Solve Newton's method for x^3 - 4x + 1 = 0 from x = 2 Step by Step

Step by step

The question

Question: estimate a root of x³ − 4x + 1 = 0 with Newton's method, starting at x = 2.

x³ − 4x + 1 = 0 =

1 The formula

Newton's idea: draw the tangent line at your guess.

Where it meets the x-axis is the next guess.

That gives this formula.

x_n+1 = x_n − f(x_n)f′(x_n)

See it on the graph.

Start at x₀ = 2.

Follow the tangent line down to the x-axis.

It lands at x₁: much closer to the root.

2 Find f′(x)

The slope: f′(x) = 3x² − 4, by the power rule.

f′(x) = 3x² − 4

Careful!

f goes on top, and f′ at the bottom.

Not the other way round.

Careful: not x₀ − f′(x₀)f(x₀) ✗

3 Round 1

f at 2 is 1.

f(2) = 2³ − 4(2) + 1 = 8 − 8 + 1 = 1

f′ at 2 is 8.

f′(2) = 3(2)² − 4 = 12 − 4 = 8

So x₁ is 2 minus 1/8: 1.875.

x₁ = 2 − 18 = 1.875

4 Round 2

f at 1.875 is about 0.0918.

f(1.875) ≈ 0.0918

f′ at 1.875 is about 6.54688.

f′(1.875) ≈ 6.54688

So x₂ is 1.875 minus 0.0918/6.54688: about 1.86098.

x₂ = 1.875 − 0.09186.54688 ≈ 1.86098

After 2 rounds, the estimate is 1.86098.

root ≈ 1.86098

5 Check

Let's check: put 1.86098 into f.

We get about 0.001103: almost 0.

Right!

f(1.86098) ≈ 0.001103

Let's look back.

First: the formula.

Then: find f′(x).

Then: round 1.

Then: round 2.

Last: check.

Answer