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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.

So after choosing the initial value of $x_{1}$ the subsequent approximations will

alternate between $-x_{1}$ and $+x_{1},$ which means that they will never approach

the root.

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question 34. I am Effort X, the groom's ex to the 1/2 when X is greater than zero or positive in f of X equals negative Negative X to the 1/2 1 X is negative or less than zero. Now you do my crime of eggs. You have 1/2 X to the negative 1/2 in my F crime here. Good. I'm gonna have negative 1/2 over negative X to the negative 1/2 So now I'm gonna in purple Just worry about my next to the end Plus one equals x minus x to the 1/2 over 1/2 Invert and multiply. I get to X so I have X minus two X, which gives me a negative X And that's the end, of course. And that's when X is positive. And now I do the same thing when X is negative. So I have negative square root of negative X except then minus. And then I have negative 1/2 negative. Yeah, invert and multiply. And I'm gonna get next to the end minus two x, which is gonna give me a negative X to the end as well. When exes Lutz, then zero so negative exit the end will not converge

University of Houston