Let $$f(x) = \begin{cases} \dfrac{x^3-1}{\sqrt{x}-1}, & x > 1\\ \arccos(x-1)-x^2, & x \leq 1\end{cases}$$
Use Newton's method with $x_{0} =1$, compute the second iterate to approximate value $c$ where $c$ is a stationary value that lies in the x-axis for some $0<c<1$
Since the Newton's method is as follows:
$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$
$$x_{1}=x_{0}-\frac{acos(0)-1}{-asin(0)-2}$$
Is this correct? What should I proceed on from here?