# $f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$

$f(x)=?$

If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?

Hint:

$$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}=\frac{1}{\frac{x^2-x+1}{x}}=\frac{1}{\frac{x^2+x+1-2x}{x}}=\frac{1}{\frac{x^2+x+1}{x}-2}$$

Note, that we have to assume $x\neq 0$ in this process. You will have to check if that is problematic.

HInt: you can take $$x+\frac 1x =u$$ and turn all the expression into $u$ ,or do like below $$\quad{f(\frac{x}{x^2+x+1})=\frac{x}{x^2-x+1}\\ \frac{x}{x^2+x+1}=u\\\frac{x^2+x+1}{x}=\frac 1u\\x+\frac{1}x +1=\frac 1u \\\to x+\frac{1}x =\frac 1u -1\\ \frac{x}{x^2-x+1}=\dfrac{1}{\dfrac{x^2-x+1}{x}}=\\\dfrac{1}{x+\dfrac{1}{x}-1}=\dfrac{1}{(\frac 1u -1)-1}}$$can you go on ?

• Is it mean $$f(u)=\frac{u}{1-2u}$$ ? – george.h Sep 29 '17 at 17:41
• @george.h :yes . this is final answer – Khosrotash Sep 29 '17 at 17:43

Let $y = \frac x{x^2+x+1}$ and $z = \frac x{x^2-x+1}$. Then we have

\begin{align} y(x^2+x+1) = x &\implies y(x^2-x+1)=x -2xy\\ &\implies y=z(1-2y)\\ &\implies z=\frac y{1-2y}\\ \end{align}

which gives us $f(y) = \frac{y}{1-2y}$.

• @george.h, note that there is no issue with $x = 0$ anywhere along the way. – Ennar Sep 29 '17 at 18:04

$$f\left( \dfrac{x}{x^{2}+x+1}\right)=\dfrac{x}{x^{2}-x+1}\rightarrow f\left( \dfrac{1}{x+\frac{1}{x}+1}\right)=\dfrac{1}{x+\frac{1}{x}-1}\rightarrow$$ $$f\left( \dfrac{1}{z+1}\right)=\dfrac{1}{z-1}$$ $$\dfrac{1}{z+1}=t\rightarrow \dfrac{1}{t}=z+1\rightarrow z=\dfrac{1}{t}-1\rightarrow z=\dfrac{1-t}{t}\rightarrow$$ $$f(t)=\dfrac{1}{\frac{1-t}{t}-1}=\dfrac{1}{\frac{1-2t}{t}}=\dfrac{t}{1-2t}\rightarrow f(x)=\dfrac{x}{1-2x}$$ Testing: $$f\left( \frac{x}{x^{2}+x+1}\right)=\dfrac{\frac{x}{x^{2}+x+1}}{1-2\left( \frac{x}{x^{2}+x+1}\right)}=\dfrac{\frac{x}{x^{2}+x+1}}{\frac{x^{2}+x+1-2x}{x^{2}+x+1}}=\dfrac{x}{x^{2}-x+1}$$