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To solve for the next approximation \( x_1 \) in Newton's method, we use the formula:

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

Given:
- \( f(x) = x^3 + x^2 + 4x - 4 \)
- \( x_0 = 2 \)

### Step 1: Calculate \( f(2) \)
Substitute \( x = 2 \) into \( f(x) \):

\[
f(2) = 2^3 + 2^2 + 4 \cdot 2 - 4 = 8 + 4 + 8 - 4 = 16
\]

### Step 2: Find \( f'(x) \)
The derivative \( f'(x) \) is calculated as follows:

\[
f'(x) = 3x^2 + 2x + 4
\]

Now, calculate \( f'(2) \):

\[
f'(2) = 3 \cdot 2^2 + 2 \cdot 2 + 4 = 3 \cdot 4 + 4 + 4 = 12 + 4 + 4 = 20
\]

### Step 3: Apply Newton's Method Formula
Now substitute \( x_0 = 2 \), \( f(2) = 16 \), and \( f'(2) = 20 \) into the Newton's method formula:

\[
x_1 = 2 - \frac{16}{20} = 2 - 0.8 = 1.2
\]

So, the next approximation \( x_1 \), accurate to two decimal places, is:
\[
\boxed{1.20}
\]

Let me know if you'd like further details on any steps, or if you have questions!

Here are some related questions to explore:

1. How does Newton's method converge for different initial values?
2. What are the potential drawbacks of Newton's method?
3. Can we apply Newton's method to functions with no real roots?
4. What happens if the derivative \( f'(x) \) is zero at an approximation?
5. How does the convergence speed of Newton's method compare with other root-finding methods?

**Tip:** Newton's method works best when the initial guess is close to the actual root, as it can diverge otherwise.

Let f(x) = x^3 + x^2 + 4x - 4 = 0 be given the function. Given the first approximation at x_0 = 2, the value of the next approximation x_1 in Newton's method, which is accurate to two decimal places, is?

Explore how to find the next approximation in Newton's method for the function f(x) = x^3 + x^2 + 4x - 4 with an initial approximation x_0 = 2. Detailed steps include calculating f(2), f'(2), and applying Newton's formula to obtain x_1 accurate to two decimal places.

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