Posted by Adam on Saturday, May 29, 2010 at 8:36pm.
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, Saturday, May 29, 2010 at 9:10pm
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, Sunday, May 30, 2010 at 1:27am
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, Thursday, June 3, 2010 at 5:41am
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
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