Question

12. Compute the positive root of x3 – X – 0.1 = 0, by Newton-Raphson Method
correct to six significant figures. (Hint: one of the root lie between 1 and 1.1)​

Answer 1

1.046681273628291927

Answer 2

Given : x³ - x - 0.1

one of the root lie between 1 and 1.1

To Find : one root by  by Newton-Raphson Method

correct to six significant figures.

Solution:

f(x) =  x³ - x - 0.1

f'(x) = 3x ² - 1

x₀  = 1  

xₙ₊₁  = xₙ ​−f(xₙ​)​./ f′(xₙ​)

x₁  =  1 -  (1³ - 1 - 0.1)/  (3*1² - 1)

x₁  =   1 -  (-0.1)/2

x₁  =  1.05

x₂ = 1.05 -  (1.05³ - 1.05 - 0.1)/  (3*1.05² - 1)

=> x₂ = 1.046695558

x₃ =  1.046680532

x₄ =  1.046680532

x = 1.046680532

x=  1.046681

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