case 1:
for
2x-1 >=0
=> x >=1/2
f(2x-1)< f(1/3)
=> 2x-1 < 1/3
x < 2/3
solution for case 1
x >=1/2 and x<2/3
ie 1/2 <= x < 2/3
case 2: x < 1/2
f(2x-1) = f(-(1-2x))) = f(1-2x) < f(1/3)
=> 1-2x < 1/3
=> x > 1/3
solution for case 2
x<1/2 and x> 1/3
ie 1/3< x < 1/2
solution for f（2x-1)
1/2 <= x < 2/3 or 1/3< x < 1/2
ie 1/3< x < 2/3
2x-1≥0
x≥1/2
2x-1<1/3
x<2/3
1/2≤x<2/3
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