Functional Cycle 0Db3B6

1. **Yes — that is the key idea.** We start with the functional equation $f(x)+f\left(1-\frac{1}{x}\right)=8x$. The trick is to choose inputs that make the inside expression cycle back to earlier values. 2. **First substitution: let $x=2$.** $$f(2)+f\left(1-\frac{1}{2}\right)=8(2)$$ $$f(2)+f\left(\frac{1}{2}\right)=16$$ So we get $f(2)+f\left(\frac{1}{2}\right)=16$. 3. **Second substitution: let $x=\frac{1}{2}$.** $$f\left(\frac{1}{2}\right)+f\left(1-\frac{1}{\frac{1}{2}}\right)=8\left(\frac{1}{2}\right)$$ Simplify the inside carefully: $$1-\frac{1}{\frac{1}{2}}=1-2=-1$$ So $$f\left(\frac{1}{2}\right)+f(-1)=4$$ 4. **Third substitution: let $x=-1$.** $$f(-1)+f\left(1-\frac{1}{-1}\right)=8(-1)$$ Simplify the inside: $$1-\frac{1}{-1}=1-(-1)=2$$ So $$f(-1)+f(2)=-8$$ 5. **Why three times?** Because the expression $1-\frac{1}{x}$ sends values in a cycle: $2 \to \frac{1}{2} \to -1 \to 2$. That gives three equations with the three unknown values $f(2)$, $f\left(\frac{1}{2}\right)$, and $f(-1)$. 6. **Now solve the system.** $$f(2)+f\left(\frac{1}{2}\right)=16$$ $$f\left(\frac{1}{2}\right)+f(-1)=4$$ $$f(-1)+f(2)=-8$$ Add them all: $$\big(f(2)+f\left(\frac{1}{2}\right)\big)+\big(f\left(\frac{1}{2}\right)+f(-1)\big)+\big(f(-1)+f(2)\big)=16+4-8$$ $$2f(2)+2f\left(\frac{1}{2}\right)+2f(-1)=12$$ Divide by $2$: $$\frac{\cancel{2}f(2)+\cancel{2}f\left(\frac{1}{2}\right)+\cancel{2}f(-1)}{\cancel{2}}=\frac{\cancel{12}}{\cancel{2}}$$ $$f(2)+f\left(\frac{1}{2}\right)+f(-1)=6$$ 7. **Finish the calculation.** From the first equation, $f\left(\frac{1}{2}\right)=16-f(2)$. From the second equation, $f(-1)=4-f\left(\frac{1}{2}\right)=4-(16-f(2))=f(2)-12$. Substitute into $f(2)+f\left(\frac{1}{2}\right)+f(-1)=6$: $$f(2)+\big(16-f(2)\big)+\big(f(2)-12\big)=6$$ $$f(2)+4=6$$ $$f(2)=2$$ **Final answer:** $2$