Hard Continuity problem

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I am having great, great difficulties in solving this problem, its asking me to find a function that is continuous everywhere which takes each of its values exactly 3 times(like give an example of a function, no proving). This part, i have a little imagination of my own to start, but the second part is the hardest one, which i can't even think about.
It says to prove that there is no such function if there are two values instead of three.
 

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Let f(x) be such a function and a be some member of its domain. Let its value at a be denoted by k. Then the question requires you to find f(x) such that the equation f(a)-k = 0 has exactly 3 real roots (or 3 real values of a). If f(x) is a polynomial then it must be a cubic polynomial. More generally f(x) could be a nth order polynomial with (n-3) complex roots. But this would mean that (n-3) is an even number (because complex roots occur in conjugates). Hence n is odd. So f(x) can be cubic, fifth degree, and so on. Of course all this doesn't help you solve the problem explicitly. Intuitively f(x) cannot be something like sin(x) because sin(x) attains each value twice in one period but not thrice and extending the interval to 1.5 periods, we won't get each value thrice. Do you think we can do this by a suitable modification of the interval chosen?

I can't think how we can "find" the function. If I come up with something else I'll certainly post it here.
 

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