Answer
$$
x^{3}-3x+6=0
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ x_{n}^{3}-3x_{n}+6 }{3x_{n}^{2}-3 }
\end{aligned}
$$
If we choose $x_{1} = 1$ then $f^{'}(x_{1})=f^{'}(1)= 3(1)^{2}-3=0$,
so it is impossible to find $x_{2} $ by using Formula 2 (Newton’s method) because we will be dividing by zero.
Therefore, Newton’s method doesn’t work for finding the
root of the given equation.
Work Step by Step
$$
x^{3}-3x+6=0
$$
Therefore we let
$$
f(x)= x^{3}-3x+6=0
$$
Then
$$
f^{'}(x)= 3x^{2}-3
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ x_{n}^{3}-3x_{n}+6 }{3x_{n}^{2}-3 }
\end{aligned}
$$
If we choose $x_{1} = 1$ then $f^{'}(x_{1})=f^{'}(1)= 3(1)^{2}-3=0$
so impossible to find approximating $x_{2} $ by using Formula 2 (Newton’s method) because will dividing by zero.
Therefore, Newton’s method doesn’t work for finding the
root of the given equation.
