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This problem requires using **Newton's Method** to approximate the root of the given function \( f(x) = 6x^2 - \frac{1}{x} \).

### Newton's Method Formula:
Newton's Method formula for approximating roots is:

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

Where:
- \( x_n \) is the current approximation,
- \( x_{n+1} \) is the next approximation,
- \( f(x) \) is the function,
- \( f'(x) \) is the derivative of the function.

### Step 1: Compute the derivative of \( f(x) \)
Given \( f(x) = 6x^2 - \frac{1}{x} \), we can calculate the derivative:

\[
f'(x) = \frac{d}{dx} \left( 6x^2 - \frac{1}{x} \right) = 12x + \frac{1}{x^2}
\]

### Step 2: Apply Newton's Method for the first iteration
We are given \( x_1 = 1 \).

1. **Compute \( f(x_1) \)**:
\[
f(1) = 6(1)^2 - \frac{1}{1} = 6 - 1 = 5
\]

2. **Compute \( f'(x_1) \)**:
\[
f'(1) = 12(1) + \frac{1}{(1)^2} = 12 + 1 = 13
\]

3. **Calculate \( x_2 \)**:
\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1 - \frac{5}{13} = 1 - 0.384615 = 0.615385
\]

### Step 3: Apply Newton's Method for the second iteration
Now we use \( x_2 = 0.615385 \) to find \( x_3 \).

1. **Compute \( f(x_2) \)**:
\[
f(0.615385) = 6(0.615385)^2 - \frac{1}{0.615385} \approx 6(0.378553) - 1.625 \approx 2.27132 - 1.625 = 0.64632
\]

2. **Compute \( f'(x_2) \)**:
\[
f'(0.615385) = 12(0.615385) + \frac{1}{(0.615385)^2} \approx 7.38462 + 2.6376 \approx 10.0222
\]

3. **Calculate \( x_3 \)**:
\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.615385 - \frac{0.64632}{10.0222} \approx 0.615385 - 0.0645 = 0.55087
\]

### Final Answer:
The value of \( x_3 \) is approximately **0.55087**.

---

Would you like a detailed explanation of any specific step, or do you have any further questions?

Here are 5 related questions to extend your understanding:
1. How does Newton's Method converge to the root of a function?
2. What are the conditions for Newton's Method to work effectively?
3. Can you always rely on Newton's Method to find a root?
4. What happens if you choose a bad initial approximation for Newton's Method?
5. How does Newton's Method differ from other root-finding methods like the Bisection Method?

**Tip:** Newton's Method requires a good initial guess for fast convergence; otherwise, it may diverge or fail.

You wish to use Newton's Method to approximate the root of f(x) = 6x^2 - 1/x. If you begin with an initial value of x1 = 1, what will be the value of the second iteration of Newton's Method? That is, what will be the value of x3?

Use Newton's Method to approximate the root of f(x) = 6x^2 - 1/x, starting with an initial value of x1 = 1. This step-by-step solution includes the derivative calculations and Newton's Method formula to find the second iteration x3, yielding an approximation of 0.55087.

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