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Thread: Continuity

  1. #1
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    Continuity

    Prove: If f(x) is a continuous function on a closed interval [a, b] and x1, x2, ..., xn are n points in the interval, then there exists c such that:
    $\displaystyle f(c)=\frac{f(x1)+f(x2)+...+f(xn)}{n}$

    Any ideas?
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  2. #2
    Super Member girdav's Avatar
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    Re: Continuity

    Hint: intermediate value theorem, showing that the average of $\displaystyle f(x_j)$ is between $\displaystyle \min_{[a,b]} f$ and $\displaystyle \max_{[a,b]}f$.
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  3. #3
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    Re: Continuity

    Quote Originally Posted by loui1410 View Post
    Prove: If f(x) is a continuous function on a closed interval [a, b] and x1, x2, ..., xn are n points in the interval, then there exists c such that:
    $\displaystyle f(c)=\frac{f(x1)+f(x2)+...+f(xn)}{n}$
    Note how to use subscripts, [TEX]f(c)=\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}[/TEX] gives $\displaystyle f(c)=\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}$.

    Let $\displaystyle M=\max\{f(x_n)\}\text{ and }m=\min\{f(x_n)\}$. Do you see that $\displaystyle m\le\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}\le M~?$.

    Can you use the intermediate value theorem?
    Thanks from loui1410
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