If $f(3x+1) + f(3x-10) = 10$, find the period of $f(x)$.
Step-by-Step Solution
Key Concept: Functional equations involving specific transformations can be manipulated through substitution to reveal periodicity relationships.
From $f(3x+1) + f(3x-10) = 10$, substituting $x \to x + 3$ gives $f(3(x+3)+1) + f(3(x+3)-10) = 10$, or $f(3x+10) + f(3x-1) = 10$. Subtracting the original equation from this, we get $f(3x+10) - f(3x-10) + f(3x-1) - f(3x+1) = 0$. Now replace $x$ with $x + 1$: $f(3x+1) - f(3x+19) = 0$, so $f(3x+1) = f(3x+19)$. This means $f(u) = f(u+18)$ for all $u$, so the period is 18.
Correct Answer: 18