Newtons Law problem: root of cos (x^2 + 2) = x^3

[domo256]

Problem- Use Newton's method to approximate a root of the equation cos (x^2 + 2) = x^3 as follows.
Let X1 be = to 2 the initial approximation.
The second approximation x2 =?

I think its the equal sign with the x3 that is messing me up... This is how set the problem up

x2 = x1 - [p(x1)/p'(x1)]

x2 = 2 - [cos (2^2 +2) - 2^3/ -sin (2x) - 3x^2)]

x2 = 2 -[cos (2^2 +2) - 2^3/ -sin (2(2)) - 3(2^2))]

my answer was 1.3738 but this is not the correct answer.. can someone help me with what I did wrong?

 

[skeeter]

Re: Newtons Law problem

Problem- Use Newton's method to approximate a root of the equation cos (x^2 + 2) = x^3 as follows.
Let X1 be = to 2 the initial approximation.
The second approximation x2 =?

I think its the equal sign with the x3 that is messing me up... This is how set the problem up

x2 = x1 - [p(x1)/p'(x1)]

x2 = 2 - [cos (2^2 +2) - 2^3/ -sin (2x) - 3x^2)] your derivative is incorrect

p(x) = cos(x² + 2) - x³
p'(x) = -2xsin(x² + 2) - 3x²

x2 = 2 -[cos (2^2 +2) - 2^3/ -sin (2(2)) - 3(2^2))]

my answer was 1.3738 but this is not the correct answer.. can someone help me with what I did wrong?

 

[domo256]

thank you

thank you very much... I think i've just been at this too long today...