To apply Newton's Method to estimate the zeros of the function $f(x) = x^4 - 2x - 2$, we use the iterative formula:

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

Step 1: Find the derivative of $f(x)$

The function is $f(x) = x^4 - 2x - 2$. The derivative is:

$f'(x) = 4x^3 - 2$

Step 2: Apply Newton's Method to estimate $x_2$ for $x_0 = -1$

1. Starting with $x_0 = -1$:

   • $f(-1) = (-1)^4 - 2(-1) - 2 = 1 + 2 - 2 = 1$
   • $f'(-1) = 4(-1)^3 - 2 = -4 - 2 = -6$

2. Next, calculate $x_1$: $x_1 = -1 - \frac{f(-1)}{f'(-1)} = -1 - \frac{1}{-6} = -1 + \frac{1}{6} = -\frac{6}{6} + \frac{1}{6} = -\frac{5}{6}$

3. Now, calculate $x_2$:

   • $f\left(-\frac{5}{6}\right) = \left(-\frac{5}{6}\right)^4 - 2\left(-\frac{5}{6}\right) - 2$
   • $f'\left(-\frac{5}{6}\right) = 4\left(-\frac{5}{6}\right)^3 - 2$

We'll compute this exactly in a moment:

Step 3: Apply Newton's Method to estimate $x_2$ for $x_0 = 1$

1. Starting with $x_0 = 1$:

   • $f(1) = 1^4 - 2(1) - 2 = 1 - 2 - 2 = -3$
   • $f'(1) = 4(1)^3 - 2 = 4 - 2 = 2$

2. Next, calculate $x_1$: $x_1 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{-3}{2} = 1 + \frac{3}{2} = \frac{5}{2}$

3. Now, calculate $x_2$:

   • $f\left(\frac{5}{2}\right) = \left(\frac{5}{2}\right)^4 - 2\left(\frac{5}{2}\right) - 2$
   • $f'\left(\frac{5}{2}\right) = 4\left(\frac{5}{2}\right)^3 - 2$

Let me calculate both values and find the exact $x_2$ values for both cases. sympy import symbols, Rational, diff

Define the variable and function

x = symbols('x') f = x**4 - 2*x - 2 f_prime = diff(f, x)

Define the values for x0 = -1 and x0 = 1

x0_1 = -1 x0_2 = 1

Define the function evaluations

f_x0_1 = f.subs(x, x0_1) f_prime_x0_1 = f_prime.subs(x, x0_1)

f_x0_2 = f.subs(x, x0_2) f_prime_x0_2 = f_prime.subs(x, x0_2)

Calculate x1 for both cases

x1_1 = x0_1 - f_x0_1 / f_prime_x0_1 x1_2 = x0_2 - f_x0_2 / f_prime_x0_2

Calculate x2 for both cases

f_x1_1 = f.subs(x, x1_1) f_prime_x1_1 = f_prime.subs(x, x1_1)

f_x1_2 = f.subs(x, x1_2) f_prime_x1_2 = f_prime.subs(x, x1_2)

Calculate x2

x2_1 = x1_1 - f_x1_1 / f_prime_x1_1 x2_2 = x1_2 - f_x1_2 / f_prime_x1_2

x2_1.simplify(), x2_2.simplify()The values of $x_2$ for each starting point are as follows:

• For $x_0 = -1$, $x_2 = \frac{-1489}{1864}$.
• For $x_0 = 1$, $x_2 = \frac{1907}{968}$.

If you need more details or further calculations, feel free to ask!

Here are 5 questions to explore related concepts:

1. How does the initial guess $x_0$ affect the convergence of Newton's method?
2. How can Newton's method be applied to more complex functions, like those with multiple variables?
3. What are the potential issues with Newton's method, such as divergence or slow convergence?
4. How do you determine the number of iterations needed to achieve a desired level of accuracy in Newton's method?
5. Can Newton's method be applied to functions that are not differentiable?

Tip: In Newton's method, if the derivative $f'(x)$ is very small (close to zero), the method can behave unpredictably, so be careful with such values.