Solve the inequality 1/x<or=|x-2| for real numbers?

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Answer 1 · 2018-05-22T18:23:11

$S:x in ]-oo;0[uu[1+sqrt2;+oo[$

$1/x<=|x-2|$

$D_f:x in RR^"*"$

for $x<0$ :

$1/x<=-(x-2)$

$1> -x²-2x$

$x²+2x+1>0$

$(x+1)²>0$

$x in RR^"*"$

But here we have the condition that $x<0$ , so:

$S_1:x in RR_"-"^"*"$

Now, if $x>0$ :

$1/x<=x-2$

$1<=x²-2x$

$x²-2x-1>=0$
$Δ=8$

$x_1=(2+sqrt8)/2=1+sqrt2$

$cancel(x_2=1-sqrt2)$ ( $<0$ )

So $S_2:x in [1+sqrt2;+oo[$

Finally $S=S_1uuS_2$

$S:x in ]-oo;0[uu[1+sqrt2;+oo[$

\0/ here's our answer!