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Matlab code. Need function and code to call your function example below. The Newton-Raphson method can be used to solve the root finding problem f(x)

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Matlab code. Need function and code to call your function example below.
 VII - 1 function x-rfNewtonR(fun,x, Tolx, itmax) 2% Find the root of f(x) - 9 using the Newton-Raphson 

The Newton-Raphson method can be used to solve the root finding problem f(x) = 0. This is an open method, so it starts with a single initial estimate for the root. Given a current estimate for the root, x, the next estimate for the root is found using the tangent line to the function curve at x,. Thus, the Newton-Raphson formula for the next estimate of the root is given by the root of the tangent line: Xi+1=Xi- This iteration process repeats until the root is found to within a given tolerance. However, because this is an open method, the iteration is not guaranteed to converge Your function should return the following output: x = computed root estimate Write a function rf NewtonR that implements the Newton-Raphson method. Your function should accept the following inputs: fun = a handle to a MATLAB function of the form [y, dydx] = fun(x) that returns f(x) and f'(x) as the first and second outputs respectively xe = initial guess for the root Tolx = desired relative error for the root given by itmax = maximum number of iterations f(xi) f'(x) ++1-24 || Xi S O Your solution should do the following: 1. Implement the Newton Raphson method to find a root of f(x) = 0 and print the iteration history. Do not use the MATLAB solver functions like fzero, solve.__ roots, etc. 2. Terminate the search when the relative error tolerance is reached or when the iteration limit is reached (whichever occurs first). 3. Use the MATLAB function error with a msgID to detect if the iteration does not converge before the iteration limit is reached: msgID= 'rfNewtonR: Iteration LimitExceeded' A test case has been provided in the 'Code to call your function' box. If the solution seems reasonable, submit your function for assessment. 35

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