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Explain why Newton's method doesn't work for finding the root of the equation $x^{3}-3 x+6=0$ if the initial approximation is chosen to be $x_{1}=1$
Newton's method fails
Calculus 1 / AB
Chapter 4
APPLICATIONS OF DIFFERENTIATION
Section 6
Newton's Method
Derivatives
Differentiation
Applications of the Derivative
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this question asks us to explain when Newton's back. That doesn't work for funding the root of this equation. What we know is that this equation X Cube managed three ex post sex has a local minimum axe is one. Therefore, the tangent Over here, the slope of the tension is zero. What this means is that exit to will not give us any other point but satisfies one of the given solutions and other words Exes. One does not work as an approximation. Therefore, this is the reason why Newton's method will not work is because X equals one doesn't work. And this is what we have is our local minima.
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