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The problem asks to approximate the zero(s) of the function \( f(x) = x^3 + x + 3 \) using **Newton's Method** accurate to three decimal places.

### Steps for Newton's Method:
Newton's method is an iterative process defined by the formula:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
Where:
- \( f(x) = x^3 + x + 3 \)
- \( f'(x) \) is the derivative of \( f(x) \).

### Step 1: Compute \( f'(x) \)
The derivative of \( f(x) = x^3 + x + 3 \) is:
\[
f'(x) = 3x^2 + 1
\]

### Step 2: Select an Initial Guess \( x_0 \)
We need to choose a reasonable initial guess. Since we are looking for a root where \( f(x) = 0 \), we can examine the graph or use trial and error to find a starting point. Let's try \( x_0 = -1 \) (since \( f(x) = 0 \) typically happens around negative values based on the function structure).

### Step 3: Apply Newton's Iteration
We now apply Newton's iteration formula to refine the root approximation.

#### Iteration 1:
For \( x_0 = -1 \):
\[
f(x_0) = (-1)^3 + (-1) + 3 = -1 - 1 + 3 = 1
\]
\[
f'(x_0) = 3(-1)^2 + 1 = 3 + 1 = 4
\]
\[
x_1 = -1 - \frac{1}{4} = -1 - 0.25 = -1.25
\]

#### Iteration 2:
For \( x_1 = -1.25 \):
\[
f(x_1) = (-1.25)^3 + (-1.25) + 3 = -1.953125 - 1.25 + 3 = -0.203125
\]
\[
f'(x_1) = 3(-1.25)^2 + 1 = 3(1.5625) + 1 = 4.6875 + 1 = 5.6875
\]
\[
x_2 = -1.25 - \frac{-0.203125}{5.6875} = -1.25 + 0.0357 = -1.2143
\]

#### Iteration 3:
For \( x_2 = -1.2143 \):
\[
f(x_2) = (-1.2143)^3 + (-1.2143) + 3 = -1.7906 - 1.2143 + 3 = -0.0049
\]
\[
f'(x_2) = 3(-1.2143)^2 + 1 = 3(1.4746) + 1 = 4.4238 + 1 = 5.4238
\]
\[
x_3 = -1.2143 - \frac{-0.0049}{5.4238} = -1.2143 + 0.0009 = -1.2134
\]

#### Iteration 4:
For \( x_3 = -1.2134 \):
\[
f(x_3) = (-1.2134)^3 + (-1.2134) + 3 = -1.7844 - 1.2134 + 3 = -0.0000
\]
\[
f'(x_3) = 3(-1.2134)^2 + 1 = 5.4203
\]
\[
x_4 = -1.2134
\]

The process converges, so the approximate zero is \( x = -1.213 \) to three decimal places.

### Final Answer:
The zero of \( f(x) = x^3 + x + 3 \) is approximately \( x = -1.213 \).

---

Do you want further clarification or details on this method? 

Here are 5 related questions for deeper exploration:
1. What happens if you choose a different initial guess for Newton's Method?
2. How does the convergence of Newton's Method depend on the choice of the initial guess?
3. Can Newton's Method fail, and if so, under what conditions?
4. What are other numerical methods for finding roots besides Newton's Method?
5. How is Newton's Method applied in higher dimensions for systems of equations?

### Tip:
Always graph the function before using Newton's Method to get a better idea of where the roots might be located.

Use Newton's Method to approximate the zero(s) of the function f(x) = x^3 + x + 3 accurate to three decimal places.

Use Newton's Method to approximate the zero(s) of the function f(x) = x^3 + x + 3 to three decimal places. This problem involves calculating the derivative and applying Newton's iterative formula. Ideal for students in undergraduate calculus or high school AP calculus.

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