# Problemes funcions_Solució exercici 37 (2009)

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restart : f d x/ 4\$sin x C1 Kx; x/4 sin x C1 Kx (1) plot 4\$sin x C1, x , x =KPi ..Pi gráficas: y=4sin(x)+1 , y=x 5 4 3 2 1 K3 K2 0 K1 1 K1 2 3 x K2 K3 Existen tres soluciones de f(x)=0: αe(-3,-2), βe(-1,0) y γe(2,3).
BISECCIÓN a) αe(-3,-2) f K3 K4 sin 3 C4 (2) evalf % 3.435519968 f K2 K4 sin 2 C3 (4) K0.637189707 (5) evalf % f(-3)>0 y f(-2)<0 5 2 [a0,b0]=[-3,-2]0c0 dK 5 2 K (6) f c0 : evalf % 1.106111424 (7) 5 f(c0) > 0 0 [a1,b1]=[K ,-2]0 2 9 4 c1 dK 9 4 K (8) f c1 : evalf % ; 0.137707212 (9) 9 f(c1)>0 0 [a2,b2]=[K ,-2] 4 17 c2 dK 8 17 8 K (10) f c2 : evalf % K0.276279159 (11) 9 17 f(c2)<0 0[a3,b3]=[K ,K ] 4 8 35 c3 dK 16 35 16 K (12) f c3 : evalf % K0.075657253 f(c3)<0 0 9 35 [a3,b3]=[K ,K ] 4 16 (13) 9 35 K K 4 16 c4 d 2 71 32 K (14) f c4 : evalf % 0.029467840 (15) f c4 O 0 0 71 35 [a4,b4]=[K ,K ] 32 16 71 35 K K 32 16 c5 d 2 141 64 K (16) f c5 : evalf % K0.023488572 (17) 71 141 ,K ] 32 64 f c5 ! 0 0 a5, b5 = [K 71 141 K 32 64 2 K c6 d 283 128 K (18) 283 128 α zK 283 128 αz K (19) K2.210937500 (20) 283 128 evalf K NEWTON 5 2 a)αe(-3,-2)0x0 dK 5 2 K (21) diff f x , x 4 cos x K1 (22) x/4 cos x K1 (23) df d x/4 cos x K1 x1 d x0 K f x0 df x0 : evalf % ; x2 d x1 K f x1 df x1 K2.236926666 (24) K2.210412561 (25) K2.210083995 (26) K2.210083944 (27) K2.210083944 (28) K2.210083944 = K2.210083944 (29) : evalf % ; x3 d x2 K f x2 df x2 : evalf % ; x4 d x3 K f x3 df x3 : evalf % ; x5 d x4 K f x4 df x4 : evalf % ; α =K2.210083944 1 2 b) βe(-1,0) 0x0 dK 1 2 (30) 5 2 (31) K K x1 d x0 K f x0 df x0 : evalf % ; x2 d x1 K f x1 df x1 K0.3336066921 (32) K0.3421675710 (33) K0.3421850529 (34) : evalf % ; x3 d x2 K f x2 df x2 : evalf % ; x4 d x3 K f x3 df x3 : evalf % ; K0.3421850529 (35) K0.3421850529 (36) 5 2 (37) 2.712599059 (38) 2.702081446 (39) 2.702061373 (40) 2.702061373 (41) γ = 2.702061373 (42) β =K0.3421850529 c) γe(2,3) 0x0 d x1 d x0 K f x0 df x0 5 2 : evalf % ; x2 d x1 K f x1 df x1 : evalf % ; x3 d x2 K f x2 df x2 : evalf % ; x4 d x3 K f x3 df x3 : evalf % ; γ = 2.702061373 α=K2.210083944 β =K0.3421850529 γ = 2.702061373 4 3 2 f := x → x - 4 x + 2 x - 8 10 5 -6 -4 y -2 x 0 2 0 -5 -10 -15 -20 48 -8 -16 24 4 6 8 3 2 4 x - 12 x + 4 x 3 2 df := x → 4 x - 12 x + 4 x x0 := -1 -1.050000000 -1.047687892 -1.047682733 -1.047682733 -1.047682733 x0 := 3 4.416666667 3.894756026 3.664034639 3.617843989 3.616108079 3.616105685 3.616105685 3.616105685 f := x → (x + 1) e (x - 1 ) -1 10 8 6 y 4 2 0 -20 -15 -10 -5 x 0 5 -2 -4 -1 -0.6321205588 7.154845484 e (x - 1) + (x + 1) e df := x → (x + 2) e (x - 1 ) (x - 1) x0 := 1 0.6666666667 0.5650213261 0.5571885588 0.5571456001 0.5571455989 0.5571455990 0.5571455990 0.5571455990 2 f := x → 3 x + tan(x) 100 x -10 -5 0 0 -100 -200 y -300 -400 5 10 5 x -1,5 -1 -0,5 0 0 -5 -10 y -15 -20 -7.35141995 3 - tan(1) 1.442592275 0,5 1 1,5 6 x + 1 + tan(x) 2 df := x → 6 x + 1 + tan(x) x0 := -1.25 -1.906079770 4.352994780 2.613125739 1.443805381 1.677930326 x0 := -1.375 -1.409668586 -1.403386007 -1.403061226 2 -1.403060421 -1.403060421 ...