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Let, $f:\rm I\!R \rightarrow \rm I\!R$ be some function that is equal to a linear combination of itself and its inverse. Is is possible write an explicit formula for $f(x)$? $$ f(x) = af(x) + bf^{-1}(x) \\ $$ $$ a \ne 0,1 \\ b \ne 0,1 $$ Solve for $f(x)$.

Thank you for all your responses. Based on your comments, I am adding some restrictions on a and b

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    $\begingroup$ Letting $x=f(y)$, your condition implies $f(f(y)) = by/(1-a)$. Certainly there are many wild solutions to such a functional equation. This seems hopeless in the level of generality you're asking. $\endgroup$
    – Joshua P. Swanson
    Commented Sep 11, 2019 at 23:49
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    $\begingroup$ When $b=0$ and $a=1$ this equation holds for any bijective function $f$. $\endgroup$
    – Kavi Rama Murthy
    Commented Sep 11, 2019 at 23:55
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    $\begingroup$ It seems like it can always be rewritten as $f(x) = c f^{-1}(x)$, where $c=\frac{b}{1-a}$, so we’re asking what functions are nonzero multiples of their inverses. One simple function would be $f(x)=\sqrt{c}x$, $f^{-1}(x)=\frac{1}{\sqrt{c}}x$. $\endgroup$
    – Joe
    Commented Sep 12, 2019 at 0:31

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