**Use Newton's method to approximate a root of the equation 5x^7 + 3x^4 + 3 = 0 as follows.**

**Let x1 = 2 be the initial approximation.**

**The second approximation x2 is**

**and the third approximation 23**

Newton's method is an iterative process used to approximate the roots of a function. The formula for Newton's method is:

\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]

Given the equation \( f(x) = 5x^7 + 3x^4 + 3 \), we need to find its derivative \( f'(x) \) and use it to calculate the subsequent approximations.

### Step 1: Find the derivative \( f'(x) \)
The function is:
\[
f(x) = 5x^7 + 3x^4 + 3
\]
The derivative \( f'(x) \) is:
\[
f'(x) = 35x^6 + 12x^3
\]

### Step 2: First Approximation (x1 = 2)
Given \( x_1 = 2 \), we need to calculate \( f(2) \) and \( f'(2) \).

\[
f(2) = 5(2)^7 + 3(2)^4 + 3 = 5(128) + 3(16) + 3 = 640 + 48 + 3 = 691
\]

\[
f'(2) = 35(2)^6 + 12(2)^3 = 35(64) + 12(8) = 2240 + 96 = 2336
\]

Now, use Newton's method to find the second approximation \( x_2 \):

\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{691}{2336}
\]

Calculate the value:

\[
x_2 = 2 - 0.2957 \approx 1.7043
\]

### Step 3: Second Approximation (x2 ≈ 1.7043)
Now, using \( x_2 \approx 1.7043 \), we need to calculate \( f(1.7043) \) and \( f'(1.7043) \).

\[
f(1.7043) \approx 5(1.7043)^7 + 3(1.7043)^4 + 3 
\]

\[
f(1.7043) \approx 5(24.27) + 3(8.42) + 3 \approx 121.35 + 25.26 + 3 = 149.61
\]

\[
f'(1.7043) \approx 35(1.7043)^6 + 12(1.7043)^3 
\]
\[
f'(1.7043) \approx 35(14.24) + 12(4.96) \approx 498.40 + 59.52 = 557.92
\]

Now, use Newton's method to find the third approximation \( x_3 \):

\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.7043 - \frac{149.61}{557.92}
\]
\[
x_3 = 1.7043 - 0.2682 \approx 1.4361
\]

### Conclusion
- The second approximation \( x_2 \) is approximately **1.7043**.
- The third approximation \( x_3 \) is approximately **1.4361**.

Would you like further details or have any questions?

Here are some related questions:

1. How does Newton's method work with different initial approximations?
2. Why does Newton's method sometimes fail to converge?
3. Can Newton's method be used to find complex roots?
4. How does the rate of convergence of Newton's method compare to other root-finding methods?
5. What happens if the derivative \( f'(x) \) is zero during the iteration process?
6. Can we apply Newton's method to functions with multiple roots?
7. How does the choice of the initial guess affect the outcome?
8. What is the geometric interpretation of Newton's method?

**Tip:** Always check the behavior of the function near your initial approximation to ensure that Newton's method will converge.

Learn how to approximate a root of the equation 5x^7 + 3x^4 + 3 = 0 using Newton's method. Follow a step-by-step solution with initial approximations and subsequent iterations. Suitable for undergraduate students studying numerical methods.

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