The range of the function f(x) = frac{x^2+x+1 | vedclass

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(A) Let $y = \frac{x^2+x+1}{x^2-x+1}$.$y(x^2-x+1) = x^2+x+1$$yx^2 - yx + y = x^2 + x + 1$$(y-1)x^2 - (y+1)x + (y-1) = 0$.For $x$ to be real,the discri

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

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(A) Let $y = \frac{x^2+x+1}{x^2-x+1}$.$y(x^2-x+1) = x^2+x+1$$yx^2 - yx + y = x^2 + x + 1$$(y-1)x^2 - (y+1)x + (y-1) = 0$.For $x$ to be real,the discriminant $D \geq 0$.$D = (-(y+1))^2 - 4(y-1)(y-1) \geq 0$$(y+1)^2 - 4(y-1)^2 \geq 0$$(y+1 - 2(y-1))(y+1 + 2(y-1)) \geq 0$$(y+1 - 2y + 2)(y+1 + 2y - 2) \geq 0$$(3-y)(3y-1) \geq 0$$(y-3)(3y-1) \leq 0$.This inequality holds for $y \in \left[\frac{1}{3}, 3\right]$.

The range of the function $f(x) = \frac{x^2+x+1}{x^2-x+1}$ is

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