Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 6

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

Using Approximation Formula

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


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\begin{equation}
\begin{aligned}

f'(x) =& \frac{1}{3} \frac{d}{dx} (x^3) + \frac{1}{2} \frac{d}{dx} (x^2) + \frac{d}{dx} (3)
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f'(x) =& \left(\frac{1}{\cancel{3}} \right) (\cancel{3}) (x^2) + \left( \frac{1}{\cancel{2}}\right) (\cancel{2})(x) + 0
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f'(x) =& x^2 + x
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x_2 =& x_1 - \frac{\displaystyle \frac{1}{3} x_1^3 + \frac{1}{2} x_1^2 + 3 }{x_1^2 + x_1}
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x_2 =& - 3 - \frac{\displaystyle \frac{1}{3} (-3)^3 + \frac{1}{2} (-3)^2 + 3 }{(-3)^2 + (-3)}
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x_2 =& -3 - \frac{\displaystyle \frac{1}{3} (-27) + \frac{1}{2} (9) + 3 }{9 - 3}
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x_2 =& -3 - \frac{\displaystyle (-9) + \frac{9}{2} + 3 }{6}
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x_2 =& -3 - \frac{\displaystyle \frac{9}{2} - 6}{6}
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x_2 =& -3 - \frac{9 - 12}{(6)(2)}
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x_2 =& - 3 - \frac{(-3)}{12}
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x_2 =& -3 + \frac{1}{4}
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x_2 =& \frac{-12 + 1}{4}
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x_2 =& \frac{-11}{4}
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x_3 =& x_2 - \frac{\displaystyle \frac{1}{3}x^3_2 + \frac{1}{2} x^2_2 + 3}{x^2_2 + x_2}
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x_3 =& \frac{-11}{4} - \frac{\displaystyle \frac{1}{3} \left( \frac{-11}{4} \right)^3 + \frac{1}{2} \left( \frac{-11}{4} \right) ^2 + 3 }{\displaystyle \left( \frac{-11}{4} \right)^2 + \left( \frac{-11}{4}\right) }
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x_3 =& \frac{-11}{4} - \frac{\displaystyle \left( \frac{1}{3} \right) \left( \frac{-1331}{64} \right) + \left( \frac{1}{2} \right) \left( \frac{121}{16} \right) + 3 }{\displaystyle \frac{121}{16} - \frac{11}{4}}
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x_3 =& \frac{-11}{4} - \frac{\displaystyle \frac{-1331}{192} + \frac{121}{32} + 3 }{\displaystyle \frac{121 - 44}{16}}
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x_3 =& \frac{-11}{4} - \frac{\displaystyle \frac{-1331 + 726 + 576}{192}}{\displaystyle \frac{77}{16}}
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x_3 =& \frac{-11}{4} - \frac{ (-29)(16) }{ (77)(192) }
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x_3 =& \frac{-11}{4} + \frac{464}{14784}
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x_3 =& \frac{-11}{4} + \frac{29}{924}
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x_3 =& \frac{-2541 + 29}{924}
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x_3 =& \frac{-2512}{924}
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\end{aligned}
\end{equation}
$


$\quad \boxed{x_3 \approx -2.7186}$