Can You Find a Continuous Function That Takes Each Value Exactly Three Times?

[matrix_204]

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.

 

[maverick280857]

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.

 

[wais]

The hard continuity problem that you are facing is indeed a challenging one. It requires a deep understanding of continuity and function behavior. Let me try to provide some insight and guidance to help you solve this problem.

Firstly, let's start with the easier part - finding an example of a function that is continuous everywhere and takes each of its values exactly 3 times. One possible function that satisfies this condition is f(x) = x^3. This function is continuous everywhere and takes each value exactly three times. For example, f(1) = 1, f(2) = 8, f(3) = 27, and so on.

Now, let's move on to the harder part - proving that there is no such function if there are two values instead of three. To prove this, we need to show that it is impossible for a function to be continuous everywhere and take only two values. This can be done by contradiction.

Assume that there exists a function f(x) that is continuous everywhere and takes only two values, let's say a and b. Without loss of generality, let a < b. Since f(x) is continuous everywhere, it must also be continuous at the point (a + b)/2. This means that for any ε > 0, there exists a δ > 0 such that if |x - (a + b)/2| < δ, then |f(x) - f((a + b)/2)| < ε.

Now, let's consider the points a, (a + b)/2, and b. Since f(a) = f((a + b)/2) = f(b), by the intermediate value theorem, there exists a point c between a and (a + b)/2 such that f(c) = a. Similarly, there exists a point d between (a + b)/2 and b such that f(d) = b.

But this means that f(x) takes the value a at two different points, c and (a + b)/2, violating the condition that f(x) takes each value exactly three times. This contradiction proves that there is no such function that is continuous everywhere and takes only two values.

In conclusion, the hard continuity problem can be solved by providing an example of a function that is continuous everywhere and takes each value exactly three times, and then proving that it is impossible for a function to be continuous everywhere and take only two values. I