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Calculus IChapter 4: Application of DerivativesSection 4-13

Newton’s Method — Question 2

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Question 2

Use Newton’s Method to approximate the root of the equation x3−2x−5=0x^3 - 2x - 5 = 0 starting with the initial guess x1=2x_1 = 2. Perform three iterations to compute x2x_2, x3x_3, and x4x_4. Round your answers to four decimal places.

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Question 2 - Solution

Newton’s iteration is

xn+1=xn−f(xn)f′(xn),f(x)=x3−2x−5,f′(x)=3x2−2.x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=x^3-2x-5,\quad f\prime(x)=3x^2-2.

Keeping full precision internally gives

x2=2.0000000000−−1.000000000010.0000000000≈2.1000x3=2.1000000000−0.061000000011.2300000000≈2.0946x4=2.0945681211−0.000185723211.1616468418≈2.0946\begin{aligned}x_{2}&=2.0000000000-\frac{-1.0000000000}{10.0000000000}\approx\boxed{2.1000}\\[6pt]x_{3}&=2.1000000000-\frac{0.0610000000}{11.2300000000}\approx\boxed{2.0946}\\[6pt]x_{4}&=2.0945681211-\frac{0.0001857232}{11.1616468418}\approx\boxed{2.0946}\end{aligned}

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