Answer:
The first and second iteration of Newton's Method are 3 and [tex]\frac{11}{6}[/tex].
Step-by-step explanation:
The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form [tex]f(x) = 0[/tex] based on the following formula:
[tex]x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}[/tex]
Where:
[tex]x_{i}[/tex] - i-th Approximation, dimensionless.
[tex]x_{i+1}[/tex] - (i+1)-th Approximation, dimensionless.
[tex]f(x_{i})[/tex] - Function evaluated at i-th Approximation, dimensionless.
[tex]f'(x_{i})[/tex] - First derivative evaluated at (i+1)-th Approximation, dimensionless.
Let be [tex]f(x) = x^{2}-8[/tex] and [tex]f'(x) = 2\cdot x[/tex], the resultant expression is:
[tex]x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}[/tex]
First iteration: ([tex]x_{1} = 2[/tex])
[tex]x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}[/tex]
[tex]x_{2} = 2 + \frac{4}{4}[/tex]
[tex]x_{2} = 3[/tex]
Second iteration: ([tex]x_{2} = 3[/tex])
[tex]x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}[/tex]
[tex]x_{3} = 2 - \frac{1}{6}[/tex]
[tex]x_{3} = \frac{11}{6}[/tex]
