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2. Four different methods were used to solve f(x) = 0, and the computed values for x₁, x₂, x₃, ? ? ? are shown in Table below.(a) One of them is Newton?s method. Which of the four is most likely Newton?s method, and why?(b) One of them is the bisection method. Which of the four is most likely the bisection method, and why?(c) Suppose someone claimed they computed the solution using the secant method, and they obtained the results given for Method 3. Why would you tell them that they are most likely mistaken (i.e., they are wrong)?

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In the derivation of Newton's method, to determine the formula for xi+1, the function f(x) is approximated using a first-order Taylor approxi- mation centered at r;. This problem investigates what happens when you try to use a second-order Taylor approximation. (a) Approximating f(r) using a second-order Taylor approximation centered at xį, what is the resulting formula for xi+1? Note your formula will have a in it. (b) In theory, a second-order Taylor approximation should be more accurate than a first-order Taylor approximation (at least when you are close to the solution). However, the formula in part (a) has several unpleasant complications that Newton's method doesn't have. Identify two of them. (c) Given that i+1 is close to xi, what choice should be made for the in part (a)? (d) One way to avoid the complications considered in part (b) is to note that the Taylor approximation used in part (a) contains a term of the form (x+1- x)2. Explain why this can be approximated with -(x₁+1- xi)f(xi)/f'(xi). If this is done, what is the resulting formula for +1? Note that the formula you are deriving is known as Halley's method, and it is an example of a third-order method. In the derivation of Newton's method, to determine the formula for xi+1, the function f(x) is approximated using a first-order Taylor approxi- mation centered at r;. This problem investigates what happens when you try to use a second-order Taylor approximation. (a) Approximating f(r) using a second-order Taylor approximation centered at xį, what is the resulting formula for xi+1? Note your formula will have a in it. (b) In theory, a second-order Taylor approximation should be more accurate than a first-order Taylor approximation (at least when you are close to the solution). However, the formula in part (a) has several unpleasant complications that Newton's method doesn't have. Identify two of them. (c) Given that i+1 is close to xi, what choice should be made for the in part (a)? (d) One way to avoid the complications considered in part (b) is to note that the Taylor approximation used in part (a) contains a term of the form (x+1- x)2. Explain why this can be approximated with -(x₁+1- xi)f(xi)/f'(xi). If this is done, what is the resulting formula for +1? Note that the formula you are deriving is known as Halley's method, and it is an example of a third-order method.