PLEASE PLEASE ONLY GIVE ME THE CODE THAT YOU WILL PUT ON MATLAB TO RUN THIS SOLUTION. I WILL TRULY APPRECIATE IT. THANK YOU!

FIGURE 5.7 An M-file to implement the bisection method. function [root, fx, ea, iter]=bisect(func, x1,xu, es, maxit,varargin) % bisect: root location zeroes % [root, fx, ea, iter]=bisect(func,x1,xu, es, maxit,p1.p2,...): % uses bisection method to find the root of func % input: % func = name of function % x1, xu = lower and upper guesses % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1, p2, ... = additional parameters used by func % output: % root = real root % fx = function value at root % ea = approximate relative error (%) % iter = number of iterations if nargin0,error('no sign change'), end if nargin 0 x1 = xr; else ea = 0; end if ea = maxit, break, end end root = xr; fx = func(xr, varargin{:}); If End is the desired error, this equation can be solved for? log(Ax/Eu) log 2 "" = log, (4x) (5.6) Let's test the formula For Example 5.4, the initial interval was Axo = 200 50 = 150. After eight iterations, the absolute error was until the app 52 Develop your own ion to Fig. 5.7. H. iterations and Eq. (5 criterion. Make sure next highest integer (Hir way to do this). The our own M-file for bisection in a similar fash- However, rather than using the maximum a Eq. (5.5), employ Eq. (5.6) as your stopping ve sure to round the result of Eq. (5.6) up to the Integer (Hint: the ceil function provides a handy this). The first line of your function should be on froot, Ea, ea, n] = bisectnew(func, x1, xu, Ead, function [root. Ea varargin) FIGURE 5.7 An M-file to implement the bisection method. function [root, fx, ea, iter]=bisect(func, x1,xu, es, maxit,varargin) % bisect: root location zeroes % [root, fx, ea, iter]=bisect(func,x1,xu, es, maxit,p1.p2,...): % uses bisection method to find the root of func % input: % func = name of function % x1, xu = lower and upper guesses % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1, p2, ... = additional parameters used by func % output: % root = real root % fx = function value at root % ea = approximate relative error (%) % iter = number of iterations if nargin0,error('no sign change'), end if nargin 0 x1 = xr; else ea = 0; end if ea = maxit, break, end end root = xr; fx = func(xr, varargin{:}); If End is the desired error, this equation can be solved for? log(Ax/Eu) log 2 "" = log, (4x) (5.6) Let's test the formula For Example 5.4, the initial interval was Axo = 200 50 = 150. After eight iterations, the absolute error was until the app 52 Develop your own ion to Fig. 5.7. H. iterations and Eq. (5 criterion. Make sure next highest integer (Hir way to do this). The our own M-file for bisection in a similar fash- However, rather than using the maximum a Eq. (5.5), employ Eq. (5.6) as your stopping ve sure to round the result of Eq. (5.6) up to the Integer (Hint: the ceil function provides a handy this). The first line of your function should be on froot, Ea, ea, n] = bisectnew(func, x1, xu, Ead, function [root. Ea varargin)