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