The minimum value of f(x) = frac{x^2-2x+3}{x^ | vedclass

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(B) Let $f(x) = y = \frac{x^2-2x+3}{x^2-4x+7}$.$y(x^2-4x+7) = x^2-2x+3$$(y-1)x^2 + (2-4y)x + (7y-3) = 0$.Since $x$ is real,the discriminant $D \geq 0$

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$\frac{3-\sqrt{3}}{3}$

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(B) Let $f(x) = y = \frac{x^2-2x+3}{x^2-4x+7}$.$y(x^2-4x+7) = x^2-2x+3$$(y-1)x^2 + (2-4y)x + (7y-3) = 0$.Since $x$ is real,the discriminant $D \geq 0$:$(2-4y)^2 - 4(y-1)(7y-3) \geq 0$$4(1-2y)^2 - 4(7y^2 - 3y - 7y + 3) \geq 0$$(1 - 4y + 4y^2) - (7y^2 - 10y + 3) \geq 0$$-3y^2 + 6y - 2 \geq 0$$3y^2 - 6y + 2 \leq 0$.Solving $3y^2 - 6y + 2 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:$y = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$.Thus,$1 - \frac{\sqrt{3}}{3} \leq y \leq 1 + \frac{\sqrt{3}}{3}$.The minimum value is $1 - \frac{\sqrt{3}}{3} = \frac{3-\sqrt{3}}{3}$.

The minimum value of $f(x) = \frac{x^2-2x+3}{x^2-4x+7}$ is

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