Consider the following nonlinear equation: 𝑓(𝑥)=𝑥^3−2𝑥^2−2𝑥+1=0 Solve for the root using interval halving with 𝑥=0 and 𝑥=1 Solve for the root using false position with 𝑥=0 and 𝑥=1 Solve for the root using fixed point with 𝑥_0=0 Solve for the root using Newton’s method with 𝑥_0=1

Question 2 5 MarksFind a root of the following equation in the interval (0,1) using Newton-Raphson Method after threeiterations𝑥𝑒𝑥 − cos 𝑥 = 0Take Initial value 0.5.Note: Accuracy up to four decimal places is required. Here is a transcendental equation all thecalculation should be done in the radians mode.

The basic Newton method is used for the solution of nonlinear problems

Implement Newton-Raphson method to find all the possible roots of the given function and verify it with built-in functions scipy.optimize.root() in the SciPy library. Given Functions are: y= f(x)=x^2-x-1 y= f(x)=x^3-x^2-2x+1 follow the template please : # Root Finding Method import math import numpy as np import scipy as sp import matplotlib.pyplot as plt def plot_function(func, a, b): """ This function plot the graph of the input func within the given interval [a,b). """ # Your code goes here def newton_method(func, grad, x0, tol=1e-6, max_iter=100): '''Approximate solution of f(x)=0 by Newton-Raphson's method. Parameters ---------- func : function Function value for which we are searching for a solution f(x)=0, grad: function Gradient value of function f(x) x0 : number Initial guess for a solution f(x)=0. tol : number Stopping criteria is abs(f(x)) < tol. max_iter : integer Maximum number of iterations of Newton's method. Returns ------- xn : root Example -------- >>> fun = lambda x: x**2 - x - 1 >>> grad = lambda x: 2*x - 1 >>> root = newton_method(fun, grad, 1, max_iter=20) ''' # Main Loop starts here iter_count = 1 while iter_count <= max_iter: # Your code goes here iter_count += 1 print("Warning! Exceeded the maximum number of iterations.") return root # Main Driver Function: if __name__ == "__main__": # Define the 1st Function for which the root is to be found func = lambda x: x**2 - x - 1 # Define the gradient of the Function grad = lambda x: 2*x -1 # Uncomment the next two lines to use the 2nd Function #func = lambda x: x**3 - x**2 - 2*x + 1 #grad = lambda x: 3*x**2 - 2*x -2 # Call plot_function to plot graph of the function # Your code goes here x0 = 0 # Initial guess for 1st (change the value as required) # Call the Newton's method for 1st root our_root_1 = # Your code goes here # Call SciPy method (reference method) for 1st root sp_result_1 = sp.optimize.root(func, x0) sp_root_1 = sp_result_1.x.item() # Call the Newton's method for 2nd root x0 = 0 # Initial guess for 2nd root (change the value as required) our_root_2 = # Your code goes here # Call SciPy method (reference method) for 2nd root sp_result_2 = sp.optimize.root(func, x0) sp_root_2 = sp_result_2.x.item() # Print the result print("1st root found by Newton's Method = {:0.8f}.".format(our_root_1)) print("1st root found by SciPy = {:0.8f}".format(sp_root_1)) print("2nd root found by Newton's Method = {:0.8f}.".format(our_root_2)) print("2nd root found by SciPy = {:0.8f}".format(sp_root_2))

Change sqrt (including its Javadoc comments) so it also works when x = 0. Note: if your code from Newton1 appears to work without any changes, but it is such that it might execute a division by 0, then it is not correct. Division by 0, in general, is undefined and you should not write code that attempts to compute it.

Compute two iterations for the function f(x) = x3 – 5x + 1 = 0 using the secant method, in which the real roots of the equation f(x) lies in the interval (0, 1).