Newton-Raphson for finding roots on a concave slope.

Question

How does the Newton-Raphson method work for a concave curve?

If I am trying to find the root and use $x_0 = -2$ then the tangent at $-2$ would cross the $x$ axis and $x_1$ would be past the root and you would have to follow down from $x_1$ until you meet the graph and then draw a new tangent. How would this locate the root graphically (I missed this in the first edit)? This is the graph of $y=-x^2+10$ by the way. Thank you in advance.

Answer 1

It works just fine. Here are the first few $x$'s from Newton-Raphson

$$x \to x-\frac{f(x)}{f'(x)} = x-\frac{10-x^2}{-2x}$$

n:             x_n    f(x_n)
----------------------------
0: -2.000000000000   6.0e+00
1: -3.500000000000  -2.2e+00
2: -3.178571428571  -1.0e-01
3: -3.162319422151  -2.6e-04
4: -3.162277660444  -1.7e-09
5: -3.162277660168  -1.8e-15
6: -3.162277660168   1.8e-15

From $x_1$ on, the $x$-values are smaller than $-\sqrt{10}$ and approximate "from the left", i.e. the $x$'s increase towards $-\sqrt{10}$ and $f(x_n)<0$.

Note: The last value at $n=6$ in the table above is already at the numerical resolution of IEEE double, so it's just fluctuating around the real solution. This is due to the numerical resolution and not due to the very dynamic of the iteration.