The range of X in the following expression is . #abs(abs(x+1)+1)>=1# ?

1 Answer
Sep 27, 2017

All #x# or

#{x inRR}#

Explanation:

We don't need to try and remove absolute bars to solve this problem.

Notice in #||x+1|+1|>=1# that the value of #|x+1|>=0# for any real #x# since the absolute value is always positive.

So even at the minimum value of #0#

#||0| +1|>= 1#