## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$, so the given function is the inverse of f.
If a function $f$ is one-to-one, then $f^{-1}$ is the unique function for which $(f^{-1}\circ f)(x)=f^{-1}(f(x))=x$ and $(f\circ f^{-1})(x)=f(f^{-1}(x))=x.$ --- $(f^{-1}\circ f)(x)=f^{-1}(f(x))=(f(x))^{3}+4$ $=(\sqrt[3]{x-4})^{3}+4$ $=x-4+4$ $=x$ $(f\circ f^{-1})(x)=f(f^{-1}(x))=\sqrt[3]{f^{-1}(x)-4}$ $=\sqrt[3]{x^{3}+4-4}$ $=\sqrt[3]{x^{3}}$ $=x$ $(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$, so the given function is the inverse of f.