For real values of x,the range of frac{x^2+2x | vedclass

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

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$(-\infty, 0] \cup (1, \infty)$

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(A) Let $y = \frac{x^2+2x+1}{x^2+2x-1}$.$y(x^2+2x-1) = x^2+2x+1$$yx^2 + 2xy - y = x^2 + 2x + 1$$(y-1)x^2 + 2(y-1)x - (y+1) = 0$.For $x$ to be real,the discriminant $D \geq 0$ (if $y 
eq 1$):$D = [2(y-1)]^2 - 4(y-1)(-(y+1)) \geq 0$$4(y-1)^2 + 4(y-1)(y+1) \geq 0$$4(y-1)[(y-1) + (y+1)] \geq 0$$4(y-1)(2y) \geq 0$$8y(y-1) \geq 0$.This inequality holds for $y \in (-\infty, 0] \cup [1, \infty)$.However,if $y=1$,the equation becomes $0 = -2$,which is impossible,so $y 
eq 1$.Thus,the range is $y \in (-\infty, 0] \cup (1, \infty)$.

For real values of $x$,the range of $\frac{x^2+2x+1}{x^2+2x-1}$ is

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