. Using Newton's Method, find a numerical approximation to the Zero of

F(x) = e^(-x/3) - 0.05471*x

on [2.5,2.9] starting with the endpoint (k = 0) having the smallest value of |f|,

keeping track of the number of ALL function evaluations kfe, the current

Change in sign interval [a_k,b_k], and tabulating

K kfe a_k b_k x_k f_k f_k' x_{k+1} |x_{k+1}-x_k|

0

... ...

until |x_{k+1}-x_k| < 0.5e-2.

 

|for k = 0 to 3 iterations

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1 Answer

x=4.324341

....................
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