Answer
$x_{3}\approx-0.6825$
Work Step by Step
We have:
$f(x)=2x^{3}-3x^{2}+2$
Then, $f'(x)= 6x^{2}-6x$
With $x_{1}=-1$, apply Newton's Method:
$x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}=-1-\frac{2\times(-1)^{3}-3\times(-1)^{2}+2}{6\times(-1)^{2}-6(-1)}=-1-\frac{-3}{12}=\frac{-3}{4}$
Similarly,
$x_{3}=x_{2}-\frac{f(x_{2})}{f'(x_{2})}\approx-0.6825$
