Answer:
[tex]x_{2} = 0.0000[/tex]
Step-by-step explanation:
The formula for the Newton's method is:
[tex]x_{i+1} = x_{i} + \frac{f(x_{i})}{f'(x_{i})}[/tex]
Where [tex]f' (x_{i})[/tex] is the first derivative of the function evaluated in [tex]x_{i}[/tex].
[tex]x_{i+1} = x_{i} + \frac{x_{i}^{4}-x_{i}-3}{4\cdot x_{i}^{3}-1}[/tex]
Lastly, the value of [tex]x_{2}[/tex] is determined by replacing [tex]x_{1}[/tex] with its numerical value:
[tex]x_{2} = x_{1} + \frac{x_{1}^{4}-x_{1}-3}{4\cdot x_{1}^{3}-1}[/tex]
[tex]x_{2} = 1.0000 + \frac{1.0000^{4}-1.0000-3}{4\cdot (1.0000)^{3}-1}[/tex]
[tex]x_{2} = 0.0000[/tex]
