1. Construct a function f (x) so that Newton's method gets 'hanging' in an infinite cycle xn = (-1)n x0 , no matter how the start value x0 is chosen. 2. Relevant Equations: xn+1 = xn - f(xn) / f'(xn) 3. The attempt at a solution xn+1 = xn - f(xn) / f'(xn) = (-1)n+1x0 = (-1)nx0 - f(xn) / f'(xn) [itex]\Rightarrow[/itex] f(xn) / f'(xn) = 2(-1)nx0 But, I don't know if that's what I wanna do or what to do with it. Any ideas?