Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 18

Use Newton's Method to find all roots of $x^5 = 5x - 2$ correct to six decimal places.

We rewrite the equation in standard form

$x^5 - 5x + 2 = 0$

So we let $f(x) =x^5 - 5x + 2$, then


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (x^5) - 5 \frac{d}{dx} (x) + \frac{d}{dx} (2)
\\
\\
f'(x) =& 5x^4 - 5

\end{aligned}
\end{equation}
$


Using Approximation Formula


$
\begin{equation}
\begin{aligned}

x_{n + 1} =& x_n - \frac{f(x_n)}{f'(x_n)}
\\
\\
x_{n + 1} =& x_n - \frac{x^5_n - 5x_n + 2}{5x^4_n - 5}

\end{aligned}
\end{equation}
$








To find the initial approximation, we graph $y = x^5$ and $y = 5x - 2$. Based from the graph, they have three intersection points in $x$-coordinate where they are very close to $-1.6, 0.4$ and $1.4$. So we have $x_1 = -1.6, x_1 = 0.4$ and $x_1 = 1.4$

So we get


$
\begin{equation}
\begin{aligned}

& x_2 = x_1 - \frac{x_1^5 - 5x_1 + 2}{5x^4_1 - 5}
&&
&&&
\\
\\
& x_2 = -1.6 - \frac{(-1.6)^5 - 5 (-1.6) + 2}{5 (-1.6)^4 - 5}
&& x_2 = 0.4 - \frac{(0.4)^5 - 5 (0.4) + 2}{5 (0.4)^4 - 5}
&&& x_2 = 1.4 - \frac{(1.4)^5 - 5 (1.4) + 2}{5 (1.4)^4 - 5}
\\
\\
& x_2 \approx -1.582506
&& x_2 \approx 0.402102
&&& x_2 \approx 1.373378
\\
\\
& x_3 = -1.582506 - \frac{f(-1.582506)}{f'(-1.582506)}
&& x_3 = 0.402102 - \frac{f(0.402102)}{f'(0.402102)}
&&& x_3 = 1.373378 - \frac{f(1.373378)}{f'(1.373378)}
\\
\\
& x_3 \approx -1.582036
&& x_3 \approx 0.402102
&&& x_3 \approx 1.371886
\\
\\
& x_4 = -1.582036 - \frac{f(-1.582036)}{f'(-1.582036)}
&&
&&& x_4 = 1.371886 - \frac{f(1.371886)}{f'(1.371886)}
\\
\\
& x_4 \approx -1.582036
&&
&&& x \approx 1.371882
\\
\\
&
&&
&&& x_5 = 1.371882 - \frac{f(1.371882)}{f'(1.371882)}
\\
\\
&
&&
&&& x_5 \approx 1.371882

\end{aligned}
\end{equation}
$


Therefore, the roots are $x \approx -1.582036, x \approx 0.402102$ and $x \approx 1.371882$