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Let's solve this step-by-step using Newton’s Method. 

### Problem:
Given the function:
\[
f(x) = 6x^2 - \frac{1}{x}
\]
You need to find the value of \(x_3\) (the second iteration) using Newton's method, starting from an initial value of \(x_1 = 1\). Newton's formula is:

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

---

### Step 1: Compute the derivative \(f'(x)\)
Using basic differentiation rules:

\[
f'(x) = 12x + \frac{1}{x^2}
\]

---

### Step 2: Write down Newton’s iteration formula
\[
x_{n+1} = x_n - \frac{6x_n^2 - \frac{1}{x_n}}{12x_n + \frac{1}{x_n^2}}
\]

---

### Step 3: First iteration \(x_2\)
Using the initial value \(x_1 = 1\):

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

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

3. Use the Newton formula:
   \[
   x_2 = 1 - \frac{5}{13} = 1 - 0.3846 \approx 0.6154
   \]

---

### Step 4: Second iteration \(x_3\)
Now, use \(x_2 \approx 0.6154\) to find \(x_3\).

1. Compute \(f(0.6154)\):
   \[
   f(0.6154) = 6(0.6154)^2 - \frac{1}{0.6154}
   \]
   \[
   (0.6154)^2 \approx 0.3787, \quad 6 \cdot 0.3787 \approx 2.272
   \]
   \[
   \frac{1}{0.6154} \approx 1.625 \quad \Rightarrow f(0.6154) \approx 2.272 - 1.625 = 0.647
   \]

2. Compute \(f'(0.6154)\):
   \[
   f'(0.6154) = 12(0.6154) + \frac{1}{(0.6154)^2}
   \]
   \[
   12 \cdot 0.6154 \approx 7.385, \quad (0.6154)^2 \approx 0.3787 \quad \Rightarrow \frac{1}{0.3787} \approx 2.641
   \]
   \[
   f'(0.6154) \approx 7.385 + 2.641 = 10.026
   \]

3. Use the Newton formula:
   \[
   x_3 = 0.6154 - \frac{0.647}{10.026} \approx 0.6154 - 0.0645 \approx 0.5509
   \]

---

### Final Answer:
The value of \(x_3\) (the second iteration) is approximately:
\[
x_3 \approx 0.5509
\]

---

Do you have any further questions or need more details?

---

### Related Questions:
1. What is the convergence criterion for Newton's method?
2. How do we know if Newton's method will converge for this function?
3. What happens if the derivative \(f'(x_n)\) is close to zero during the iteration?
4. Can Newton's method be modified for systems of nonlinear equations?
5. How does the choice of the initial value \(x_1\) affect convergence?

**Tip:** If Newton’s method is slow to converge, consider using an alternative like the Secant method!

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?

Learn how to apply Newton's Method to approximate the root of f(x) = 6x^2 - 1/x, starting with an initial guess of x1 = 1. This step-by-step solution includes calculations for the first and second iterations, showing the value of x3 as approximately 0.5509. Suitable for calculus students.

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