Question

Find the range of $f\left(x\right)=\frac{x^2-x+1}{x^2+x+1}$.

• A. $[\frac{1}{3},3]$
• B. $[\frac{-1}{3},3]$
• C. $[\frac{2}{3},3]$
• D. $[\frac{-2}{3},3]$

Answer 1

Answer: (A)

$[\frac{1}{3},3]$

Let $y=\frac{x^2-x+1}{x^2+x+1}$
$\Rightarrow (1-y)x^2-(1+y)x+1-y=0$
Now x is real, then $D\ge 0$
$\Rightarrow (1+y{)}^2-4(1-y{)}^2\ge 0$
$\Rightarrow (1+y-2+2y)(1+y+2-2y)\ge 0$
$\Rightarrow (2y-1)(3-y)\ge 0$
$\Rightarrow 3(y-\frac{1}{3})(y-3)\le 0$
$\Rightarrow \frac{1}{3}\le y\le 3\Rightarrow$ The range is $[\frac{1}{3},3]$.