Question

Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal places.) f(x) = x2 − 5, x1 = 2n xn f(xn) f '(xn) f(xn)/f '(xn) xn − f(xn)/f '(xn)1 2

Answer 1

Answer:

Step-by-step explanation:

Given that:

$\mathsf{f(x) = x^2 -5 } \\ \\ \mathsf{x_1 = 2}$

The derivative of the first function of (x) is:

$\mathsf{f'(x) =2x }$

According to Newton's Raphson method for function formula:

${\mathrm{x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}}$

where;

$\mathbf{x_1 =2}$

The first iteration is as follows:

$\mathtt{f(x_1) = (2)^2 - 5} \\ \\ \mathbf{f(x_1) = -1}$

$\mathtt{f'(x_1) = 2(2)} \\ \\ \mathbf{ = 4}$

$\mathtt{\dfrac{f(x_1)}{f'(x_1)}} = \dfrac{-1}{4}}$

$\mathbf{\dfrac{f(x_1)}{f'(x_1)} =-0.25}$

$\mathtt{x_1 - \dfrac{f(x_1)}{f'(x_1)}} = \mathtt{2 - (-0.25)}}$

$\mathbf{x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2.25}$

Therefore;

$\mathbf{x_2 = 2.25}$

For the second iteration;

$\mathtt f(x_2) = (2.25)^2 -5}$

$\mathtt f(x_2) = 5.0625-5}$

$\mathbf{ f(x_2) =0.0625}$

$\mathtt{f'(x_2)= 2(2.25)}$

$\mathbf{f'(x_2)= 4.5}$

$\mathtt{ \dfrac{f(x_2)}{f'(x_2)}} = \dfrac{0.0625}{4.5}}$

$\mathbf{ \dfrac{f(x_2)}{f'(x_2)} = 0.01389}$

$\mathtt{x_2 - \dfrac{f(x_2)}{f'(x_2)}} = \mathtt{2.25 -0.01389}}$

$\mathbf{x_2 - \dfrac{f(x_2)}{f'(x_2)} = 2.2361}}$

Therefore, $\mathbf{x_3 = 2.2361}$