%
% This example shows that some functions can be very difficult (numerically
% speaking) to work with.  In this case, there is a triple root at x=2/3
% (that is, f is the expansion of (x-2/3)^3)

f=inline('x.^3-2*x.^2+(4/3)*x-(8/27)');

%Use a modified bisection algorithm to give all the iterates (See below)
format long e
xx=bisection1(f,0,1,25);

(xx-(2/3))'  %List all the errors
xx(end)      %Give the final entry
f(xx(end))   %Is it really?