If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are? If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are?
I thought that $(x+1)/(x+2)=-x$.But I'm getting only two values by solving this.How do I get the other two values?
And if i put $(x+1)/(x+2)=x$ then the answer is not matching.I dont know why.

Don't forget the possibility $(x+1)/(x+2)=+x$.
Setting $\frac{x+1}{x+2}=-x$ leads to $x+1=-x^2-2x$, i.e., $x^2+3x+1=0$, $x=\frac{-3\pm\sqrt{5}}2$. Setting $\frac{x+1}{x+2}=+x$ leads to $x+1=x^2+2x$, i.e., $x^2+x-1=0$, $x=\frac{-1\pm\sqrt{5}}2$.
All four values are well within $(-5,5)$.
