- #1
SomeRandomGuy
- 55
- 0
1.) Suppose f:[a,b]->R and g:[a,b]->R are continuous such that f(a)<=g(a) and f(b)=>g(b). Prove that f(c)=g(c) for some c in [a,b].
I started out by using the Intermediate Value Property for some c1 and c2 with f(c1)=L1 and g(c2)=L2. I am trying to conclude that L1=L2. This was one approach. Then, I also tried setting up a ratio where f(a)/g(a) > 1 and f(b)/g(b) < 1 by assumption. These are continuous if g !=0, so there must exist a point where f/g = 1. But, what if g(x)=0. This method seems to fall apart as well, and now I'm stuck.
2.) Let f:[0,1]->[0,1] be a continuous function. Prove that f must have a fixed point; that is, show f(x)=x for some x in [0,1]. I'm trying to show that f must intersect the graph g(x)=x by using the IVP again. Don't quite know how, though. Seems that if I figure out the first problem, this one would be very similiar.
Thanks for any help, I appreciate it.
I started out by using the Intermediate Value Property for some c1 and c2 with f(c1)=L1 and g(c2)=L2. I am trying to conclude that L1=L2. This was one approach. Then, I also tried setting up a ratio where f(a)/g(a) > 1 and f(b)/g(b) < 1 by assumption. These are continuous if g !=0, so there must exist a point where f/g = 1. But, what if g(x)=0. This method seems to fall apart as well, and now I'm stuck.
2.) Let f:[0,1]->[0,1] be a continuous function. Prove that f must have a fixed point; that is, show f(x)=x for some x in [0,1]. I'm trying to show that f must intersect the graph g(x)=x by using the IVP again. Don't quite know how, though. Seems that if I figure out the first problem, this one would be very similiar.
Thanks for any help, I appreciate it.