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1 solution
y = ⎩ ⎪ ⎨ ⎪ ⎧ 4 0 3 1 − 2 x 1 2 x − 4 0 3 1 if x ≤ 2 0 1 5 if 2 0 1 5 ≤ x ≤ 2 0 1 6 if x ≥ 2 0 1 6 When we split cases, we can determine the minimum value to be 1 over all cases, hence m i n ( y ) = 1