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If $f:R\rightarrow R$ is a continuous function with the properties: $f(2017)=2016$ and $f(x)\cdot f(f(x))=1$ find the value of $f(2015)$

I replaced $x$ with $2017$ and got $f(2017)\cdot f(f(2017))=1$

$2016\cdot f(2016)=1$ which means $f(2016)=\frac1{2016}$, but I don't know how to get to $f(2015)$

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You have that for every $y$ in the image of $f$ $f(y)=\frac 1y$. Furthermore 2016 and $\frac 1{2016}$ stay in the image, so since $f$ is continous also 2015 stays in the image. It follows that $f(2015)=\frac 1{2015}$

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  • $\begingroup$ If $\displaystyle f(y)=\frac1y$, how can we have $f(2017)=2016\not=\frac1{2017}$? $\endgroup$
    – Dr. Mathva
    Feb 11 '20 at 10:56
  • $\begingroup$ @Dr.Mathva it means that 2017 doesn’t stay in the range of $f$! $\endgroup$ Feb 11 '20 at 10:59
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$f(t)=\frac 1 t$ for any $t$ in the range of $f$. Since $\frac 1 {2016}$ and $2016$ are both in the range and $f$ is continuous it any number between these two is in the range. So $f(2015)=\frac 1 {2015}$.

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