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.
The point is, if there's more than one root, then the function has to turn somewhere. But we have shown that it doesn't turn, so there is at most one root.
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.