Questions: Use a calculator or program to compute the first 10 iterations of Newton's method for the given function and initial approximation. f(x)=4 sin x+5 x-1, x0=1.4 Complete the table. (Do not round until the final answer. Then round to six decimal places as needed.) k xk k xk 1 6 2 7 3 8 4 9 5 10

Use a calculator or program to compute the first 10 iterations of Newton's method for the given function and initial approximation.
f(x)=4 sin x+5 x-1, x0=1.4

Complete the table.
(Do not round until the final answer. Then round to six decimal places as needed.)

k xk k xk
1 6
2 7
3 8
4 9
5 10

Transcript text: Use a calculator or program to compute the first 10 iterations of Newton's method for the given function and initial approximation. \[ f(x)=4 \sin x+5 x-1, x_{0}=1.4 \] Complete the table. (Do not round until the final answer. Then round to six decimal places as needed.) \begin{tabular}{|c|c|c|c|} \hline $\mathbf{k}$ & $\mathbf{x}_{\mathbf{k}}$ & $\mathbf{k}$ & $\mathbf{x}_{\mathbf{k}}$ \\ \hline 1 & $\square$ & 6 & $\square$ \\ \hline 2 & $\square$ & 7 & $\square$ \\ \hline 3 & $\square$ & 8 & $\square$ \\ \hline 4 & $\square$ & 9 & $\square$ \\ \hline 5 & $\square$ & 10 & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

To solve this problem using Newton's method, we need to iteratively apply the formula for Newton's method:

\[ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \]

where $ f(x) = 4 \sin x + 5x - 1 $. First, we need to compute the derivative $ f'(x) $, which is $ f'(x) = 4 \cos x + 5 $. Starting with the initial approximation $ x_0 = 1.4 $, we will compute the next 10 iterations of $ x_k $ using the formula above.

Step 1: Define the Function and Derivative

We start with the function $ f(x) = 4 \sin x + 5x - 1 $ and its derivative $ f'(x) = 4 \cos x + 5 $.

Step 2: Initial Approximation

The initial approximation is given as $ x_0 = 1.4 $.

Step 3: Apply Newton's Method

Using the formula for Newton's method:

\[ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \]

we compute the first 10 iterations:

For $ k = 1 $: \[ x_1 = 1.4 - \frac{f(1.4)}{f'(1.4)} \approx -0.3504 \]
For $ k = 2 $: \[ x_2 \approx 0.1207 \]
For $ k = 3 $: \[ x_3 \approx 0.1112 \]
For $ k = 4 $: \[ x_4 \approx 0.1112 \]
For $ k = 5 $: \[ x_5 \approx 0.1112 \]
For $ k = 6 $: \[ x_6 \approx 0.1112 \]
For $ k = 7 $: \[ x_7 \approx 0.1112 \]
For $ k = 8 $: \[ x_8 \approx 0.1112 \]
For $ k = 9 $: \[ x_9 \approx 0.1112 \]
For $ k = 10 $: \[ x_{10} \approx 0.1112 \]

Final Answer

The results of the first 10 iterations of Newton's method are as follows:

\[ \begin{align_} x_1 & \approx -0.3504 \\ x_2 & \approx 0.1207 \\ x_3 & \approx 0.1112 \\ x_4 & \approx 0.1112 \\ x_5 & \approx 0.1112 \\ x_6 & \approx 0.1112 \\ x_7 & \approx 0.1112 \\ x_8 & \approx 0.1112 \\ x_9 & \approx 0.1112 \\ x_{10} & \approx 0.1112 \\ \end{align_} \]

Thus, the final boxed answers for the first three iterations are:

\[ \boxed{x_1 \approx -0.3504, \quad x_2 \approx 0.1207, \quad x_3 \approx 0.1112} \]

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