To approximate the solution to the equation $f(x) = x^3 + x^2 - 2x + 4 = 0$ using Newton's Method, we follow these steps:

Step 1: Define the function and its derivative

The equation we are working with is:

$f(x) = x^3 + x^2 - 2x + 4$

To apply Newton's Method, we also need the derivative of this function. The derivative $f'(x)$ is:

$f'(x) = 3x^2 + 2x - 2$

Step 2: Newton's Method formula

Newton's Method provides a recursive way to approximate the root of a function. The formula for the next approximation is:

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

where $x_n$ is the current approximation and $x_{n+1}$ is the next approximation.

Step 3: Apply the formula iteratively

We are given the initial guess $x_0 = -2.5$. Now, we apply Newton's Method iteratively:

Iteration 1:

1. Calculate $f(x_0)$: $f(x_0) = (-2.5)^3 + (-2.5)^2 - 2(-2.5) + 4$ $f(x_0) = -15.625 + 6.25 + 5 + 4 = -0.375$

2. Calculate $f'(x_0)$: $f'(x_0) = 3(-2.5)^2 + 2(-2.5) - 2$ $f'(x_0) = 3(6.25) - 5 - 2 = 18.75 - 5 - 2 = 11.75$

3. Apply the Newton's Method formula: $x_1 = -2.5 - \frac{-0.375}{11.75} = -2.5 + 0.03191 = -2.46809$

Iteration 2:

1. Calculate $f(x_1)$: $f(x_1) = (-2.46809)^3 + (-2.46809)^2 - 2(-2.46809) + 4$ $f(x_1) \approx -15.01818 + 6.08657 + 4.93618 + 4 = 0.00457$

2. Calculate $f'(x_1)$: $f'(x_1) = 3(-2.46809)^2 + 2(-2.46809) - 2$ $f'(x_1) \approx 3(6.087) - 4.93618 - 2 = 18.261 - 4.93618 - 2 = 11.32482$

3. Apply the Newton's Method formula: $x_2 = -2.46809 - \frac{0.00457}{11.32482} = -2.46809 - 0.00040 = -2.46849$

Iteration 3:

1. Calculate $f(x_2)$: $f(x_2) = (-2.46849)^3 + (-2.46849)^2 - 2(-2.46849) + 4$ $f(x_2) \approx -15.01885 + 6.08745 + 4.93698 + 4 = 0.00000$

Since the value of $f(x_2)$ is very close to zero, the approximation has converged.

Final Answer:

The root of the equation, accurate to five decimal places, is approximately:

$x \approx -2.46849$

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Would you like any additional details or have any further questions? Here are some related questions you might find interesting:

1. How do you determine the convergence rate of Newton's Method for this equation?
2. What happens if the initial guess is too far from the true root?
3. Can Newton's Method fail for some functions? If so, in what situations?
4. How can you apply Newton's Method to solve nonlinear equations in multiple variables?
5. What is the geometrical interpretation of Newton's Method?

Tip: Always check the behavior of the function and its derivative before starting iterations to avoid division by zero or slow convergence.