Question

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

Answer 1

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`