Students' answers to review problems for the second exam
in Math 311, spring 2003

#1 Benson #2 #3 #4 #5 #6 #7
#8 #9 #10 #11 Oleynick #12 #13 #14

2.Let f:[0,3]->R be a continuous function. Suppose f(0)-f(1)=f(2)-f(3). Prove that there exists x in the interval [0,1] such that f(x)+f(x+2)=2f(x+1).

Let g(x)=f(x)+f(x+2)-2f(x+1). We will try to show that g has a root in the interval [0,1], which will show that for some x in [0,1] 0=f(x)+f(x+2)-2f(x+1) which implies f(x)+f(x+2)=2f(x+1). Note that g(0)=f(0)+f(2)-2f(1) and g(1)=f(1)+f(3)-2f(2). But from our given information, f(3)=f(2)+f(1)-f(0). Thus g(1)=f(1)+[f(2)+f(1)-f(0)]-2f(2) so that g(1)=-f(0)-f(2)+2f(1). We see that g(0)=-g(1). By trichotomy either g(0) is positive, negative, or zero. If g(0) is 0, then 0 is a root of g and we are done. If g(0) is postive, we have that g(0)>0>g(1), and if g(0) is negative we have that g(0)<0Sent by Michael Benson on Sat, 12 Apr 2003 21:42:20

12.Let c be a cluster point of set A. Let L be a real number. Let f: A -> R be a function such that lim x->c f(x) = L and f(x) not = L for each x in A. Prove that L is a cluster point of { f(x) : x in A }.

To show that L is a cluster point of { f(x) : x in A }, we must show that for all deltaB>0, there exists a y in { f(x) : x in A }, y != L, such that |y - L| < delta-sub-B. We will demonstrate how to find a value y, given any value of deltaB>0.

Because c is a cluster point of A and lim x->c f(x) = L, by the definition of the limit of a function, for all epsilon>0, there exists a deltaF(epsilon)>0 such that |f(x) - L|<epsilon when x in A and 0<|x - c|<deltaF(epsilon). Because c is a cluster point of A, by the definition of cluster point, for all deltaA>0, there exists an x in A, x not= c, such that |x - c|<deltaA. For any value deltaB, if we take deltaF(deltaB) as the value of deltaA, then we know that there exists a z in A such that z not= c and |z - c|<deltaF(deltaB). But, because lim x->c f(x) = L, and z in A, and 0<|z - c|<deltaF(deltaB), we also know that |f(z) - L|<deltaB. Because f(x) not= L for all x in A, we also know that f(z) not= L. So, given any deltaB, take the value f(z) as y. y = f(z) so y is in { f(x) : x in A }, y not= L, and |y - L|<deltaB. Therefore, L is a cluster point of { f(x) : x in A }.

Sent by John Oleynick on Sat, 12 Apr 2003 18:52:11