Re: First Year Uni Calculus Marathon
(A bit tricky)
For which positive integers k is it possible to find a continuous function f:R->R that such that
f(x)=y
has exactly k solutions x for every real number y?
Also unanswered. I will get people started:
For k=2 it is not possible.
Proof:
Assume f is a cts function taking every value exactly twice.
There must exist two zeros a < b of the function f.
Between a and b f must have fixed sign (otherwise the intermediate value theorem would provide the existence of a third zero).
Without loss of generality, we assume f is positive in (a,b).
Then f must attain a positive maximum m at some p in (a,b) (Extreme value theorem applied to [a,b]).
Now this means that f(x)=2m must have its solutions OUTSIDE the interval [a,b], without loss of generality we may assume that one such solution q is greater than b. Then by the intermediate value theorem, the equation f(x)=m/2 must have:
-at least one solution in (a,p)
-at least one solution in (p,b)
-at least one solution in (b,q).
This contradiction completes the proof.
As for k=3, it IS actually possible.
Picture the square [0,1]x[0,1] and draw line segments between O, (1/3,1), (2/3,0), (1,1). (Looks like a zigzag.)
Now replicate this curve in each square [k,k+1]x[k,k+1].
The resulting curve is the graph of a continuous function with the sought property.
What about for higher k?