USe Newtons method to approx imate the zevo of the equahen f

USe Newton\'s method to approx imate the zevo of the equahen f(x)= 3x22x+2 Correct to ten decimal places Make sure fo write the general formule which You will plug in the inihial Cquessed) value

Solution

Newton’s Method is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. Let f(x) is a function.

Here The Newton–Raphson method in one variable is defined as follows

X: f(x) = 0

Given a function ƒ defined over the reals x, and its derivative ƒ\',

Hence, The Formula

x(n+1) = x(n) –    f(n)                                                                   (Let Eqn. – 1)

                               f’(x0)   

Now, we begin with a first guess x₀ for a root of the function f. provided the function satisfies all the assumptions made in the derivation of the formula, a better approximation x₁ is

x1 =      x0   -       f(x0)    

                            f’(x0)

Your Function is f(x) = 3x^2 – 2x+2

So, f’(x) = 6x-2

As per detail above, we can simply tell as

f(0) = 3*(0)^2-2*(0)+2= 2

f’(0)= 6*(0)-2 = - 2

So, Let us guest the initial value of x0 = 1.

Now, x1 =   x0 -    f(x0)

                              f’(x0)

Or, (a)   x1 = 1 –      (2)       = 1+1 = 2 or 2.0000000000

                                 (-2)

Similarly

(b) X2 =x1 - f(x1)     = 2-        {3*(2) ^2 – 2*(2) +2}                    

                       f’(x)                             6*(2) -2

          

=     2-        10             =   2-1 = 1    or 1.0000000000

                    10

In this way, we can calculate,

(c) x3 = x2 –      f(x2)                      1-          6                      = 1-1.5 =0.5 or -0.5000000000        

                         f’(x2)                                    4

(d) x4 = x3-   f(x3)               = 0.50 -    1.75        = -1.25 or -1.2500000000

                      f’(x3)                                 1

=================

Quetion No. 2

Wire of Circle is used to take

( e) x5 = x4 – f(x4) =   (-1.25) -    (-4.1875)        = (-1.25) – 0.7613636363

                        f’(x)                            (-5.50)

= (- 2.0113636363)

(f) x6 = x5-        f(x5) =      (-2.011363636) –        (-10.1140237603)

                        f’(x5)                                                (-10.0681818182)

= (-2.011363636) – 1.0045531500 = -3.0150167860