Question

The equation f(x) is given as x3+4x+1=0. Considering the initial approximation atx=1 then the
value of x is given as​

Answer 1

Given : f(x) =  x³ +4x+1  = 0

initial approximation at x=1

To Find :  value of  x₁

Solution:

f(x) =  x³ +4x+1

f'(x) = 3x ² +4

x₀  = 1  

xₙ₊₁  = xₙ ​−f(xₙ​)​./ f′(xₙ​)

x₁  =  1 -  (1³ + 4 + 1)/  (3*1² + 4)

x₁  =   1 -  0.8571

x₁  =  0.1429

value of  x₁  is 0.1429

x₂  =  -0.2448

x₃  =  -0.24627

x₄  =  -0.24627

-0.24627  is one of the solution of  x³ +4x+1  = 0

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