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Use Newton's method with the specified initial approximation $x_{1}$ to find $x_{3},$ the third approximation to the root of the given equation. (Give your answer to four decimal places.)$\frac{1}{3} x^{3}+\frac{1}{2} x^{2}+3=0, \quad x_{1}=-3$

$-2.7186$

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

Campbell University

Harvey Mudd College

Idaho State University

Lectures

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in this function. Where in this pub away Rusty's Newton's method to find the third approximation of the boot today function that where our issue dez is negative free. So let's remember the intense method. It's a practice is dark. The next up information could've rude ISS plus one, the previous of pollution minus the function relates that the previous point divided by their toe, functional. It's at the service or waiting to find the derrick of functional, which is, um, express Plus. So the apartment, your guess is negative. Three. Then our next test of the extra is swollen. Miners have fallen over X prime one and this is negative. Three minus nine negative time bus then plus three divided by very ministry. And this gives us negative 2.75 Then the burger Pricks mission to read off dysfunction this text to minus that true over a crime of two. And this is the Linus native to form 75 minus. No one heard O negative 2.7 Fire Jew boasts one or 2.75 Spring plus three What were, um, negative 2.7 Fire squared minus 247 and he did this cognition. This gives us the third approximation to Group B negative. 2.7186

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