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