Let the function $ f : [a,b] \to \mathbb{R} $ be continous, such that $ f(a) \neq f(b)$.
For $ n \in \mathbb{N}$ let $ x_1,x_2,...,x_n $ be points in $[a,b]$.
Prove that there is a point $z \in [a,b]$ such that $f(z)= {f(x_1)+ \dots + f(x_n)\over n} $
Give an example in which $f(a) = f(b)$ where the conclucion shown before is not true.
Sketch: let $m=\min_i f(x_i)$ and $M=\max_i f(x_i)$ both of which are attained since $f$ is bounded and the sequence is finite. Let $A$ denote the average of $f$ over the points. Then $m\leq A\leq M$. By the intermediate value theorem there is a point $z$ that achieves the average.
Now figure out why you need $f(a)\neq f(b)$ to make the above precise. I have a feeling you meant to say $z\in (a,b)$.
Proof
$f:[a,b]\to \mathbb{R}$ continuous such that $f(a)\ne f(b)$. Points $x_1,\dots,x_n\in [a,b]$ where $f(x_1),\dots,f(x_n)$ are in the range of $f$.
Let $\alpha = \min\{f(x_1),\dots,f(x_n)\},\beta = \max\{f(x_1),\dots,f(x_n)\}$. Then the average value must lie between the minimm and maximum. By Intermediate Value Theorem, $\exists \zeta\in (a,b)$ such that
$$ f(\zeta) = \frac{f(x_1)+\dots+f(x_n)}{n} $$
Suppose $f(a)=f(b)$. Let $n=2$, and $x_1=a,x_2=b$. Consider $f(x) = -(x+1)^2+1$ defined on $[-2,0]$. Notice that $\nexists \zeta \in (-2,1)$ such that $f(\zeta) = \frac{f(-2)+f(0)}{2}=0$.
This situation arises because if the function $f$ has the same value for its two end-points, but $f$ is not a constant and also not a function with oscillations, then we can take $n=2$ and the average of the value of the end-points is just the value of the end-point itself, which is not attained elsewhere in the function. If the function is a constant then the case is simple.
