Let f(x) = 4 - (x - 7)³ be an invertible function, then find f⁻¹(x).

Let $y = 4 - (x-7)^3$. Then $(x-7)^3 = 4 - y$, so $x = 7 + (4-y)^{1/3}$. Interchanging $x$ and $y$: $f^{-1}(x) = 7 + (4-x)^{1/3}$.