# Newton - (1 Use Newtons method to nd critical points of the function y = ex 2x2 Solution The critical points are located at x values for which y =

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(1) Use Newton’s method to ﬁnd critical points of the function y = e x - 2 x 2 . Solution: The critical points are located at x values for which y 0 = f ( x )= e x - 4 x =0.If there exists any root for f ( x ) = 0, there should normally be two. Let’s start from x 0 =0 to ﬁnd the smaller root. x 1 = x 0 - f ( x 0 ) /f 0 ( x 0 )=0 . 333333, x 2 = x 1 - f ( x 1 ) /f 0 ( x 1 . 357246, x 3 = x 2 - f ( x 2 ) /f 0 ( x 2 . 357402, x 4 = x 3 - f ( x 3 ) /f 0 ( x 3 . 357402. Let start from x 0 = 5 to ﬁnd the larger root. x 1 = x 0 - f ( x 0 ) /f 0 ( x 0 )=4 . 110793, x 2 = x 1 - f ( x 1 ) /f 0 ( x 1 )=3 . 329113, x 3 = x 2 - f ( x 2 ) /f 0 ( x 2 )=2 . 718701, x 4 = x 3 - f ( x 3 ) /f 0 ( x 3 . 334689, x 5 =2 . 178594, x 6 . 153872, x 7 . 153292, x 8 . 153292. (2) Solution: (3) Use Newton’s method to ﬁnd an approximate value for 8. (Hint: First think of a function, f ( x ), such that f ( x ) = 0 has the solution x = 8). Solution: x = 8 is a solution of the equation f ( x x 2 - 8 = 0. Thus, f 0 ( x x .W e know also that 9=3and 8 must be close to this value. We start from x 0 =3. x 1 = x 0 - f ( x 0 ) /f 0 ( x 0 - (9 - 8) / 2 × 3=3 - 1 / 6=2 5 6 2 . 83 x 2 = x 1 - f ( x 1 ) /f 0 ( x 1 . 83 - (2 . 83 2 - 8) / 2 × 2 . 83 x 2 . 83 - 0 . 0089 / 5 . 66 = 2 . 83 - 0 . 0016 2 . 83 Therefore, x = 8 2 . 83.

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## This note was uploaded on 01/12/2010 for the course STAT 100 taught by Professor Denissjerve during the Winter '06 term at San Jose State University .

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Newton - (1 Use Newtons method to nd critical points of the function y = ex 2x2 Solution The critical points are located at x values for which y =

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