Answer
The root of the equation
$$
f(x)=x^{5}-x^{4}+3 x^{2}-3 x-2
$$
in $[1,2]$ is 1,297383, to six decimal places.
Work Step by Step
We apply Newton’s method with
$$
f(x)=x^{5}-x^{4}+3 x^{2}-3 x-2
$$
and
$$
f^{\prime}(x)=5 x^{4}-4 x^{3}+6 x-3
$$
Newton himself used this equation to illustrate his method and he chose $x_{1}=1$ so,
$$
\begin{aligned}
x_{n+1} &=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\ &=x_{n}-\frac{x_{n}^{5}-x_{n}^{4}+3 x_{n}^{2}-3 x_{n}-2}{5 x_{n}^{4}-4 x_{n}^{3}+6 x_{n}-3}
\end{aligned}
$$
With $n = 1$ we have
$$
\begin{aligned}
x_{2} &=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}\\ &=x_{1}-\frac{x_{1}^{5}-x_{1}^{4}+3 x_{1}^{2}-3 x_{1}-2}{5 x_{1}^{4}-4 x_{1}^{3}+6 x_{1}-3}\\
&=(1)-\frac{(1)^{5}-(1)^{4}+3 (1)^{2}-3 (1)-2}{5 (1)^{4}-4 (1)^{3}+6 (1)-3}\\
&=\frac{3}{2}=1.5
\end{aligned}
$$
With $n = 2$ we have
$$
\begin{aligned}
x_{3} &=x_{2}-\frac{f\left(x_{2}\right)}{f^{\prime}\left(x_{2}\right)}\\ &=x_{2}-\frac{x_{2}^{5}-x_{2}^{4}+3 x_{2}^{2}-3 x_{2}-2}{5 x_{2}^{4}-4 x_{2}^{3}+6 x_{2}-3}\\
&=(1.5)-\frac{(1.5)^{5}-(1.5)^{4}+3 (1.5)^{2}-3 (1.5)-2}{5 (1.5)^{4}-4 (1.5)^{3}+6 (1.5)-3}\\
&\approx 1.343860
\end{aligned}
$$
With $n = 3$ we have
$$
\begin{aligned}
x_{4} &=x_{3}-\frac{f\left(x_{3}\right)}{f^{\prime}\left(x_{3}\right)}\\ &=x_{3}-\frac{x_{3}^{5}-x_{3}^{4}+3 x_{3}^{2}-3 x_{3}-2}{5 x_{3}^{4}-4 x_{3}^{3}+6 x_{3}-3}\\
&=(1.343860)-\frac{(1.343860)^{5}-(1.343860)^{4}+3 (1.343860)^{2}-3 (1.343860)-2}{5 (1.343860)^{4}-4 (1.343860)^{3}+6 (1.343860)-3}\\
&\approx 1.300320
\end{aligned}
$$
With $n = 4$ we have
$$
\begin{aligned}
x_{5} &=x_{4}-\frac{f\left(x_{4}\right)}{f^{\prime}\left(x_{4}\right)}\\ &=x_{4}-\frac{x_{4}^{5}-x_{4}^{4}+3 x_{4}^{2}-3 x_{4}-2}{5 x_{4}^{4}-4 x_{4}^{3}+6 x_{4}-3}\\
&=(1.300320)-\frac{(1.300320)^{5}-(1.300320)^{4}+3 (1.300320)^{2}-3 (1.300320)-2}{5 (1.300320)^{4}-4 (1.300320)^{3}+6 (1.300320)-3}\\
&\approx 1,297396\\
&\approx x_{6}
\end{aligned}
$$
so the root in $[1,2]$ is 1,297383, to six decimal places.
