>> my_fun = @(x) 5*x^4 - 2.7*x^2 - 2*x + .5;
low = .1;
high = 0.5;
tolerance = .00001;
x = bisection(my_fun, low, high, tolerance);
Bisection Method
Iter    low        high          x0
 0 	 0.100000 	 0.500000 	 0.300000
Root at x = 0.200000




CODE


function  m = bisection(f, low, high, tol, maxit)
disp('Bisection Method');

% Evaluate both ends of the interval
j = maxit;
y1 = feval(f, low);
y2 = feval(f, high);
i = 0;

% Display error and finish if signs are not different
if y1 * y2 > 0
   disp('Have not found a change in sign. Will not continue...');
   m = 'Error'
   return
end

% Work with the limits modifying them until you find
% a function close enough to zero.
disp('Iter    low        high          x0');
while (abs(high - low) >= tol) && (maxit<j)




    j = j + 1;
    i = i + 1;
    % Find a new value to be tested as a root
    m = (high + low)/2;
    y3 = feval(f, m);
    if y3 == 0
        fprintf('Root at x = %f \n\n', m);
        return
    end
    fprintf('%2i \t %f \t %f \t %f \n', i-1, low, high, m);

    % Update the limits
    if y1 * y3 > 0
        low = m;
        y1 = y3;
    else
        high = m;
    end
end



% Show the last approximation considering the tolerance
w = feval(f, m);
fprintf('\n x = %f produces f(x) = %f \n %i iterations\n', m, y3, i-1);
fprintf(' Approximation with tolerance = %f \n', tol);