Answer
$f(x)=2(x+1)^3$, domain $(-\infty,\infty)$, range $(-\infty,\infty)$;
$f^{-1}(x)=\sqrt[3] {\frac{x}{2}}-1$, domain $(-\infty,\infty)$, range $(-\infty,\infty)$
Work Step by Step
1. Given $f(x)=2(x+1)^3$, we can identify it is one-to-one with domain $(-\infty,\infty)$ and range $(-\infty,\infty)$
2. Rewrite the function as $y=2(x+1)^3$
3. Exchange $x,y$ to get $x=2(y+1)^3$
4. Solve for $y$ to get $y=\sqrt[3] {\frac{x}{2}}-1$
5. Replace $y$ with $f^{-1}(x)$ to get $f^{-1}(x)=\sqrt[3] {\frac{x}{2}}-1$
6. For $f^{-1}(x)$, we can find its domain $(-\infty,\infty)$ and range $(-\infty,\infty)$
