If $f(x)=x^3-8$, what is $f^{-1}(f(f^{-1}(19)))$?
I need to find the inverse function of $f(x)$ first. To do that, I switch the roles of $x$ and $y$ and solve for $y$. So $y=x^3-8$ becomes $x=y^3-8$. Adding $8$ to both sides, I get $x+8=y^3$. Taking the cube root of both sides, I get $\sqrt[3]{x+8}=y$. This is the inverse function of $f(x)$, so I can write it as $f^{-1}(x)=\sqrt[3]{x+8}$. Now I can plug in $f^{-1}(19)$ into $f(x)$. That means $f(f^{-1}(19))=f(\sqrt[3]{19+8})$. Simplifying the radical, I get $f(f^{-1}(19))=f(\sqrt[3]{27})$. I know that $\sqrt[3]{27}=3$, so $f(f^{-1}(19))=f(3)$. Plugging in $3$ into $f(x)$, I get $f(3)=(3)^3-8=27-8=19$. So $f(f^{-1}(19))=19$. Finally, I need to find $f^{-1}(f(f^{-1}(19)))$. That means I need to plug in $19$ into $f^{-1}(x)$. So $f^{-1}(f(f^{-1}(19)))=f^{-1}(19)$. But I already know that $f^{- {1}}(19)=\sqrt[3]{19+8}$. So I just need to simplify that radical. I see that $19+8=27$ and $\sqrt[3]{27}=3$. So $f^{-1}(f(f^{-1}(19)))=3$. 