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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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    $\begingroup$ If you use the Mean Value Theorem you can get one point $x_1$, since $f(b)-f(a)=b-a$. Does it help? $\endgroup$ – Sigur Feb 11 '13 at 18:31
  • $\begingroup$ @Sigur: What about the another distinct one? $\endgroup$ – mrs Feb 11 '13 at 18:32
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    $\begingroup$ @BabakSorouh, I don't know. So I asked if the first one could be useful. $\endgroup$ – Sigur Feb 11 '13 at 18:33
  • $\begingroup$ See another excellent answer at math.stackexchange.com/questions/1373408/… $\endgroup$ – Paramanand Singh Jul 26 '15 at 4:33
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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.

Added:

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.

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  • $\begingroup$ Why does $x_1\neq f(x_1)$? $\endgroup$ – Potato Feb 11 '13 at 18:43
  • $\begingroup$ Ditto ${}{}{}{}{}{}$. $\endgroup$ – copper.hat Feb 11 '13 at 18:43
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    $\begingroup$ Not quite done. But, if $f(x_1)=x_1$, then one may apply the MVT on the intervals $[a,x_1]$ and $[x_1,b]$ to obtain the desired result. $\endgroup$ – David Mitra Feb 11 '13 at 18:50
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    $\begingroup$ If $f(x) \neq x$ for all $x \in [a, b]$, then there must be two fixed points $x_1 \neq x_2$ (perhaps just $a$ and $b$) such that no $x$ satisfying $x_1 \leq x \leq x_2$ is a fixed point for $f$. Proceed with $x_1$ and $x_2$ in MVT. $\endgroup$ – Shaun Ault Feb 11 '13 at 18:53

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