Use Newton's Method to approximate the zero(s) of the function.

Question

Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations differ by less than 0.001. Then find the zero(s) to three decimal places using a graphing utility and compare the results.

f(x) = 2 - x^3

Newton's method:

X=

Graphing utility:

X=

Answer 1

Zero = x1 = x0 -f(x)/f'(x) where x1 is the new approximation and x0 is the old approximation.

f(x)=2-x^3

f'(x) = -3x^2

Start with x = 1

x1 = 1 - f(1)/f'(1) = 1 - (2-1^3)/(-3(1)^2) = 1 - 1/-3 = 1 + 1/3 = 4/3

x2 = 1 - f(4/3)/f'(4/3) = 1.263888888888888...

x3 = 1 - f(1.2638888888...)/f'(2.638888888...) = about 1.2599335 (difference is still more than .001)

x4 = 1-f(1.2559933...)/f'(1.259933...) = about 1.2599211 (difference is less than .001)

Graphed on Desmos the zero is shown at 1.260