Posted by Adam on .
Use Newton's method to approximate a root of the equation x3+x+3=0 as follows.
Let x1=–1 be the initial approximation.
The second approximation x2 is _____?
and the third approximation x3 is _____?

CALC 
Damon,
y = x^3 + x + 3
Xn+1 = Xn  Yn/y'
where here
y' = 3Xn^2 +1
X1 = 1
Y1 = 11+3 = 1
y'= 3+1 = 4
X2 = X1  Y1/(3X1^2+1)
X2 = 1  1/4 = 1.25
now do X3 
CALC 
Adam,
im not very clear on the steps of newton's method because my professor didn't explain it very well. Do you just take the second derivative of y and then plug in 1.25 to find x3. I don't quite understand the concept...thanks for the help in advance.

CALC 
gurbz,
You're right Adam, just plug 1.25 to find X3.
So...you know...
y=x^3+x+3
y'=3x^2+1
Xn=1
The equation you use is:
Xn+1 = Xn (f(Xn)/f'(Xn))
So for:
X2=(1) ( ((1)^3+(1)+3)/ (3(1)^2+1))
X2= (1) (1/4)
X2= 1.25
Then...
X3=(1.25) ( ((1.25)^3+(1.25)+3)/ (3(1.25)^2+1))
X3=(1.25)(1/28)
X3=1.214285714