For x in mathbb{R},the least value of frac{x^ | vedclass

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(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in \mathbb{R}$.Let $y = \frac{x^2-6x+5}{x^2+2x+1}$.$y(x^2+2x+1) = x^2-6x+5$.$yx^2 + 2yx + y = x^2 -

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$-\frac{1}{3}$

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(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in \mathbb{R}$.Let $y = \frac{x^2-6x+5}{x^2+2x+1}$.$y(x^2+2x+1) = x^2-6x+5$.$yx^2 + 2yx + y = x^2 - 6x + 5$.$(y-1)x^2 + (2y+6)x + (y-5) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = (2y+6)^2 - 4(y-1)(y-5) \geq 0$.$4(y+3)^2 - 4(y^2 - 6y + 5) \geq 0$.$(y^2 + 6y + 9) - (y^2 - 6y + 5) \geq 0$.$12y + 4 \geq 0$.$12y \geq -4$.$y \geq -\frac{4}{12} = -\frac{1}{3}$.Thus,the least value is $-\frac{1}{3}$.

For $x \in \mathbb{R}$,the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is

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