Question #1cf68

1 Answer
Oct 31, 2016

Answer:

#x in [-1, 1]#

Explanation:

The absolute value function #|x|# is defined by

#|x| = {(x if x>=0), (-x if x<0):}#

As such, we will consider three cases:

Case 1: #x < -1#

#=> x+1 < 0 and x-1 < 0#

#=>|x+1| = -(x+1) and |x-1| = -(x-1)#

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

#=> -x-1-x+1 <= 2#

#=> -2x <= 2#

#=> 2x => 2#

#=> x>= 1#

As this contradicts the premise that #x < -1#, there are no solutions on #(-oo, -1)#

Case 2: #-1 <= x <= 1#

#=> x+1 >= 0 and x-1<= 0#

#=> |x+1| = x+1 and |x-1| = -(x-1)#

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

#=> x+1-x+1 <=2#

#=> 2 <= 2#

As this is true in all cases, every value in #[-1, 1]# is a solution.

Case 3: #x > 1#

#=> x+1 > 0 and x-1 > 0#

#=> |x+1| = x+1 and |x-1| = x-1#

#=> x+1+x-1 <= 2#

#=> 2x <= 2#

#=> x <= 1#

This contradicts the premise that #x>1#, meaning there are no solutions on #(1, oo)#.


Taken together, we have the solution set as #[-1, 1]#. The inequality will hold for any #x# in that interval, and will not for any #x# outside of it.