Question

The equation f(x) is given as x2_4=0. Considering the initial approximation at

Answer 1

Answer:

The value of $x_{1}$ is given by $\frac{10}{3}$. 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 $f(x) = 0$. The expression is now described:

$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}}$

Where:

$x_{n}$ - Current approximation.

$x_{n+1}$ - New approximation.

$f(x_{n})$ - Function evaluated in current approximation.

$f'(x_{n})$ - First derivative of the function evaluated in current approximation.

If $f(x) = x^{2} - 4$, then $f'(x) = 2\cdot x$. Now, given that $x_{0} = 6$, the function and first derivative evaluated in $x_{o}$ are:

$f(x_{o}) = 6^{2} - 4$

$f(x_{o}) = 32$

$f'(x_{o})= 2 \cdot 6$

$f'(x_{o}) = 12$

$x_{1} = x_{o} - \frac{f(x_{o})}{f'(x_{o})}$

$x_{1} = 6 - \frac{32}{12}$

$x_{1} = 6 - \frac{8}{3}$

$x_{1} = \frac{18-8}{3}$

$x_{1} = \frac{10}{3}$

The value of $x_{1}$ is given by $\frac{10}{3}$. Hence, the answer is A.