source: #youtube https://www.youtube.com/watch?v=XwsNp3WS6mI tags: #iterate course: [[Algebra 2]] status: #solved ### Problem Given $f(x)+f\left(\frac{x-1}{x}\right)=1+x$ For all $x \neq 0,1$ Find $f(x)$ #### Solution Let $g(x)=\frac{x-1}{x}$ for all $x \neq 0,1$ then $\begin{aligned} (g\circ g)(x)&=&\frac{\left(\frac{x-1}{x}\right)-1}{\left(\frac{x-1}{x}\right)}=\frac{x-1-x}{x-1}&=&\frac{1}{1-x}\\ (g\circ g\circ g)(x)&=&\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}=\frac{1-(1-x)}{1}&=&x\\ \end{aligned}$ So we use adjacent pairs of these as inputs into the original equation $\begin{aligned} {\textcolor{lightgreen}{(1)}}\quad&f(x)&+&f\left(\frac{x-1}{x}\right)&=&1+x\\ {\textcolor{pink}{(2)}}\quad&f\left(\frac{x-1}{x}\right)&+&f\left(\frac{1}{1-x}\right)&=&1+\frac{x-1}{x}&=&\frac{2x-1}{x}\\ {\textcolor{lightblue}{(3)}}\quad&f\left(\frac{1}{1-x}\right)&+&f\left(x\right)&=&1+\frac{1}{1-x}&=&\frac{2-x}{1-x}\\ \end{aligned}$ Combining ${\textcolor{lightgreen}{(1)}}-{\textcolor{pink}{(2)}}+{\textcolor{lightblue}{(3)}}$ to get cancellation $\begin{aligned} &\textcolor{lightgreen}{\left(f(x)+\cancel{f\left(\frac{x-1}{x}\right) }\right)}&&\textcolor{lightgreen}{1+x}\\ -&\textcolor{pink}{\left(\cancel{f\left(\frac{x-1}{x}\right)}+\cancel{f\left(\frac{1}{1-x}\right)}\right)}&&\textcolor{pink}{-\frac{2x-1}{x}}\\ +&\textcolor{lightblue}{\left(\cancel{f\left(\frac{1}{1-x}\right)}+f\left(x\right)\right)}&=&\textcolor{lightblue}{\frac{2-x}{1-x}}\\ \\ &2\cdot f(x)&=& 1+x-\frac{2x-1}{x}+\frac{2-x}{1-x}\\ & f(x)&=&\frac{(1+x)(x)(1-x)-(2x-1)(1-x)+(2-x)(x)}{2x(1-x)} \end{aligned}$ Now to simplify $\begin{aligned} f(x)&=\frac{(1+x)(x)(1-x)-(2x-1)(1-x)+(2-x)(x)}{2x(1-x)}\\ &=\frac{-x^3+x+2x^2-3x+1+2x-x^2}{2x(1-x)}\\ &=\frac{-x^3+x^2+1}{2x(1-x)}\\ &=\boxed{\frac{x^3-x^2-1}{2x(x-1)}} \end{aligned}$