Use the Intermediate Value Theorem to prove $f:[0,1]\to [0,1]$ continuous and $C\in[0,1]$, there is some $c \in [0,1]$ such that $f(c) = C$.
Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an $f(x) = C$, but I am having trouble proving that a $f(c) = C$.
This is what I have:
Since $f$ is continuous on $[0,1]$ there exists $f(a) = 0$ and $f(b) = 1$.
Let $g(x) = F(x) - C$. We assume that $f(a) < C < f(b)$ $\to$ $0 < C < 1$
when $x = a$, $g(a)$ is negative when $x = b$, $g(b)$ is positive
Therefore, $g(a) < 0 < g(b)$, and since $f$ is continuous on $[0,1]$, so is $g$.
Therefore there exists a $g(x) = 0$, and $f(x) = C$.
How can I prove there is a $f(c) = C$ ? Is this a rule for IVT?