Math, asked by sathishvms002, 7 months ago

solve for a postive root of f(x)=2x3-3x-6=0 newton raphson method correct to four decimal places k3 co1

Answers

Answered by schhn2021
4

Answer:

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Step-by-step explanation:

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Answered by amitnrw
4

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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