Questions: Use Newton's method with the specified initial approximation (x1) to find (x3), the third approximation to the solution of the given equation [ x^3-frac2x-x^2+1=0, quad x1=2 ]

Solution Steps

To solve the equation using Newton's method, we need to:

1. Define the function \( f(x) = \frac{2}{x} - x^2 + 1 \).
2. Compute its derivative \( f'(x) \).
3. Use the iterative formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to find the next approximation.
4. Start with the initial approximation \( x_1 = 2 \) and compute up to \( x_3 \).

We start with the function defined as: \[ f(x) = \frac{2}{x} - x^2 + 1 \] The derivative of this function is: \[ f'(x) = -\frac{2}{x^2} - 2x \]

The initial approximation is given as: \[ x_1 = 2 \]

Using Newton's method, we compute the next approximations. The iterative formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

For \( n = 1 \): \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{f(2)}{f'(2)} \] Calculating \( f(2) \) and \( f'(2) \): \[ f(2) = \frac{2}{2} - 2^2 + 1 = 1 - 4 + 1 = -2 \] \[ f'(2) = -\frac{2}{2^2} - 2 \cdot 2 = -\frac{2}{4} - 4 = -0.5 - 4 = -4.5 \] Thus, \[ x_2 = 2 - \frac{-2}{-4.5} = 2 - \frac{2}{4.5} = 2 - \frac{4}{9} = \frac{18}{9} - \frac{4}{9} = \frac{14}{9} \approx 1.5556 \]

For \( n = 2 \): \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \] Calculating \( f\left(\frac{14}{9}\right) \) and \( f'\left(\frac{14}{9}\right) \): \[ f\left(\frac{14}{9}\right) = \frac{2}{\frac{14}{9}} - \left(\frac{14}{9}\right)^2 + 1 = \frac{18}{14} - \frac{196}{81} + 1 \] Calculating \( f'\left(\frac{14}{9}\right) \): \[ f'\left(\frac{14}{9}\right) = -\frac{2}{\left(\frac{14}{9}\right)^2} - 2 \cdot \frac{14}{9} \] After performing the calculations, we find: \[ x_3 \approx 1.5215 \]

Final Answer