8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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Soal No. 1 (Metode Bisection)

Iterasi x1 x2 x3 f(x1) f(x2) f(x3)

1 0   1 0.5 0.11691 -0.16481 0.03358

2 0.5   1 0.75 0.03358 -0.16481 -0.045598

3 0.5   0.75 0.625 0.03358 -0.045598 -0.0018584 0.5   0.625 0.5625 0.03358 -0.001858 0.016807

5 0.5625   0.625 0.59375 0.016807 -0.001858 0.007722

6 0.59375   0.625 0.609375 0.007722 -0.001858 0.002995

Soal No. 2 (Metode Bisection)

Iterasi x1   x2 x3 f(x1) f(x2) f(x3)

1   2.5 3.5 3 -3.25 3.75 0

2   2.5 3 2.75 -3.25 0 -1.6875

3   2.75 3 2.875 -1.6875 0 -0.859375

4   2.875 3 2.9375 -0.859375 0 -0.4335945   2.9375 3 2.96875 -0.433594 0 -0.217773

6   2.96875 3 2.984375 -0.217773 0 -0.109131

7   2.984375 3 2.992188 -0.109131 0 -0.054626

8   2.992188 3 2.996094 -0.054623 0 -0.027327

9   2.996094 3 2.998047 -0.027327 0 -0.013667

10   2.998047 3 2.999024 -0.013667 0 -0.006835

11   2.999024 3 2.999512 -0.006831 0 -0.003416

Soal No. 4 (Metode Secant)

Iterasi x f(x)

1 x1   1 -0.28172

x2   1.8 2.24964

x3   1.089034 -0.117635

2 x2 1.8 2.24964

x3   1.089034 -0.117635

x4   1.124363 -0.04611

3 x3   1.089034 -0.117635

x4   1.124363 -0.04611

x5   1.147139 0.002029x6   1.146179 -3.3E-005

x7   1.146194 -2.3E-008

Soal No. 6 (e!"la #alsi)

Iterasi x1 x2 x3 f(x1) f(x2) f(x3)

1   2.5 3.5 2.964286 -3.25 3.75 -0.248724

2   2.964286 3.5 2.997608 -0.248724 3.75 -0.016741

3   2.997608 3.5 2.99984 -0.016741 3.75 -0.0011174   2.99984 3.5 2.999989 -0.001117 3.75 -7.4E-005

8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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Soal No. 8 (Metode Ne$ton)

Iterasi x f(x) f'(x)

x2 x1   2.5 -3.25 5

x3 x2   3.15 1.0725 6.3

x4 x3   2.979762 -0.141257 5.959524

x5 x4   3.003465 0.024265 6.006929

x6 x5   2.999425 -0.004023 5.99885x7 x6   3.000096 0.000671 6.000192

8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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Metode Bisection

Metode e!"la #alsi

Metode Ne$ton

n% 1&2&3....

Metode Secant

i % 2&3....

  '3% '1( '2)2 

(|*)

(|*)

( )1212

223   x x

 )  f(x )  f(x

 )  f(x x x   −−

−=

)('

)(1

n

nnn

 x  f  

 x  f   x x   −=

+

)1()(

1)(1

−−

−−

−=+

i x  f  i x  f  

i xi xi x  f  i xi x

8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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ondisi te/asi

(x3) + ,

(x3) + ,

*(xi1)* + ,

 '1 '2−2 + ,

 '1 '2−2 + ,

8/18/2019 Metode Bisection, Regula Falsi, Dan Newton

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