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If $f$ is continuous real function such that $f(11) = 10$ and $\forall x, f(x)f(f(x)) = 1$, then $f(9) =$ ?

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$f(11)=10$, so $f(11)f(10)=1$ and hence $f(10)=1/10$. But $f(x)$ is continuous, so $f(x_0)=9$ for some $10<x_0<11$. Hence $f(x_0)f(9)=1$, so $f(9)=1/9$.

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  • $\begingroup$ Clear and concise. +1 $\endgroup$
    – Mark Viola
    Apr 11 '16 at 19:44

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