Figure 1: Example showing that Newton's method sets mn+1 to be the value at which 6(33), the line tangent to f(5'3) at (513m f(:c,,)), goes to 0. In this example, ax) : :53 i 8 and f(a:.,.) : 0 at 1:9. = 2. For 33,, = 3, => Hm) = f'(:cn)(:e 7 5%) + an) = 27(93 7 3) + 19, and 30611\") = 0 at 1.2,,11 = 2 % z: 2.3. Question 2. Let's use Newton's method to solve rep = 2 given p and 2. To do this, we can set at) = a3\" 7 z or f(:li) = z 7 (up. Either way is ne. Use Newton's method to nd the formula for 1En+1 given run. You will get the same formula for sen+1 for either choice of f(;1:) Question 3. In the chapter titled \"Lucky Numbers\" in Surely You're Joking, Mr. Feynman, by Richard Feynman and Ralph Leighton, Nobel physicist Richard Feynman tells how he used Newton's method to solve a problem posed by an abacus salesman: nd the cube root of 1729.03. Feynman knew that 123 = 1728, because that is the number of cubic inches in a cubic foot. Therefore, he chose 33.3 = 12. In his head, Feynman found that (1729.03)\"3 \"A: 12.002 much more quickly and accurately than the man who used the abacus. Use your answer to the previous question with p = 3, z = 1729.03, and $0 = 12 to verify Feynman's calculation by showing that 171 = 12 + 103/432. With rounding, 1.03/432 s: 1/500 = 0.002, so .131 z 12.002. Next, be more accurate by using a calculator to evaluate m1 = 12 + 1.03 / 432 and then evaluate (12 + 1.03 / 432)3. \\U) 0')