Use Newton's method to approximate a unique root in [-1,1] for $f(x)=arccos(x)-2x-3$ to within $10^{-6}$ Edited for readability
I'm having some trouble with this because I've tried solving this myself through the iterations, and I've also tried Matlab and an online calculator.
The first time I tried the iterations myself I began with x=1. On the second iteration I was getting $1-\frac{0}{0}$ which did not look right to me at all. I get the same behavior (an error of division by 0 when I try Matlab).
Here is what I am doing - now I have chosen $-.5$ because it is closer to the root (which is approx $-.47$ by looking at the graph of f(x)=arccos(x)−2x−3.
$$f(x) = arccos(x)-2x-3$$
$$f'(x) = \frac{-1}{\sqrt{1-x^2}} - 2$$
Newton's method : $$x - \frac{f(x)}{f'(x)} $$
which, in this particular case $$x - \frac{ arccos(x) - 2·x - 3}{ -\sqrt{ 1-x^2 }} - 2$$
Now when I plug in x=−.5
$$p1 = (-.5) - \frac{( arccos(-.5) - 2·(-.5) - 3 )}{ -\sqrt{ 1-(-.5)^2} } - 2 = -2.3910019244$$
$$p2 = (-2.3910019244)-\frac{arccos(-2.3910019244)-2·(-2.3910019244)-3}{-\sqrt{1-(-2.3910019244)^2}}-2$$
p2=not a real number
So this is where I am stuck because I can clearly see the root on the graph but I do not seem to be getting close to it.
Thanks for the help!

Note that $f'(x)$ is undefined at $x= \pm 1$ because of division by zero.  Newton's method works well if you start close to the root.  If you graph it, the root is somewhere near $-0.4$.  Starting at $-0.4$ I converge rapidly to $-0.4699722$
