STEM-Master
Calculus IChapter 4: Application of DerivativesSection 4-13

Newton’s Method — Question 6

PDF ↗

Question 6

Use Newton’s Method to approximate a root of the equation: x3−2x−5=0x^3 - 2x - 5 = 0 Start with an initial guess of x1=2x_1 = 2, and compute the next three approximations: x2x_2, x3x_3, and x4x_4. Round answers to four decimal places.

Original worksheet page 1: question and worked solution for 4-13-006
Show solutionHide solution

Question 6 - 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}

Original worksheet page 2: question and worked solution for 4-13-006

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

← Section questionsChapter contents