y= f(x)=x^2-x-1 y = = f(x)=x^3-x^2-2x+1 Implement Bisection Method to find all the possible roots of these 2 given functions and verify it with built-in functions scipy.optimize.root() in the SciPy library. follow this given template # 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 bisection_method(func, a, b, tol=1e-6, max_iter=100): """ Bisection method to find the root of a function within a given interval. Parameters: - func: The function for which the root is to be found. - a, b: Interval [a, b] within which the root is searched for. - tol: Tolerance level for checking convergence of the method. - max_iter: Maximum number of iterations. Returns: - root: Approximation of the root. Example -------- >>> fun = lambda x: x**2 - x - 1 >>> root = bisection_method(fun, 1, 2, max_iter=20) """ # Check if the interval is valid (signs of f(a) and f(b) are different) # Your code goes here # 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 # Example usage: if __name__ == "__main__": # Define the function for which the root is to be found func = lambda x: x**2 - x - 1 # First Function # Uncomment the below line to use the Second Function # func = lambda x: x**3 - x**2 - 2*x + 1 # Second Function # Call plot_function to plot graph of the function # Your code goes here # Set the interval [a, b] for the search a_1 = 0; b_1 = 0; # For first root (change the values as required) a_2 = 0; b_2 = 0; # For second root (change the values as required) # Call the bisection method our_root_1 = # your code goes here our_root_2 = # your code goes here # Call SciPy method root, which we consider as a reference method. x0 = (a_1 + b_1)/2 sp_result_1 = sp.optimize.root(func, x0) sp_root_1 = sp_result_1.x.item() x0 = (a_2 + b_2)/2 sp_result_2 = sp.optimize.root(func, x0) sp_root_2 = sp_result_2.x.item() # Print the result print("1st root found by Bisection Method = {:0.8f}.".format(our_root_1)) print("1st root found by SciPy = {:0.8f}".format(sp_root_1)) print("2nd root found by Bisection Method = {:0.8f}.".format(our_root_2)) print("2nd root found by SciPy = {:0.8f}".format(sp_root_2))

y= f(x)=x^2-x-1 y= f(x)=x^3-x^2-2x+1 Write Python codes to plot the given function using Matplotlib library. This will give an idea where the roots of the given function are located.

Wrie a python program to use a function that computes the roots of a quadratic equation

MATLAB has only one built-in function for finding roots : fzero. (The Optimization Toolbox has more).To find out about it, start up Matlab and typehelp fzeroThis hybrid algorithm is due to Richard Brent (1946-), emeritus Professor of Computer Science, ANU.A simplified version called fzerotx can be found in the Asst2 folder. A description of the method as wellas the code is given in Sections 4.6 & 4.8 of the textbook. You can use fzerogui from the Asst2 folder to seehow the algorithm works.a. Modify fzerotx to return, in addition to the computed root, the number of iterations taken.b. Try your modified fzerotx on the following problems and see how it performs in terms of reliability andnumber of iterations.• Find the first 2 positive solutions of the equationsin x = cos(2x 2 )• try it on the function f (x) = 1/(x − π) on the interval [0,5] and comment.• try it on the function f (x) = 1 − (1 + 3x) exp(−3x) on the interval [-1,1] and comment.c. Exactly what kind of stopping criterion does fzerotx use?d. (Tricky) Answer Q4.7 in the textbook (Numerical Computing with MATLAB by Moler)

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))​