If $f$ is continuous in $[a,b]$ show that there exist $x_1,x_2,x_3,x_0$ in $[a,b]$ such that
$$ f(x_0)=\frac{f(x_1)+f(x_2)+f(x_3)}{3}.$$
I tried to use the intermediate value theorem and the fact that $f$ attains a maximum and a minimum in the interval but I am stuck.
Best Answer
I will restrict to the non-trivial case in which $x_1\neq x_2\neq x_3$. Without loss, assume $x_1<x_2<x_3$. Since $f$ is continuous in the compact set $[x_1,x_3]$, we know, from the intermediate value theorem, that for any number $v$ such that: $$ \min\{f(x_1),f(x_2),f(x_3)\}\leq v \leq \max\{f(x_1),f(x_2),f(x_3)\}, $$
there exists a value $x_0\in [x_1,x_3]$ such that:
$$ f(x_0)=v. $$
Now, pick $3v=f(x_1)+f(x_2)+f(x_3)$.