Question

Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=0 as your equation.

Answer 1

Let
`f(x) = x^3 - 2x - 2`. Then differentiating, we get

`f'(x) = 3x^2 - 2`

We approximate
`f(x)` at
`x_1=2` with the tangent line,

`f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18`

The
`x`-intercept for this approximation will be our next approximation for the root,

`10x - 18 = 0 \implies x_2 = \frac95`

Repeat this process. Approximate
`f(x)` at
`x_2 = \frac95`.

`f(x) \approx f(x_2) + f'(x_2) (x-x_2) = (193)/(25)x - (1708)/(125)`

Then

`(193)/(25)x - (1708)/(125) = 0 \implies x_3 = (1708)/(965)`

Once more. Approximate
`f(x)` at
`x_3`.

`f(x) \approx f(x_3) + f'(x_3) (x - x_3) = (6,889,342)/(931,225)x - (11,762,638,074)/(898,632,125)`

Then

`(6,889,342)/(931,225)x - (11,762,638,074)/(898,632,125) = 0 \\\\ \implies x_4 = (5,881,319,037)/(3,324,107,515) \approx 1.769292663 \approx \boxed{1.769293}`

Compare this to the actual root of
`f(x)`, which is approximately 1.769292354, matching up to the first 5 digits after the decimal place.