How do you solve #abs(x+1)+ abs(x-1)<=2#?

1 Answer
George C.
May 30, 2015

With moduli it is useful to split into cases at the point that the sign of the enclosed expression changes. #(x+1)# changes sign at #x=-1# and #(x-1)# changes sign at #x=1#....

Case (a) : #x < -1#

#x + 1 < 0# so #abs(x+1) = -(x+1)#
#x - 1 < 0# so #abs(x-1) = -(x-1)#

#2 >= abs(x+1)+abs(x-1) = -(x+1) + -(x-1)#
#=-x-1-x+1#
#= -2x#

Dividing both ends by #-2# and reversing the inequality we get:
#x >= -1#

Notice that we have to reverse the inequality, because we are dividing by a negative number.

So in Case (a) we have #x < -1# and #x >= -1#
These conditions cannot be satisfied at the same time, so Case (a) yields no solutions.

Case (b) : #-1 <= x <= 1#

#x + 1 >= 0# so #abs(x+1) = (x+1)#
#x - 1 <= 0# so #abs(x-1) = -(x-1)#

#abs(x+1)+abs(x-1) = (x+1) + -(x-1) = 1+1 = 2 <= 2#

So the target inequality is satisfied for all #x# in Case (b).
That is #-1 <= x <= 1#.

Case (c) : #x > 1#

#x + 1 > 0# so #abs(x+1) = (x+1)#
#x - 1 > 0# so #abs(x-1) = (x-1)#

#abs(x+1)+abs(x-1) = (x+1)+(x-1) = 2x > 2#
So in Case (c) the inequality is never satisfied.

So the solution is #-1 <= x <= 1#

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