Question

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

Answer 1

$\frac{x^2+x+1}{x^2-x+1}=\frac{x^2-x+1+2x}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1}$

Now, $-2\le x+\frac{1}{x}\le 2$

The minimum value of given expression occurs at $x=-1$ and maximum value occurs at $x=1$

Therefore the range is $[\frac{1}{3},3]$