Question

Consider the nonlinear function: y = f(x) = x3 cos x a. Plot y as a function of x as x is varied between -67 and 67. In this plot mark all the locations of x where y = 0. Make sure to get all the roots in this range. You may need to zoom in to some areas of the plot. These locations are some of the roots of the above equation. b. Use the fzero function in MATLAB (or equivalent function in python which is fsolve) to solve for all the roots of the above equation between -6 and 61. These should match the results that you get in 'a'. c. Determine the first root between 6 and 6t using bisection method and carry out the first four iterations. Use initial values for a and b as you find appropriate. d. Determine the first root again between 6 and 6t using Newton's method and carry out the first four iterations. Start with initial guess value as you find appropriate.

In MATLAB please

Answer 1

1 $ Consider the non-linear function clc; clear; a) Plot y asfunction of x = –6*pi():0.5:6*pi) 3 y = x.-3. fcos (x); plot (x, y, -0'); xlabel(x'); ylabel('Non-linear function title { Plot of y=if (x)=x^scos (xi); grid on: 88 b) Use fzero function in MATLAB 8 10 12 1 13 HS 15 $ There are thirteen distinct roots as evident from the plot zeros (13,1); - 6*pi() :pi() : 6 pia fzero (@(x) x^3 cos (x),XO); i = i + 1 = 16 17 1 endl disp The Loots dispir) 1

Command window The roots are: -17.2.788 -14 13 72 -10.9956 --7: 8540 -4.7124 -1.5708 1.5708 4.7124 7.8540 10.9956 14.13 72 172 788 fx >>

a and b are solved.