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?
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?