Use newton's method to find the absolute minimum value of the function f(x)=x2+sinx correct to six decimal places.

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).

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry