1,x_2 \in(a,b)$ such that $f '(x_1)f '(x_2)=1$? Thanks in advance.
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Apply the MVT for $g(x)=f(f(x)).$ Thus there exists $x_1 \in (a,b)$ s.t. $g'(x_1)=f'(x_1)f'(f(x_1))=1,$ so we're done.
If it happens $x_1=f(x_1),$ then apply the MVT for $g$ over $[a,x_1]$ therefore, there exists $x_2 \in (a,x_1)$ s.t. $g'(x_2)=f'(x_2)f'(f(x_2))=1.$ If $x_2 \neq f(x_2)$ so we're done, but if $x_2=f(x_2),$ since $(f'(x_1))^2=1=(f'(x_2))^2$ we will then have $f'(x_1)f'(x_2)=1$ or $-1.$ For the latter, we again need to apply the MTV over $[x_1,b]$ and run the argument for finding $x_3 \in (x_1,b),$ but now we can choose two points out of three ($x_1,x_2,x_3$) with the desired property.