Let's work through the problem using Newton's Method. The given function is:

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

Step 1: Newton's Method Formula

Newton's method uses the following formula to iteratively find the roots of a function:

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

We need to find the derivative $f'(x)$ of the function $f(x)$.

Step 2: Derivative of the Function

To apply Newton's method, we first calculate the derivative of $f(x)$:

$f'(x) = 3x^2 - 6x + 1$

Step 3: Iterations

Starting with $x_0 = 1$, we now compute the next values using the Newton's method formula.

1. Iteration 1:

   • $x_0 = 1$
   • $f(x_0) = 1^3 - 3(1^2) + 1 + 3 = 1 - 3 + 1 + 3 = 2$
   • $f'(x_0) = 3(1^2) - 6(1) + 1 = 3 - 6 + 1 = -2$

   $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{2}{-2} = 1 + 1 = 2$

   So, $x_1 = 2$ (which is already correct).

2. Iteration 2:

   • $x_1 = 2$
   • $f(x_1) = 2^3 - 3(2^2) + 2 + 3 = 8 - 12 + 2 + 3 = 1$
   • $f'(x_1) = 3(2^2) - 6(2) + 1 = 12 - 12 + 1 = 1$

   $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{1}{1} = 2 - 1 = 1$

   So, $x_2 = 1$.

3. Iteration 3:

   • $x_2 = 1$ (repeats as we calculated before)
   • We already know that repeating the previous process gives $x_3 = 2$.

4. Iteration 4 and 5:

   • These values will also cycle between 1 and 2.

Conclusion:

Thus, the answers for the red boxes are:

• $x_2 = 1$
• $x_3 = 2$
• $x_4 = 1$
• $x_5 = 2$

Let me know if you'd like further clarifications or have any questions!

Here are 5 related questions to deepen your understanding:

1. Why does Newton's Method fail when $f'(x_n) = 0$?
2. Can Newton's Method be applied to all types of functions? Why or why not?
3. How can the initial guess $x_0$ affect the convergence of Newton's Method?
4. How would Newton's Method behave if the function has multiple roots?
5. What are some alternatives to Newton's Method for finding roots of functions?

Tip: For functions where Newton's Method fails to converge, try using methods like the Secant Method or Bisection Method.