## University Calculus: Early Transcendentals (3rd Edition)

$$f^{-1}(x)=\sqrt[3]{x+1}$$
$$y=f(x)=x^3-1$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y = x^3-1$$ $$x^3=y+1$$ $$x=\sqrt[3]{y+1}$$ 2) Interchange $x$ and $y$: $$y=\sqrt[3]{x+1}$$ Therefore, the inverse of function $y=f(x)=x^3-1$ is the function $y=\sqrt[3]{x+1}$. In other words, $$f^{-1}(x)=\sqrt[3]{x+1}$$