METODE REGULA FALSI

C=b−f (b )(b−a)f (b )−f (a)

Soal : f ( x )=3 x5+4 x4−2 x2−2027,991018 Dimana : a=3,4 ;b=3,5 ;ε=10−5

r a c b F(a) F© F(b) Selang Baru

Lebar

0 3,4 3,455434 3,5 -153,513898 -3,751057 123,415232 [C,B] 0,044566

1 3,455434 3,456749 3,5 -3,751057 -0,086453 123,415232 [C,B] 0,043251

2 3,456749 3,456779 3,5 -0,086453 -0,002788 123,415232 [C,B] 0,043221

3 3,456779 3,456780 3,5 -0,002788 0,00000038 123,415232/2 =

61,707616

[A,C] 0,000001

PERHITUNGAN :

r=0 ; f (a )=f (3,4 )=3 (3,4 )5+¿ 4 (3,4 )4−2 (3,4 )2−2027,991018=−153,513898 ¿

f (b )= f (3,5 )=3 (3,5 )5+4 (3,5 )4−2¿

C=3,5−{ [ (123,415232 ) (3,5−3,4 ) ][ (123,415232 )−(−153,513898 ) ] }=3,455434

f ( c )=f (3,455434 )=3 (3,455434 )5+4 (3,455434 )4−2 (3,455434 )2−2027,991018=−3,751057

Lebar= (3,5−3,455434 )=0,04456 6

r=1;C=3,5−{[ (123,415232 ) (3,5−3,455434 ) ][ (123,415232 )−(−3,751057 ) ] }=3,45674 9

f ( c )=f (3,456749 )=3 (3,456749 )5+4 (3,456749 )4−2 (3,456749 )2−2027,991018=−0,0864453

Lebar= (3,5−3,456749 )=0,043251

r=2;C=3,5−{ [ (123,415232 ) (3,5−3,456749 ) ][ (123,415232 )— 0,086453 )

=3,456779

f ( c )=f (3,456779 )=3 (3,456779 )5+4 (3,456779 )4−2 (3,456779 )2−2027,991018=−0,002788

Lebar= (3,5−3,456779 )=0,043221

r=3 ;C=3,5−{[ (123,415232 ) (3,5−3,456779 ) ][ (123,415232 )−(−0,002788 ) ] }=3,45677 9

Karena C pada r 3=C pada r 2=3,456779→f (b )2

=123,4152322

=61,707616

C=3,5−{ [ (61,707616 ) (3,5−3,456779 ) ][ (61,707616 )−(−0,002788 ) ] }=3,45678 0

f ( c )=f (3,456780 )=3 (3,456780 )5+4¿

Lebar= (3,456780−3,456779 )=0,000001

Untuk nilai a, c, b, f(a), f(c), f(b), lebar pada r4 sampai r14 TIDAK DAPAT DICARI dikarenakan

nilai f(c) pada r3 sudah melebihi batas yang ditentukan pada soal.