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I am to prove that there exists $x^*\in [0,1] \cup [2,3]$ for a function $f$ from $[0,1] \cup [2,3] \to [0,1] \cup [2,3]$ s.t $f(x^*) = x*$ basically the fixed point theorem for the above disjoint interval. It is given that $f$ is continous.

My attempt : I tried going the standard route and define $g(x) = f(x) - x$ then i say that $g(0) > 0$ for $g(1)$ we have two cases it can be either positive or negative, if it is negative then proof is done by IVT, if it is positive then I check $g(2)$ which again has the same two cases, now if $g(2)$ is also positive then I check for $g(3)$ and I say that $g(3)$ has to be negative. So now I have $g(0) > 0$ and $g(3) < 0$ so by invoking IVT I could say that there exists a point s.t. $f(x^*) = x^*$ but since the interval is disjoint I am not sure that I can use IVT here( or maybe the function being continuous allows me to???).

So can someone help me out in this and let me know if there is a proof for the above problem if my proof is incorrect.

user601297
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1 Answers1

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The claim in the question is not true in general: let $f(x)=3-x$ for $x$ in its domain, then an $x^*$ does not exist.

Parcly Taxel
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