The sum of the maximum and minimum values of | vedclass

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(C) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2-x+1$, which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.For $x$ to be real, the discriminant

Vedclass · Accepted Answer

$\frac{10}{3}$

Vedclass · Answer

(C) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2-x+1$, which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.For $x$ to be real, the discriminant $D \ge 0$.$D = (y+1)^2 - 4(y-1)^2 \ge 0$.$(y+1-2(y-1))(y+1+2(y-1)) \ge 0$.$(-y+3)(3y-1) \ge 0$.$(y-3)(3y-1) \le 0$.Thus, $\frac{1}{3} \le y \le 3$.The maximum value is $3$ and the minimum value is $\frac{1}{3}$.The sum of the maximum and minimum values is $3 + \frac{1}{3} = \frac{10}{3}$.

The sum of the maximum and minimum values of the function $f(x)=\frac{x^2-x+1}{x^2+x+1}$ is

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