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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.)$x^{3}+2 x-4=0, \quad x_{1}=1$

1.1797

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

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this question asked us to use Newton's method to solve. So first things first. Let's take the derivative if f of X is x cubed plus two x minus for the derivative of X Cubed is three are exponent becomes on new coefficient. We subtract, exploited by one and then two acts. The derivative is simply too, Which means now, except to is gonna be one minus one cubed plus two times one minus four. Divide by three times one squared plus two, which is 1.2 and then exit three is gonna be 1.2 minus 1.2 cubed plus two times 1.2 minus four over three times 1.2 squared plus two, which is 1.1797

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