04 380 4(0 2.5 x=B1x0)+f= 0+3 3 =[25,3,31|x 4.924 3 8 4(2.5(2.5 4 x2)=Bx+f=-10 3+3 3 0 2 4 =[2875236361]y|x(2)-x(=2.1320                − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 x B x f = J + (1) (0)           + 3 3 2.5            0 0 0 T = [2.5, 3, 3] 4.924 (1) (0) x − x =                 − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 x B x f = J + (2) (1)           + 3 3 2.5            3 3 2.5 T = [2.875, 2.3636,1] 2.1320 (2) (1) x − x =