`1/x = 1 + x^3` Use Newton's method to find all roots of the equation correct to six decimal places.
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`1/x=1+x^3`
Set the left side equal to zero.
`0=1+x^3 -1/x`
To solve this using Newton's method, apply the formula:
`x_(n+1) = x_n - (f(x_n))/(f'(x_n))`
Let the function be:
`f(x) = 1+x^3-1/x`
Take the derivative of f(x).
`f'(x) = 3x^2 +1/x^2`
Plug-in f(x) and f'(x) to the formula of Newton's method.
`x_(n+1)=x_n-(1+x_n^3-1/x_n)/(3x_n^2+1/x_n^2)`
This simplifies to:
`x_(n+1) = x_n- (x_n^5+x_n^2-x_n)/(3x_n^4+1)`
Then, refer to the graph of the function to get the initial values x when f(x) =0. (See attached figure.)
(The entire section contains 265 words.)
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