Use Newton's method to approximate a root of the equation x3+x+3=0 as follows.

Question

Let x1=–1 be the initial approximation.
The second approximation x2 is _____?
and the third approximation x3 is _____?

Answer 1

y = x^3 + x + 3
Xn+1 = Xn - Yn/y'
where here
y' = 3Xn^2 +1
X1 = -1
Y1 = -1-1+3 = 1
y'= 3+1 = 4
X2 = X1 - Y1/(3X1^2+1)
X2 = -1 - 1/4 = -1.25
now do X3

Answer 2

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.

Answer 3

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