Continuous function from the unit interval into the unit interval has a fixed point 
Show that if $f$ is continuous on $[0,1]$ and $0\leq f(x) \leq 1$ for all $ x \in [0,1]$, then there exists one point  $c \in [0,1]$ at which $f(c)=c$.
(Hint: Apply the Intermediate Value Theorem to the function $g(x)=x−f(x)$.)

I have some idea how to do this, but the area where I am getting stumped is...the value of $f(x)$?
How can I find the value of $f(x)$?  Do I even need to find $f(x)$?
Edit:
okay I got something...
I only need f(0)>0 and f(1)<1 to show 
g(0)<0

So then what happens for f(0)=0 or f(1)=1 ?
 A: To begin, some clarification:  under the stated conditions, $f(x)$ may have more than one fixed point.  To see this, take $f(x) = \cos^2 (n \pi x)$ for sufficiently large $n$.  
This being the case:
You don't need to know the value of $c$ such that $f(c) = c$, only that such a $c$ exists.  Without more detailed knowledge of $f(x)$, $c$ cannot be determined in any event; but its existence may be affirmed.  Let $I$ denote the closed unit interval $[0, 1]$: $I  = [0, 1]$, and let $f:I \to I$ be continuous.  (Note that we need $f:I \to I$ here; to see this, consider $f:I \to \Bbb R$ given by, e.g. $f(x) = -x - 2$.)  First look at the values of $f(x)$ at $0$ and $1$.  If $f(0) = 0$ and/or $f(1) = 1$, we are done.  Assuming then that neither of these possibilities hold, since $f:I \to I$, we must have $f(0) > 0$ and $f(1) < 1$.  Next, look at $g(x) = f(x) - x$.  $g(x)$ is manifestly continuous, with $g(0) > 0$ and $g(1) < 0$.  By the intermediate value theorem, there must be a $c \in (0, 1)$ with $g(c) = 0$.  For such a $c$, we have $f(c) = c$, i.e., $c$ is the sought-for fixed point of $f(x)$.  QED.
Hope this helps.  Cheerio, 
and of course,
Fiat Lux!!!
