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Assume that $g$ is a continuously differentiable function and that the Fixed-Point Iteration $g(x)$ has exactly three fixed points, $-3, 1$ and $2$. Assume that $g '(-3) = 2.4$ and that FPI started sufciently near the fixed point $2$ converges to $2$. Find $g'(1)$.

Having trouble doing this.

Thank you for the help!

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I'll try to get you started.

Note that $g'(-3) \geq 1$. Therefore, since $g(x) \neq x$ for $x \neq -3, 1, 2$, this implies that $g(x) \geq x$ for all $x \in (-3, 1)$. In other words, the graph of $g(x)$ is above the line $y=x$ on that interval.

Since $2$ is an attractor of $g$, what can you say about $g'(2)$? Given that information, what can you say about the graph of $g(x)$ on the interval $(1,2)$? Can you finish the problem from here?

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If -3,1,2 are fixed points then why is g(x) no equal to x at thouse points? I thought that was the definition of a fixed point? – user24648 Feb 18 '12 at 23:41
I mean that g(x) is not equal to x everywhere /except/ those points, since the problem stipulates that they are the /only/ fixed points. – Lopsy Feb 19 '12 at 20:08
@Lopsy I think that $g'(2)$ is irrelevant and it is very hard to say anything useful about it. – Trismegistos Feb 23 '12 at 20:20

This discussion along with @Lopsy hints helped me to solve this interesting question. I think though that Lopsy made one mistake telling you to consider $g'(2)$. It is not important and you can not say much about it. Important thing is that this is stable fixed point.

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