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March 26, 2017

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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 - ,

    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

  • CALC - ,

    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 - ,

    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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