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

$$f^{-1}(x)=\Big(\frac{x^5-1}{2}\Big)^{1/3}$$ - Domain: $(-\infty,\infty)$ - Range: $(-\infty,\infty)$
$$y=f(x)=(2x^3+1)^{1/5}\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=(2x^3+1)^{1/5}$$ $$2x^3+1=y^5$$ $$x^3=\frac{y^5-1}{2}$$ $$x=\sqrt[3]{\frac{y^5-1}{2}}=\Big(\frac{y^5-1}{2}\Big)^{1/3}$$ 2) Interchange $x$ and $y$: $$y=\Big(\frac{x^5-1}{2}\Big)^{1/3}$$ Therefore, $$f^{-1}(x)=\Big(\frac{x^5-1}{2}\Big)^{1/3}$$ - Domain: $x$ is defined in $R$. So the domain of $f^{-1}$ is $(-\infty,\infty)$. - Range: Since $x$ is defined in $R$, $\Big(\frac{x^5-1}{2}\Big)^{1/3}$ also ranges from $-\infty$ to $\infty$. That means the range of $f^{-1}$ is $(-\infty,\infty)$. *Check: $$f(f^{-1}(x))=\Big[2\times\Big(\Big(\frac{x^5-1}{2}\Big)^{1/3}\Big)^3+1\Big]^{1/5}=[2\times\frac{x^5-1}{2}+1]^{1/5}=[x^5-1+1]^{1/5}=[x^5]^{1/5}=x$$ $$f^{-1}(f(x))=\Big[\frac{((2x^3+1)^{1/5})^5-1}{2}\Big]^{1/3}=\Big[\frac{2x^3+1-1}{2}\Big]^{1/3}=\Big[\frac{2x^3}{2}\Big]^{1/3}=\Big[x^3\Big]^{1/3}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$