# Thread: continuous on closed and bounded interval

1. ## continuous on closed and bounded interval

Let f be continuous on the closed and bounded interval [a,b] and let x1, x2, .... , xn be in [a,b]. Show that there must exist a c in [a,b] such that f(c) =
f(x1) + f(x2) + .... + f(xn)
----------------------------------------.
n

I think all I have to show is that:
minimum value of f(x) in [a,b] is less than or equal to the expression above which is less than or equal to the maximum value of f(x) in [a,b]. But how do I show that this is true? Then, I can just use the Intermediate Value Theorem to prove that a f(c) exists.

Thank you for any help (the ------------ is obviously a division sign).

2. Because f is continuous on [a,b] it has a maximum and minimum there.
Let $U = \max \left\{ {f(x):x \in [a,b]} \right\}\,\& \,L = \min \left\{ {f(x):x \in [a,b]} \right\}$.
By continuity $\left( {\exists c \in [a,b]} \right)\left[ {f(c) = L} \right]\,\& \,\left( {\exists d \in [a,b]} \right)\left[ {f(d) = U} \right]$.

Now we have $nL \le \sum\limits_{k = 1}^n {f\left( {x_k } \right)} \le nU\quad \Rightarrow \quad L \le \frac{{\sum\limits_{k = 1}^n {f\left( {x_k } \right)} }}{n} \le U$.

Using the intermediate value theorem finish.