Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 8

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Find $x_3$, the 3rd approximation to the root of $\displaystyle x^5 + 2 = 0$ using Newton's Method with the specified initial approximation $x_1 = -1$. (Give your answer to four decimal places.)

Using Approximation Formula

$\displaystyle x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}$

\begin{aligned} f'(x) =& \frac{d}{dx} (x^5) + \frac{d}{dx} (2) \\ \\ f'(x) =& 5x^4 \\ \\ x_2 =& x_1 - \frac{x_1^5 + 2}{5x_1^4} \\ \\ x_2 =& -1 - \frac{(-1)^5 + 2}{5(-1)^4} \\ \\ x_2 =& -1 - \frac{-1 + 2}{5 (-1)^4} \\ \\ x_2 =& - 1 - \frac{-1 + 2}{5} \\ \\ x_2 =& -1 - \frac{1}{5} \\ \\ x_2 =& \frac{-5 - 1}{5} \\ \\ x_2 =& \frac{-6}{5} \\ \\ \\ \\ x_3 =& x_2 - \frac{x^5_2 + 2}{5x^4_2} \\ \\ x_3 =& \frac{-6}{5} - \frac{\displaystyle \left( \frac{-6}{5} \right)^5 + 2 }{ \displaystyle 5 \left( \frac{-6}{5} \right)^4 } \\ \\ x_3 =& \frac{-6}{5} - \frac{\displaystyle \frac{-7776}{3125} + 2 }{\displaystyle 5 \left( \frac{1296}{625} \right)} \\ \\ x_3 =& \frac{-6}{5} - \frac{\displaystyle \frac{-7776 + 6250}{3125}}{\displaystyle \frac{1296}{125}} \\ \\ x_3 =& \frac{-6}{5} - \frac{(-1526)(125)}{(3125)(1296)} \\ \\ x_3 =& \frac{-6}{5} + \frac{190750}{4050000} \\ \\ x_3 =& \frac{-6}{5} + \frac{763}{16200} \end{aligned}

$\quad \boxed{x_3 \approx -1.1529}$