Use newtons method to approximate zeros

Question

Use Newton's Method to approximate the zeros of the function  to 3 decimal places and to approximate the zeros of the function  to 3 decimal places.

Answer 1

For Newton's method, you can use the equation:

1) Find the zeros of f(x) = x³ + x - 1

Taking the derivative: f '(x) = 3x² + 1

Let's start with x₀ = 0 and calculate x₁:

x₁ = 0 - f(0)/f '(0) = 0 - -1/1 = 1

Now, calculate x₂:

x₂ = 1 - f(1)/f '(1) = 1 - 1/4 = 0.75

Now calculate x₃:

x₃ = 0.75 - f(0.75)/f '(0.75) = 0.75 - 0.171875/2.6875 = 0.75 - 0.0639534884 = 0.6860465116

Now calculate x₄:

x₄ = 0.6860465116 - f(0.6860465116)/f '(0.6860465116) = 0.6860465116 - 0.0089410366/2.411979448 = 0.6860465116 - 0.003706929 = 0.6823395826

Now calculate x₅:

x₅ = 0.6823395826 - f(0.6823395826)/f '(0.6823395826) = 0.6823395826 - 0.000028230628/2.396761918 = 0.6823395826 - 0.00001177865348 = 0.6823278039

Notice that the 3rd decimal point did not change between x₄ and x₅ so our answer is 0.682

Repeat this process on the 2nd problem.

I'd begin with x₀ = -1

f '(x) = 3x²

Calculate x₁:

x₁ = -1 - f(-1)/f '(-1) = -1 - 2/3 = -5/3

Calculate x₂:

x₂ = -5/3 - f(-5/3)/f '(-5/3) = -5/3 - -1.62962963/8.3333333333 = -5/3 - -0.19555555556 = -1.471111111

Calculate x₃:

x₃ = -1.471111111 - f(-1.471111111)/f '(-1.471111111) = -1.471111111 - -0.1837314451/6.492503703 = 1.471111111 - -0.0282990128 = -1.442812098

Calculate x₄:

x₄ = -1.442812098 - f(-1.442812098)/f '(-1.442812098) = -1.442812098 - -0.0035116836/6.24512025 = -1.442812098 - -0.0000562308404 = -1.44224979

Since x₃ rounds to -1.443 and x₄ rounds to -1.442, let's do x₅

Calculate x₅:

x₅ = -1.44224979 - f(-1.44224979)/f '(-1.44224979) = -1.44224979 - -0.0000013709372/6.24025337 = -1.44224979 - -0.000000219692554 = -1.44224957

So our answer is -1.442