Question

If Newton's method is used to approximate the real root of Xˆ3+x-1=0 then a first approximation of x1=1 would lead to an x2 value of _?

Answer 1

\[f(x)=x^3+x-1\]\[f'(x)=3x^2+1\]\[f(1)=1\]\[f'(1)=4\]\[y=4(x-1)+1\implies y=4x-3\implies4x-3=0\implies x=3/4\] would be your approximation

Answer 2

\[ x_{n+1} = x_n - \frac{f(x)}{f'(x)} = 1 - \frac{1}{4} = 3/4\]