(5). (i) When using the Newton method to find solutions accurate to within 10-6 to the equation r4-2x2 + 1 = (x2-1)2-0 with initial guess po very close to the roots p 1 or p1, the Newton method fails to converge in 4 to 5 iterations. Why? (ii) Applying the modified Newton's method (the code "alg023ModifiedNewton.m") to solve the given equation with po 0.8 and 0.9, are the numbers of iterations reasonable (i.e., it only takes 4 to 5 iterations)? (3). Use Newton's method to find solutions accurate to within 10-6 for- 1 0 on [1,2] by code with Po = 0, 1, 2, and 4. (Use code). Which initial value is the best in the sense that the least number of iterations is needed? Why does the Newton's method perform poorly for Po = 0? I got the code done, at P0-1 (5 iterations-least # of iterations) This is the code at Po 0 with 21 iterations Why does the Newton method perform poorly on this one? Newtons Method F (P) 11.000000000000e+00 -1.000000000000e+00 2-5.000000000000e-01 -6.250000000000e-01 3 -3.000000000000e+00 -2.500000000000e+01 4 2.038461538462e+00-7.432009558489e+00 5 1.390282147217e+002.296972594679e+00 6 -9.116118977179e-01 -8.459706370957e-01 7-3.450284967482e-01-6.960453045687e-01 8 -1.427750704027e+00 2.482679233125e+00 9-9.424179125095e-01 -8.945919906750e-01 10 4.049493571994e-01 -6. 614558508605e-01 11 1.706904645183e+004.266202095763e+00 12 -1.155756361075e+00 1.388071510371e+00 13 6.941918133295e-01 -6.403407998936e-01 14 7.424942987207e-01 1.333158837476e+00 15 2.781295940677e+00 1.773371671200e+01 16 1.982725247044e+00 4.811763064936e+00 17 1.536927379758e+00 1.093519130134e+00 18 1.357262483188e+00 1.430341385223e-01 19 1.325663094429e+00 4.034214117256e-03 20 1.324718788615e+00 3.545492862633e-06 21 1.324717957245e+00 2.747135852132e-12 fx