How do you solve and write the following in interval notation: #absx> abs(x-1)#?

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Answer 1 · 2017-10-28T22:50:19

See below.

#absx-epsilon^2=abs(x-1)# with #epsilon ne 0# or

#absx-abs(x-1)=epsilon^2# or

#(absx-abs(x-1))^2=epsilon^4# or

#x^2+(x-1)^2-2absx abs(x-1) = epsilon^4# or

#4x^2 (x-1)^2 = (x^2+(x-1)^2-epsilon^4)^2#

#4x^2 (x-1)^2 - (x^2+(x-1)^2-epsilon^4)^2=0# or

#(1-epsilon^4)(2x-1 - epsilon^2) (2x-1 + epsilon^2)=0# or

#{(2x-1 - epsilon^2=0),(2x-1 + epsilon^2=0):}# or

#{(2x-1 gt 0),(2x-1 lt 0):}#

so finally the inequality #absx gt abs(x-1)# has not real solution and consequently the represented set is void.