Question

Write the formula for Newton's method and use the given initial approximation to compute the approximations x_1 and x_2. f(x) = x^2 + 21, x_0 = -21 x_n + 1 = x_n - (x_n)^2 + 21/2(x_n) x_n + 1 = x_n - (x_n)^2 + 21 x_n + 1 = x_n - 2(x_n)/(x_n)^2 + 21 Use the given initial approximation to compute the approximations x_1 and x_2. x_1 = (Do not round until the final answer. Then round to six decimal places as needed.)

Answer 1

Answer:

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

$x_{1} = -10$

$x_{2} = -3.95$

Step-by-step explanation:

Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:

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

We are given that:

f(x) = $x^{2} + 21$; $x_{0} = -21$

$f^{'} (x)$ = df(x)/dx = 2x

Therefore, using the formula for Newton-Raphson method to determine $x_{1}$ and $x_{2}$

$x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}$

$f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462$

$f^{'}(x_{0}) = 2*(-21) = -42$

Therefore:

$x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10$

Similarly,

$x_{2} = x_{1} - \frac{f(x_{1} )}{f^{'}(x_{1})}$

$f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121$

$f^{'}(x_{1}) = 2*(-10) = -20$

Therefore:

$x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95$