Examples of when Newton's Method will fail? I'm currently working on Newton's Method, and my instructor gave four instances where Newton's Method will fail.
(A) Newton's method converges to another solutions x=b such that f(b)=0 instead of converging to the desired solution x=a.
(B) Newton's method eventually gets into the never ending cycle, bouncing between the same two approximations $x_i$ and $x_{i+1}$.
(C) Eventually, each next approximation $x_{i+1}$  falls further from desired solution $x_a$ than the previous approximation $x_i$  determined by the Newton's method.
(D) Newton's method is not able to find the next approximation $x_{i+1}$  because f'($x_i$)=0 or f'($x_i$) Does Not Exist.
However, there aren't any examples of when this happens. Would anyone be willing to provide examples of these instances?

Example for Case (A):  $$f(x) = \frac{1}{1+x^2} - \frac{1}{2},$$ which has roots at $x \in \{-1,1\}$.  The initial choice $x_0 = 2$ converges to the negative root.
Example for Case (B):  $$f(x) = \begin{cases}\sqrt{x}, & x \ge 0 \\ -\sqrt{-x}, & x < 0 \end{cases}$$ has the peculiar property that for any initial guess $x_0 \ne 0$, the orbit is trapped in a cycle of period $2$, with $x_k = -x_{k-1}$.  This is quite easy to prove and is left as an exercise for the reader.
Example for Case (C):  $$f(x) = x^{1/3}.$$  The Newton's method recursion has no fixed point except for the initial guess $x_0 = 0$.
