solve for a postive root of f(x)=2x3-3x-6=0 newton raphson method correct to four decimal places k3 co1
Answers
Answer:
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Step-by-step explanation:
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Given : f(x)=2x³ - 3x - 6
To Find : one positive root by by Newton-Raphson Method
correct 4 decimal
Solution:
f(x) = 2x³ - 3x - 6
f'(x) = 6x² - 3
x₀ = 2
xₙ₊₁ = xₙ −f(xₙ)./ f′(xₙ)
x₁ = 2 - (2*2³ - 3*2- 6)/ (6*2² - 3)
x₁ = 2 - (4)/21
x₁ = 1.809524
Similarly
x₂ = 1.784200
x₃ = 1.783769
x₄ = 1.783769
x₃ = x₄
Hence 1.783769 is the required root
up to 4 decimal place 1.7838
hence x = 1.7838 upto 4 decimal places
Taking x₀ = 1
x₁ = 3.333333
x₂ =2.421175
x₃ = 1.951123
x₄ = 1.799821
x₅ = 1.783938
x₆ = 1.783769
x₇ = 1.783769
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