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Solve 0 lt | x-3| le 5...

Solve `0 lt | x-3| le 5`

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`0 lt | x-3| le 5`
`rArr -5 le x -3 lt 0 or 0 lt x -3 le 5`
`rArr -2 le x lt 3 or 3 lt x le 8`
`rArr x in [ -2 ,3 ) cup ( 3,8]`
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f(x) = (2x^(2) + 3)/(5), for oo lt x le 1 = 6 -5x, for 1 lt x lt 3 = x-3, for 3 le x lt 8, then