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!