Prove there is $z \in (a,b)$ such that $f(z)=1/k \sum_{n=1}^k f(x_n)$

Question

Suppose that the function $f:(a,b) \rightarrow \mathbb{R}$ is continuous for $x_1, \ldots ,x_k$ in $(a,b)$. Show there is $z \in (a,b)$ such that $f(z) = \frac{1}{k}\sum_{n=1}^k f(x_n)$

Answer 1

There are $ s\ t\in \{1\ldots k\}\ $ such that

$$ \forall_{n=1}^k\quad f(x_s)\le\ f(x_n)\le\ f(x_t) $$

hence:

$$ f(x_s)\,\ \le\,\ \frac 1k\cdot\sum_{n=1}^k\,f(x_n)\,\le\,\ f(x_t) $$

Thus, there exists $$ z\ \in\ [\min(x_s\ x_t);\ \max(x_s\ x_t)]\ \subseteq (a;b)\ $$

such that $$ f(z)\ =\ \frac 1k\cdot\sum_{n=1}^k\,f(x_n) $$