Question

Find the range of the function
$y=\frac{x^2-x+1}{x^2+x+1}$

Answer 1

$y=\frac{x^2-x+1}{x^2+x+1}$
$\therefore y(x^2+x+1)=x^2-x+1\therefore x^2-x+1=x^{2}y+xy+y\therefore x^2-x+1-x^{2}y-xy-y=0\therefore x^2(1-y)+x(-y-1)+(1-y)=0$
Discriminant $=(-y-1{)}^2-4(1-y\left)\right(1-y)$
$=(y+1{)}^2-4(1-y{)}^2$
$=y^2+2y+1-4(1-2y+y^2)$
$=y^2+2y+1-4+8y-4y^2$
Discriminant $=-3y^2+10y+1$
Range is set for discriminant $\ge$ 0
$\therefore -3y^2+10y+1\ge 0$
$\therefore \frac{1}{3}\le y\le 3$
$\therefore Range\to [\frac{1}{3},3]$.