Answer
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 $(-\infty,\infty)$.
Work Step by Step
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 $(-\infty,\infty)$.
