Find the value of `x^2` for the given values of `xdot` `x<2` (ii) `x >-1` (iii) `xgeq2` (iv) `x<-1`

Explanation

(i) When `x lt 3`, we have `x in (-oo,0)cup[0,3)`
for `x in [0,3),x^(2) in [0,9)`
for `x in (- oo,0),x^(2) in (0,oo)`
`implies "for " x lt 3, x^(2) in [0,9) cup(0,oo)`
`implies x in [0,oo)`
(ii) When `x gt -1`, we have `x in (-1,0)cup[0,oo)`
for `x in [-1,0),x^(2) in [0,1)`
for `x in [-0,oo),x^(2) in [0,oo)`
`implies "for " x gt -1, x^(2) in (0,1) cup(0,oo)`
`implies x in [0,oo)`
(iii) Here `x in [2,oo)`
`implies x^(2) in {4, oo)`
Here `x in (-oo, -1)`
`implies x^(2) in (1, oo)`