Finally, consider the following fixed point iteration xk+1 =g(xk) = arccos −1 1 + e 2x and show that finding a fixed point ofg(x) is equivalent to finding a root of f(x) = 0. Use the codefixedpt.m to try to approximate the root using an initial guess ofx0 = −3. Can you explain why your iteration behaves as it does?Hint: Plot the fixed-point function and think convergence!

Code in fixedpt.m:-

function [xfinal, niter, xlist] = fixedpt( gfunc, xguess, tol )% FIXEDPT: Fixed point iteration for x=gfunc(x).%% Sample usage:%   [xfinal, niter, xlist] = fixedpt( gfunc, xguess, tol )%% Input:%   gfunc  - fixed point function %   xguess - initial guess at the fixed point%   tol   - convergence tolerance (OPTIONAL, defaults to 1e-6)%% Output:%   xfinal - final estimate of the fixed point%   niter  - number of iterations to convergence%   xlist  - list of interates, an array of length 'niter'% First, do some error checking on parameters.if nargin < 2 fprintf( 1, 'FIXEDPT: must be called with at least two arguments' ); error( 'Usage: [xfinal, niter, xlist] = fixedpt( gfunc, xguess, [tol] )' );endif nargin < 3, tol = 1e-6; end% fcnchk(...) allows a string function to be sent as a parameter, and% coverts it to the correct type to allow evaluation by feval().gfunc = fcnchk(gfunc);x = xguess;xlist = [ x ];niter = 0;done = 0;while ~done, xnew = feval(gfunc, x); xlist = [ xlist; xnew ]; % create a list of x-values  niter = niter + 1; if abs(x-xnew) < tol,   % stopping tolerance for x only  done = 1; end x = xnew;endxfinal = xnew;