# Newton-Raphson method with $c_0=3$ to calculate $c_1$.

I have a small problem

Calculate a polynomial function $g$ with integer coefficient that has $\sqrt[3]{28}$ as a root, and then use the Newton-Raphson method with $c_0=3$ to calculate $c_1$.

I now the Newton-Raphson method$$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$$ But don't understand how to solve this using it. Any help would be appreciated.

• What are $c_0,c_1$? – Galc127 Mar 24 '17 at 6:49

Given root of polynomial function$x=\sqrt[3]{28}\\g(2)=x^3-28=0\tag*{}$Applying Newton's method $x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}\\\text{For}\;c_1\\c_1=c_0-\dfrac{g(c_0)}{g'(c_0)}\\\text{Thus,}\;c_0=3\\\text{So,}\;g'(3)=27\\g(3)=27-28=-1\\c_1=\dfrac{3-(-1)}{27}=\dfrac{81+1}{27}\\c_1=\dfrac{82}{27}\tag*{}$