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Question #4435e

1 Answer

Feb 2, 2017

$x^{2} - 2 x + 2$ $x^2-2x+2$

$= x^{2} - 2 \cdot x \cdot 1 + 1^{2} + 1$ $=x^2-2*x*1+1^2+1$

$= {(x - 1)}^{2} + 1$ $=(x-1)^2+1$

Again $0 < x < 1$ $0 < x < 1$

$⇒x→+ fraction$ $=>x->"+ fraction"$

Let $x = f > 0$ $x=f>0$

So the given expression

$= {(x - 1)}^{2} + 1 = f^{2} + 1 > 1 \neq 0$ $=(x-1)^2+1=f^2+1>1!=0$

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