the function f defined by $f(x)=(x^3+1)/3$ has three fixed points say α,β,γ where
$-2<α<-1$, $0<β<1$, $1<γ<2$.
For arbitrarily chosen $x_{1}$, define ${x_{n}}$ by setting $x_{n+1}=f(x_{n})$
If $α I think I must prove three things, but not sure: 1: if $α 2: if $α 3: if $β could you please help me?
How I can prove this? (fixed points of $f(x)=(x^3+1)/3$)
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real-analysis
2 Answers
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Hints: Since there are only those three fixed points, we either have $f(x) < x$ for all $x$ with $\alpha < x < \beta$ or $f(x) > x$ for all those $x$. Check one point $x$ in that interval to see which it is. Similarly for $\beta < x < \gamma$.
Since $f$ is an increasing function, if $x > \alpha$ then $f(x) > f(\alpha) = \alpha$, and similarly ....
You will also want to use the fact that an increasing sequence that is bounded above, or a decreasing sequence that is bounded below, has a limit.
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This is not meant to be an answer, but I can't add a picture to comments or another answer...
