How do you solve #abs((x+1)/x)>2# using a sign chart?

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Answer 1 · 2018-07-09T06:06:36

The solution is # x in (-1/3,0) uu (0, 1)#

Start by solving

#(x+1)/x=0#

#=>#, #x=-1# and #x!=0#

Let #g(x)=|(x+1)/x|#

The sign chart is as follows

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-1##color(white)(aaaaaaaa)##0##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##g(x)##color(white)(aaaa)##-(x+1)/x##color(white)(aaaa)##(x+1)/x##color(white)(aaaa)##(x+1)/x##color(white)(aaaa)#

In the interval #I_1=(-oo,-1)#

#-(x+1)/x-2>0#

#=>#, #(-x-1-2x)/x>0#

#=>#, #(-3x-1)/x>0#

#=>#, ##x>1/3

In the interval #I_2=(1,+oo)#

#(x+1)/x-2>0#

#=>#, #(x+1-2x)/x>0#

#=>#, #(1-x)/x>0#

#=>#, #x<1#

Therefore ,

The solution is # x in (-1/3,0) uu (0, 1)#

graph{|(x+1)/x|-2 [-10, 10, -5, 5]}