There exists x on closed interval such that f(x)=x

Question

If $f$ is a continuous function on a closed interval, how can I show that there exists some $x$ on $f$ that $f(x)=x$?

I know it will require the Intermediate Value Theorem. Initially I thought of letting $g(x)=f(x)-x$, and showing that there exists $x$ such that $g(x)=0$ implying $f(x)=x$, but don't have any information about $f$ on either end of the interval.

Answer 1

The function $g$ defined by $g(x)=f(x)-x$ is continuous and

$$g(0)g(1)\le0$$ so by IVT there's $x_0\in[0,1]$ s.t. $g(x_0)=0\iff f(x_0)=x_0$.

Answer 2

Since $f(0) \in [0,1]$, then $g(0)=f(0)-0\ge 0$; and Since $f(0) \in [0,1]$, then $g(1)=f(1)-1\le 0.$

Hence $g(0)\cdot g(1) \le 0$.