Complete two iterations of Newton's Method for the function using the initial guess . Round all numerical values in your answer to four decimal places.
A)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 1.0000 & 0.0068 & -3.6021 & 4.9021 \
\hline 2 & 4.9021 & 0.2675 & -0.2675 & 0.0068 & 4.8953 \
\hline
\end{array}$

B)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.2675 & -0.9636 & -0.2776 & 1.5776 \
\hline 2 & 1.5776 & 0.9636 & 0.0068 & -0.2776 & 1.8552 \
\hline
\end{array}$

C)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.9636 & -0.2675 & -3.6021 & 4.9021 \
\hline 2 & 4.9021 & 1.0000 & 0.0068 & 146.6399 & -141.7378 \
\hline
\end{array}$

D)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.2675 & -0.9636 & -0.2776 & 1.5776 \
\hline 2 & 1.5776 & -0.0068 & -1.0000 & 0.0068 & 1.5708 \
\hline
\end{array}$

E)
$\begin{array}{l}
\begin{array} { | c | c | c | c | c | c | }
\hline n & x _ { n } & f \left( x _ { n } \right) & f ^ { \prime } \left( x _ { n } \right) & \frac { f \left( x _ { n } \right) } { f ^ { \prime } \left( x _ { n } \right) } & x _ { n } - \frac { f \left( x _ { n } \right) } { f ^ { \prime } \left( x _ { n } \right) } \
\hline 1 & 1.3 & 0.9636 & - 0.9636 & 0.0068 & 1.2932 \
\hline 2 & 1.2932 & 1.0000 & - 1.0000 & - 3.6021 & 4.8953 \
\hline
\end{array}
\end{array}$

Question

Complete two iterations of Newton's Method for the function using the initial guess . Round all numerical values in your answer to four decimal places. 
A)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 1.0000 & 0.0068 & -3.6021 & 4.9021 \
\hline 2 & 4.9021 & 0.2675 & -0.2675 & 0.0068 & 4.8953 \
\hline
\end{array}$

B)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.2675 & -0.9636 & -0.2776 & 1.5776 \
\hline 2 & 1.5776 & 0.9636 & 0.0068 & -0.2776 & 1.8552 \
\hline
\end{array}$

C)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.9636 & -0.2675 & -3.6021 & 4.9021 \
\hline 2 & 4.9021 & 1.0000 & 0.0068 & 146.6399 & -141.7378 \
\hline
\end{array}$

D)
$\begin{array}{|c|c|c|c|c|c|}
\hline n & x_{n} & f\left(x_{n}\right) & f^{\prime}\left(x_{n}\right) & \frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} & x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \
\hline 1 & 1.3 & 0.2675 & -0.9636 & -0.2776 & 1.5776 \
\hline 2 & 1.5776 & -0.0068 & -1.0000 & 0.0068 & 1.5708 \
\hline
\end{array}$

E)
$\begin{array}{l}
\begin{array} { | c | c | c | c | c | c | } 
\hline n & x _ { n } & f \left( x _ { n } \right) & f ^ { \prime } \left( x _ { n } \right) & \frac { f \left( x _ { n } \right) } { f ^ { \prime } \left( x _ { n } \right) } & x _ { n } - \frac { f \left( x _ { n } \right) } { f ^ { \prime } \left( x _ { n } \right) } \
\hline 1 & 1.3 & 0.9636 & - 0.9636 & 0.0068 & 1.2932 \
\hline 2 & 1.2932 & 1.0000 & - 1.0000 & - 3.6021 & 4.8953 \
\hline
\end{array}
\end{array}$

Quizplus · Accepted Answer

Newton's Method involves iterating the formula $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. The correct table must show the application of this formula with consistent values for $f(x_n)$ and $f'(x_n)$ leading to the correct update of $x_n$. Option D correctly applies Newton's Method with consistent updates and rounding.

Complete Two Iterations of Newton's Method for the Function Using