If x is real,then the value of frac{x^2-3x+4} | vedclass

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

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$[\frac{1}{7}, 7]$

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(D) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.$y(x^2+3x+4) = x^2-3x+4$$x^2(y-1) + x(3y+3) + (4y-4) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = (3y+3)^2 - 4(y-1)(4y-4) \geq 0$.$9(y+1)^2 - 16(y-1)^2 \geq 0$.$9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$.$9y^2 + 18y + 9 - 16y^2 + 32y - 16 \geq 0$.$-7y^2 + 50y - 7 \geq 0$.$7y^2 - 50y + 7 \leq 0$.$(7y-1)(y-7) \leq 0$.Thus,$\frac{1}{7} \leq y \leq 7$.

If $x$ is real,then the value of $\frac{x^2-3x+4}{x^2+3x+4}$ lies in the interval

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