Use Newton's method to obtain an approximate solution to the equation x^3+x-1=0. Start with the guess x0=0 and continue doing iterations until you get an answer with four correct decimal places. Finally, draw a graph for this particular problem that shows geometrically how to go from x0 to x1 and from x1 to x2.
ick use a spread sheet
maybe not so bad for first couple. use \[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\]
so if \[x_1=0\] you get \[f(0)=-1,f'(x)=3x^2+1\] so your second guess will be \[0-\frac{-1}{1}=1\] oh this might be trouble
lets see what we get for the second one. \[f(1)=1, f'(1)=4\] so third one will be \[1-\frac{1}{4}=\frac{3}{4}\] ok we will get there. but you can see that the computation is going to be lousy.
if you like you can write \[x-\frac{f(x)}{f'(x)}=x-\frac{x^3+x-1}{3x^2+1}=\frac{2x^3+1}{3x^2+1}\] and work directly with that instead. keep plugging you new number into that formula repeatedly until you get the same number twice. that is why a spread sheet would be handy
but i have to draw a graph too. Can you draw for me?
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