Help Please.?

Question

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

Answer 1

the absolute minimum value will be one of the critical points which are solutions of f'(x) = 0
So you need to find the roots of
f'(x) = 2x + cos x = 0

f''(x) = 2 - sin x , which is positive( why?)
Since f'' is positive the function f is concave upward everywhere. This tells you that , if you find a critical point, this should be absolute minimum and not maximum.( why?)
Now back to our duty :
we have 2x + cos x = 0. The approximations of the root are given by Newton's method:
x(n+1) = x(n) - (2x + cos x(n))/(2 - sin x(n))
take x(0) = 1
so x(1) = 1 -(2+ cos 1)/(2 - sin1)
compute this using calculator( sorry i don't have here one)
after this compute x(2), x(3) and so on . When two consecutive approximations agree to six decimals then that one is the approximation that you search for.

Answer 2

= -0.017452. Newtons method is x (n+1) = x(n) - (f(fxn)/f ' (xn)), so f ' (xn)= 2x + cos x. for xn, use 1. so you'll have 1- (1+ sin1)/(2+cos1)= 0.660831978877. now, for x(n+1), use this number, so you'll have 0.660831978877- (1+sin(0.660831978877))/(2+cos(0.660831978877)), and you'll get another decimal. Use that decimal for the third one (keep pluggin in what you got on your last time, until you get to a point where all the iterations are equal to the same exact number. Once you do that, you will get the answer -0.017452406437