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,

$10x - 18 = 0 \implies x_2 = \dfrac95$

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

$f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}$

Then

$\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}$

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

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

Then

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

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.