Ordered Solution Sets

The first step is to limit the domain of f(x) to every third odd number, including 1 and 7. A new function will be constructed that has a domain of all real numbers, and a range of only the elements in the domain of f(x). The new function is called D(x), and its values are that of the domain of f(x), so it will replace "x" in f(x)= $x^2$ .

To start, it is a greatest integer function. I will represent this as $\boxed{D(x)=[x]}$ . The range, so far, is the set of all integers. Since an even number can always be obtained by doubling an integer, and adding 1 to an even number always yields an odd number, then $\boxed{D(x)=2[x]+1}$ only contains all odd-numbered y-values. Now, to get every third odd number, the 2[x] term must be multiplied by 3 to give 6[x]. This is because " every third odd number " translates to " every sixth integer. " The distance between adjacent steps in the function becomes 6 because 6[x]-6[x-1] always equals 6. The function $\boxed{D(x)=6[x]+1}$ finally must be tested to verify that it contains {1,7}. It does, so this is our new function.

• f(x)= $x^2$
• f(D(x))= $D(x)^2$
• f(D(x)= $(6[x]+1)^2$

It's a very strange-looking function. Counting up from f(D(x))=0, the tenth value is 841, and this is the final answer.

I could not come up with a way to algebraically determine the 10th whole number value of this function. Ideas will be appreciated!