The solution would be like this for this specific problem:
f(x) = x^2 + sin(x)
f '(x) = 2x + cos(x)
The minimum value is at f
'(x) = 0,
So, let g(x) = 2x + cos(x)
Thus, g '(x) = 2 - sin(x)
x(new) = x - g(x) / g '(x)
or
x(new) = x - [2x + cos(x)] / [2 - sin(x)]
Calculation
x1 = -0.5 - [2 * -0.5 + cos(-0.5)] / [2 - sin(-0.5)]
= -0.4506266931
x2 = -0.4501836476
x3 = -0.4501836113
x4 = -0.4501836113
This value for x, f(x) =
-0.2324655752.
After converting to 6 decimal places: the minimum point is (-0.450184,
-0.232466).
