Write the formula for Newton's method and use the given initial approximation to compute the approximations $$ x_{1} $$ and $$ x_{2} $$. $$ f(x)=2 e^{-x}-7 x, x_{0}=\ln (4) $$

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**Newton's Method Formula:**

Newton's method is an iterative technique for finding increasingly accurate approximations to the roots (or zeroes) of a real-valued function. The formula for Newton's method is:

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

**Given:**
$$
f(x) = 2e^{-x} - 7x
$$
$$
x_0 = \ln(4) \approx 1.38629436112
$$

**Step 1: Compute $$ f(x_0) $$ and $$ f'(x_0) $$**

1. **Calculate $$ f(x_0) $$:**
   $$
   f(x_0) = 2e^{-x_0} - 7x_0
   $$
   Since $$ x_0 = \ln(4) $$, $$ e^{-x_0} = \frac{1}{4} $$.
   $$
   f(x_0) = 2 \times \frac{1}{4} - 7 \times 1.38629436112 \approx 0.5 - 9.70406052784 \approx -9.20406052784
   $$

2. **Calculate $$ f'(x_0) $$:**
   $$
   f'(x) = \frac{d}{dx}(2e^{-x} - 7x) = -2e^{-x} - 7
   $$
   $$
   f'(x_0) = -2 \times \frac{1}{4} - 7 = -0.5 - 7 = -7.5
   $$

**Step 2: Compute $$ x_1 $$**

Using Newton's formula:
$$
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.38629436112 - \frac{-9.20406052784}{-7.5} = 1.38629436112 - 1.22600807 \approx 0.160286291
$$

**Step 3: Compute $$ f(x_1) $$ and $$ f'(x_1) $$**

1. **Calculate $$ f(x_1) $$:**
   $$
   f(x_1) = 2e^{-x_1} - 7x_1
   $$
   $$
   e^{-0.160286291} \approx 0.85192158
   $$
   $$
   f(x_1) = 2 \times 0.85192158 - 7 \times 0.160286291 \approx 1.70384316 - 1.12200404 \approx 0.58183912
   $$

2. **Calculate $$ f'(x_1) $$:**
   $$
   f'(x_1) = -2e^{-x_1} - 7 \approx -2 \times 0.85192158 - 7 = -1.70384316 - 7 = -8.70384316
   $$

**Step 4: Compute $$ x_2 $$**

Using Newton's formula:
$$
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.160286291 - \frac{0.58183912}{-8.70384316} \approx 0.160286291 + 0.066849 \approx 0.227135
$$

**Final Approximations:**
$$
x_1 \approx 0.1603
$$
$$
x_2 \approx 0.2271
$$

**Summary:**
Using Newton's method with the initial approximation $$ x_0 = \ln(4) $$, the first two approximations are:
$$
x_1 \approx 0.1603 \quad \text{and} \quad x_2 \approx 0.2271
$$
Using Newton's method with the initial approximation $$ x_0 = \ln(4) $$, the first two approximations are: $$ x_1 \approx 0.1603 \quad \text{and} \quad x_2 \approx 0.2271 $$

Write the formula for Newton’s method and use the given initial approximation to compute the approximations and .

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