In Matlab language

the error is 1000*maxeps(x0))

Matlab function

2) Considering Bisection method in workseet-5, there is another method to find a solution of f(x) = 0 called Newton method. In this method, we define a sequence of numbers xn which is converging to the solution of f (x) = 0 by using the following iteration formula f(xn) Xn+1 = Xn %3D f'(xm) ' where f'(xn) is the derivative of f at the point xon which f'(xn) #0. Write a function newton(f, fp,x,) on which f,fp and x, are inputs with f, fp to be input function and its derivative, x, to be starting point, and x_sol (solution of f (x) = 0) %3D to be output. Define the same error as in bisectnew (of course for just xo). Test your code for f (x) = x - 3x +1 = 0, x, = 2 and also for x, -5 and write your %3D answers in the end of the code leading by %. Copy and paste your code in your answer sheet (W_6_24)