f(x)=x3-x2+1

Question

using newton's method of approximation, find the first upto the fifth solution for the function f(x)=x3-x2+1.

Answer 1

The cubic can have up to three zeroes. In fact, it has only one real zero and two complex zeroes. f(0)=1 and f(1)=1, but f(-1)=-1, so there must be a real zero between x=0 and x=-1.

We can use Newton's method to find the real zero.

x_n+1=x_n-f(x_n)/f'(x_n).

f'(x)=3x²-2x. We need to choose a suitable x₀. We can't use x₀=0 because it would lead to division by zero, so let x₀=-1.

x₁=-1-f(-1)/f'(-1)=-4/5, x₂=-333/440, x₃=-0.7548814744, x₄=-0.7548776663, x₅=-0.7548776662, (x₆=-0.754877666247). The fifth iteration after x₀=-1 is stable so is an approximate solution for the real zero of f(x).