Newton's method is an iterative method to approximate the roots of a function. It is defined as:

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

Given Function:

$f(x) = x - x^2 + 8$

Step 1: Find $f'(x)$:

$f'(x) = 1 - 2x$

For $x_0 = -2$:

1. Calculate $f(x_0)$ and $f'(x_0)$: $f(-2) = -2 - (-2)^2 + 8 = -2 - 4 + 8 = 2$ $f'(-2) = 1 - 2(-2) = 1 + 4 = 5$

2. Compute $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -2 - \frac{2}{5} = -2 - 0.4 = -2.4$

3. Calculate $f(x_1)$ and $f'(x_1)$: $f(-2.4) = -2.4 - (-2.4)^2 + 8 = -2.4 - 5.76 + 8 = -0.16$ $f'(-2.4) = 1 - 2(-2.4) = 1 + 4.8 = 5.8$

4. Compute $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -2.4 - \frac{-0.16}{5.8} = -2.4 + 0.027586 = -2.372414$

Thus, for $x_0 = -2$, $x_2 \approx -2.372$.

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For $x_0 = 3$:

1. Calculate $f(x_0)$ and $f'(x_0)$: $f(3) = 3 - 3^2 + 8 = 3 - 9 + 8 = 2$ $f'(3) = 1 - 2(3) = 1 - 6 = -5$

2. Compute $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{2}{-5} = 3 + 0.4 = 3.4$

3. Calculate $f(x_1)$ and $f'(x_1)$: $f(3.4) = 3.4 - (3.4)^2 + 8 = 3.4 - 11.56 + 8 = -0.16$ $f'(3.4) = 1 - 2(3.4) = 1 - 6.8 = -5.8$

4. Compute $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 3.4 - \frac{-0.16}{-5.8} = 3.4 - 0.027586 = 3.372414$

Thus, for $x_0 = 3$, $x_2 \approx 3.372$.

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Final Answers:

1. For $x_0 = -2$, $x_2 \approx -2.372$.
2. For $x_0 = 3$, $x_2 \approx 3.372$.

Would you like a detailed explanation of the iterations?
Here are 5 related questions:

1. What is the significance of Newton's method in finding roots?
2. Can Newton's method fail or converge slowly? When?
3. What happens if we use a different starting value for this problem?
4. How can we improve the accuracy of $x_2$ further?
5. How does Newton's method compare to other root-finding methods like the bisection method?

Tip: Always check $f'(x) \neq 0$ when using Newton's method to avoid division by zero!