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.