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!