1. ## 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?

2. ## Re: Continuity

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

3. ## Re: Continuity

Originally Posted by loui1410
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?