Suppose $f: \mathbb R \to \mathbb R$ is smooth, that it has a root and that $f'(x)>0$ for all $x$. Does this guarantee that Newton's method will converge (quadratically) for all choices of initial points?
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These days you need to explain more about where you got the problem from and what have you tried, so your question will not be closed by the moderators. – Momo Nov 26 '21 at 16:02
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@Momo Yes, thank you. – joseMoralez1888 Nov 26 '21 at 16:05
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Also not strictly related, because it's not always increasing, but this answer from the other day might be interesting. https://math.stackexchange.com/questions/4314240/example-of-numerical-methods#4314428 – Jakob Streipel Nov 26 '21 at 16:07