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

`x_(n+1) = x_(n) - (f(x_(n) ))/(f^(')(x_(n)))`

`x_(1) = -10`

`x_(2) = -3.95`

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) - (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) - (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 - (462)/(-42) = -21 + 11 = -10`

Similarly,

`x_(2) = x_(1) - (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 - (121)/(20) = -10+6.05 = -3.95`