In Newton Raphson method, if the curve 𝑓(𝑥)f(x) is constant then ________.a)𝑓′(𝑥)=0f ′ (x)=0b)𝑓(𝑥)=0f(x)=0c)None of the mentionedd)𝑓′(𝑥)=𝑐f ′ (x)=c

The Iterative formula for Newton Raphson method is given by __________a)𝑥(1)=𝑥(0)+𝑓′(𝑥(0))𝑓(𝑥(0))x(1)=x(0)+ f(x(0))f ′ (x(0))​ b)𝑥(1)=𝑥(0)−𝑓(𝑥(0))𝑓′(𝑥(0))x(1)=x(0)− f ′ (x(0))f(x(0))​ c)𝑥(1)=𝑥(0)+𝑓(𝑥(0))𝑓′(𝑥(0))x(1)=x(0)+ f ′ (x(0))f(x(0))​ d)𝑥(1)=𝑥(0)−𝑓′(𝑥(0))𝑓(𝑥(0))x(1)=x(0)− f(x(0))f ′ (x(0))​

The Newton Raphson method is also called as ____________a)Chord methodb)Diameter methodc)Secant methodd)Tangent method

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))

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.