Task 1: Solutions
Check that f has a maximum of a/4 at x = 1/2.
Graphing f we
see that it is a convex parabola, symmetric about the line y = 1/2. To prove this, first note that the first derivative of f is
f¢ = a ·(1- 2x). Solving f¢(x) = 0 gives x = 1/2, which
must be a turning point. The second derivative of f is f¢¢(x) = -2a which is negative (everywhere) so f has a maximum at 1/2.
Substituting x = 1/2 in f(x) gives a/4.
Deduce that f maps [0,1] back onto [0,1] for 0 £ a,1] for 0 £ a £ 4.
Since a, x and 1-x are all positive for x Î [0,1], f(x) is also positive. The maximum value of f on [0,1] is a/4, which is £ 1 . And f(0) (and f(1)) equals 0. Hence f([0,1]) Í [0,1] by the Intermediate Value Theorem.
Show (xn) is unbounded if x1 is a member of (0,1) and a > 4.
Show that the 1-tail of (xn) is constantly equal to 0 if x1 is either 0 or 1.
Follows from f(0) = 0 = f(1).