- #1

peterianstaker

- 2

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Using the Newton Raphson method with X0=2 to find the root of the equation:

x^3-x-1=0 (correct to 4.d.p)

My answer is:

f'(x)= 3x^2-1

xn+1= 2-x^3-x-1/3x^2-1

xn+1= 2-2^3-2-1/3(2^2)-1

x1= 17/11

x2= 17/11-(17/11^3)-17/11-1/3x(17/11^2)-1

= 1.3596