Answer:
The value of [tex]x_{1}[/tex] is given by [tex]\frac{10}{3}[/tex]. Hence, the answer is A.
Step-by-step explanation:
This exercise represents a case where the Newton-Raphson method is used, whose formula is used for differentiable function of the form [tex]f(x) = 0[/tex]. The expression is now described:
[tex]x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}}[/tex]
Where:
[tex]x_{n}[/tex] - Current approximation.
[tex]x_{n+1}[/tex] - New approximation.
[tex]f(x_{n})[/tex] - Function evaluated in current approximation.
[tex]f'(x_{n})[/tex] - First derivative of the function evaluated in current approximation.
If [tex]f(x) = x^{2} - 4[/tex], then [tex]f'(x) = 2\cdot x[/tex]. Now, given that [tex]x_{0} = 6[/tex], the function and first derivative evaluated in [tex]x_{o}[/tex] are:
[tex]f(x_{o}) = 6^{2} - 4[/tex]
[tex]f(x_{o}) = 32[/tex]
[tex]f'(x_{o})= 2 \cdot 6[/tex]
[tex]f'(x_{o}) = 12[/tex]
[tex]x_{1} = x_{o} - \frac{f(x_{o})}{f'(x_{o})}[/tex]
[tex]x_{1} = 6 - \frac{32}{12}[/tex]
[tex]x_{1} = 6 - \frac{8}{3}[/tex]
[tex]x_{1} = \frac{18-8}{3}[/tex]
[tex]x_{1} = \frac{10}{3}[/tex]
The value of [tex]x_{1}[/tex] is given by [tex]\frac{10}{3}[/tex]. Hence, the answer is A.
