Please show your work clearly.. Thanks As discusscd in class (and in the textboo

Question

Please show your work clearly.. Thanks

As discusscd in class (and in the textbook in section 11.4), to solve the equation f(x) = 0 by Newton's Method we start with a good initial guess .To and then run the iteration until we get an approximation xn+i that is good enough for our purposes. Suppose you want to compute the cube root of 4 by solving the equation x3-4=0. Since 13 = 1 and 23 = 8 let's start with x0 = 1.5 Then To check your answer compute x33= Enter x1, x2 and x3 with at least 6 correct digits beyond the decimal point.

Explanation / Answer

x1 = 1.5 - [(1.5^3-4)] / 3. (1.5^2)

=> 1.5 +0.625/6.75

=> 1.592592

x2 =>1.592592- [(1.592592^3-4)] / 3. (1.592592^2)

  => 1.592592 - 0.039374/7.609053

=> 1.587417

x3 = 1.587417 - [(1.587417^3-4)] / 3. (1.587417^2)

=> 1.587417- 0.000123/7.559678

=>1.587401

now x3^3 = 3.999999 ~ 4

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