Question

How do you use Newton's method to find the approximate solution to the equation `x^3-10x+4=0, 0<x<1`?

Answer 1

`x ~~.4067`

Explanation:

Given: `x^3 - 10x + 4 = 0`.

Problem: Use Newton's method to find the approximate zero `c`, where `f(c) = 0`.

Process:
The function must be differentiable between `0 < x < 1`

1. Make an initial estimate `x`: Let `x_1 = 0.5`

2. Determine a new approximation using `x_(n+1) = x_n - (f(x_n))/(f'(x_n))`

3. If `|x_n - x_(n+1)|` is within the desired accuracy, let `x_(n+1)`approximate `c`.

4. Iterate until you get the desired accuracy.

Find the first derivative of the function: `f' (x) = 3x^2 - 10`

TIP: The easiest way to make the least amount of mistakes is to graph both functions and use your calculator to get the values. Always round the numbers to the accuracy value, e.g. `.0001`

Create a table:

`n|" "x_n" "|f(x_n)" "|f'(x_n)" "|(f(x_n))/(f'(x_n))" "|x_n - (f(x_n))/(f'(x_n))|`

`1|" "0.5000|-0.8750|-9.2500|" "0.9459|" "0.5410|`

`2|" "0.5410|-1.2517|-9.1220|" "0.1372|" "0.4038|`

`3|" "0.4038|" "0.2784|-9.5108|-0.0293|" "0.4331|`

`4|" "0.4331|-0.2498|-9.4373|" "0.0265|" "0.4066|`

`5|" "0.4066|" "0.0012|-9.5040|-0.0001|" "0.4067|`

`6|" "0.4067|" "0.0003|-9.5038|-0.0000|" "0.4067|`

The zero with `.0001` accuracy is `0.4067`