As mentioned in the comments, the reason is that the cost functions mentioned might not have any zeroes at all, in which case Newton's method will fail to find the minima.
I have created a visualization to show this:
As you can see, the method is not converging at all for this particular case.
The code used to create this is stored here.
Adding the relevant portion of the code here itself for convenience:
% Dummy statement to avoid writing function in the first line and making it a 'function file' instead of a 'script file'
% The function to find zeroes of.
% The function is specifically chosen to not have any zeroes
% so as to show the weakness of Newton's method.
function y = f(x)
y = (x - 5).^2 + 5;
% The derivative of f(x)
function y = fd(x)
y = 2 * (x - 5);
% Initial guess
x0 = 1.5;
% Max number of iterations
itermax = 20;
% Epsilon value initialized to a very large value
eps = 1;
% A vector for storing the history of the approximate roots
xvals = x0;
% Number of iterations done
itercount = 0;
% Required for plotting f(x) vs x
x = linspace(0, 10, 100);
% Create a figure whose output is not rendered on the screen
% Not working currently; supposedly a bug in Octave
% A workaround is to use gnuplot instead of qt - `graphics_toolkit gnuplot`
% but this is very slow.
% Uncomment the following to activate the feature once the bug is fixed
% The main loop
while eps >= 1e-5 && itercount <= itermax
% x1 = New value of root
% x0 = Current value of root
x1 = x0 - f(x0) / fd(x0);
% Plot f(x)
% Plot a line passing through points [x0, f(x0)] and [x1, 0]
% Plot a line passing through points [x1, 0] and [x1, f(x1)]
% Plot a line passing through points [x0, 0] and [x0, f(x0)]
plot(x, f(x), ";f(x);", [x0 x1], [f(x0) 0], "-r;f'(x);", [x1 x1], [0 f(x1)], ":r", [x0 x0], [0 f(x0)], ":r");
title('f(x) = (x-5)^2 + 5');
% Set limits for the axes shown in the plots
% Label the two consecutive zeroes on the X-axis
text(x0, -2, sprintf('x%d', itercount), 'color', 'red');
text(x1, -2, sprintf('x%d', itercount+1), 'color', 'red');
% Print the plot to a file
filename = sprintf('output/%05d.jpg', itercount);
% Append the zero to the array of zeroes calculated so far
xvals = [xvals; x1];
% Calculate the epsilon value
eps = abs(x1-x0);
x0 = x1;
itercount = itercount+1;
% Print the result of the iteration
f_zero = f(xvals(end))