Determine o zero da função,
, utilizando o Método de Newton, no intervalo
para 5 iterações com arredondamento padrão.
Passo 1
O enunciado pede para encontrar o zero da função f(x)=2x+3 cos(x)−e^x no intervalo [−1,0] usando o método de Newton em 5 iterações. Vamos precisar da derivada f’(x)=2−3 sin(x)−e^x e escolho x_0=−1 como chute inicial.
Passo 2
Em x_0=−1:
f(x_0)=2(−1)+3 cos(−1)−e^{−1}=−2+3·0.540302−0.367879≈−0.746973
f’(x_0)=2−3 sin(−1)−e^{−1}=2+3·0.841471−0.367879≈4.156533
Logo
x_1=x_0−f(x_0)/f’(x_0)=−1−(−0.746973)/4.156533≈−0.820358
f(x_0)=2(−1)+3 cos(−1)−e^{−1}=−2+3·0.540302−0.367879≈−0.746973
f’(x_0)=2−3 sin(−1)−e^{−1}=2+3·0.841471−0.367879≈4.156533
Logo
x_1=x_0−f(x_0)/f’(x_0)=−1−(−0.746973)/4.156533≈−0.820358
Passo 3
Em x_1=−0.820358:
f(x_1)=2(−0.820358)+3 cos(−0.820358)−e^{−0.820358}≈−0.038810
f’(x_1)=2−3 sin(−0.820358)−e^{−0.820358}≈3.754569
Logo
x_2=−0.820358−(−0.038810)/3.754569≈−0.810021
f(x_1)=2(−0.820358)+3 cos(−0.820358)−e^{−0.820358}≈−0.038810
f’(x_1)=2−3 sin(−0.820358)−e^{−0.820358}≈3.754569
Logo
x_2=−0.820358−(−0.038810)/3.754569≈−0.810021
Passo 4
Em x_2=−0.810021:
f(x_2)=2(−0.810021)+3 cos(−0.810021)−e^{−0.810021}≈0.008254
f’(x_2)=2−3 sin(−0.810021)−e^{−0.810021}≈3.726990
Logo
x_3=−0.810021−0.008254/3.726990≈−0.812237
f(x_2)=2(−0.810021)+3 cos(−0.810021)−e^{−0.810021}≈0.008254
f’(x_2)=2−3 sin(−0.810021)−e^{−0.810021}≈3.726990
Logo
x_3=−0.810021−0.008254/3.726990≈−0.812237
Passo 5
Em x_3=−0.812237:
f(x_3)=2(−0.812237)+3 cos(−0.812237)−e^{−0.812237}≈0.000673
f’(x_3)=2−3 sin(−0.812237)−e^{−0.812237}≈3.723020
Logo
x_4=−0.812237−0.000673/3.723020≈−0.812418
f(x_3)=2(−0.812237)+3 cos(−0.812237)−e^{−0.812237}≈0.000673
f’(x_3)=2−3 sin(−0.812237)−e^{−0.812237}≈3.723020
Logo
x_4=−0.812237−0.000673/3.723020≈−0.812418
Passo 6
Em x_4=−0.812418:
f(x_4)=2(−0.812418)+3 cos(−0.812418)−e^{−0.812418}≈−0.000055
f’(x_4)=2−3 sin(−0.812418)−e^{−0.812418}≈3.722547
Logo
x_5=−0.812418−(−0.000055)/3.722547≈−0.812403
f(x_4)=2(−0.812418)+3 cos(−0.812418)−e^{−0.812418}≈−0.000055
f’(x_4)=2−3 sin(−0.812418)−e^{−0.812418}≈3.722547
Logo
x_5=−0.812418−(−0.000055)/3.722547≈−0.812403
Resposta
x≈−0.812403