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If $ f(x) = \left\{ \begin{array}{ll} \sqrt{x} & \mbox{if} x \geqslant 0\\ -\sqrt{-x} & \mbox{if} x < 0\\ \end{array} \right. $
then the root of the equation $ f(x) = 0 $ is $ x = 0 $. Explain why Newton's method fails to find the root no matter which initial approximation $ x_1 \not= 0 $ is used. Illustrate your explanation with a sketch.
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Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 8
Newton's Method
Derivatives
Differentiation
Volume
Missouri State University
Harvey Mudd College
Baylor University
Idaho State University
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