If $f(x)=2x^3+4$, find $f^{-1}(58)$.
I need to find the inverse function of $f$, which means I need to switch the roles of $x$ and $y$ and then solve for $y$. So, if $f(x)=2x^3+4$, then $y=2x^3+4$, and swapping $x$ and $y$ gives $x=2y^3+4$. Now I need to isolate $y$ on one side of the equation. First, I'll subtract $4$ from both sides to get $x-4=2y^3$. Next, I'll divide both sides by $2$ to get $\frac{x-4}{2}=y^3$. Finally, I'll take the cube root of both sides to get $y=\sqrt[3]{\frac{x-4}{2}}$. This is the inverse function of $f$. To find $f^{-1}(58)$, I just need to plug in $x=58$ into the inverse function and simplify. So, $f^{-1}(58)=\sqrt[3]{\frac{58-4}{2}}=\sqrt[3]{27}=3$. 