- #1
Treadstone 71
- 275
- 0
"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.
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.