Prove: If f(x) is a continuous function on a closed interval [a, b] and x_{1}, x_{2}, ..., x_{n}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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- Jul 11th 2012, 01:52 AMloui1410Continuity
Prove: If f(x) is a continuous function on a closed interval [a, b] and x

_{1}, x_{2}, ..., x_{n}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? - Jul 11th 2012, 02:52 AMgirdavRe: Continuity
Hint: intermediate value theorem, showing that the average of $\displaystyle f(x_j)$ is between $\displaystyle \min_{[a,b]} f$ and $\displaystyle \max_{[a,b]}f$.

- Jul 11th 2012, 02:52 AMPlatoRe: Continuity
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*?