Answer:
Step-by-step explanation:
Given that:
[tex]\mathsf{f(x) = x^2 -5 } \\ \\ \mathsf{x_1 = 2}[/tex]
The derivative of the first function of (x) is:
[tex]\mathsf{f'(x) =2x }[/tex]
According to Newton's Raphson method for function formula:
[tex]{\mathrm{x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}}[/tex]
where;
[tex]\mathbf{x_1 =2}[/tex]
The first iteration is as follows:
[tex]\mathtt{f(x_1) = (2)^2 - 5} \\ \\ \mathbf{f(x_1) = -1}[/tex]
[tex]\mathtt{f'(x_1) = 2(2)} \\ \\ \mathbf{ = 4}[/tex]
[tex]\mathtt{\dfrac{f(x_1)}{f'(x_1)}} = \dfrac{-1}{4}}[/tex]
[tex]\mathbf{\dfrac{f(x_1)}{f'(x_1)} =-0.25}[/tex]
[tex]\mathtt{x_1 - \dfrac{f(x_1)}{f'(x_1)}} = \mathtt{2 - (-0.25)}}[/tex]
[tex]\mathbf{x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2.25}[/tex]
Therefore;
[tex]\mathbf{x_2 = 2.25}[/tex]
For the second iteration;
[tex]\mathtt f(x_2) = (2.25)^2 -5}[/tex]
[tex]\mathtt f(x_2) = 5.0625-5}[/tex]
[tex]\mathbf{ f(x_2) =0.0625}[/tex]
[tex]\mathtt{f'(x_2)= 2(2.25)}[/tex]
[tex]\mathbf{f'(x_2)= 4.5}[/tex]
[tex]\mathtt{ \dfrac{f(x_2)}{f'(x_2)}} = \dfrac{0.0625}{4.5}}[/tex]
[tex]\mathbf{ \dfrac{f(x_2)}{f'(x_2)} = 0.01389}[/tex]
[tex]\mathtt{x_2 - \dfrac{f(x_2)}{f'(x_2)}} = \mathtt{2.25 -0.01389}}[/tex]
[tex]\mathbf{x_2 - \dfrac{f(x_2)}{f'(x_2)} = 2.2361}}[/tex]
Therefore, [tex]\mathbf{x_3 = 2.2361}[/tex]
