problem: Let \[f(x) =
  \begin{cases}
  |\lfloor{x}\rfloor| &\text{if }x\text{ is rational}, \\
  \lceil{x}\rceil^2 &\text{if }x\text{ is irrational}.
  \end{cases}
  \] Find $f(\sqrt[3]{-8})+f(-\pi)+f(\sqrt{50})+f\left(\frac{9}{2}\right)$.
solution: Since we know that $\sqrt[3]{-8}=-2$ is a rational number, $$f(\sqrt[3]{-8})=|\lfloor{-2}\rfloor|=2.$$Continuing from here, we know that $-\pi$ is irrational, thus $$f(-\pi)=\lceil{-\pi}\rceil^2=(-3)^2=9.$$Because 50 is not a perfect square, $\sqrt{50}$ must be irrational as well, so $$f(\sqrt{50})=\lceil{\sqrt{50}}\rceil^2=8^2=64.$$Finally, we know that $\frac{9}{2}$ is a rational number, so $$f\left(\frac{9}{2}\right)=\left|\left\lfloor{\frac92}\right\rfloor\right|=4.$$Therefore $$f(\sqrt[3]{-8})+f(-\pi)+f(\sqrt{50})+f\left(\frac{9}{2}\right)=2+9+64+4=\boxed{79}.$$