Show that the equation x^3 + x = 3 has exactly one root, ∝. Find an integer n such that ∝ lies between n and n + 1. Starting with x0 = n as a first approximation, use Newton's Method to find 2 further approximations to ∝.
f(x) = x^3 + x - 3, [n, n + 1]
f'(x) = 3x^2 + 1
x2 = ( x1 + x1^3 + x1 - 3 ) / ( 3x1^2 + 1 )
x2 = ( n + n^3 + n - 3 ) / ( 3n^2 + 1 )
x2 = ( n^3 + 2n - 3 ) / ( 3n^2 + 1 )
x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 )
I dont you what to do from here.
To prove that there is a root, substitute some values in and see if you get both positive and negative values. If so then it must have crossed the axis somewhere.