1
$\begingroup$

Let $f:(a,b) \rightarrow \mathbb{R}$ be a continuous function. Let $x_1,x_2,x_3,\dots,x_n \in (a,b)$. Prove that there exists a point $c \in (a,b)$ such that $$f(c) = \dfrac{f(x_1)+f(x_2)+......+f(x_n)}{n} $$

$\endgroup$
1
$\begingroup$

Since $x_1, x_2, \ldots, x_n \in (a,b)$, there is a closed interval $[a',b'] \subset (a,b)$ which contains all of them.

Let $$ \alpha = \frac{f(x_1) + f(x_2) + \ldots + f(x_n)}{n} $$ and write $$ M = \max\{f(x) : x\in [a',b']\}, \text{ and } m = \min\{f(x) : x\in [a',b']\} $$ Then $$ m \leq \alpha \leq M $$ and hence by the Intermediate Value Theorem, there exists a $c \in [a',b']$ such that $$ f(c) = \alpha $$

$\endgroup$
5
  • $\begingroup$ The function $f$ is defined on $(a,b)$, rather than $[a,b]$; your argument can be easily adjusted, though. $\endgroup$
    – egreg
    Oct 24 '13 at 13:23
  • $\begingroup$ @egreg : Thanks for pointing that out. Argument fixed :) $\endgroup$ Oct 24 '13 at 13:27
  • $\begingroup$ The intermediate value theorem requires that the function is continuous at the interval specified. How do we prove that the function is continuous at $[a',b']$ ? $\endgroup$
    – Papul
    Oct 24 '13 at 13:37
  • $\begingroup$ Its continuous on $(a,b)$ which contains $[a',b']$ $\endgroup$ Oct 24 '13 at 14:31
  • $\begingroup$ How do we prove that $m\leq \alpha \leq M$? $\endgroup$
    – Papul
    Oct 24 '13 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.