 # Suppose the function f : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struc Caitlyn Cole 2022-04-30 Answered
Suppose the function $f:\mathbb{R}\to \mathbb{R}$ is continuous. For a natural number $k$, let ${x}_{1},{x}_{2},...,{x}_{k}$ be points in $\left[a,b\right]$. Prove there is a point $z$ in $\left[a,b\right]$ at which
$f\left(z\right)=\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k$
So I'm thinking about applying the intermediate value theorem:

If
$a<{x}_{1}
then
$f\left(a\right)
or
$k.f\left(a\right)
$f\left(a\right)<\left(f\left({x}_{1}\right)+f\left({x}_{2}\right)+...+f\left({x}_{k}\right)\right)/k
But I couldn't think of any way to prove that $f\left(a\right) or is it even true?

EDIT: Thanks everyone for your effort.
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Hint:

Take $z\in \left\{{x}_{1},\cdots {x}_{k}\right\}$ such that $f\left(z\right)\le f\left({x}_{j}\right)$ for any ${x}_{j}$.

Take $w\in \left\{{x}_{1},\cdots {x}_{k}\right\}$ such that $f\left(w\right)\ge f\left({x}_{j}\right)$ for any ${x}_{j}$.

Then
$f\left(z\right)\le \frac{f\left({x}_{1}\right)+\cdots +f\left({x}_{k}\right)}{k}\le f\left(w\right)$
Now, use the intermediate value theorem.
###### Not exactly what you’re looking for? tswe0uk
$m\le \frac{1}{k}\sum _{i=1}^{k}f\left({x}_{i}\right)\le M$ where $M=maxf\left(x\right),m=minf\left(x\right)$ on $\left[a,b\right]$, Now apply IVT