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Let $F (x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $F (x) + F ((x − 1)/x) = 1 + x$. Find $F (x)$.

This functional equation looks like I could do an inverse substitution. Meaning, let $x = \dfrac{x-1}{x}$ then we have $F \left(\dfrac{x-1}{x} \right)+F\left( \dfrac{1}{x-1}\right) = \dfrac{2x-1}{x}$. Thus, $1+x-F(x) = \dfrac{2x-1}{x}-F\left( \dfrac{1}{x-1}\right)$. I am not sure how to proc

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By solving the system of questions:

$$ \begin{align} F(x) + F\left(\frac{x-1}{x}\right) &= 1+x \\ F\left(\frac{x-1}{x}\right) + F\left(\frac{1}{1-x}\right) &= \frac{2x-1}{x} \\ F\left(\frac{1}{1-x}\right) + F(x) &= \frac{2-x}{1-x} \end{align} $$

letting $A=F(x)$, $B=F\left(\frac{x-1}{x}\right)$, $C=F\left(\frac{1}{1-x}\right)$, we find via linear algebra we have

$$ A=F(x)=\frac{x^{3} - x^{2} - 1}{2 \, {\left(x^{2} - x\right)}} $$

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  • $\begingroup$ What do you get for $F(x)$? $\endgroup$
    – Puzzled417
    Feb 5, 2016 at 21:23

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