Precalculus (6th Edition) Blitzer

a. $f^{-1}(x)=(x-1)^3$ b. see graph c. $f(x)$: domain $(-\infty,\infty)$, range $(-\infty,\infty)$; $f^{-1}(x)$: domain $(-\infty,\infty)$, range $(-\infty,\infty)$.
a. Given $f(x)=\sqrt[3] x+1$, we follow the steps below: (i) write $y=\sqrt[3] x+1$; (ii) exchange $x,y$ to get $x=\sqrt[3] y+1$; (iii) solve for $y$, $\sqrt[3] y=x-1$ Thus: $y=(x-1)^3$; (iv) replacing $y$ with $f^{-1}(x)$, we have $f^{-1}(x)=(x-1)^3$ b. We can graph both functions together with $y=x$ as shown in the figure. c. We can find the domain and range for each function as: for $f(x)$, domain $(-\infty,\infty)$, range $(-\infty,\infty)$; for $f^{-1}(x)$, domain $(-\infty,\infty)$, range $(-\infty,\infty)$.