use newton's method to estimate the two zeros of the function f(x) = x^4 - 2x - 2. Start with x0 = 1 for the zero on the right. then in each case, find x2.

Determine x2 when x0 = -1

x2= ____ type an integer or a simplified fraction


Then determine x2 when x0 = 1

x2 = ____ type an integer or a simplified fraction





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.

Use Newton's method to estimate the two zeros of the function f(x) = x^4 - 2x - 2. Start with x0 = 1 for the zero on the right. Then in each case, find x2. Determine x2 when x0 = -1. Then determine x2 when x0 = 1.

Learn how to use Newton's method to approximate the zeros of the polynomial function f(x) = x^4 - 2x - 2. Step-by-step calculations are provided, starting from initial guesses x0 = -1 and x0 = 1, with detailed evaluations of x2 for both cases. Suitable for high school calculus or college-level math students.

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