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The graph of y = f(x) is given below. 2 1 y= f(x) 2 3 4 6 -1+ If Newton's method is applied to the function f, starting with an initial guess of ₁ = 1, what happens to the iterates 22, 23, 24,...? 22=1.5, 33, and z = 6.75. O26 and 3 cannot be calculated because f'(x2)=0. 4 O The iterates 22, 23, 24,... oscillate and do not approach any zero. O The iterates #2, #3, #4,... will approach the zero z=5
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Transcribed Image Text:The graph of y = f(x) is given below. 2 1 y= f(x) 2 3 4 6 -1+ If Newton's method is applied to the function f, starting with an initial guess of ₁ = 1, what happens to the iterates 22, 23, 24,...? 22=1.5, 33, and z = 6.75. O26 and 3 cannot be calculated because f'(x2)=0. 4 O The iterates 22, 23, 24,... oscillate and do not approach any zero. O The iterates #2, #3, #4,... will approach the zero z=5
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