Let f be a function defined on [a,b] and suppose that z is a point in (a,b) such that f(z) = z. Further suppose that there is a number alpha < 1 such that f ' (x) < alpha for all x contained in (a,b) and that 0 < f ' (x) for all x contained in (a, b).

a.) Prove that if z

b.) Let x0 > z and set xn = f ( x n-1) for every natural number n > or = 1. Prove that 0 < xn - z < alphan (x0- z). You may use the result of part a.

c.) Short answer - this does not require a formal proof - just an indication of why it may be so. How could the result of part b be used to argue that if x0 > z, then the sequence of iterates xn = f ( x n-1) tends towards z.