Y= (667.38./x)*(1-e (-x/68.1)/10 ) - 40 Akar Iterasi XL XU f(XL) f(XU) XR f(XR) 1 10.0000 16.0000 11.3691 -2.2688 13.0000 3.7270 2 13.0000 16.0000 3.7270.

Presentasi berjudul: "Y= (667.38./x)*(1-e (-x/68.1)/10 ) - 40 Akar Iterasi XL XU f(XL) f(XU) XR f(XR) 1 10.0000 16.0000 11.3691 -2.2688 13.0000 3.7270 2 13.0000 16.0000 3.7270."— Transcript presentasi:

1 y= (667.38./x)*(1-e (-x/68.1)/10 ) - 40 Akar Iterasi XL XU f(XL) f(XU) XR f(XR) 1 10.0000 16.0000 11.3691 -2.2688 13.0000 3.7270 2 13.0000 16.0000 3.7270 -2.2688 14.5000 0.5523 3 14.5000 16.0000 0.5523 -2.2688 15.2500 -0.8993 4 14.5000 15.2500 0.5523 -0.8993 14.8750 -0.1841 5 14.5000 14.8750 0.5523 -0.1841 14.6875 0.1814 6 14.6875 14.8750 0.1814 -0.1841 14.7812 -0.0020 7 14.6875 14.7812 0.1814 -0.0020 14.7344 0.0895 Iterasi XL XU f(XL) f(XU) XR f(XR) 1 10.0000 16.0000 11.3691 -2.2688 15.0019 -0.4284 2 10.0000 15.0019 11.3691 -0.4284 14.8202 -0.0779 3 10.0000 14.8202 11.3691 -0.0779 14.7874 -0.0141 4 10.0000 14.7874 11.3691 -0.0141 14.7815 -0.0025 5 10.0000 14.7815 11.3691 -0.0025 14.7804 -0.0005 Metode Bagi Dua Metode Posisi Palsu

2 Iterasi Xi 0 0 1 0.5000 2 0.5663 3 0.5671 4 0.5671 5 0.5671 Metode Newton Raphson Iterasi Xi 0 0 0.5000 1 0.5596 2 0.5671 3 0.5671 Metode Secant Akar y= e -x - x

3 Iterasi Xi 0 0 1 -0.5000 2 -0.8772 3 -1.0917 4 -0.7387 5 -1.0463 6 -0.8974 7 -1.0889 Metode Newton Raphson Akar y= x 10 - 1 Iterasi XL XU f(XL) f(XU) XR f(XR) 1 0 1.3000 -1.0000 12.7858 0.6500 -0.9865 2 0.6500 1.3000 -0.9865 12.7858 0.9750 -0.2237 3 0.9750 1.3000 -0.2237 12.7858 1.1375 2.6267 4 0.9750 1.1375 -0.2237 2.6267 1.0563 0.7285 Metode Bagi Dua Iterasi XL XU f(XL) f(XU) XR f(XR) 1 0 1.3000 -1.0000 12.7858 0.0943 -1.0000 2 0.0943 1.3000 -1.0000 12.7858 0.1818 -1.0000 3 0.1818 1.3000 -1.0000 12.7858 0.2629 -1.0000 4 0.2629 1.3000 -1.0000 12.7858 0.3381 -1.0000 5 0.3381 1.3000 -1.0000 12.7858 0.4079 -0.9999 6 0.4079 1.3000 -0.9999 12.7858 0.4726 -0.9994 7 0.4726 1.3000 -0.9994 12.7858 0.5326 -0.9982 Metode Posisi Palsu

4 Iterasi Xi 0 10 1 7.0667 2 6.1246 3 6.0021 4 6.0000 Metode Newton Raphson Akar y= x 2 – 5x - 6 Iterasi XL XU f(XL) f(XU) XR f(XR) 1 0 10 -6 44 5 -6 2 5.0000 10.0000 -6.0000 44.0000 7.5000 12.7500 3 5.0000 7.5000 -6.0000 12.7500 6.2500 1.8125 4 5.0000 6.2500 -6.0000 1.8125 5.6250 -2.4844 5 5.6250 6.2500 -2.4844 1.8125 5.9375 -0.4336 6 5.9375 6.2500 -0.4336 1.8125 6.0938 0.6650 Metode Bagi Dua Iterasi XL XU f(XL) f(XU) XR f(XR) 1 0 10.0000 -6.0000 44.0000 1.2000 -10.5600 2 1.2000 10.0000 -10.5600 44.0000 2.9032 -12.0874 3 2.9032 10.0000 -12.0874 44.0000 4.4327 -8.5149 4 4.4327 10.0000 -8.5149 44.0000 5.3354 -4.2108 5 5.3354 10.0000 -4.2108 44.0000 5.7428 -1.7345 6 5.7428 10.0000 -1.7345 44.0000 5.9042 -0.6613 Metode Posisi Palsu Iterasi Xi 11 0 10 1 7.2500 2 6.4082 3 6.0589 4 6.0032 5 6.0000 Metode Secant

5 Metode Bisection : XL XU f (XL) f(XU) XR f(XR) 1.0000 8.0000 -5.3000 1.2800 4.5000 -3.7250 4.5000 8.0000 -3.7250 1.2800 6.2500 -1.6512 6.2500 8.0000 -1.6512 1.2800 7.1250 -0.2928 7.1250 8.0000 -0.2928 1.2800 7.5625 0.4668 7.1250 7.5625 -0.2928 0.4668 7.3438 0.0803 7.1250 7.3438 -0.2928 0.0803 7.2344 -0.1079 7.2344 7.3438 -0.1079 0.0803 7.2891 -0.0142 7.2891 7.3438 -0.0142 0.0803 7.3164 0.0329 7.2891 7.3164 -0.0142 0.0329 7.3027 0.0093 7.2891 7.3027 -0.0142 0.0093 7.2959 -0.0025 7.2959 7.3027 -0.0025 0.0093 7.2993 0.0034 Metode False position : XL XU f(XL) f(XU) XR f(XR) 1.0000 8.0000 -5.3000 1.2800 6.6383 -1.0749 6.6383 8.0000 -1.0749 1.2800 7.2598 -0.0644 7.2598 8.0000 -0.0644 1.2800 7.2953 -0.0035 7.2953 8.0000 -0.0035 1.2800 7.2972 -0.0002 7.2972 8.0000 -0.0002 1.2800 7.2973 0.0000 Contoh: Tentukan akar real positif dari persamaan y= 0.14x 2 – 0.32x – 5.12 Dengan batas-batas: X L = 1, X U = 8