Find F2 D57E4E

1. **State the problem clearly:** We are given the functional equation $$f(x)+f\left(1-\frac{1}{x}\right)=8x$$ and we want to find $$f(2)$$. 2. **Important idea:** We will plug in a few values of $$x$$ that make the inside expressions cycle back to earlier inputs. 3. **Find one equation by letting $$x=2$$:** $$f(2)+f\left(1-\frac{1}{2}\right)=8(2)$$ $$f(2)+f\left(\frac{1}{2}\right)=16$$ So we get $$f(2)+f\left(\frac{1}{2}\right)=16$$ 4. **Find a second equation by letting $$x=\frac{1}{2}$$:** $$f\left(\frac{1}{2}\right)+f\left(1-\frac{1}{\frac{1}{2}}\right)=8\left(\frac{1}{2}\right)$$ First simplify the inside carefully: $$1-\frac{1}{\frac{1}{2}}=1-2=-1$$ So $$f\left(\frac{1}{2}\right)+f(-1)=4$$ 5. **Find a third equation by letting $$x=-1$$:** $$f(-1)+f\left(1-\frac{1}{-1}\right)=8(-1)$$ Now simplify the inside: $$1-\frac{1}{-1}=1-(-1)=2$$ So $$f(-1)+f(2)=-8$$ 6. **Now we have a system of three equations:** $$f(2)+f\left(\frac{1}{2}\right)=16$$ $$f\left(\frac{1}{2}\right)+f(-1)=4$$ $$f(-1)+f(2)=-8$$ 7. **Add all three equations together:** $$\big(f(2)+f\left(\frac{1}{2}\right)\big)+\big(f\left(\frac{1}{2}\right)+f(-1)\big)+\big(f(-1)+f(2)\big)=16+4-8$$ This becomes $$2f(2)+2f\left(\frac{1}{2}\right)+2f(-1)=12$$ Now divide both sides by $$2$$: $$\frac{\cancel{2}f(2)+\cancel{2}f\left(\frac{1}{2}\right)+\cancel{2}f(-1)}{\cancel{2}}=\frac{\cancel{12}}{\cancel{2}}$$ So $$f(2)+f\left(\frac{1}{2}\right)+f(-1)=6$$ 8. **Use the earlier equations to replace the unknowns:** From $$f(2)+f\left(\frac{1}{2}\right)=16$$, we get $$f\left(\frac{1}{2}\right)=16-f(2)$$ From $$f\left(\frac{1}{2}\right)+f(-1)=4$$, we get $$f(-1)=4-f\left(\frac{1}{2}\right)$$ Substitute the first into the second: $$f(-1)=4-(16-f(2))$$ Simplify: $$f(-1)=4-16+f(2)=f(2)-12$$ 9. **Substitute into the summed equation:** $$f(2)+\big(16-f(2)\big)+\big(f(2)-12\big)=6$$ Now simplify step by step: $$f(2)+16-f(2)+f(2)-12=6$$ $$f(2)+4=6$$ 10. **Solve for $$f(2)$$:** $$f(2)=6-4=2$$ **Final answer:** $$2$$