Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 14

Use Newton's Method to approximate the root of $2.2x^5 - 4.4x^3 + 1.3x^2 - 0.9x - 4.0 = 0$ in the interval $[-2, -1]$ correct to six decimal places.







We apply Newton's Method with


$
\begin{equation}
\begin{aligned}

f(x) =& 2.2x^5 - 4.4x^3 + 1.3x^2 -0.9x - 4 \text{ and }
\\
\\
f'(x) =& 2.2 \frac{d}{dx} (x^5) - 4.4 \frac{d}{dx} (x^3) + 1.3 \frac{d}{dx} (x^2) - 0.9 \frac{d}{dx} (x) - \frac{d}{dx} (4)
\\
\\
f'(x) =& (2.2)(5x^4) - (4.4)(3x^2) + (1.3)(2x ) - 0.9(1) - 0
\\
\\
f'(x) =& 11x^4 - 13.2x^2 + 2.6x - 0.9

\end{aligned}
\end{equation}
$


Based from the graph, we choose $x = 1.4$

Using Approximation Formula


$
\begin{equation}
\begin{aligned}

x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}
\\
\\
x_2 =& x_1 - \frac{2.2x_1^5 - 4.4x_1^3 + 1.3x_1^2 - 0.9x_1 - 4}{11x_1^4 - 13.2x_1^2 + 2.6x_1 - 0.9}
\\
\\
x_2 =& -1.4 - \frac{2.2(-1.4)^5 - 4.4(-1.4)^3 + 1.3(-1.4)^2 - 0.9(-1.4) - 4}{11(-1.4)^4 - 13.2(-1.4)^2 + 2.6(-1.4) - 0.9}
\\
\\
x_2 \approx & -1.4
\\
\\
\\
\\
x_3 =& -1.404176 - \frac{f(-1.404176)}{f'(-1.404176)}
\\
\\
x_3 \approx & -1.404118
\\
\\
x_4 =& -1.404118 - \frac{f(-1.404118)}{f'(-1.404118)}
\\
\\
x_4 \approx & -1.404118


\end{aligned}
\end{equation}
$


Since $x_3$ and $x_4$ agree to six decimal decimal places, therefore $x \approx -1.404118$