Theorem: Let $f$ be a continuous function on $[a,b] \in \Bbb R$. Prove that there exists a $c \in [a,b]$ so that $ f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx $
I cannot use derivatives or any other integral theorems, I am doing these proofs using Riemann Sums and continuity.
My Thoughts:
I am planning on approaching this problem using the following theorem:
Theorem: Let $f$ be a continuous function on the interval $[a,b]$. Then $\left| \int_a^bf(x)\,dx \right| \ \le \ (b-a)\sup\limits_{[a,b]}|f(x)| $
Or I could show that this $f(c)$ exists, but then I am unsure as to how I would prove that $c \in [a,b]$. Hints would be appreciated.
Edit: So I have realized that I have the following inequality for $m := \inf\limits_{[a,b]}f(x)(b-a)$, $M := (b-a)\sup\limits_{[a,b]}|f(x)|$: $ m \le \frac{1}{b-a}\left| \int_a^bf(x)\,dx\right| \le M $
So now how do I relate this to $f(c)$?
Edit 2: I changed the title as upon considering the existing MVT, I don't think this is the same thing.