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Let $f$ continuous at $[0,2]$ with $f(0)=f(2)$.Check if there are $x_{1},x_{2} \in [0,2]$ such that $x_{1}-x_{2}=1$ and $f(x_{1})=f(x_{2})$ . How can I do this?

evinda
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1 Answers1

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Consider $g\colon[0,1]\to\mathbb R$ with $g(x)=f(x)-f(x+1)$. What can you say about $g(0)$ and $g(1)$?

Hagen von Eitzen
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  • $g(0)=-g(1)$ ,right? – evinda Feb 09 '14 at 12:56
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    @evinda: Consider the signs of $g(0)$ and $g(1)$, and the fact that $g$ is continuous. – MPW Feb 09 '14 at 13:04
  • We could use the Bolzano theorem,because $g$ is continuous at $[0,1]$ and $g(0)g(1)<0$.So we conclude that $g$ has at least one root $w$ at $[0,1]$,so $g(w)=0 \Rightarrow f(w)=f(w+1)$ ,right?But,could I also use the Intermediate value theorem? – evinda Feb 09 '14 at 13:06