Will upvote for complete help, Thanks

Problem 2 (16 points) (CCO 3) / MatlabGrader Develop a Matlab function that finds a root of a function f(x) starting from the given initial estimates c(-1) and 2 (0) with a tolerance in function of at least cok using the secant method. Name the function my Secant using as input the anonymous function f, the initial estimates xml and x0, and the required tolerance in function epsok. As output, the function shall return four scalar variables: the numerical solution x, its tolerance in function tolFunc, its estimated relative error ere, and the number of iterations performed n. If the requested tolerance in function cannot be reached within N = 80 iterations, the function shall display an error message to the screen and return a value of x = realmax(). Other than this potential error message, do not print out any results to screen or do any plotting within the function. Minimize as much as possible the number of function calls required per iteration, by storing function call results in temporary storage variables. You should require only 1 function call to f to update the solution per iteration. Problem 2 (16 points) (CCO 3) / MatlabGrader Develop a Matlab function that finds a root of a function f(x) starting from the given initial estimates c(-1) and 2 (0) with a tolerance in function of at least cok using the secant method. Name the function my Secant using as input the anonymous function f, the initial estimates xml and x0, and the required tolerance in function epsok. As output, the function shall return four scalar variables: the numerical solution x, its tolerance in function tolFunc, its estimated relative error ere, and the number of iterations performed n. If the requested tolerance in function cannot be reached within N = 80 iterations, the function shall display an error message to the screen and return a value of x = realmax(). Other than this potential error message, do not print out any results to screen or do any plotting within the function. Minimize as much as possible the number of function calls required per iteration, by storing function call results in temporary storage variables. You should require only 1 function call to f to update the solution per iteration