If x is real,then the range of frac{x^2+2x+1} | vedclass

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(A) Let $y = \frac{x^2+2x+1}{x^2+2x+7}$.$y(x^2+2x+7) = x^2+2x+1$$yx^2 + 2yx + 7y = x^2 + 2x + 1$$(y-1)x^2 + 2(y-1)x + (7y-1) = 0$.If $y=1$,then $0x^2

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$[0,1)$

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(A) Let $y = \frac{x^2+2x+1}{x^2+2x+7}$.$y(x^2+2x+7) = x^2+2x+1$$yx^2 + 2yx + 7y = x^2 + 2x + 1$$(y-1)x^2 + 2(y-1)x + (7y-1) = 0$.If $y=1$,then $0x^2 + 0x + 6 = 0$,which is impossible. Thus,$y 
eq 1$.For $x$ to be real,the discriminant $D \geq 0$:$D = [2(y-1)]^2 - 4(y-1)(7y-1) \geq 0$$4(y-1)^2 - 4(y-1)(7y-1) \geq 0$$(y-1)[(y-1) - (7y-1)] \geq 0$$(y-1)(-6y) \geq 0$$6y(y-1) \leq 0$This implies $y \in [0, 1]$.Since $y 
eq 1$,the range is $[0, 1)$.

If $x$ is real,then the range of $\frac{x^2+2x+1}{x^2+2x+7}$ is

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