Proving Existence of x for Continuous Identity Function f[a,b]->[a,b]

[Treadstone 71]

"Let f[a,b]->[a,b] be continuous. Prove that there exists at least 1 x in [a,b] such that f(x)=x."

This seems simple geometrically since if we consider the identity function g(x)=x, if f(x) is continuous, then if you "draw" the graw of f, it must intersect g at some point. At that point, f(x)=x. But I have no idea how to translate this intuition into analytical lingo.

 

[Muzza]

Use the intermediate value theorem.

 

[math-chick_41]

and what can you say about g(x) = f(x)-x on [a,b]?

 

[HallsofIvy]

In particular, what is f(a)- a? What is f(b)-b?
(Remember that f(a) and f(b) is in [a,b]?)

 

[Treadstone 71]

I got it! Thanks for the help.