We are tasked with applying Newton's Method to approximate $x_3$ for the given function $f(x) = 3x^7 + 5x^4 + 4$. Newton's Method formula is:

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

Step 1: Compute the derivative of $f(x)$

To apply Newton's method, we need both $f(x)$ and its derivative $f'(x)$.

$f(x) = 3x^7 + 5x^4 + 4$ $f'(x) = 21x^6 + 20x^3$

Step 2: Start with $x_1 = 2$

We will perform two iterations to find $x_3$.

First iteration to find $x_2$:

1. Evaluate $f(2)$: $f(2) = 3(2)^7 + 5(2)^4 + 4 = 3(128) + 5(16) + 4 = 384 + 80 + 4 = 468$

2. Evaluate $f'(2)$: $f'(2) = 21(2)^6 + 20(2)^3 = 21(64) + 20(8) = 1344 + 160 = 1504$

3. Apply Newton's method to find $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{468}{1504} \approx 2 - 0.31117 = 1.68883$

Second iteration to find $x_3$:

1. Evaluate $f(1.68883)$: $f(1.68883) = 3(1.68883)^7 + 5(1.68883)^4 + 4 \approx 3(28.826) + 5(8.119) + 4 \approx 86.478 + 40.595 + 4 \approx 131.073$

2. Evaluate $f'(1.68883)$: $f'(1.68883) = 21(1.68883)^6 + 20(1.68883)^3 \approx 21(17.072) + 20(4.817) \approx 357.512 + 96.34 \approx 453.852$

3. Apply Newton's method to find $x_3$: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.68883 - \frac{131.073}{453.852} \approx 1.68883 - 0.28878 = 1.40005$

Final Answer:

The third approximation for the x-intercept is $x_3 \approx 1.40005$.

So the ordered pair is approximately $(1.40005, 0)$.

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. How would Newton's method behave if the derivative of the function at the starting point were very small?
2. Why does Newton's method require the function to be differentiable?
3. Can Newton's method always find the exact root of a function, or does it have limitations?
4. How does the choice of the initial guess $x_1$ affect the convergence of Newton's method?
5. What happens if the function has a horizontal tangent near the root?

Tip: When using Newton's method, if the values of $f'(x)$ are close to zero, the method might fail to converge or take significantly more iterations.