#### Answer

$$
f(x)=x^{3}-3 x+6
$$
the tangent line used for approximating $x_{2}$ is horizontal. Attempting to find $x_{2}$ results in trying to divide by zero.
Therefore, Newton’s method doesn’t work for finding the
root of the given equation .

#### Work Step by Step

$$
f(x)=x^{3}-3 x+6
$$
We apply Newton’s method with
$$
f(x)=x^{3}-3 x+6
$$
and
$$
f^{\prime}(x)=3 x^{2}-3
$$
If
$$x_{1}=1$$
then
$$
f^{\prime}(x_{1})=3 (1)^{2}-3=0
$$
since
$$
x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}
$$
With $n= 1$ we have
$$
x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}
$$
the tangent line used for approximating $x_{2}$ is
horizontal. Attempting to find $x_{2}$ results in trying to divide by zero.
Therefore, Newton’s method doesn’t work for finding the
root of the given equation .