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

$$f^{-1}(x)=\sqrt[4] x$$ - Domain: $[0,\infty)$. - Range: $[0,\infty)$.
$$y=f(x)=x^4\hspace{1cm}x\ge0$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^4$$ $$|x|=\sqrt[4]y$$ Since $x\ge0$, we have $|x|=x$, therefore: $$x=\sqrt[4]y$$ 2) Interchange $x$ and $y$: $$y=\sqrt[4]x$$ Therefore, $$f^{-1}(x)=\sqrt[4] x$$ - Domain: $x$ is defined where $x\ge0$. So the domain of $f^{-1}$ is $[0,\infty)$. - Range: We see here that $\sqrt[4]{x}\ge0$ for all $x\in[0,\infty)$ and it will range to $\infty$. In other words, the range of $f^{-1}$ is $[0,\infty)$ *Check: $$f(f^{-1}(x))=(\sqrt[4]x)^4=x$$ $$f^{-1}(f(x))=\sqrt[4]{x^4}=|x|$$ Again, since $x\ge0$, $|x|=x$: $$f^{-1}(f(x))=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$