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\\(x_0=\\var{x0}\\)

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\\(x_1=\\) [[0]]

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\\(x_2=\\) [[1]]

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\\(x_3=\\) [[2]]

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\\(x_4=\\) [[3]]

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\\(x_{n+1}=x_n-\\frac{\\var{a}x_n^3-\\var{b}x_n^2+\\var{c}x+\\var{d}}{3(\\var{a})x_n^2-2(\\var{b})x+\\var{c}}\\)

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\\(x_0=\\var{x0}\\)

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\\(x_1=\\var{x0}-\\frac{(\\var{a})*(\\var{x0})^3-\\var{b}*(\\var{x0})^2+\\var{c}*(\\var{x0})+\\var{d}}{3(\\var{a})(\\var{x0})^2-2(\\var{b})(\\var{x0})+\\var{c}}=\\var{x1}\\)

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\\(x_2=\\var{x1}-\\frac{(\\var{a})*(\\var{x1})^3-\\var{b}*(\\var{x1})^2+\\var{c}*(\\var{x1})+\\var{d}}{3(\\var{a})(\\var{x1})^2-2(\\var{b})(\\var{x1})+\\var{c}}=\\var{x2}\\)

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\\(x_3=\\var{x2}-\\frac{(\\var{a})*(\\var{x2})^3-\\var{b}*(\\var{x2})^2+\\var{c}*(\\var{x2})+\\var{d}}{3(\\var{a})(\\var{x2})^2-2(\\var{b})(\\var{x2})+\\var{c}}=\\var{x3}\\)

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\\(x_4=\\var{x3}-\\frac{(\\var{a})*(\\var{x3})^3-\\var{b}*(\\var{x3})^2+\\var{c}*(\\var{x3})+\\var{d}}{3(\\var{a})(\\var{x3})^2-2(\\var{b})(\\var{x3})+\\var{c}}=\\var{x4}\\)

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\\(x_5=\\var{x4}-\\frac{(\\var{a})*(\\var{x4})^3-\\var{b}*(\\var{x4})^2+\\var{c}*(\\var{x4})+\\var{d}}{3(\\var{a})(\\var{x4})^2-2(\\var{b})(\\var{x4})+\\var{c}}=\\var{x5}\\)

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Perform four iterations of the Newton-Raphson method, taking  \\(x_0=\\var{x0}\\)  as your initial estimate to find a root of the function:

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\\(f(x)=\\var{a}x^3-\\var{b}x^2+\\var{c}x+\\var{d}\\)

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Give your answers correct to three decimal places.

", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}