To solve this problem using Newton's Method, we need to iteratively calculate $x_n$ using the formula:

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

Step 1: Define $f(x)$ and $f'(x)$

The function given is:

$f(x) = e^x - 2x - 4$

The derivative is:

$f'(x) = e^x - 2$

Step 2: Start with $x_0 = 0$

We are given the initial guess $x_0 = 0$. Now we compute the next approximations.

Iteration 1: Compute $x_1$

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

Substitute $x_0 = 0$ into $f(x)$ and $f'(x)$:

$f(0) = e^0 - 2(0) - 4 = 1 - 4 = -3$ $f'(0) = e^0 - 2 = 1 - 2 = -1$

$x_1 = 0 - \frac{-3}{-1} = 0 - 3 = -3$

Iteration 2: Compute $x_2$

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

Substitute $x_1 = -3$ into $f(x)$ and $f'(x)$:

$f(-3) = e^{-3} - 2(-3) - 4 = e^{-3} + 6 - 4 = e^{-3} + 2$ $f'(-3) = e^{-3} - 2$

Using a calculator:

$e^{-3} \approx 0.0498$

$f(-3) \approx 0.0498 + 2 = 2.0498$ $f'(-3) \approx 0.0498 - 2 = -1.9502$

$x_2 = -3 - \frac{2.0498}{-1.9502} \approx -3 + 1.0513 = -1.9487$

Iteration 3: Compute $x_3$

$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}$

Substitute $x_2 = -1.9487$ into $f(x)$ and $f'(x)$:

$f(-1.9487) = e^{-1.9487} - 2(-1.9487) - 4$ $f'(-1.9487) = e^{-1.9487} - 2$

Using a calculator:

$e^{-1.9487} \approx 0.1420$

$f(-1.9487) \approx 0.1420 + 2(-1.9487) - 4 = 0.1420 - 3.8974 - 4 \approx -1.7554$ $f'(-1.9487) \approx 0.1420 - 2 \approx -1.858$

$x_3 = -1.9487 - \frac{-1.7554}{-1.858} \approx -1.9487 + 0.9447 = -1.0040$

Final Answers:

$x_1 = -3.00000, \quad x_2 = -1.94870, \quad x_3 = -1.00400$

Would you like detailed explanations for these steps or assistance with other related concepts?

5 Questions to Explore Further:

1. What are the theoretical foundations of Newton's Method?
2. How does the choice of the initial guess ($x_0$) affect convergence?
3. Can Newton's Method fail to converge? If so, under what conditions?
4. How does this method compare with other root-finding algorithms (e.g., bisection)?
5. Can we use Newton's Method for systems of equations? How?

Tip: Always ensure the derivative $f'(x) \neq 0$ at any point during Newton's Method to avoid division by zero.