Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. x5 − x − 6 = 0,    x1 = 2

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Newton's method is numerical method for solving nonlinear equations of type `f(x)=0.` If you have `n`-th approximation you can calculate `(n+1)`-st approximation using the formula:

`x_(n+1)=x_n-(f(x_n))/(f'(x_n)).`

For further explenation see the link below.

Lets calculate derivation: `f'(x)=5x^4-1.` So you have:

`x_2 = 2 - (24)/(79) = 1.6962`

`x_3 = 1.6962 - (6.34439)/(40.38836) = 1.53912` 

Your final solution would be something like `x_n = 1.49612.` Of course that's only one of 5 solutions to this equation. Other 4 solutions are complex numbers. If you want to get complex solutions, your initial approximation must be complex.

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