## Calculus: Early Transcendentals (2nd Edition)

a. $f^{-1}(x) = \sqrt[3] x/\sqrt[3]3$ b. $f(f^{-1}(x)) = 3(\sqrt[3] x/\sqrt[3]3)^{3}$ $f(f^{-1}(x)) = 3x/3$ $f(f^{-1}(x)) = x$ $f^{-1}(f(x)) = \sqrt[3] 3x^{3}/\sqrt[3]3$ $f^{-1}(f(x)) = \sqrt[3]x^{3}$ $f^{-1}(f(x)) = x$
a. $f(x) = 3x^{3}$ $x = 3f^{-1}(x)^{3}$ $x/3 = f^{-1}(x)^{3}$ $\sqrt[3] x/\sqrt[3]3 = f^{-1}(x)$ b. $f(f^{-1}(x)) = 3(\sqrt[3] x/\sqrt[3]3)^{3}$ $f(f^{-1}(x)) = 3x/3$ $f(f^{-1}(x)) = x$ $f^{-1}(f(x)) = \sqrt[3] 3x^{3}/\sqrt[3]3$ $f^{-1}(f(x)) = \sqrt[3]x^{3}$ $f^{-1}(f(x)) = x$