For x in R, the least value of frac{x^2-6x+5} | vedclass

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(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in R$. Set $y = \frac{x^2-6x+5}{x^2+2x+1}$. Then $y(x^2+2x+1) = x^2-6x+5$, which simplifies to $(y-

Vedclass · Accepted Answer

$-\frac{1}{3}$

Vedclass · Answer

(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in R$. Set $y = \frac{x^2-6x+5}{x^2+2x+1}$. Then $y(x^2+2x+1) = x^2-6x+5$, which simplifies to $(y-1)x^2 + (2y+6)x + (y-5) = 0$. Since $x$ is a real number, the discriminant $D$ must be greater than or equal to $0$. $D = (2y+6)^2 - 4(y-1)(y-5) \geq 0$. $4(y+3)^2 - 4(y^2-6y+5) \geq 0$. $(y^2+6y+9) - (y^2-6y+5) \geq 0$. $12y + 4 \geq 0$. $12y \geq -4$. $y \geq -\frac{1}{3}$. Thus, the least value of the expression is $-\frac{1}{3}$.

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

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