Question

Find the range of rational expression $\frac{x^2-x+1}{x^2+x+1}$ if $x$ is real.

Answer 1

$y=\frac{x^2-x+1}{x^2+x+1}\Rightarrow (y-1)x^2+(y+1)x+y-1=0$ (i)
case I: If $y\ne 1,$ then equation (i) is quadratic in $x$
and $\because$ $x$ is real.
$\therefore$ $D\ge 0$
$\Rightarrow$ ${(y+1)}^2-4{(y-1)}^2\ge 0$
$\Rightarrow$ $(y-3)(3y-1)\le 0$
$\therefore$ $yϵ[\frac{1}{3},3]-\left\{1\right\}$
case II: if $y=1,$ then the equation become
$2x=0$ $\Rightarrow$ $x=0$ which is possible as $x$ is real.
$\therefore$ Range $[\frac{1}{3},3]$