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$g$ be a continuous function on $[a,b]$, suppose $\xi_1,\xi_2\in [a,b]$, I need to show $\exists \xi\in [a,b]\ni g(\xi)=\frac{g(\xi_1)+g(\xi_2)}{2}$. Thanks

Myshkin
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  • Why would $g(\xi)$ need to be in $\left[a,b\right]$? – Tucker Jun 15 '17 at 00:28
  • In your question, I see $g(\xi)\in \left[a,b\right]$. – Tucker Jun 15 '17 at 00:32
  • Not to be rude, but this is astounding that with that much reputation you post so low quality questions and unconstructive comments! – C. Falcon Jun 15 '17 at 00:33
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    @C.Falcon sadly most of their questions are of similar quality, but some have upwards of 30, even 60 votes. – Dando18 Jun 15 '17 at 00:35
  • @Urgent Sure I am the finest mathematician ever born in the universe. Seriously, I think you don't get my point, I do not judge the difficulty of your question nor your level, this is meaningless to me. However, you do not respect in any way the rules of this site. You must provide some thoughts on the problem. – C. Falcon Jun 15 '17 at 00:38

1 Answers1

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If $g(\xi_1)=g(\xi_2)$. Then we may take $\xi=\xi_1$.

If $g(\xi_1)\ne g(\xi_2)$. Let $\displaystyle f(x)=g(x)-\frac{g(\xi_1)+g(\xi_2)}{2}$. Then $\displaystyle f(\xi_1)=\frac{g(\xi_1)-g(\xi_2)}{2}$ and $\displaystyle f(\xi_2)=\frac{g(\xi_2)-g(\xi_1)}{2}$.

By Bolzano's theorem, there exists $\xi\in[a,b]$ such that $f(\xi)=0$.

CY Aries
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