Roots of a Polynomial Earlier in the class we discussed the problems associated with finding the roots of functions specifically polynomials but other functions, e.g. logarithms and exponentials, have similar issues. Specifically, there isn't a formula you can use to find the exact roots for these functions. In the earlier class, we talked about using the midpoint method for nding an approximate value. In this project, we're going to look at a more efficient procedure called the NewtonRaphson method. The Newton-Raphson Method The idea behind the method is that the tangent line can be used to approximate the roots of an equation. You can see this illustrated in the graph to the right. If you start with x = 3 as an approximate value then find the tangent line's x-intercept, that value is closer to the actual root which makes it a better approximation. The methods procedure is an iterative one like the mid point method was, i.e. you start by finding an approximate value, use the formula to get a better , approximate value then substitute that value into the r formula to get an even better approximation. J, Here's how the method works: First find an approximate value for the root, call itpa. The next approximation, p1, is calculated using the formula flpc} f'lpc} [51:23 Dnce you've calculated p1, 1,rou would substitute it into the right hand side ofthe equation to getpz, etc. The process continues until the difference between two successive approximations is less than whatever level of accuracy you need. Using the Method 1. Use Newton's method to approximate the roots 0er3 + 3x2- 1 = 0 between -4 and U to within 10'\". Use Newton's method to approximate the root of 312 e" = 0 between D and 3 to within 10'\". Use the mid-point method to find the same root and compare the number of steps required by each method to get the required accuracy. If? While ('Idrw | rltnsaiinn, )(l'l': Compute an approximate value of ME to within 10\" using Newton's Method. Newton's Method has its drawbacks. Consider the functionx} =xI4. What happens if you try to find the root between D and 3 using x = 0 as your first approximation? If you use x = .25 as your first approximation is the second approximation better or worse than .25? {The actual root is at x: 2.} Explain the issues graphically by looking at the graph and its tangent line