Explanation:
−1,
2
−1
]∪[0,
2
1
]∪{1}
For f to be defined 2{x}
2
−3{x}+1≥0
({x}−1)(2{x}−1)≥0
⇒{x}∈(−∞,
2
1
)∪[1,∞)
But we know, {x}∈[0,1)
Thus {x}∈[0,
2
1
)→(1)
Also given that x∈[−1,1]
Hence required domain is [−1,−
2
1
]∪[0,
2
1
],∪{1}
Note {−x}=1−{x}