Trouble with newton's method.

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This is the problem I'm doing and the start of the explanation of how to do it:

Use Newton's Method to approximate to $4$ decimal place: $$\sqrt[3]{11}$$

This would be the equivalence of finding the zero for $x^3-14=0$. So that would make our function $f(x)=x^3-14$. $x_1=3$ will be our starting point since it is closest perfect cube root.

Based on other example problems, I thought it should be $x^3-11 = f(x)$. Trying that, using the formula $x_{n+1} = x_{1} - \frac{x^3-11}{3x^3}$ I just kept getting negative numbers that just bounced back and forth, eventually, between -1.something and -2.something. So with my understanding of how I get f(x) is wrong, or the example problem is wrong again I somehow either screwed up the overly simple math of setting the cube root of a 11 to zero, or (despite checking and adding plenty of parentheses) I'm failing at my calculator input. Or some combination of those. I need held figuring what.

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1
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It must have been some typo in the formula that you have used. If you do everything correctly you will get something like this

3.0000
2.4074
2.2376
2.2241
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
2.2240
1
On

It looks like there's a typo in the explanation, it should be $x^3 - 11$ not $x^3 - 14$. Then the approach you've described attempting is correct except that you've got the wrong $f'$ in your correction term. It should be:

$$x_{n+1} = x_n - \frac{x_n^3-11}{3x_n^2}$$