Question

If $x$ be real then, find range of values of $\frac{x^2-2x+1}{x+1}$.

Answer 1

$y=\frac{x^2-2x+1}{x+1}$

$\Rightarrow xy+y=x^2-2x+1$

$\Rightarrow x^2-2x-xy+1-y=0$

$\Rightarrow x^2-(2+y)x+1-y=0$

As $x\in R$

$\Delta \ge 0$

$\Rightarrow (2+y{)}^2-4(1-y)\ge 0$

$\Rightarrow 4+y^2+4y-4+4y\ge 0$

$\Rightarrow y^2+8y\ge 0$

$\Rightarrow y(y+8)\ge 0$

$\therefore y\in (-\infty ,-8]\cup [0,\infty )$