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

$$f^{-1}(x)=\frac{1}{\sqrt[3] x}$$ - Domain: $(-\infty,0)\cup(0,\infty)$ - Range: $(-\infty,0)\cup(0,\infty)$
$$y=f(x)=\frac{1}{x^3}\hspace{1cm}x\ne0$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{1}{x^3}$$ $$x^3=\frac{1}{y}$$ $$x=\frac{1}{\sqrt[3] y}$$ 2) Interchange $x$ and $y$: $$y=\frac{1}{\sqrt [3]x}$$ Therefore, $$f^{-1}(x)=\frac{1}{\sqrt[3] x}$$ - Domain: $x$ is defined where $x\ne0$. So the domain of $f^{-1}$ is $(-\infty,0)\cup(0,\infty)$. - Range: Since $\sqrt[3]{x}\gt0$ ranges from $-\infty$ to $\infty$ in the defined $x$, $\frac{1}{\sqrt[3]x}$ also ranges from $-\infty$ to $\infty$, except there is no value of $x$ so that $f^{-1}(x)=0$ In other words, the range of $f^{-1}$ is $(-\infty,0)\cup(0,\infty)$ *Check: $$f(f^{-1}(x))=\frac{1}{\Big(\frac{1}{\sqrt[3] x}\Big)^3}=\frac{1}{\frac{1}{x}}=x$$ $$f^{-1}(f(x))=\frac{1}{\sqrt[3]{\frac{1}{x^3}}}=\frac{1}{\frac{1}{x}}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$